CORRESPONDENCE BETWEEN PROF. BARTOCCI

AND ARDESHIR MEHTAThe following is the exact text of e-correspondence between

Prof. Umberto Bartocci of the University of Perugia, Italy,

and myself, regarding Relativity (especially my "Simple

Challenge"). The correspondence is complete and unedited,

translated by me when necessary from Italian into English;

and the only other changes that have been made are the

elimination of typographical errors, and reformatting to fit all

messages into the same margins (for the sake of legibility).From the following, any rational person should be able to

judge between Prof. Bartocci and myself, and see who is right

from a logical and mathematical point of view.Comments regarding the correspondence may be sent via e-

mail to us at:ardeshirmehta@myself.com

and

bartocci@dipmat.unipg.it

_____________________________________________________

Subject: Argomento Contro la Teoria della Relativita Speciale

[Translation: Argument Against the Special Theory of Relativity]

Date: Thu, 14 Jun 2001 17:26:19 -0400

From: "Ardeshir Mehta, N.D." <ardeshir@sympatico.ca>

To: bartocci@dipmat.unipg.it

[TRANSLATED FROM THE ITALIAN BY ARDESHIR MEHTA]

Dear Professor:

Having read your Web page:

<http://www.dipmat.unipg.it/~bartocci/quest.htm>

... I have decided to write my argument against the Special Theory

of Relativity -- at least the one I consider to be the best and simplest.You can read it at my Home Page, and at the following URL:

<http://homepage.mac.com/ardeshir/RelativityDebunked.html>

Regards,

Ardeshir Mehta

Ottawa, CanadaHome Page: <http://homepage.mac.com/ardeshir/education.html>

______________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

[Translation: Argument Against the Special Theory of Relativity]

Date: Fri, 15 Jun 2001 08:59:57 +0200

From: umberto bartocci <bartocci@dipmat.unipg.it>

To: "Ardeshir Mehta, N.D." <ardeshir@sympatico.ca>

References: 1

Dear Dr Mehta,

thank you very much for your attention. I shall read your argument with

great interest, and possibly I shall then send to you some comments.Best wishes,

Umberto Bartocci

(How it happens that you know so well Italian?)

(Are you a relative of the famous Indian conductor Zubin Mehta?)

--

Umberto Bartocci

Dipartimento di Matematica

Universita' di Perugia

06100 - Italy

http://www.dipmat.unipg.it/~bartocci

______________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

[Translation: Argument Against the Special Theory of Relativity]

Date: Sat, 16 Jun 2001 13:49:26 -0400

From: "Ardeshir Mehta, N.D." <ardeshir@sympatico.ca>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2

Hello Professor:

You wrote:

> Dear Dr Mehta,

> thank you very much for your attention. I shall read your

> argument with great interest, and possibly I shall then send

> to you some comments.I shall await your comments and read them with interest.

I am in touch with Dr. Al Kelly of Ireland, and Prof. Paul

Marmet of Ottawa, both of whom, I understand, you know

also. (In fact I became acquainted with your Web site after

reading Prof. Marmet's work, and then read about Dr Kelly

on your site, whereupon I contacted him.)By the way: Please call me "Ardeshir"! The "N.D." after

my name stands for "No Degrees". :-) I am so ashamed

of the way the academic world has distorted twentieth

century physics and mathematics, that I would personally

take no pride at all in being called "Doctor."> (How it happens that you know so well Italian?)

I lived in Italy three years, from 1964 to '67. Infatti, sono

stato anche a' Perugia, nell'anno '65, ed a volte ho mangiato

nella mensa degli studenti all'università. (In quei giorni ero

molto più giovane!) Per di più, ricordo benissimo il mio

soggiorno a' Perugia, e mi piacerebbe molto ritornarci un di'.> (Are you a relative of the famous Indian conductor Zubin

> Mehta?)No, I am afraid not; however, both he and I belong to a very

small ethnic group called "Parsis", less than 80,000 in number,

who fled Iran after Islam became the dominant religion there,

and found refuge in India. We are Zoroastrians, followers of

the Persian Prophet Zarathushtra.As a result of intermarriage within such a small community,

which took place over many centuries, it is quite possible

that a few of my genes and the conductor's are from the same

source!Saluti,

Ardeshir <http://homepage.mac.com/ardeshir/education.html>

___________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

[Translation: Argument Against the Special Theory of Relativity]

Date: Fri, 27 Jul 2001 13:42:18 +0200

From: umberto bartocci <bartocci@dipmat.unipg.it>

To: "Ardeshir Mehta, N.D." <ardeshir@sympatico.ca>

References: 1

Dear Mehta,

I have at last found some time for reading your "best

argument". I do not know if you are interested in my

comments/help, in case let me know. Anyway, I must tell you

in short that it is "obvious" that once more I did not find any

good argument against relativity. As a matter of fact, it is

simply NOT true that "the Lorentz transformation requires that

the stop watch should show a lesser time for this event if it is

calculated under the assumption that A is moving and B is

stationary, than it would if the time were calculated under the

assumption that B is moving and A is stationary. Thus the

Lorentz transformation requires the readings on the stop watch

to be calculated to be different, depending on whether A is

assumed to be moving or stationary...". If you do the

computation correctly, you find a unique answer to your

question.In any case, you should be much more clear since the

beginning, in the description of the physical situation you

propose to analyze. I presume that you imagine to have A and

B BOTH moving with respect to a THIRD (inertial) observer

C. Which is then their speed with respect to a Lorentzian

coordinate system associated with C? And when you ask "how

much time should it take for A to pass by the spot X marked

on B, as measured by a stop watch carried on board A?", you

should specify carefully which is the STARTING POINT of

the time interval you intend to compute (in A "proper time",

namely, with regard to the time marked by a watch carried by

A).If you do all things correctly, all "troubles" will disappear.

Unfortunately, relativity cannot be defeated with such simple

arguments: one needs some (strong) experimental evidence, for

instance of the non-validity of the "principle of relativity" for

electromagnetic phenomena...Molti cordiali saluti da Perugia,

dal suo Umberto Bartocci

--

Umberto Bartocci

Dipartimento di Matematica

Universita' di Perugia

06100 - Italy

http://www.dipmat.unipg.it/~bartocci

_______________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

[Translation: Argument Against the Special Theory of Relativity]

Date: Mon, 30 Jul 2001 11:38:49 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2

Caro Professore:

With the utmost respect, I think you are *completely* wrong.

And there is a *simple* test to prove it. Try to take my

challenge, given towards the end of my paper as follows:"I hereby challenge *anyone* to give a consistent answer

to the above question -- one that does not contradict

either the Principle of Relativity or the Lorentz transfor-

mation equations."You will see that you cannot give any *consistent* answer to

the question, "exactly what should the snapshots show?" No

one has ever been able to do so, not even the most respected

physicists in the world.You write:

If you do the computation correctly, you find a unique

answer to your question.Well, here is your chance to prove that you mean what you

say! Please do the computation, omitting no detail, and try

your best to prove yourself right. I am 100% sure I will be

able to find *specific* mathematical and / or logical errors in

any such computation.BTW: There is no need for a "third (inertial) observer C". It is

the *snapshots* taken of the stop watches carried on board A

and B that count. All observers, wherever they are, and even if

they are not in inertial frames, will see the *same* snapshots -

- it would be impossible for them to see different ones!You see, I do not even *need* to "defeat relativity" -- I only

have to show that the relativists cannot *support* relativity by

computations of their own, when I am allowed to set up the

*Gedankenexperimente* myself! That suffices to show that

they have not proved their own case.

Auguri e saluti,

Ardeshir <http://homepage.mac.com/ardeshir/education.html>

************************************************************

PS: Shall we consult with my good cyber-friend Al Kelly of Ireland? Or

Dr Christoph von Mettenhem, author of the book *Popper vs. Einstein*?

************************************************************

________________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

[Translation: Argument Against the Special Theory of Relativity]

Date: Mon, 13 Aug 2001 19:23:43 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

CC: "Ardeshir Mehta, N.D." <ardeshir@sympatico.ca>

References: 1 , 2

Caro professore:

Thinking that perhaps you did not understand my paper, I have condensed

my argument into a one-page challenge, which can be found at:<http://homepage.mac.com/ardeshir/SimpleChallenge-Relativity.html>

Se vuole, potrei anche tradurre la mia sfida in italiano!

[If you wish, I can even translate my challenge into Italian!]

Molti saluti,

Ardeshir <http://homepage.mac.com/ardeshir/education.html>

________________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

[Translation: Argument Against the Special Theory of Relativity]

Date: Tue, 14 Aug 2001 07:25:33 +0200

From: umberto bartocci <bartocci@dipmat.unipg.it>

To: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

References: 1 , 2 , 3

Dear Mehta,

I did not forget you, and I was preparing a detailed answer (no need to

translate), but I have got too many things to carry on simultaneously (either

in the classical or in the relativistic sense!), and I am getting old and with

less resources every day which goes...Always best wishes, UB

Umberto Bartocci

Dipartimento di Matematica

Universita' di Perugia

06100 - Italy

http://www.dipmat.unipg.it/~bartocci

_______________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

[Translation: Argument Against the Special Theory of Relativity]

Date: Wed, 15 Aug 2001 23:32:22 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2 , 3 , 4

Caro Prof. Bartocci,

Thank you for your response!

You wrote:

> Dear Mehta,

>

> I did not forget you, and I was preparing a detailed answer (no need to

> translate), but I have got too many things to carry on simultaneously (either

> in the classical or in the relativistic sense!), and I am getting old and with

> less resources every day which goes...

>

> Always best wishes, UBI think, from a picture of you I saw on the Web, that you

must be quite a bit older than I am (I am 58 now.) So I shall

understand if you cannot reply immediately -- just so long as

you reply eventually!I would like you to recall your own words in the Web page

entitled <ANSWERS TO "BARTOCCI INQUIRY">:Remember as scientists the duty to answer to criticism.

I find that you had also written there:

There are no mathematical contradictions in SR, which

is quite a coherent theory.It is to counter this second statement that I have devised a

simple one-page mathematical challenge to supporters of

Special Relativity, available at:<http://homepage.mac.com/ardeshir/SimpleChallenge-Relativity.html>

Of course I realise that you are not a supporter of SR, but

perhaps you still think that SR is *mathematically* correct. It

is definitely not.If you *do* the mathematics, which should not take you more

than twenty minutes at the most, you will find that there are

absolutely NO correct mathematical answers to the problem I

have posed in my challenge -- i.e., answers which are ALSO

consistent with the Special Theory of Relativity.No one to whom I have put the challenge has been able to

answer me. And indeed I think no one ever will, because SR,

despite your claim, is both logically and mathematically

flawed!Remember that Einstein was not a good mathematician. But

you are, so you should be able to see the mathematical flaw --

that is, if you actually *do* the mathematics.I suspect that the problem with SR is that the flaws are TOO

SIMPLE, and that is why they has not been widely discovered

till now. I have written yet another short *reductio-ad-

absurdum* refutation of SR -- this time logical in nature -- at:<http://homepage.mac.com/ardeshir/ShortestRefutatnRelativity.html>.

This is so short that it takes almost no time at all to read! It

is, as I say, a logical argument, not mathematical; but it is

still devastating to SR.But of course the mathematical argument is even more

conclusive in this regard.And naturally, it is impossible to properly answer a

mathematical challenge unless the mathematical calculations

are actually *done*!Tanti saluti,

Ardeshir <http://homepage.mac.com/ardeshir/education.html>

_______________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

[Translation: Argument Against the Special Theory of Relativity]

Date: Sun, 19 Aug 2001 10:24:35 +0200

From: umberto bartocci <bartocci@dipmat.unipg.it>

To: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

References: 1 , 2 , 3 , 4 , 5

Dear Mehta,

here it is at last my answer, which I shall divide in various

steps. I shall be VERY FRANK with you, after all it is you

who makes a "challenge", and I always believe that "amicus

Plato, sed magis amica veritas" (I hope that you already know

these words, or that you can be able to understand their

meaning in force of the Italian you can remember).A - First of all, let me say that I had to change this message at

least twice, since in the meantime I received two more mails

from you! For instance, I was trying to argue [1 - footnote]

that I AM NOT A SUPPORTER OF RELATIVITY, but then

I acknowledge with pleasure that you wrote in your last mail:> Of course I realise that you are not a supporter of SR, but

> perhaps you still think that SR is *mathematically* correct.Yes, I am definitively NOT a supporter of relativity, since I

believe (and I hope it, too) that it can be shown wrong in the

EXPERIMENTAL SIDE, but of course I still obviously think

that SR is *mathematically* correct.I would be indeed the HAPPIEST man in the world if it was

true that your argument (just to quote one between many

similar, like Dingle's, or Kelly's, etc.) is correct, and that I am

*completely* wrong"! UNFORTUNATELY IT IS NOT SO,

and there is no doubt about it.I can say it, I must say it, with no arrogance at all, but in

mathematics THERE IS NO FREEDOM OF OPINIONS: an

assertion is either right or wrong, and we must consider (both

special and general) relativity as "simple" mathematical

theories (which does NOT mean that they are in agreement

with the ORDINARY CATEGORIES of space and time!).

Special relativity, for instance, is just the theory of the

Minkowski space-time (a flat space-time). In this space the

"principle of relativity" for Lorentzian coordinate systems

(inertial observers - observers are not "people", but precisely

defined mathematical entities!) simply holds in force of the

definition of the space itself; the Lorentz transformations are

just the automorphisms of this space, no problems at all. If it

was true, as you claim, that:> SR, despite your claim, is both logically and

> mathematically flawed!then the WHOLE MATHEMATICS would be affected by

contradictions, and I hope that you will agree that this is

rather impossible, unbelievable (Goedel theorems apart)!I really cannot understand how is it possible that there is

always a lot of people (the most famous was perhaps Dingle,

with his notorious, but "stupid", "Dingle syllogism") which

believes that simple "paradoxes" can show that relativity is

wrong! As I wrote in a whole paper about

"Misunderstandings..." (which is available in point N. 2 of

my web page on the Foundations of Physics), the origin of

ALL of them is in some MISUNDERSTANDING of the

theory. As a matter of fact, these misconceptions are

sometimes "interesting", because to find where is the

"mistake" can be amusing, or fruitful for deeper understanding

of the theory, sometimes they are of no value at all. They

show in any case how ordinary space and time (which means

even ORDINARY LANGUAGE) differ from the relativistic

correspondents!Of course, SR is even a "physical theory", which means that

the mathematical theory is equipped with a code, a set of

rules, which allow to transform real situations into

mathematical ones, and conversely. That is to say, relativity

pretends to describe in some sense REALITY, at least in

certain well defined conditions, and we could indeed contest

THIS claim, but nothing more.This is why I can assert what I assert, and why I am able to

"answer" to all questions you ask, even if I am not a supporter

of relativity. The fact is that I understand relativity, which is

one thing that many "classical physicists" seem unable to

achieve.B - After this general introduction, I come to the special case

of your "challenge". There are many ways to make

"comments" to it, but there is no way to give a "strict answer"

to your question, since it is ILL POSED (in other words, it is

not well formulated; in Italian: MAL POSTA)! I am not

surprised at all that:> No one to whom I have put the challenge has been able to

> answer me.I have just stated which is the first possible reason for this

fact. It is as if you said: two men are going together for a

walk, one of them is tall x meters, the weight of the other is y

kilo, which is their age?I shall PROVE immediately that this is (of course only in

some sense) the case with your "question", but I wish to add

two more remarks.B1 - I am well aware that to show where are the "flaws" in the

simple statement of your question does not finish the

discussion, and that it could be the "duty" of a teacher to do

something more. For instance, a good teacher should even

show how one could correct the flaws, posing then a good

question, and then answer to it. I shall try to do even this for

you, but only for friendship, and not because I think that this

discussion can have some scientific interest, I think sincerely

that this is not the caseB2 - What I have just said could be another reason why you

did not get until now a reply to your question. Did really any

"competent" physicist assert that your problem was difficult,

and that he was not able to give an answer? I do not believe

that. I just believe that other people to whom you have sent

your challenge simply did not reply because they did not

intend to waste their time in a discussion which they

immediately judged unproductive.B3 - As far as the previous point is concerning, let me tell

you something which has been said to me by a colleague here

in Perugia - one of the most competent expert in relativity, I

believe. He told me that he cannot understand why people

understands that they have to pay lawyers, physicians, etc., in

order to have their opinion, even for matters which require less

knowledge, time and commitment than scientific ones, but

then the same people suppose instead that scientists should

loose their time in answering to amateurish ill-posed

questions. He added that a teacher could do this work only for

the benefit of his own students, but if a student insists in not

understanding then he gets fired (in Italian: BOCCIATO).C - I see with pleasure that you have changed, simplified,

your previous question*, seeming to have appreciated the

objections contained in one of my last mails. Now the

physical situation about which you require an explanation is

defined rather more clearly, but all the same it needs some

correction, if you want to formulate a question which has

some meaning from a relativistic point of view, and then "can

be answered".You say that one has a spaceship A in empty space, L metres

in length (why do you introduce numbers instead than simple

letters, parameters?! There is no need at all of numbers in

mathematics, even if I understand which is your strategy, as I

shall soon argue!), and first of all:- what do you mean by this length?

You know very well that in relativity there is not the

possibility to define an absolute length for any "object", so

you could say for instance that L is the proper length of the

spaceship, namely the length with respect to some coordinate

system in which the ship stands still. Or otherwise you could

say that L is the length of the ship with respect to the buoy,

but then you have to deal with DIFFERENT lengths, as far as

relativity is concerning. You must decide which is the

meaning of this L . The same objection holds for the

definition of the buoy's length, which I shall call M . I shall

assume from now on that both L and M are PROPER

LENGTHS. You then assume that the relative speed between

the spaceship A and the buoy B is some percentage v of the

speed of light (in relativity, one says that in this case he is

measuring speeds in "geometrical unities", c = 1, OK), etc.,

and here we have not problems.D - In your words this spaceship is "passing rectilinearly just

centimetres past a small spherical capsule or buoy B" (avoid

please that "small", it has not an absolute meaning at all!).Then you claim:

> that some suitable mechanism (which can easily be devised)

> causes both the stop watches to measure the time interval

> taken for the spaceship A to pass the buoy B.THIS IS EXACTLY THE POINT WHICH HAS LESS

MEANING, for more than one reason, and which requires

greater attempts of correction, in comparison with what I did

before in the case of the lengths!D1 - First of all, what does it precisely mean this "time

interval"? You did not define it enough well in order to

identify it exactly. You use a generic expression which could

have a meaning in rough ordinary language, but not in a

physical problem like the one we are now dealing with. I shall

try to explain the situation with more detail.A time interval measured by some SINGLE clock can be

defined as the DIFFERENCE of two values t2 - t1 marked

by this clock, where t2 is the time corresponding to some

SECOND EVENT in the clock's "life", in some given space-

time coordinate system, and t2 the time corresponding to

some FIRST EVENT. Which are precisely the EVENTS you

wish to introduce for defining your time interval?Let us assume for instance THAT THE BUOY IS

POINTLIKE, namely that M = 0 (I understand very well that

you have said 1 meter in length, with respect to 259,627,884 ,

exactly because you were looking for a "clever" way to avoid

complications, and then to say that 1 is "almost" zero with

respect to 259,627,884 ! The point is that one must be very

precise in mathematics, no ad hoc approximations at all, when

they are not needed, and above all when one is trying to hide

with them exactly the point which on the contrary one should

have to bring to the light!), and that there is only one clock in

the buoy. Then we could ask indeed the following question:- the clock which is in the buoy marks t1 when the "bow" of

the spaceship is in front of it; this same clock marks t2 when

the "stern" of the spaceship is in front of it (let us introduce

bow and stern with respect to the buoy, first the bow, then the

stern are passing in front of the buoy), and then we take t2 - t1

(which, after all, does not depend from the choice of the origin

of times for this clock). Well, we could then ask: how much

is this difference?The answer is very easy: it is L'/v , where L' is the length of

the spaceship WITH RESPECT TO THE BUOY, which in

relativity is NOT the same L we introduced before. But let

me draw a picture, as I can, with the purpose of being more

clear:S ____________ B -----> (vector velocity v) °

the spaceship has a bow the pointlike

and a stern S with respect buoy, with a

to the buoy clock B

If you ask for a comparison which is the time which is needed

in this case for the buoy to pass the spaceship, from the point

of view of clocks which are in the spaceship, it is obviously

L/v , and it is a time interval GREATER than the other I have

indicated before:L/v > L'/v .

D2 - Having said that, what happens if the buoy is not

supposed pointlike, and it has too a bow and a stern with

respect to the spaceship (it would be much more "honest" to

speak of TWO SPACESHIPS, with possibly different

lengths!)?S ____________ B -----> (vector velocity v) B° ___S°

spaceship buoy

Do you wish to compute the time interval which is needed for

the two external points of the spaceship in order to pass in

front of the SAME POINT in the buoy, a point in which there

is a clock? Or do you suppose to want to compute for instance

the time which goes from the bow B being in front of the

other bow B° of the buoy, and then the stern S being in front

of the other stern S° of the buoy? (if you object that these

remarks are too much "pedantic", since you have introduced a

buoy of only 1 meter, and a much "longer" spaceship, think

what happens when you try to do the converse: namely, to

measure this investigated time interval from the point of view

of the spaceship! In any case, I repeat it, in mathematics there

are not long and small, there are only precise computations,

and possibly, AT THE END, when we are doing physics, we

can make APPROXIMATIONS, but not at the very

beginning).D3 - In truth, I understand that you have foreseen this

objection, and that you have tried to cleverly avoid it, by

introducing that "suitable mechanism (which can easily be

devised)" etc.! But where is this mechanism? In B, or in S, or

in B°, in S°, or elsewhere? And where is the clock (or better,

where are the clocks) you use for the measure of time?If they are at some distance from the "mechanism", in any case

you have to deal with the time which is needed for a signal

sent by this instrument in order to arrive to the clock. All

your efforts notwithstanding, you cannot avoid the relativistic

speculations in this matter, which claim that no signal can

travel faster than light, etc., and then assuming that exactly c

is the "better speed" (faster) which is required for

"communications" between your "mechanism" and the clock!D4 - I try to explain the problems in yet another way. At last

you propose to compare TWO measures, with apparently only

two clocks, but this is simply impossible. If it is possible to

suppose (or to approximate, if you prefer) the buoy as

pointlike, and to make the measurement you require with only

one clock in this case, how can you do the same measurement

from the spaceship? If there is only one clock in the spaceship,

say for instance in the bow, then this clock shows some time

when the bow of the spaceship is in front of the buoy (or of

the clock which is the buoy), but how can you get a second

value for the time measured by this clock, the second value

which is needed in order to compute the required difference?YOU NEED FOR INSTANCE ANOTHER CLOCK IN THE

STERN, or even - if you think that it is a different thing (but

it isn't!) - you need a signal going from the stern to the bow,

in order to know where the stern WAS in front of the buoy.

Here it is exactly where relativity comes in, with all his

stories about the synchronization of distant clocks etc.. If the

clock you are talking about was placed instead in the stern of

the spaceship, it could register only the time when the stern is

in front of the buoy, but he cannot "know" when the bow was

in front of the buoy! Summing up, you need TWO

synchronized clocks in order to do the measurement you wish,

both in the spaceship and in the buoy, and you must be more

precise in specifying which is the time interval you wish to

compute, and most of all HOW...E - Now that it has become clear, I hope, how your question

SHOULD HAVE BEEN FORMULATED, let us go on,

choosing (between many), one possibility, namely ask to

measure the time which passes - either from the point of view

of the buoy, or of the spaceship (in both cases you need a

whole space-time coordinate system associated with the two

objects: there is no difference at all with the attempt of

introducing only TWO objects, clock and mechanism!) - from

the two events which correspond to the coincidences:

bow/bow, stern/stern. In other words, I say that you have to

pose your question for instance in the following way, and that

then there is an easy not contradictory answer to it:- One has an inertial spaceship, with some proper length L,

and ANOTHER ONE, of some proper length M , the relative

speed is v. One has synchronized clocks on both spaceships [I

repeat that it is really enough to have TWO of them for each

spaceship, so FOUR in all], and one asks how much time is

needed, from the point of view of both spaceships, for

completely passing one the other.Now we have at last a WELL POSED question, and an easy

answer. The two time intervals are:L/v + M'/v (from the point of view of the first spaceship -

with obvious notation)L'/v + M/v (from the point of view of the second spaceship,

the "old" buoy) ,where of course it is L' = L*sqr(1-v^2/c^2) , and M' =

M*sqr(1-v^2/c^2) (in force of the so called length contraction

- we do not even need to introduce Lorentz transformations,

even if of course the length contraction is one of their

consequences).In order to decide which time interval is bigger than the other,

one has just to decide which length is bigger between L and

M . If one supposes that the first spaceship is longer than the

second (the ex-buoy), then one has:(L+M')/v > (L'+M)/v ,

no doubt at all about that, the second time interval is smaller

than the first.If one supposes instead COMPLETE SYMMETRY, namely

L = M (in other words, TWO EQUAL SPACESHIPS), one

gets at last TWO EQUAL TIME INTERVALS.As you see, everything is very simple, much more simple for

instance than introducing a coordinate system with respect to an

accelerated observer, like one must do when discussing the twin

paradox (see for instance the paper in the point 4 in the above

quoted web page, which is unfortunately only written in Italian,

but there is some comment to this problem even in the fifth

section of the aforesaid paper about "Misunderstandings...").The question is always the same, which after all is not even

too much difficult to overcome: it is to understand the

counterintuitive (but mathematically precise, and then

"logically possible") relativistic treatment of space and time

(not to have an "intuition" of it with the common thought

space and time categories, which would be of course rather

impossible)...F - I have talked too much - and with my bad English this is

perhaps harmful for a better understanding - yet I have not said

everything I could have said. For instance, in some lecture to

students, I could have shown even how one could have

connected length contraction and time dilation in the proposed

exercise.In any case I hope:

F1 - first of all, to not have made mistakes (which should

have to be considered my personal mistakes, and not

mistakes/contradictions in the theory!);F2 - that you have understood where are the mistakes in your

posing the question;F3 - that you have understood how could one pose a good

question, between many possible of the "same nature";F4 - that you will not require to me endless discussions, as it

happened many times to me with people not willing to

understand relativity (there is an example of such useless

discussions in my web site, with Percival, or Galeczki, etc.,

all people who do not know enough well relativity);F5 - to have been useful to you at least from an ethical point

of view, showing that one could indeed answer to all people,

even when one believes that the required effort of writing will

not produce anything "useful".I am at your disposal with sincere friendship for a possible

next round, but please only one, no more (and please spare to

me a comment about your new "logical" "Short Reductio-ad-

Absurdum Refutation of Special Relativity", which is no

better "idea" than the one I have discussed until now); time is

indeed a precious thing, mostly at the end of one's life, and

there are better, most productive, ways to spend it...Once again best wishes, from yours most sincerely

correspondent perugino UB

[1] There were even other things which I started to discuss in

my preliminary attempts of answer, for instance about what

you have said in one of yours intermediate mails:> There is no need for a "third (inertial) observer C"...

I do agree about that, and as a matter of fact you avoided to do

it in the second formulation of your "challenge", but it was

you that in the first formulation of it had written:> Imagine two objects, A and B, in rectilinear motion past

> one another.In order to "imagine" this situation, one needed to introduce

the point of view of a third observer! The fact is that there are

always many different manners to describe in mathematical

terms the "same" physical situation, since the "code" which

translates from "reality" to mathematics is rather flexible, but

in any case the answers must always be the same!

--

Umberto Bartocci

Dipartimento di Matematica

Universita' di Perugia

06100 - Italy

http://www.dipmat.unipg.it/~bartocci

__________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

Date: Fri, 24 Aug 2001 14:24:44 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2 , 3 , 4 , 5 , 6

Caro Prof. Bartocci:

Saluti!

Thank you for your detailed response. I am especially glad to

see that you have given philosophical arguments against my

views also!But I am very, very sorry to see that you have made a BIG

MISTAKE -- namely, you have *completely* ignored time

dilation!As you yourself have pointed out, in mathematics one must

be ABSOLUTELY PRECISE; and as you yourself also say,

"with no arrogance at all, ... in mathematics THERE IS NO

FREEDOM OF OPINIONS: an assertion is either right or

wrong".So if we get *one* answer using length contraction *alone*,

and *another* answer using both length contraction *and*

time dilation, then *both* answers cannot possibly be correct!

Only ONE of them can be correct.And then the question arises: Which one? Obviously, if one

accepts the Lorentz transformations, the right answer can only

be the one using BOTH length contraction AND time

dilation!SO YOUR ANSWER, CALCULATED *WITHOUT* USING

TIME DILATION, CANNOT POSSIBLY BE CORRECT!There is no freedom of opinion: you HAVE to be wrong!

Do you see my point? Am I not right? And are you not

wrong? Be honest, now.Also, as you would have seen had you carefully read my first

article "The Single Best Argument Against Relativity", my

entire "Challenge" is based on Dingle's observation that time

dilation, on the one hand, and the principle of all rectilinear

motion being relative, on the other, are contradictory to one

another.Thus time dilation is AN INTEGRAL PART of my

Challenge! By not mentioning time dilation you have not

answered my Challenge at all.I am moreover also sorry to see that you did not read my two

articles *carefully*. All the points you have raised have

already been answered in them. None of the objections you

have raised are new to me, and I have foreseen and taken care

of *all* of these objections.(Did you imagine you were entering into a Challenge with

some first-year university student? My dear Sir, you are

carrying on a mathematical discussion with someone who has

written a 110-page book entitled "Critique of Goedel's

Theorem"! Trust me, if I can find mistakes as subtle as those

made by Kurt Goedel in his most celebrated theorem, I can

find mistakes in anything YOU can write -- that is, if any

mistakes exist in what you write, and especially if they are

*blatant* mistakes.)Now I shall describe your mistakes in detail. I shall reply in

the beginning to the mistake in mathematical part of your

answer, and then the mistakes arising from the fact that you

did not read my articles carefully; and I shall tackle the

philosophical part at the end.You wrote:

> You say that one has a spaceship A in empty space, L metres in

> length (why do you introduce numbers instead than simple letters,

> parameters?! There is no need at all of numbers in mathematics,

> even if I understand which is your strategy, as I shall soon argue!),

> and first of all:

>

> - what do you mean by this length?I meant, of course, the so-called "rest length". Even in

discussions concerning Relativity, when one speaks of

"length" *without* any qualifications, one normally means

"rest length" or "length as measured by a rod which is at rest

with respect to the object being measured." If I remember

correctly, Einstein himself in his book *Relativity: the

Special and General Theory* does this.(But I think you know this fact already!)

By the way, I could easily have used "length L" instead of

precise numbers; but I used precise numbers so that the

calculations become easy. The numbers I have used allow the

Lorentz <gamma> factor to be exactly 2.00, which simplifies

calculations, especially for precise numbers. This is for those

who are not professional mathematicians, like many

philosophers I know, who are interested in Relativity from a

philosophical point of view.Also, in physics, measurements cannot be performed with

ABSOLUTE perfection. With every measurement there is

associated a MARGIN OF ERROR. If we take the margin of

error to be +/- 10^(-10) -- which is reasonable -- then my

numbers make perfect sense from the point of view of physics.

(After all, it is the APPLICATION of mathematics that we are

discussing, not PURE mathematics!)> You know very well that in relativity there is not the

> possibility to define an absolute length for any "object", so

> you could say for instance that L is the proper length of the

> spaceship, namely the length with respect to some coordinate

> system in which the ship stands still. Or otherwise you could

> say that L is the length of the ship with respect to the buoy,

> but then you have to deal with DIFFERENT lengths, as far as

> relativity is concerning. You must decide which is the

> meaning of this L . The same objection holds for the

> definition of the buoy's length, which I shall call M . I shall

> assume from now on that both L and M are PROPER

> LENGTHS.Right. (But the length M is not necessary, as I shall show

below.)> You then

> assume that the relative speed between the spaceship A and the

> buoy B is some percentage v of the speed of light (in relativity,

> one says that in this case he is measuring speeds in "geometrical

> unities", c = 1, OK), etc., and here we have not problems.Right again.

> D - In your words this spaceship is "passing rectilinearly just

> centi-metres past a small spherical capsule or buoy B" (avoid

> please that "small", it has not an absolute meaning at all!).Okay: one can assume that there is a point marked "X" on the

buoy which is the point past which the spaceship A must

pass. (I already wrote that in my longer, first article entitled

"Single Best Argument Against Relativity", which, if you had

read carefully, you would have seen clearly. Indeed it is

mentioned in the Abstract at the very *beginning* of the

article.)And as I said also, if we take the margin of error to be +/-

10^(-10), as specified by me in my first article, then even this

"X" is not necessary.So the buoy's length M is NOT NECESSARY AT ALL.

> Then you claim:

>

> > that some suitable mechanism (which can easily be devised)

> > causes both the stop watches to measure the time interval

> > taken for the spaceship A to pass the buoy B.

>

> THIS IS EXACTLY THE POINT WHICH HAS LESS

> MEANING, for more than one reason, and which requires

> greater attempts of correction, in comparison with what I did

> before in the case of the lengths!

>

> D1 - First of all, what does it precisely mean this "time

> interval"? You did not define it enough well in order to

> identify it exactly. You use a generic expression which could

> have a meaning in rough ordinary language, but not in a

> physical problem like the one we are now dealing with. I shall

> try to explain the situation with more detail.

>

> A time interval measured by some SINGLE clock can be

> defined as the DIFFERENCE of two values t2 - t1 marked

> by this clock, where t2 is the time corresponding to some

> SECOND EVENT in the clock's "life", in some given space-

> time coordinate system, and t2 the time corresponding to

> some FIRST EVENT. Which are precisely the EVENTS you

> wish to introduce for defining your time interval?I thought it was obvious that the two events are:

t1 = the time corresponding to the event of the "bow" of the

spaceship A coming to its closest possible distance compared

to the spot marked "X" on the buoy B, andt2 = the time corresponding to the event of the "stern" of the

spaceship A coming to its closest possible distance compared

to the spot marked "X" on the buoy B.> Let us assume for instance THAT THE BUOY IS POINTLIKE,

> namely that M = 0 (I understand very well that you have said 1 meter

> in length, with respect to 259,627,884, exactly because you were

> looking for a "clever" way to avoid complications, and then to say

> that 1 is "almost" zero with respect to 259,627,884!)Yes. The difference would have been negligible, indeed zero

when calculated to ten decimal places (which is the level of

accuracy I have specified in my first article entitled "Single

Best Argument Against Special Relativity".)I also did not say the buoy was pointlike because then there

would be a problem of how, from a *practical* point of view,

one could put a clock *inside* the buoy.But both these problems can be avoided by saying there is a

spot on the buoy marked "X" past which the spaceship passes.

(As I said above, in my earlier article entitled "Single Best

Argument Against Relativity" I have already mentioned this.)> The point is that one must be very precise in mathematics, no ad

> hoc approximations at all, when they are not needed, and above all

> when one is trying to hide with them exactly the point which on the

> contrary one should have to bring to the light!), and that there is only

> one clock in the buoy.Yes, I agree fully. But as I said, I already explained this in

my first article.And in physics there is ALWAYS a margin of error.

> Then we could ask indeed the following question:

>

> - the clock which is in the buoy marks t1 when the "bow" of

> the spaceship is in front of it; this same clock marks t2

> when the "stern" of the spaceship is in front of it (let us

> introduce bow and stern with respect to the buoy, first the

> bow, then the stern are passing in front of the buoy), and

> then we take t2 - t1 (which, after all, does not depend from

> the choice of the origin of times for this clock). Well, we

> could then ask: how much is this difference?

>

> The answer is very easy: it is L'/v , where L' is the length

> of the spaceship WITH RESPECT TO THE BUOY, which in

> relativity is NOT the same L we introduced before.

>

> But let me draw a picture, as I can, with the purpose of being

> more clear:

>

> S ____________ B -----> (vector velocity v) °

>

> the spaceship has a bow the pointlike

> and a stern S with respect buoy, with a

> to the buoy clock B

>

> If you ask for a comparison which is the time which is needed in

> this case for the buoy to pass the spaceship, from the point of view

> of clocks which are in the spaceship, it is obviously L/v , and it is

> a time interval GREATER than the other I have indicated before:

>

> L/v > L'/v.Now do you not SEE your mistake? You have

COMPLETELY IGNORED TIME DILATION in your above

calculation!In that case, HOW can your calculation be correct? If the

answer *using* time dilation *and* length contraction is

different from your above answer, then your answer cannot

possibly be correct!And there is NO WAY the answer *using* time dilation

*and* length contraction is NOT going to be different from

your above answer!If I have understood you correctly, you say that the stop watch

on the buoy would show t2 - t1 = L'/v, while the stop watch

in the spaceship would show t2 - t1 = L/v where L/v > L'/v.

Right?So WHERE is the time dilation calculation here? It does not

show up ANYWHERE in your calculations! ALL YOU

HAVE APPLIED IS THE LENGTH CONTRACTION.Maybe you are trying to side-step my Challenge -- which is

based, as I wrote in my first article, on Dingle's argument

about two clocks each ticking slower than the other -- by

*avoiding* all mention of time dilation? If so, then your

calculations contain an error -- and a very simple, elementary

and blatant error at that!You see, you are NOT applying the Lorentz transformation

equations properly. The Lorentz transformations absolutely

and categorically *require* time dilation. Length contraction

alone, *without* time dilation, IS NOT PERMITTED BY

THE LORENTZ TRANSFORMATIONS!If two DIFFERENT answers are obtained -- one when using

*only* length contraction and another when using *both*

length contraction *and* time dilation -- then only ONE of

the answers can be correct: namely, the one calculated using

both length contraction *and* time dilation.It is not a matter of opinion: YOU ARE

MATHEMATICALLY WRONG!(But why is it that I have to EXPLAIN all this to you, who

are a senior professor of mathematics? Surely this should have

been obvious to you from the start!)And as for time dilation, one CANNOT have two clocks, each

of which ticks slower than the other. That is *logically*

impossible.Suppose however we DO take time dilation into account.

Then we would have to say that the MOVING stop watch

ticks SLOWER than the one which is NOT moving.(N.B.: It is IMPOSSIBLE to calculate the time dilation

without specifying WHICH stop watch is moving and which

one is NOT.If you think it IS possible, please show exactly HOW -- for I

do not think you will be able to do so, ever!)Let me finish doing your calculations the way you SHOULD

have finished doing them.Suppose we assume the buoy to be MOVING. Then the time

recorded by its stop watch would *not* be L'/v, but

(L'/v)/<gamma> (where <gamma> = 1/sqrt.[1-(v^2/c^2)], and

which as a consequence must always be greater than 1)!And since L' = L/<gamma>, this equals

[(L/<gamma>)/v]/<gamma>, which is (L/v)/<gamma>^2

(and again, <gamma> is greater than 1)!And so it is clear that (L'/v)/<gamma> =/= L'/v !

Or suppose we take the SPACESHIP to be moving. Then the

time recorded by *its* stop watch would *not* be L/v, but

(L/v)/<gamma> (where <gamma> is once again greater than

1) !And again it is clear that (L/v)/<gamma> =/= L/v !

So your answers are wrong, wrong, WRONG!!! It is NOT a

matter of opinion. ANY mathematical judge will say so.Now remember the words of my Challenge: "exactly what

should the snapshots [of the stop watches' readouts] show?"Suppose you say that the snapshots would show that the

buoy's stop watch would show (L'/v)/<gamma> and the stop

watch on the spaceship would show L/v.Aha! But then WE CAN DETERMINE THAT THE BUOY

WAS MOVING AND THE SPACESHIP WAS

STATIONARY!For if the buoy was NOT moving, the readout of its stop

watch would NOT be (L'/v)/<gamma>, but rather, according

to your calculations, L'/v, which is not the same as

(L'/v)/<gamma> !Or suppose you say that the snapshots would show that the

spaceship's stop watch would show (L/v)/<gamma> and the

stop watch on the spaceship would show L'/v.Then WE CAN DETERMINE THAT THE SPACESHIP

WAS MOVING AND THE BUOY WAS STATIONARY!For if the spaceship was NOT moving, the readout of its stop

watch would NOT be (L/v)/<gamma>, but rather, according to

your calculations, L/v, which is not the same as

(L/v)/<gamma> !So depending on the snapshots before us, we can determine

which of the two -- buoy or spaceship -- was stationary and

which was moving.But according to Relativity there should be NO WAY to

determine which of the two -- spaceship or buoy -- was

moving! For Relativity (both Special and General) denies that

there is such a thing as absolute motion.Thus even *your own answer* above, when PROPERLY

calcu- lated, disproves Relativity -- both the Special and

General Theories, both of which require that there can be *no*

such thing as absolute motion!On the other hand, if you claim that MY CORRECTION IS

WRONG, and that your calculations are correct AS THEY

STAND, then the Lorentz time dilation CANNOT have

occurred in either of the stop watches!This TOO would disprove the Theory of Relativity, for

Special Relativity absolutely and categorically DEMANDS

time dilation.Again, it is NOT a matter of opinion. Either Relativity

demands time dilation or it does not! And ALL the books on

Relativity demand time dilation: it is NOT optional!So EITHER WAY the Theory of Relativity is

*mathematically* disproved ... whether by YOUR OWN

answers above, or by MY CORRECTIONS to them.

Do you now admit that this is so? Am I not right? HAVE I

NOT WON MY CHALLENGE?

If you still claim to be right, prove it! Prove that the time

dilation *has* been applied by you, or else prove that time

dilation does *not* have to be applied in Special Relativity!Or prove that even if you apply the time dilation, you will get

the SAME results you gave me above.But I am virtually certain you can NEVER do any of the

above.So here I rest my case. Any impartial mathematical judge

would rule in my favour!(My wife is a lawyer, so I am used to arguing my case -- and

let me tell you that I often end up winning, even against her!)The rest of this e-mail is now superfluous; but for the sake of

being thorough, I shall reply in detail to your other specific

comments.You have added all the following material to your e-mail,

which is absolutely and completely unnecessary, since if you

had read my articles carefully you will have seen that I have

taken care of all the objections you raise:> D2 - Having said that, what happens if the buoy is not supposed

> pointlike, and it has too a bow and a stern with respect to the

> spaceship (it would be much more "honest" to speak of TWO

> SPACESHIPS, with possibly different lengths!)?

>

> S ____________ B -----> (vector velocity v) B° ___ S°

>

> spaceship buoyThis needlessly complicates the problem, since all that is

required is for the spaceship to pass a specific point *on* the

buoy.And as I said, it we have a buoy of 1 m diameter, and accept

the margin of error to be +/- 10^(-10), then the calculations in

*precise numbers* (i.e., in arithmetic and not in algebra) also

remain the same.> Do you wish to compute the time interval which is needed for

> the two external points of the spaceship in order to pass in

> front of the SAME POINT in the buoy, a point in which there

> is a clock? Or do you suppose to want to compute for instance

> the time which goes from the bow B being in front of the

> other bow B° of the buoy, and then the stern S being in front

> of the other stern S° of the buoy? (if you object that these

> remarks are too much "pedantic", since you have introduced a

> buoy of only 1 meter, and a much "longer" spaceship, think

> what happens when you try to do the converse: namely, to

> measure this investigated time interval from the point of view

> of the spaceship! In any case, I repeat it, in mathematics there

> are not long and small, there are only precise computations,

> and possibly, AT THE END, when we are doing physics, we

> can make APPROXIMATIONS, but not at the very

> beginning).Of COURSE in mathematics one must be precise; but all that

needs to be done is to make it clear that the spaceship is

required to pass a specified point *on* the buoy.And it is necessary to specify the margin of error in

PRACTICE.I can re-word the "Challenge" appropriately -- though as I said,

I have already mentioned both in my earlier, longer article: the

point marked "X" is mentioned, in fact, right in the first para-

graph of the Abstract.> D3 - In truth, I understand that you have foreseen this objection,

> and that you have tried to cleverly avoid it, by introducing that

> "suitable mechanism (which can easily be devised)" etc.! But

> where is this mechanism? In B, or in S, or in B°, in S°, or else-

> where? And where is the clock (or better, where are the clocks)

> you use for the measure of time?I have, as you say, already foreseen this difficulty, and have

*described in some detail* all the above, including the

mechanism, in my first article "Single Best Challenge to

Special Relativity".But I see that you have not read the article carefully enough to

understand what I have written (or perhaps I should have

translated the article?)In any case, here are the details for you, here in this very e-

mail, and expressed simply and in words that are easy to

understand:1. The stop watch in the spaceship is exactly at the

*midpoint* of the spaceship. (Nota bene: it is *not* a "clock"

but a *stop watch*, which can be made to *start* and *stop*

by signals. The reason for this will be explained below.)2. The stop watch in the buoy is at the spot marked "X" on

the buoy.3. The mechanism for activating the stop watches is as

follows:A blue laser light shines from the spot "X" on the buoy B, the

direction of this light being at *right angles* to the direction

of relative motion between A and B.This light is so aimed that when the bow of A passes by B,

the light will begin to shine on A.As soon as the blue laser light impinges upon the hull of A,

light sensors located all over the hull of A sense this blue

laser light as long as it continues to shine on A.As soon as it begins shining on A, these sensors send a signal

to activate a stop watch carried on board A -- the stop watch

being located, as mentioned above, exactly at the mid-point of

A.(The exact nature of the signal will also be described below).

And as soon as the sensors detect that NO blue laser light is

shining on A any more, they send ANOTHER signal to the

stop watch to STOP ticking.All the sensors send all their signals to the stop watch carried

on board A *at a fixed and known speed*.This could be accomplished -- just as an example -- by a

system of electrical wires, as follows:An electrical wire of rest length L/2 connects the sensor at the

bow of the spaceship to the stop watch at the mid-point of the

spaceship, and an identical electrical wire of identical rest

length connects the sensor at the stern of the spaceship to the

same stop watch. (Nota bene: there is only ONE stop watch

on board the spaceship!)And electrical wires of rest lengths shorter than L/2 connect all

the other sensors along the hull of the spaceship to the stop

watch.Note that the wires are all *inside* the spaceship, and thus

they are all *stationary with respect to the spaceship*. If the

spaceship contracts, they contract along with it! But they all

contract in equal proportion, so the time taken (in the IFR of

the spaceship) for the signal to travel the length of the wire

from the bow to the stop watch is still identical to the time

taken for the signal to travel the length of the wire from the

stern to the stop watch.(N.B.: "IFR" = "Inertial Frame of Reference".)

And the time taken for the signals to travel from the other

sensors to the stop watch is always less than the time taken

for the signal to travel from the bow and the stern to the stop

watch.Thus while the spaceship is passing the buoy, one or another

sensor on the spaceship is always sending a signal to the stop

watch, and the stop watch is always receiving it, and therefore

continues to tick until the blue light stops shining on A.So when A moves past B, the stop watch located at the

midpoint of A records a time interval *exactly equal* to the

time interval t2 - t1.Of course the stop watch carried at the mid point of A starts

ticking with a fixed *delay* after the bow of A passes by the

spot "X" marked on B, because the sensor at the bow of A

takes a precise amount of time to send its signal, at the speed

at which electrical signals travel, to the mid-point of A, so as

to get the stop watch starting to tick.But then again, this is *exactly compensated* by the fact that

the signal from the sensor at the stern of A takes exactly an

*identical* amount of time to reach the stop watch, and

inform it that the laser light from B has *stopped* shining on

A's hull, and thereby to get the stop watch to stop ticking!And thus the amount of time recorded by the stop watch on A

will be *exactly* equal to the time it takes for A to pass by

the spot "X" marked on B.All this, of course, is to be taken as in the IFR of the

spaceship.(It is because of this *delay* that I changed the timepieces in

my Challenge from "clocks" to *stop watches*. What matters

here is to record the specific time *interval*, and not the

*time*!)And furthermore, as the front end of A passes by the spot "X"

marked on B, an AMBER laser light emanating from the front

end of A, pointed at right angles to the direction of relative

motion between A and B, shines on a light sensor on B

located at the spot marked "X", activating a stop watch on

board B, which is also located at the spot marked "X", just

behind the sensor.And as the rear end of A passes by the spot "X" marked on B,

a GREEN laser light, similarly pointed at right angles to the

direction of the relative motion between A and B, shines on a

light sensor carried aboard B at the spot marked "X", and as

soon as this sensor on B detects this green laser light, it sends

a signal causing the stop watch on B to stop ticking.Thus both the stop watches record the very *same* time

interval, namely t2 - t1. The only difference is that the stop

watch on the spaceship records it with a slight *delay*, that

delay being exactly measurable and even calculable; and thus

for this delay, allowances can be *precisely* made.You see, there is NO PROBLEM AT ALL.

> If they are at some distance from the "mechanism", in any case

> you have to deal with the time which is needed for a signal sent by

> this instrument in order to arrive to the clock. All your efforts not-

> withstanding, you cannot avoid the relativistic speculations in this

> matter, which claim that no signal can travel faster than light, etc.,

> and then assuming that exactly c is the "better speed" (faster) which

> is required for "communications" between your "mechanism" and

> the clock!Ah, but if the *delay* for the signal to reach the stop watch is

*identical*, regardless of whether the signal is sent from the

bow or the stern, the time interval t2 - t1 will *still* be

accurately recorded by the stop watch on the spaceship!(Did you seriously imagine that I had not already thought of

this problem?)Note also that the signals can be of *any* kind, however slow

or fast -- within reason, of course. (In my earlier article I have

described an OPTICAL mechanism to perform the same

function.)Even SOUND TRAVELLING IN A RIGID ROD can attain

the same objective. As long as the rods are carried *inside*

the spaceship, and as long as the time it takes for the signal to

reach from sensor to stopwatch is the same in both the rods, it

would still be possible to have the stop watch record a time

interval EQUAL EXACTLY to the time interval t2 - t1.(Note that it is not necessary to record the time interval t2 - t1

ITSELF, but only a time interval exactly EQUAL to it!)And *even* if the time it takes for the signal to reach from

sensor to stopwatch is *not* the same in both the rods, as

long as the time to for the signal to travel in *each* rod is

known, it is still possible to make the necessary adjustments,

by CALCULATION! The calculations would be a little more

complex, of course, but it is still possible to perform them

fairly easily.Thus it is ALWAYS POSSIBLE to know the time interval t2

- t1, REGARDLESS of the kind of signal used to trigger the

stop watch on board the spaceship. Even if the signal has

different speeds in different parts of the spaceship, as long as

what the speed is, and where, is known with precision, it is

possible to precisely compen- sate for the signal's delay in

reaching the stop watch.Indeed we may locate the stop watch *anywhere* in the

spaceship: as long as the speed of the signal each way is

known with precision, allowances can be precisely made for

the delay(s)!I did not insult your intelligence by explaining all this,

thinking that you, being a professor, would easily understand

that this can be done. I am surprised, indeed, that you raise

this objection!> D4 - I try to explain the problems in yet another way. At last

> you propose to compare TWO measures, with apparently only

> two clocks, but this is simply impossible. If it is possible to

> suppose (or to approximate, if you prefer) the buoy as

> pointlike, and to make the measurement you require with only

> one clock in this case, how can you do the same measurement

> from the spaceship? If there is only one clock in the spaceship,

> say for instance in the bow, then this clock shows some time

> when the bow of the spaceship is in front of the buoy (or of

> the clock which is the buoy), but how can you get a second

> value for the time measured by this clock, the second value

> which is needed in order to compute the required difference?

>

> YOU NEED FOR INSTANCE ANOTHER CLOCK IN THE

> STERN, or even - if you think that it is a different thing (but

> it isn't!) - you need a signal going from the stern to the bow, in

> order to know where the stern WAS in front of the buoy.As I have explained above, measuring the time interval t2 - t1

with one single stop watch is NOT AT ALL IMPOSSIBLE,

as you claim, and has been taken care of by the above

mechanism.And I can describe many other mechanisms to do the same

thing. (I am also an engineer, with over thirty registered

inventions to my credit, many of them in aerospace

technology: so it is very easy for me to do this!)Perhaps you are not an engineer: in which case, please take my

description to any competent engineer and ask him or her!

ANY competent engineer -- and not necessarily one working

in aerospace technology -- will definitely confirm what I say.I think that Einstein, not being an engineer, did not

understand how the measurement of an exact time interval

between events that occur in different widely-separated

locations using one single clock could be done. That is HIS

mistake. But WE, having good engineering skills (or at least

having access to good engineers in our midst), do not have to

make the same mistake HE made!> Here it is exactly

> where relativity comes in, with all his stories about the synchro-

> nization of distant clocks etc. If the clock you are talking about was

> placed instead in the stern of the spaceship, it could register only the

> time when the stern is in front of the buoy, but he cannot "know"

> when the bow was in front of the buoy! Summing up, you need

> TWO synchronized clocks in order to do the measurement you wish,

> both in the spaceship and in the buoy, and you must be more precise

> in specifying which is the time interval you wish to compute, and

> most of all HOW...The absolute and utter nonsense about synchronisation of

clocks in Relativity is just that: ABSOLUTE AND UTTER

NONSENSE!ANY competent engineer knows how to synchronise clocks

that are merely separated (but not MOVING relative to one

another): *he simply allows for the time a signal takes to go

from one to the other* !!!And it can be ANY kind of signal, as long it has a measurable

and constant speed!Perhaps, as I said, you are not an engineer, but I assure you it

can easily be done. Please ask any competent engineer in Italy:

he or she will confirm what I say.As I wrote to a friend of mine only yesterday, criticising Max

Born's famous book "Einstein's Theory of Relativity":[QUOTE]

On pages 228 and 229, Max Born writes:

"From this it follows that absolute simultaneity can likewise

be ascertained in no way whatsoever."... and:

"THERE IS NO SUCH THING AS ABSOLUTE

SIMULTANEITY." [His emphasis, no less!]I mean, *come ON*. Did we not read *Born's own* words on page

225 of his book -- just a few pages before! -- that "the velocity of

light is independent of the state of motion of the observer and has

always the same value c"? If this is indeed the case, what is the

difficulty in making the necessary correction, using the equation

{time = distance / velocity}?There are *numerous* ways to measure distance [in any particular

IFR]: for example, one can use a ruler, or use triangulation, or use

the time taken to send a signal -- such as electrical, or sound

travelling in a rigid rod -- whose speed is actually known. Once the

distance between two objects is known, and the speed of the signal

is also known, one can calculate the *time* it takes for the signal

to get from one clock to the other using the above formula! One

expects even Grade 6 students to be able to do this.Just *what* is Max Born's -- and Einstein's -- difficulty here?

(Maybe they didn't pass their mathematics exams in grade school?)And even if the speed of light were *not* constant, but changed

depending on the conditions (such as for example the direction in

which it travels), as long as the different velocities were known for

the different conditions, the requisite adjustments could *still* be

made! (Maybe not in Grade 6, but at least in Grade 9.)...

For instance, even if the time it takes for the signal to go one way is

different from the time it takes to go the other way, the difference

can be measured by sending the signal both ways along the very

same route.Heck, astronomers have known since long before Max was born

that they can calculate the exact positions of *all* the planets --

and their satellites as well -- at *any* given instant. Not, mind you,

where the planets *appear* to be, but where they actually *are* at

that instant. It is *so-o-o-o-o-o* very easy: *one simply allows for

the time it takes for the signal (which in this case is light) to reach

us from them!*[END QUOTE]

As you can see, the very fact that the planets' actual positions

can be calculated proves that there *can be* simultaneity all

throughout the Solar System. (This was known even

BEFORE Relativity!)I simply do not understand why Relativists keep on repeating

*ad nauseam* that simultaneity is impossible, when

astronomers have known for so long that not only it IS

possible, but is ROUTINELY USED in their calculations.

(Otherwise we would never calculate when eclipses, or

alignments of the planets, would occur!)You see now, your objection to measuring the time interval t2

- t1 with the help of a single stop watch carried on the

spaceship is completely cleared up by simple but competent

reasoning.> E - Now that it has become clear, I hope, how your question

> SHOULD HAVE BEEN FORMULATED, let us go on,

> choosing (between many), one possibility, namely ask to

> measure the time which passes - either from the point of view

> of the buoy, or of the spaceship (in both cases you need a

> whole space-time coordinate system associated with the two

> objects: there is no difference at all with the attempt of

> introducing only TWO objects, clock and mechanism!) - from

> the two events which correspond to the coincidences:

> bow/bow, stern/stern. In other words, I say that you have to

> pose your question for instance in the following way, and that

> then there is an easy not contradictory answer to it:

>

> - One has an inertial spaceship, with some proper length L, and

> ANOTHER ONE, of some proper length M , the relative speed is

> v. One has synchronized clocks on both spaceships [I repeat that it

> is really enough to have TWO of them for each spaceship, so

> FOUR in all], and one asks how much time is needed, from the

> point of view of both spaceships, for completely passing one the

> other.All this is not needed at all. (I do wish you had read my

articles more carefully! And I do hope that you will read THIS

e-mail carefully: I select all my words with great care, and

seldom use a word when a more precise one is available.)As I have shown in my articles and also shown above, *two*

stop watches -- one in the buoy B and one in the spaceship A

-- are QUITE sufficient to measure the time interval t2 - t1;

and a single spot marked "X" on the buoy is ALSO quite

sufficient to represent the point past which the spaceship

passes.And if the margin of error in measurement is 10^(-10), then

the length M is also not needed.And as explained also, it IS possible to precisely measure the

interval t2 - t1 by using a *single* stop watch carried on the

spaceship.So all the rest of your so-called "WELL POSED question" is

totally unnecessary!> Now we have at last a WELL POSED question, and an easy

> answer. The two time intervals are:

>

> L/v + M'/v (from the point of view of the first spaceship -

> with obvious notation)

>

> L'/v + M/v (from the point of view of the second spaceship,

> the "old" buoy) ,

>

> where of course it is L' = L*sqr(1-v^2/c^2) , and M' =

> M*sqr(1-v^2/c^2) (in force of the so called length contraction

> - we do not even need to introduce Lorentz transformations,

> even if of course the length contraction is one of their

> consequences).

>

> In order to decide which time interval is bigger than the other,

> one has just to decide which length is bigger between L and

> M . If one supposes that the first spaceship is longer than the

> second (the ex-buoy), then one has:

>

> (L+M')/v > (L'+M)/v ,

>

> no doubt at all about that, the second time interval is smaller

> than the first.

>

> If one supposes instead COMPLETE SYMMETRY, namely

> L = M (in other words, TWO EQUAL SPACESHIPS), one

> gets at last TWO EQUAL TIME INTERVALS.

>

> As you see, everything is very simple, much more simple for

> instance than introducing a coordinate system with respect to

> an accelerated observer, like one must do when discussing the

> twin paradox (see for instance the paper in the point 4 in the

> above quoted web page, which is unfortunately only written in

> Italian, but there is some comment to this problem even in the

> fifth section of the aforesaid paper about

> "Misunderstandings...").

>

> The question is always the same, which after all is not even

> too much difficult to overcome: it is to understand the

> counterintuitive (but mathematically precise, and then

> "logically possible") relativistic treatment of space and time

> (not to have an "intuition" of it with the common thought

> space and time categories, which would be of course rather

> impossible)...I hope you have understood by now that all this is COM-

PLETELY unnecessary.> F - I have talked too much - and with my bad English this is

> perhaps harmful for a better understanding - yet I have not said

> everything I could have said. For instance, in some lecture to stu-

> dents, I could have shown even how one could have connected

> length contraction and time dilation in the proposed exercise.But it is not an *optional* matter to connect length

contraction and time dilation! In Relativity it is absolutely

*necessary* to do so.The Lorentz transformations are nothing WITHOUT time

dilation. By ignoring time dilation you have ignored the

Lorentz transfor- mation equations.Please excuse my being blunt, but in science one can be a

respecter only of the Truth, and not of persons. If anything

anyone tells me is untrue -- even if it be you who are older

than I am -- then I shall not mince words in condemning what

is said!For like you, I also believe that "amicus Plato, sed magis

amica veritas".But I have great respect for all persons *qua* persons, as also

for you. If I am blunt and contemptuous, it is only towards

the untruth: not towards the person in whose mind it may

temporarily reside ... because, of course, everyone is at liberty

to change his or her mind!And please feel free to reciprocate: for I am never afraid or

reluctant to change my mind when the truth is clearly shown

to me!> In any case I hope:

>

> F1 - first of all, to not have made mistakes (which should

> have to be considered my personal mistakes, and not

> mistakes/contradictions in the theory!);

>

> F2 - that you have understood where are the mistakes in your

> posing the question;

>

> F3 - that you have understood how could one pose a good

> question, between many possible of the "same nature";

>

> F4 - that you will not require to me endless discussions, as it

> happened many times to me with people not willing to

> understand relativity (there is an example of such useless

> discussions in my web site, with Percival, or Galeczki, etc.,

> all people who do not know enough well relativity);

>

> F5 - to have been useful to you at least from an ethical point

> of view, showing that one could indeed answer to all people,

> even when one believes that the required effort of writing will

> not produce anything "useful".Firstly, and most importantly, I hope you understand that you

have made a BIG MISTAKE in your calculations in ignoring

the time dilation. I hope that you now UNDERSTAND and

ACCEPT your mistake in answering the question -- as the

honest person I am sure you are.Or else can you show me with what justification time dilation

*may* be left out of the calculations when calculating a

problem in Special Relativity? I doubt very much that you can

do so.Also I hope that you now understand and accept that your

criticism of my posing of my question is totally unfounded,

because you did not read my articles carefully before

answering me.But as I said, even though you did not read my two articles

carefully enough, YOUR OWN WORDS ABOVE HAVE

DISPROVED SPECIAL RELATIVITY: for you did not

apply the Lorentz time dilation, which is *absolutely

REQUIRED* by Special Relativity.And even if the time dilation IS applied, it *still* disproves

Relativity, for then it becomes possible to determine which of

the two -- spaceship and buoy -- must have been moving and

which must have been at rest! And this contradicts the

principle according to which all rectilinear motion must be

relative and not absolute.So either way your answer -- as it stands, OR as corrected by

me -- disproves the Special Theory of Relativity.I hope you see this clearly now?

If you still do not see, I can explain it in even greater detail.

As for an ongoing discussion, there is never a need for

discussion between NON-scientists, all of whom are entitled

to their own views, however irrational; but as you yourself

have written, it is the absolute obligation of any *scientist* to

answer criticism -- for science *must* be rational and logical.Thus in science, discussion cannot, unfortunately, be avoided.

No one who claims to be a scientist can avoid answering to

criticism.But if you do have the courage of your convictions to discuss

Relativity with me further, let us have no more silly mistakes,

okay? Please make sure that you have thoroughly gone over

everything I have written, and also over your reply; and

remove from your reply *all* the mistakes that you are able to

remove.Please also do not insult my intelligence: I am not a first-year

university student, but a person almost 60 years old, well-

travelled and well read, and extremely highly educated, with

several first-rate books and many papers and inventions -- and

that too, in several different disciplines -- to my credit.> I am at your disposal with sincere friendship for a possible next

> round, but please only one, no more (and please spare to me a com-

> ment about your new "logical" "Short Reductio-ad-Absurdum Refu-

> tation of Special Relativity", which is no better "idea" than the one

> I have discussed until now); time is indeed a precious thing, mostly

> at the end of one's life, and there are better, most productive, ways to

> spend it...Surely there is no better or more productive way to spend

one's time on earth -- at whatever stage in one's life one may

be -- than seeking the Truth, and clearing one's mind of

untruth?If even on my death-bed I am able to learn something new, or

to find out that something I had thought of as truth was

untruth, I would be very, very happy.And as to these following words of yours:

> ... let me tell you something which has

> been said to me by a colleague here in Perugia - one of the most com-

> petent expert in relativity, I believe. He told me that he cannot

> understand why people understands that they have to pay lawyers,

> physicians, etc., in order to have their opinion, even for matters which

> require less knowledge, time and commitment than scientific ones, but

> then the same people suppose instead that scientists should loose their

> time in answering to amateurish ill-posed questions. He added that a

> teacher could do this work only for the benefit of his own students, but

> if a student insists in not understanding then he gets fired (in Italian:

> BOCCIATO).... direi ch'è *Lei* chi sta bocciato qui! [Translation: "I'd say

it is YOU who have failed the exam here!]In any case, I have approached you, NOT as a student, but as a

Challenger: an adversary. I would approach a lawyer in a law

court likewise! I do not have to PAY my opposing lawyer in a

law court -- on the contrary, it is expected either that my

challenge be fully and satisfactorily answered, or else that the

opponent concede defeat -- that is, if my opponent has any

self-respect and honour at all (as I hope and believe you have!)So: give me your counter-arguments against my arguments

here -- if you have any -- OR ELSE CONCEDE DEFEAT!(But please, no more silly mistakes like you have made

above! And no more misreading or non-reading of my clearly-

written words!)>Once again best wishes, from yours most sincerely

>

> correspondent perugino UB

>

> [1] There were even other things which I started to discuss in my

> preliminary attempts of answer, for instance about what you have

> said in one of yours intermediate mails:

>

> > There is no need for a "third (inertial) observer C"...

>

> I do agree about that, and as a matter of fact you avoided to do it

> in the second formulation of your "challenge", but it was you that

> in the first formulation of it had written:

>

> > Imagine two objects, A and B, in rectilinear motion past one another.Yes, this was a small mistake in my first article. I shall

correct it. (In English, one sometimes says "Imagine" when

one intends to say "Suppose that there is". But as you

indicate, this is a bit imprecise.)> In order to "imagine" this situation, one needed to introduce the point

> of view of a third observer! The fact is that there are always many different

> manners to describe in mathematical terms the "same" physical situation,

> since the "code" which translates from "reality" to mathematics is rather

> flexible, but in any case the answers must always be the same!Right.

But as you have seen from the above, it does not matter in the

present case, because the calculations above have shown that if

length contraction *and* time dilation are accepted as valid, it

*is* possible to distinguish between the two -- spaceship or

buoy -- as to which of them was moving and which was

stationary.In other words, time dilation, on the one hand, and the

principle of all rectilinear motion being relative, on the other,

are contradictory to one another. This, essentially, is what

Dingle says too. What I have done is to provide a

mathematical PROOF of Dingle's argument.Philosophically, however, there is an EVEN MORE FUNDA-

MENTAL objection to the Lorentz transformation: namely,

that they CANNOT be derived from the axioms of

mathematics and the propositions and postulates of Euclidean

geometry!The *only* transformations that can be derived for *Euclidean

space* from the axioms of mathematics -- such as those of

Peano, or those expounded by Zermelo and Fraenkel (later

extended by John von Neumann) -- and the propositions and

postulates of Euclidean geometry, are the GALILEAN

transformations. The Lorentz transformations CONTRADICT

the Galilean, and therefore *cannot* be a part of mathematics

and Euclidean geometry.(Nota bene: the Lorentz transformations refer to *Euclidean*

space, and so there is no need to include non-Euclidean

geometry in this argument!)You being a mathematician should understand that *every

mathematical theorem must be constructed from previously-

proven theorems, or from the axioms.* And *every

geometrical theorem in Euclidean geometry must be

constructed from the propositions and postulates of Euclidean

geometry, or from previously proven geometrical theorems*.

These are absolute and SINE-QUA-NON conditions in

mathematics and Euclidean geometry.But there is *no* way the Lorentz-Fitzgerald transformations

can be constructed from either previously-proven mathematical

and geometrical theorems, or from the axioms of mathematics

and the propositions and postulates of Euclidean geometry.

But there *is* a way to construct the Galilean transformations

therefrom!Therefore the Lorentz-Fitzgerald transformations MUST be

*mathematically* and *geometrically* invalid, at least from

the point of view of Euclidean geometry (and that is the only

kind of geometry used in Special Relativity, strictly

speaking).This, too, would disprove Special Relativity from a mathe-

matical point of view -- contrary to your claim that Special

Relativity is "mathematically quite a consistent theory".Indeed I intend to write a second Challenge -- of course not a

"Simple" one, and understandable only by mathematicians --

viz., to try and derive the Lorentz-Fitzgerald transformations

from the axioms of mathematics and the definitions, axioms

and postulates of Euclidean geometry. This is *impossible*,

because the Galilean transformations *can* be derived

therefrom, and they *contradict* the Lorentz transformations!By the way, this would not, as you say, result in the

conclusion that "then the WHOLE MATHEMATICS would

be affected by contradictions" -- it would just mean that the

LORENTZ TRANSFORMATIONS must affected by

contradictions!Yours always in sincere friendship,

Ardeshir <http://homepage.mac.com/ardeshir/education.html>.

____________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

[Translation: Argument Against the Special Theory of Relativity]

Date: Sun, 26 Aug 2001 09:25:03 +0200

From: umberto bartocci <bartocci@dipmat.unipg.it>

To: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7

Dear Mehta,

thanks for your long and warm ("appassionata") answer, to

which I answer immediately, at least in connection with the

most important part, just one piece at a time. But let me say

that your mail shows that we have got a kind of intellectual

friendship, since we share a common dislike for the work of

some "great" scientists which the whole world, quite on the

contrary, worships.SO, MAY I PLEASE ASK YOU TO LET ME HAVE A

COPY OF YOUR "CRITIQUE OF GOEDEL'S

THEOREM"?I am very interested about it, because I think that one MUST

criticise Goedel, even if until now I did not know of any one

who attempted to do it, and I would not even know where to

start from in order to do that!As far as the case against relativity is concerning, from a quite

HONEST point of view, there is unfortunately no doubt at all

about who is right and who is wrong. After all, if you were

right, there would be no question to discuss between each

other, and you should have to discuss with the greatest experts

in the field of theoretical physics (and, believe me, I would be

very happy to see relativity down, even if I had to consider

myself "bocciato"!).As a matter of fact, your objection:

> But I am very, very sorry to see that you have made a BIG

> MISTAKE -- namely, you have *completely* ignored

> time dilation!is wrong. I must say sincerely that these words show to me

that you have not got a full understanding of relativity, as I

said does indeed happen with ALL the pseudo-arguments

against relativity (and this does not mean that one is not

clever enough, but just that unfortunately he has to study

more).As you know very well, Minkowski said since the beginning,

that space and time "are doomed to fade away into mere

shadows, and only a kind of UNION of the two will preserve

an independent reality". In your criticism to my simple answer

to your easy challenge, you "separate" in some sense space and

time, while in relativity there exists ONLY ONE space-time

interval, and the SAME phenomenon could be explained - as I

shall soon prove - FROM A RELATIVE POINT OF VIEW

once with a length contraxtion, another with a time dilation.I shall try to explain better all these assertions, after all I am a

professor, and I have explained these things many times to

students, even if writing them, and in English, is rather

difficult for me, and it takes too much time. I shall analyze

your objection, splitting it in order to find that it is grounded

on the following assumptions:1 - you point out the existence in relativity of one "length

contraction" and of one "time dilation" (which is in some

sense correct);2 - you pretend to see BOTH of them at work in the situation

we are studying (which is in some sense correct);3 - you see that in "my" computation there is only one of the

two, WRONG, and you pretend to see even the other, RIGHT.The fact is that it is you who sees only one of it, and not the

two together! In the computation that your challenge required

from me, and which I have fully satisfied, THERE ARE

BOTH length contraction and time dilation, DEPENDING

FROM THE POINTS OF VIEW. One must simply be able

to interpret the result which is the only one possibly correct. I

shall try to show you where you have to look for finding the

time dilation in the same and one formula.Now that we have made a complete understanding of the

physical situation, we must say that we have only one

phenomenon to study in Minkowski space-time, namely we

have three different world lines, two of which are parallel,

which intersect, something of the following kind:

/ / !

/ / !

/ / !

/ / !

/ / !stern S bow B buoy X

(the time orientation is from top to down).

The world lines of the bow of the spaceship and of the buoy,

let us call it in the order B and X , thought of as single

points (no worry for instance about the fact that a "real clock"

is not pointlike, we are doing simple "theory"!), intersect in

some event E1 ; the lines of the stern S of the spaceship and

of the buoy X intersect in some event E2 . We do not need

to introduce more than that.Then one can ask which is the proper time elapsed WITH

RESPECT TO X from E1 to E2 . The situation is like that

from X's point of view:

______________ -------> °

S B vector velocity X

the spaceship the buoy is "still"

is "moving"and the computation is very easy, I repeat it. If the proper

length of the spaceship is L, then L' = L*sqr(1-v^2/c^2)

(the star * stands for product) is the length of the spaceship

with respect to X , and the required time is (MUST BE!):(1) L'/v

(here v is the scalar speed, the "module" of the vector velocity).

Conversely, from the point of view of the spaceship, we

have a physical situation like the following one:______________ <------- °

S B vector velocity X

the spaceship the buoy is "moving"

is "still"and WE CANNOT ASK EXACTLY the same question as

before, since E1 is an event in the life of B , and E2 is an

event in the life of S ! So, there is no possibility to ask a

difference between proper times of anybody!!But we can of course suppose to have synchronized clocks

aboard of the spaceship (this is exactly the meaning of the

words in my last mail, about the role of your "mechanism"),

and we can ask which is the time which is needed for the buoy

to pass the spaceship, FROM THE POINT OF VIEW OF

THE SPACESHIP, which means from the point of view of a

coordinate system associated with the spaceship (we could

better think of X as a meteorite). Well, since the length of

the ship is L , and the module of the velocity does not

change, this time MUST BE now equal to:(2) L/v .

Let me remark that relativity is not so stupid to give different

answers from these ones which I have given to you!In conclusion, once again, you have got the two times that

you challenged me to compute. The second time, (2), is surely

GREATER than the first time (1), at least if v is not zero.

Now you have got all the answers to all your questions. Try

to ask to do these easy computations to any other good

physicist, and you will obviously get the same answers (if

not, he is not a good physicist - or at least he does not know

enough relativity).Now you object: I see length contraction at work in (1) -

CORRECT - so you must be wrong, because you forgot the

time dilation. This is WRONG, since in (1) there is even a

time dilation. As a matter of fact, I said it already at the end

of my previous mail:"I have talked too much - and with my bad English this is

perhaps harmful for a better understanding - yet I have not said

everything I could have said. For instance, in some lecture to

students, I COULD HAVE SHOWN EVEN HOW ONE

COULD HAVE CONNECTED LENGTH CONTRACTION

AND TIME DILATION IN THE PROPOSED EXERCISE."In order to show that, let us put the thing in the following

terms. The people aboard of the spaceship measure the time

needed for the meteorite to pass, and they claim that it is L/v ,

good. Somebody aboard of the meteorite says that this time is

not L/v , but it is LESS than that, because he measures L'/v ,

good once again. Well, this assertion from the point of view of

the time elapsed for the meteorite is exactly what the crew of

the spaceship would interpret as the time dilation occurring

"aboard of the meteorite"!The proper time of some phenomenon (between any two

events E1 and E2 in the world line of some "observer") is

defined as the LEAST value of this time with respect to all

inertial observer, in the same way as the proper length is, on

the contrary, the GREATEST value etc. etc.. This proper time

is connected to the measure of the "same time-interval" with

respect to other coordinate times by the well known relation:proper time interval = sqr(1-v^2/c^2) * coordinate time interval

while the corresponding formula for lengths is the "inverse":

proper length * sqr(1-v^2/c^2) = coordinate (or "apparent") length.

Well, in (1) you have got exactly the first formula at work:

proper time of the phenomenon we are studying, namely the

passing of the meteorite X = sqr(1-v^2/c^2) * coordinate time

interval of the same event = sqr(1-v^2/c^2) * (L/v) .As a matter of fact, the coordinate time interval of this

phenomenon is L/v , as we said in formula (2) , and if you

write the previous formula "associating" differently (a shifting

of parentheses), you find exactly (L*sqr(1-v^2/c^2))/v , and

the numerator of this fraction (ratio) is L' , the "contracted

length"...That is to say, in (1) time dilation and length contraction

THERE ARE BOTH PRESENT, I hope that you will be

satisfied by that, since this is the simple easy truth.I could say even more, showing how one could explain even

the "time dilation" which happens (must happen!) in the time

of spaceship as "seen" from the point of view of the meteorite

(until now we had only a "time dilation" in the meteorite from

the point of view of the spaceship).This is just a bit more

difficult exercise for students, which requires a little attention,

and an explicit use of Lorentz transformations. I avoid to do

that, thinking that I have answered enough to your criticism,

but if you want I can send to you the complete "solution" - it

is one argument that I have experienced most amateur people

do not know precisely, and then they get in other "paradoxes",

because they are not able (for instance Dingle was unable to

see it, and he was not even an amateur!) to see the

COMPLETE SYMMETRY which is now at work (of course,

there is no possibility to see even the length contraction of X

, which we have supposed pointlike).Ciao, I HOPE TO READ SOON YOUR OBJECTIONS

AGAINST GOEDEL, perhaps they are more useful than the

ones against relativity (I am sympathetic with these attempt to

show that relativity is wrong, at least they show that one has

understood that to do this is a most important task, but one

must not underestimate the problems), one never knows,I return the expression of a sincere friendship,

UB

Umberto Bartocci

Dipartimento di Matematica

Universita' di Perugia

06100 - Italy

http://www.dipmat.unipg.it/~bartocci

________________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

[Translation: Argument Against the Special Theory of Relativity]

Date: Mon, 27 Aug 2001 20:53:22 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8

Caro Professore:

Grazie tanto per la Sua risposta!

Però potrà vedere dal sottoscritto che ANCORA una volta Lei sbaglia.

[Translation:

Thank you very much for your answer.

However, you will be able to see from what is written below

that ONCE AGAIN you are mistaken.](By the way: If you prefer to write in Italian, and if you accept my answers in

English, I am quite agreeable, if it will make your life easier!)You wrote:

> Dear Mehta,

>

> thanks for your long and warm ("appassionata") answer, to

> which I answer immediately, at least in connection with the

> most important part, just one piece at a time. But let me say

> that your mail shows that we have got a kind of intellectual

> friendship, since we share a common dislike for the work of

> some "great" scientists which the whole world, quite on the

> contrary, worships.Indeed! The more "dogmatic" the theory, the more it SHOULD

be criticised, because if it cannot meet strict and logical criticism,

it must have been false!> SO, MAY I PLEASE ASK YOU TO LET ME HAVE A

> COPY OF YOUR "CRITIQUE OF GOEDEL'S

> THEOREM"?

>

> I am very interested about it, because I think that one MUST

> criticise Goedel, even if until now I did not know of any one

> who attempted to do it, and I would not even know where to

> start from in order to do that!Anyone can download my book in PDF format (i.e., *Adobe

Acrobat* format) from my Home Page<http://homepage.mac.com/ardeshir/education.html>,

... and print it out on a printer.

The same file is also available at my Web site from the following

URL: <http://homepage.mac.com/ardeshir/Critique_Of_Goedel.pdf>.Unfortunately it is not possible to publish this book in

HTML format because of the way Goedel has written his

formulae, which requires elaborate type- setting. (Even then I

have had to modify the formulae for typesetting with

Microsoft Word, which does not allow them to be typeset

EXACTLY the way Goedel does in his original German

Paper.)I warn you, however, that since it is a full book, it is a LONG

download -- almost 1 MB. And it takes a long time to read

also. (Part 1 is fairly short and easy, however, and contains

most of the more important arguments against the Theorem.

But the entire book is one SINGLE download, unfortunately.)One reason why the book is long is that I have quoted almost

all of Goedel's *own words* verbatim, and criticised them,

sometimes sentence by sen- tence, and often paragraph by

paragraph, so as to make sure that there is NO

misunderstanding as to WHAT I am criticising and WHY I

am criticising it. So my book includes almost the FULL text

of Section 1 and Section 2 of Goedel's 1931 paper entitled

"On Formally Undecidable Propositions of Principia

Mathematica and Related Systems", which is more popularly

known as "Goedel's Theorem" or sometimes as "Goedel's First

Incomplete- ness Theorem".But I have to admit that the ideas were not ORIGINALLY

mine. I first read them on the Web, written in not very good

English, by Ferdinand Romero, who at present lives in

Argentina, and who later became my colleague and good

friend. He does not write English very well, and although he

does write it, it is somewhat hard for most English-speaking

readers to understand from his English words exactly what he

is trying to say.Still, as a LOGICIAN he is superb: indeed the very BEST I

have ever known.And since I know Italian, and for that reason can to some

extent understand Spanish also, I re-wrote, for my own

benefit, Ferdinand's arguments, using better English, trying to

understand what he might have been thinking in Spanish as he

tried to write in English. And from re-reading his arguments

in better English, I easily understood that he MUST be right.So we two collaborated to write our book, using good English

-- I have had many years of experience as a writer and editor in

English, and of course Ferdinand checked to see that there

were no logical mistakes in what I wrote.The book took about 8 months to write, and was finished -- or

almost finished -- this past winter. (There are still a few

mistakes of mine that need to be corrected, according to

Ferdinand's comments, but they are comparatively minor. You

may say that the presently-available edition is the "pre-press"

edition!)In fact originally I thought I would do the same thing with

Relativity, quoting Einstein's OWN words and disproving

them. Perhaps I shall do so some day.But Special Relativity (not General) is TEN TIMES MORE

EASY to disprove than Goedel's Theorem! So I thought I

would simply write a few short articles against Relativity

first, and publish them on my Home Page.But I see that many people -- including your good self -- are

not able to understand even my simple arguments. So one day

I shall DEFINITELY disprove Special Relativity IN EVERY

SINGLE DETAIL, quoting EVERY WORD of Einstein's

1905 paper and disproving what he says sentence by sentence

or at least paragraph by paragraph, as Ferdinand and I have

done with Goedel.(For his information, I am separately sending to Ferdinand a

copy of this first part of my e-mail to you.)So now on to Relativity. As I said, you are AGAIN mistaken,

as follows:> As far as the case against relativity is concerning, from a quite

> HONEST point of view, there is unfortunately no doubt at all

> about who is right and who is wrong. After all, if you were

> right, there would be no question to discuss between each

> other, and you should have to discuss with the greatest experts

> in the field of theoretical physics (and, belive me, I would be

> very happy to see relativity down, even if I had to consider

> myself "bocciato"!).

>

> As a matter of fact, your objection:

>

> > But I am very, very sorry to see that you have made a BIG

> > MISTAKE -- namely, you have *completely* ignored

> > time dilation!

>

> is wrong. I must say sincerely that these words show to me

> that you have not got a full understanding of relativity, as I

> said does indeed happen with ALL the pseudo-arguments

> against relativity (and this does not mean that one is not

> clever enough, but just that unfortunately he has to study

> more).

>

> As you know very well, Minkowski said since the beginning,

> that space and time "are doomed to fade away into mere

> shadows, and only a kind of UNION of the two will preserve

> an independent reality". In your criticism to my simple answer

> to your easy challenge, you "separate" in some sense space and

> time, while in relativity there exists ONLY ONE space-time

> interval, and the SAME phenomenon could be explained - as I

> shall soon prove - FROM A RELATIVE POINT OF VIEW

> once with a length contraction, another with a time dilation.

>

> I shall try to explain better all these assertions, after all I am a

> professor, and I have explained these things many times to

> students, even if writing them, and in English, is rather

> difficult for me, and it takes too much time. I shall analyze

> your objection, splitting it in order to find that it is grounded

> on the following assumptions:

>

> 1 - you point out the existence in relativity of one "length

> contraction" and of one "time dilation" (which is in some

> sense correct);

>

> 2 - you pretend to see BOTH of them at work in the situation

> we are studying (which is in some sense correct);

>

> 3 - you see that in "my" computation there is only one of the

> two, WRONG, and you pretend to see even the other, RIGHT.

>

> The fact is that it is you who sees only one of it, and not the

> two together! In the computation that your challenge required

> from me, and which I have fully satisfied, THERE ARE

> BOTH length contraction and time dilation, DEPENDING

> FROM THE POINTS OF VIEW.Parenthetically, this is a mere ASSERTION on your part: there

is no DEMONSTRATION of it anywhere in your e-mail.(See also below.)

One must simply be able

> to interpret the result which is the only one possibly correct. I

> shall try to show you where you have to look for finding the

> time dilation in the same and one formula.

>

> Now that we have made a complete understanding of the

> physical situation, we must say that we have only one

> phenomenon to study in Minkowski space-time, namely we

> have three different world lines, two of which are parallel,

> which intersect, something of the following kind:

>

> / / !

> / / !

> / / !

> / / !

> / / ! >

> stern S bow B buoy X

>

> (the time orientation is from top to down).

>

> The world lines of the bow of the spaceship and of the buoy,

> let us call it in the order B and X , thought of as single

> points (no worry for instance about the fact that a "real clock"

> is not pointlike, we are doing simple "theory"!), intersect in

> some event E1 ; the lines of the stern S of the spaceship and

> of the buoy X intersect in some event E2 . We do not need

> to introduce more than that.Now here you have cunningly shifted the argument from the

Lorentz transformation to Minkowski space-time, which is

NOT a part of Einstein's ORIGINAL 1905 formulation of the

Special Theory of Relativity -- for note that Minkowski

published his first paper on Relativity only in 1908, three

years LATER. (You are playing the part of the *advocatus

diaboli* very diabolically!)But never mind: I can show you how you are mistaken HERE

TOO.I am indeed sorry to see that it is YOU who do not have a full

understanding of Relativity. If you had, you would know that

for the scenario described in my challenge, there is not just

ONE possible Minkowski diagram, but at least TWO (or more

accurately, one can have an INFINITE NUMBER of

Minkowski diagrams, depending on which IFR is used for the

orthogonal axes -- x horizontal and ct vertical -- in the

diagram!)In particular, you should also know that in a Minkowski

diagram the PRIMED axes -- i.e., those SLANTED format,

for ct' and x' respectively -- are ALWAYS those of the

MOVING IFR, and the UN-PRIMED axes -- i.e., those in in

exact ORTHOGONAL format -- are ALWAYS those of the

STATIONARY IFR. (If this is NOT done the time dilation

solutions come out wrong -- the moving clocks tick *faster*

than the stationary ones!)So ALTHOUGH it is true that in a Minkowski diagram one

can show BOTH time dilation AND length contraction

SIMULTANEOUSLY, because we are using space-time

combined, and not space and time separately, the primed co-

ordinates once gain indicate WHICH of the two IFRs was

moving!So the question STILL remains: WHICH Minkowski diagram

is the one according to which the two stop watches will read?

Because depending on which diagram you use, YOU WILL

GET DIFFERENT TIME DILATION RESULTS!If you draw a Minkowski diagram using the IFR of the

BUOY for the ortho- gonal x and t axes, then YOUR diagram

is certainly ONE of the possible correct ones.But if you draw a Minkowski diagram using the IFR of the

SPACESHIP for the orthogonal x and t axes, then the

diagram CHANGES, and becomes something like this:! ! \

! ! \

! ! \

! ! \

! ! \ <

stern S bow B buoy X direction of velocity vectorNote that according to *this* diagram, if you WORK OUT

THE TIME INTERVALS which should be indicated by the

readouts of the TWO snapshots, THOSE two readouts would

be DIFFERENT from the two readouts computed from the

Minkowski diagram given by you!Indeed in the two diagrams even the DIRECTION of the

velocity vector is reversed.So AGAIN we can determine, if we agree as to which diagram

is the one which will determine the readouts of the two stop

watches, which of the two -- spaceship or buoy -- was

moving and which was not -- and also IN WHICH

DIRECTION the movement was taking place! But this, as I

said, contradicts the Principle of Relativity.So one way or the other -- <Lei sta "bocciato">.

Have I not made you very HAPPY, though? For this

PROVES that Relativity IS mathematically flawed!***

And one more thing: you have not actually USED the

Minkowski diagram to demonstrate clearly just WHAT the

two readouts should show!Instead you have used the LORENTZ TRANSFORMATIONS

for calculating this -- and you have NOT shown the LINK

between the Lorentz transformations and the Minkowski

diagram for doing this!So in the matter of THOROUGHNESS of showing DETAILS

of your mathematics also, you are very, VERY lacking.Indeed if a STUDENT of yours showed you an answer like

you have shown me, WITHOUT giving details of his

calculations, you would definitely have him "bocciato"!So you are TWICE "bocciato": first, for giving the wrong

answer, and the second time for giving an answer without

giving FULL DETAILS as to how you reached your

conclusions!You then added:

> Then one can ask which is the proper time elapsed WITH

> RESPECT TO X from E1 to E2 . The situation is like that

> from X's point of view:

> ______________ -------> °

> S B vector velocity X

> the spaceship the buoy

> is "still" is "moving"

>

>

> and the computation is very easy, I repeat it. If the proper

> length of the spaceship is L, then L' = L*sqr(1-v^2/c^2) (the star *

> stands for product) is the length of the spaceship with respect to X ,

> and the required time is (MUST BE!):

>

> (1) L'/v

>

> (here v is the scalar speed, the "module" of the vector

> velocity).Now you RE-INTRODUCE the Lorentz transformations into

the picture. But then WHY did you give the MINKOWSKI

DIAGRAM above, if as you say about your Minkowski

diagram: "We do not need to introduce more than that"?You see, you HAVE NOT shown how a SINGLE Minkowski

diagram is used to compute the TWO time interval readouts

which should be photo- graphed by the cameras -- the one in

the spaceship and the one in the buoy.Although it is possible, certainly, you have not shown it --

maybe because to show it in an e-mail is very hard: we do not

have graph paper.But since the same results can be obtained using the Lorentz

transformations, which are easy to show without graph paper,

why should we not use them?So forget the Minkowski diagrams: they were never a part of

the ORIGINAL Theory of Special Relativity anyway.

Remember that in the ORIGINAL 1905 Special Theory of

Relativity ONLY the Lorentz-Fitzgerald transformation is

used. Don't confuse the issue by using in one single argument

both the Lorentz-Fitzgerald transformations *and* the

Minkowski world-lines!You should be CONSISTENT in your argument. You should

use the Minkowski world-lines *alone*, OR the Lorentz-

Fitzgerald transformations *alone*. But you should use

ONLY ONE OF THEM. Otherwise you will be confusing the

issue: not only in MY mind, but more importantly, ALSO IN

YOURS.(And THAT, I suspect, must have been your mistake ALL

THESE YEARS.)That is because ANY Minkowski diagram shows BOTH time

dilation AND length contraction, whereas EACH Lorentz

transformation equation shows only ONE of the two:

EITHER length contraction OR time dilation!You added:

> Conversely, from the point of view of the spaceship, we have

> a physical situation like the following one:

>

>______________ <------- °

> S B vector velocity X

> the spaceship the buoy is "moving"

> is "still"

>

> and WE CANNOT ASK EXACTLY the same question as before,

> since E1 is an event in the life of B , and E2 is an event in the life

> of S ! So, there is no possibility to ask a difference between proper

> times of anybody!!Now here is ANOTHER one of your mistakes: because even

though E1 is an event in the life of B and E2 is an event in

the life of S, since B and S are NON-MOVING WITH

RESPECT TO ONE ANOTHER, it IS possible to know the

exact TIME DIFFERENCE (in the IFR of the spaceship)

between the "life" of the bow B and the "life" of the stern S!So if the time difference between the "life" of the bow B and

the "life" of the stern S is, say, t seconds -- i.e., if any event

occurs in the "life" of the stern S t seconds after it occurs in

the life of the bow B, and vice versa -- then if the event E2

occurred in the "life" of the stern S occurred at a time of T

seconds, then in the IFR of the spaceship, that event will have

occurred in the "life" of the bow B at a time of (T-t) seconds!Thus it IS possible to ask what is the difference between E2

and E1 in the life of the bow B compared with the same

interval in the life of the stern S.Do you STILL not see how this is accomplished? If not, I can

explain even further. (Or else, as I wrote earlier, you could ask

an engineer.)O forse sarà meglio se scriverò in italiano?

Anche se E1 è un evento nella "vita" di B, ed E2 è un evento

nella "vita" di S, per causa che B ed S non stanno muovendo

fra loro, è certamente POSSIBILE di sapere qual'è l'intervallo

temporale (nel'IFR dello 'spaceship') fra la "vita" di B e la

"vita" di S!E' vero o no, questo che ho scritto?

E si è vero, possiamo sempre SAPERE l'intervallo temporale

fra E2 ed E1, nell'IFR dello 'spaceship'!Infatti, come Lei ha scritto, è possibile avere due orologi

sincronizzati nel IFR dello spaceship, l'uno a B e l'altro a S --

il quale risolve tutti i problemi.[TRANSLATION:

Or perhaps it might be better if I were to write in Italian?

Even if E1 is an event in the "life" of B, and E2 is an event in

the "life" of S, since B and S are not moving relative to one

another, it is certainly POSSIBLE to know the time interval

(in the IFR of the spaceship) between the "life" of B and the

"life" of S!Is this not true?

And if it is, we can always KNOW the time interval between

E1 and E2, in the IFR of the spaceship!In fact, as you have written, it is possible to have two

synchronised clocks in the IFR of the spaceship, one at B and

the other at S -- which resolves all problems.END OF TRANSLATION]

> But we can of course suppose to have synchronized clocks

> aboard of the spaceship (this is exactly the meaning of the

> words in my last mail, about the role of your "mechanism"),

> and we can ask which is the time which is needed for the buoy

> to pass the spaceship, FROM THE POINT OF VIEW OF

> THE SPACESHIP, which means from the point of view of a

> coordinate system associated with the spaceship (we could

> better think of X as a meteorite). Well, since the length of

> the ship is L , and the module of the velocity does not

> change, this time MUST BE now equal to:

>

> (2) L/v .

>

> Let me remark that relativity is not so stupid to give different

> answers from these ones which I have given to you!(I see you have changed the nomenclature, saying "meteorite

X" instead of "buoy B", and have used the letter "B" for

"Bow" -- i.e., of the spaceship A. Let it be so from now on --

though it would be best in a SINGLE argument to stick to a

SINGLE nomenclature and a SINGLE set of alphabetic letters

throughout.)Now as I pointed out there must be at least TWO Minkowski

diagrams, depending on which IFR is used, and each of them

give two DIFFERENT answers to what the snapshots should

show!Indeed, as I said, there must be an INFINITE NUMBER of

solutions to the challenge -- and this is logically impossible.And EVEN if you use the Lorentz transformations ALONE,

what you claim above, namely (2) L/v as being the interval E2

- E1 in the IFR of the spaceship, is correct ONLY if the

spaceship is considered to be NON-MOVING.Note: L is the *rest-length* of the spaceship. And v is always

the same. So the rest-length divided by v gives the time

between E2 and E1 as observed by a stop watch in a NON-

MOVING IFR!But the spaceship is MOVING relative to the "meteorite"!

If the spaceship is considered on the contrary to be MOVING,

the time given by its stop watch CANNOT be L/v, as you

claim, but LESS than that, because of the TIME DILATION

which that movement causes to appear on the readout of its

stop watch, and also because its LENGTH will have

contracted!But note that a meteorite moving past a stationary spaceship

cannot affect the stop watch on board the meteorite! It is only

if the spaceship is *actually* moving (i.e., moving in an

*absolute* sense) that its stop watch will show a dilated time.So your remarks are ONCE AGAIN IN ERROR. (Why is it

you cannot see this simple error you have made? Perhaps you

are not paying attention?)As for the following, namely:

> In conclusion, once again, you have got the two times that

> you challenged me to compute. The second time, (2), is surely

> GREATER than the first time (1), at least if v is not zero.

> Now you have got all the answers to all your questions. Try

> to ask to do these easy computations to any other good

> physicist, and you will obviously get the same answers (if

> not, he is not a good physicist - or at least he does not know

> enough relativity).

>

> Now you object: I see length contraction at work in (1) -

> CORRECT - so you must be wrong, because you forgot the

> time dilation. This is WRONG, since in (1) there is even a

> time dilation. As a matter of fact, I said it already at the end

> of my previous mail:

>

> "I have talked too much - and with my bad English this is

> perhaps harmful for a better understanding - yet I have not said

> everything I could have said. For instance, in some lecture to

> students, I COULD HAVE SHOWN EVEN HOW ONE

> COULD HAVE CONNECTED LENGTH CONTRACTION

> AND TIME DILATION IN THE PROPOSED EXERCISE."

>

> In order to show that, let us put the thing in the following

> terms. The people aboard of the spaceship measure the time

> needed for the meteorite to pass, and they claim that it is L/v ,

> good. Somebody aboard of the meteorite says that this time is

> not L/v , but it is LESS than that, because he measurs L'/v ,

> good once again. Well, this assertion from the point of view

> of the time elapsed for the meteorite is exactly what the crew

> of the spaceship would interpret as the time dilation occurring

> "aboard of the meteorite"!Now PLEASE: why do you EQUIVOCATE?

At first you say that according to the IFR of the "meteorite"

X, the LENGTH of the spaceship has CONTRACTED, and

THAT is why the elapsed time is shorter: L'/v rather than L/v

(as required by the Galilean transformation.)And you also specify that "L' = L*sqr(1-v^2/c^2) (the star *

stands for product)"!And NOW you say that the shorter time recorded by the stop

watch on board the "meteorite" is due to the TIME

DILATION !!!P L E A S E --- M A K E U P Y O U R

M I N D ! ! !Just WHAT is the lesser time recorded by the stop watch on

the meteorite DUE TO: time dilation or length contraction?I am beginning to think that you do not even UNDERSTAND

my objection.Let me explain in more detail.

Note that if the meteorite is NOT considered to be moving,

there can be NO time dilation of ITS stop watch! But the

LENGTH CONTRACTION OF THE SPACESHIP, as

observed from the IFR of the meteorite, remains unaffected!Whereas if the meteorite IS considered to be moving, then it

observes the LENGTH of the spaceship as CONTRACTED

(because each is moving relativistically compared with the

other), AND it observes the TIME of its OWN stop watch --

i.e., the one on the METEORITE -- to be DILATED!If, in other words, the meteorite is considered to be moving,

BOTH length contraction AND time dilation have to be

applied to obtain the final readout of its stop watch: the length

contraction is applied to the SPACESHIP, and the time

dilation to its OWN STOP WATCH.Whereas if the meteorite is considered to be stationary, ONLY

the length contraction of the spaceship is applied to obtain the

final readout of its stop watch, since its stop watch can NOT

show a dilated time!Why is it that you cannot see this SIMPLE mistake you have

made?I think it is because you do not give full DETAILS of your

argument.Or perhaps because you confuse in your own mind the two

methods of calculation: Minkowski diagrams or Lorentz

transformations.That, I think, must be the reason why YOU DO NOT

UNDERSTAND RELATIVITY FULLY!If you wish to prove that you are NOT wrong, show me the

EXACT DETAILS of your calculations, USING ONLY

O N E OF THE TWO METHODS: Minkowski

diagrams or Lorentz transformation equations (preferably the

latter).You will see if you do the calculations IN DETAIL, you

WILL get more than one set of two readouts -- whichever

method you use: Minkowski diagrams or Lorentz

transformations.(But of course if you use BOTH Minkowski diagrams and

Lorentz transformation equations TOGETHER, you will

CONFUSE everyone, including yourself; so restrict yourself

to only ONE of the two: Minkowski diagrams or Lorentz

transformation equations.Select one, but select only ONE, of the two methods.

Preferably, select the Lorentz transformation equations,

because they are the ONLY ones used by Einstein in his

original 1905 Paper on Special Relativity.)Then also, you have added all the following nonsense about

"proper times". Please note: I have NOT ASKED about

"proper times": I have only asked, "What should the

SNAPSHOTS show?" Proper or improper, ALL that is

required is the time intervals recorded by the two stop watches

as photo- graphed by the two cameras.So all this you have written below is irrelevant:

> The proper time of some phenomenon (between any two

> events E1 and E2 in the world line of some "observer") is

> defined as the LEAST value of this time with respect to all

> inertial observer, in the same way as the proper lenght is, on

> the contrary, the GREATEST value etc. etc.. This proper time

> is connected to the measure of the "same time-interval" with

> respect to other coordinate times by the well known relation:

>

> proper time interval = sqr(1-v^2/c^2) * coordinate time interval

>

> while the corresponding formula for lengths is the "inverse":

>

> proper length * sqr(1-v^2/c^2) = coordinate (or "apparent") length .

>

> Well, in (1) you have got exactly the first formula at work:

>

> proper time of the phenomenon we are studying, namely the

> passing of the meteorite X = sqr(1-v^2/c^2) * coordinate time

> interval of the same event = sqr(1-v^2/c^2) * (L/v) .

>

> As a matter of fact, the coordinate time interval of this

> phenomenon is L/v , as we said in formula (2) , and if you

> write the previous formula "associating" differently (a shifting

> of parentheses), you find exactly (L*sqr(1-v^2/c^2))/v , and

> the numerator of this fraction (ratio) is L' , the "contracted

> length"...And as for the following:

> That is to say, in (1) time dilation and length contraction

> THERE ARE BOTH PRESENT, I hope that you will be

> satisfied by that, since this is the simple easy truth.Again, it is the simple and easy-to-understand FALSEHOOD,

and a mathematical JUDGE will easily be able to see that this

is the case.And in addition, in your referred to (1) above, you have NOT

introduced the *Lorentz* time dilation. Let me repeat (1) for

your convenience, and for the convenience of any

mathematical JUDGE who might happen in future to see

copies of this correspondence between us:[QUOTE]

> Then one can ask which is the proper time elapsed WITH

> RESPECT TO X from E1 to E2 . The situation

> is like that from X's point of view:

>

> ______________ ------- > °

> S B vector velocity X

> the spaceship the buoy is "still"

> is "moving"

>

> and the computation is very easy, I repeat it.

> If the proper length of the spaceship is L, then

> L' = L*sqr(1-v^2/c^2) (the star * stands for product)

> is the length of the spaceship with respect to

> X , and the required time is (MUST BE!):

>

> (1) L'/v

>

> (here is the scalar speed, the "module" of the vector velocity).[END QUOTE]

Now show me just WHERE is the time dilation calculation

here ? ? ? ? ? ? ?Indeed, if as you write, the buoy is "still", the time of the

stop watch carried at X *cannot* be dilated!And even you yourself admit that the time interval between

E2 and E1 in the IFR of the SPACESHIP is L/v, not L'/v!THERE IS NO LORENTZ TIME DILATION IN (1). Any

mathematically competent judge will be able to see this.> I could say even more, showing how one could explain even

> the "time dilation" which happens (must happen!) in the time

> of spaceship as "seen" from the point of view of the meteorite

> (until now we had only a "time dilation" in the meteorite from

> the point of view of the spaceship).This is just a bit more

> difficult exercise for students, which requires a little attention,

> and an explicit use of Lorentz transformations. I avoid to do

> that, thinking that I have answered enough to your criticism,

> but if you want I can send to you the complete "solution" - it

> is one argument that I have experienced most amateur people

> do not know precisely, ...It is precisely the COMPLETE SOLUTION that will prove to

you that you are wrong. So please DO send me the complete

solution, omitting absolutely no detail! In the very process of

doing so, you will PROVE to YOURSELF that you are

wrong.Actually, I have an even better idea, if you will agree to it: let

us publish the FULL AND UNEDITED TEXT of our

correspondence on the Web -- at my Home Page if you wish,

or at your Web site, or better still, at both of them -- for

public review! Then ANY competent mathematician will be

able to judge who is right and who is wrong: you or I.Do you agree to this?

If so, then I shall definitely claim -- and that too with

COMPLETE justification -- that you have NOT answered my

Challenge satisfactorily! And ANYONE with competence in

the field of physics and mathematics who reads our

correspondence will be able to SEE that this is so.In other words, I will have proved not only to YOU, but to

THE ENTIRE WORLD OF SCIENCE AND

MATHEMATICS, that you are wrong and I am right.For as I answer to Del Larson when he says:

[QUOTE]

> Dear Mr. Mehta,

>

> I read the first half of the lengthy correspondence you had with

> Umberto Bartocci. It was, unfortunately, quite clear from your

> writings that you simply do not understand the theory of relativity.

>

> There are many in the alternative space time community that share

> your disability. Relativity is not the easiest thing to understand.

> Many make claims about the "common sense" inaccuracies in the special

> theory, and think from those claims alone they have disproved it.

> But that is wrong. Special relativity is not illogical. Rather it

> has a different logic than what prevails as "common sense".

>

> All of this is highly unfortunate. Since there is a small army

> making inaccurate claims without sufficient prior study and

> understanding, it makes it even harder for the few competent relativity

> critics to be heard.

>

> Please take the time to study and understand relativity. I would

> suggest contacting your closest quality university and seeking out

> a mentor. Once you understand relativity I would hope that you could

> join the battle against it.

>

> But fighting an intellectual battle without an understanding of its

> underpinnings is simple folly.

>

> Sincerely, and with Best Hopes and Intentions,

>

> Del Larson[END QUOTE]

... to this I answer very simply:

[QUOTE]

Dear Sir:

I am sorry to say that your answer clearly shows that YOU do

not understand SCIENCE.If you did, you would know that science does not consist of

mere ASSERTIONS such as "You do not understand

Relativity", but rather demands logical, and if possible also

mathematical, ARGUMENTS!Your complete lack of any ARGUMENT against my words

shows that either you HAVE none, or else that you do not

know even the first principles of science.In fact, do you HAVE THE COURAGE OF YOUR OWN

CONVICTIONS to agree to my publishing the full and

unedited text of our brief correspondence on the Web for all to

see and judge who is a true scientist: you or I?For I think your present response would indicate to ALL who

read it that your words in our correspondence do not even

allow you to merit the title "scientist".Sincerely, and with Best Hopes and Intentions of turning a

promising human being from a dogmatist -- even if an anti-

Relativity dogmatist -- to a TRUE scientist,Ardeshir Mehta.

(Let me add that I am not so much anti-Relativity as anti-

humbug, no matter what its source.)[END QUOTE]

> and then they get in other "paradoxes",

> because they are not able (for instance Dingle was unable to

> see it, and he was not even an amateur!) to see the

> COMPLETE SYMMETRY which is now at work (of course,

> there is no possibility to see even the length contraction of X

> , which we have supposed pointlike).Your argument becomes completely INVALID as soon as you

are reminded of the SECOND Minkowski diagram!And indeed, as I said, there are an infinite number of such

diagrams possible for the scenario of my Challenge ... which

proves that there is NO mathematically consistent solution

possible for the question I have posed which is ALSO

consistent with the Theory of Special Relativity.Anyway, you should realise by now that you have NOT been

able to satisfactorily answer my Challenge, whether with the

help of the Lorentz transformation equations OR with the help

of the Minkowski diagrams -- and ANY competent

mathematical judge should be able to see that too.> Ciao, I HOPE TO READ SOON YOUR OBJECTIONS

> AGAINST GOEDEL, perhaps they are more useful than the

> ones against relativity (I am sympathetic with these attempt to

> show that relativity is wrong, at least they show that one has

> understood that to do this is a most important task, but one

> must not underestimate the problems), one never knows,I hope you have seen from the above at least that you have

YOURSELF not been able to understand Relativity until now

-- !Or else, if you do understand my arguments, you should

realise by now that it is VERY EASY to refute Relativity -- at

any rate if one restricts oneself to Special Relativity alone!

(General Relativity is not being addressed here by me

properly, except for being mentioned once in a while, mostly

in passing.)Goedel is a much more difficult matter than Special

Relativity. If you cannot even understand my arguments

against Special Relativity I don't think you will be able to

understand the arguments against Goedel, whose own

arguments in his 1931 Paper are MUCH more subtle than

those in Special Relativity.I also don't think you are up to the job of tackling Goedel's

Theorem because you have not given me an answer to my

PHILOSOPHICAL argument against the Lorentz

transformations: namely that they contradict the Galilean ones,

and thus they *cannot* possibly be deduced in a step-by-step

mathematical fashion from the axioms of mathematics and the

propositions and postulates of Euclidean geometry, whereas

the Galilean transformations CAN be so deduced.(Why did you not answer my philosophical argument? Maybe

you did not read my e-mail carefully enough ... AGAIN?

Don't say I didn't warn you.)This philosophical argument is MUCH stronger than the one

in my Simple Challenge ... and of course it is not a SIMPLE

one either. If you are not a philosopher versed in the

philosophy of mathematics, however -- as for example

Bertrand Russell was -- it may be best for you to avoid

responding to my argument: you may not even understand it.(Of course if you DO claim to understand it, I shall willingly

debate your counter-arguments -- and that too, very

thoroughly.)But by all means TRY to understand my Critique of Goedel's

Theorem: and who knows? Maybe I shall be proven mistaken

in my opinion, and you WILL understand it after all! -- In

which case no one will be happier than I.In any case, I remain always your sincere friend in our

common battle against dogma and humbug, whether in

mathematics or in science.Tanti saluti,

Ardeshir.

Home Page: <http://homepage.mac.com/ardeshir/education.html

_______________________________________________________

Subject: Re: Argomento Contro la Teoria della Relativita Speciale

(addendum molto breve)

[Translation: Argument Against the Special Theory of

Relativity (very brief addendum)]

Date: Tue, 28 Aug 2001 02:56:22 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8

Caro Prof. Bartocci:

Here is an absolutely clinching argument that you are

DEFINITELY wrong.You had written:

> But we can of course suppose to have synchronized clocks

> aboard of the spaceship (this is exactly the meaning of the

> words in my last mail, about the role of your "mechanism"),

> and we can ask which is the time which is needed for the buoy

> to pass the spaceship, FROM THE POINT OF VIEW OF

> THE SPACESHIP, which means from the point of view of a

> coordinate system associated with the spaceship (we could

> better think of X as a meteorite). Well, since the length of

> the ship is L , and the module of the velocity does not

> change, this time MUST BE now equal to:

>

> (2) L/v .Now let us give some hard figures, and let us exaggerate

GREATLY (and you will see how useful this is!).Let us say, just hypothetically, that L = 1 light-second, and v

= 0.9999 ... 999 c (with ONE BILLION 9's after the decimal

point).Then according to your calculation, which as you say MUST

be right, the time which is needed for the buoy to pass the

spaceship, from the point of view of a co-ordinate system

associated with the spaceship, ACCURATE TO FIFTY

DECIMAL PLACES, should be L/v = (1/0.9999 ... 999)

seconds or 1.0000 ... 000 second (with fifty 0's after the

decimal point). Right?And let us assume that it is impossible to measure

ANYTHING with an accuracy GREATER than fifty decimal

places -- i.e., that this is the allowable margin of error in our

measuring instruments.But according to Relativity it should make NO

DIFFERENCE to the readout of the stop watch (or of the stop

watches, if there are more than one) on board the spaceship,

whether the spaceship passes the buoy, or the buoy passes the

spaceship!Right?

So according to you, the time which is needed for the

SPACESHIP to pass the BUOY, again from the point of view

of a co-ordinate system associated with the spaceship, and

again accurate to fifty decimal places, should ALSO be

1.0000 ... 000 second.Right?

Now let us suppose that this spaceship is passing, not a buoy,

but the axis of the Sun -- i.e., the axis around which the Sun

rotates every so many hours -- so that this axis lies at right

angles to the relative motion between the spaceship and the

Sun's axis of rotation. (The axis has only one dimension, and

this takes care of the problem created by the finite size of a

buoy or a meteorite.)(And for the purposes of THIS calculation, a stop watch on

the buoy is not at all required: only the stop watch(es) in the

spaceship is/are taken into consideration.)And let us suppose the spaceship passes the Sun at a distance

of 200 million kilometres from the Sun's centre, so it misses

the Sun's heat by a great distance.Now if Relativity is correct, according to the TIME

DILATION which the stop watch in the spaceship must be

undergoing, time should be running SO very slowly on board

the spaceship that its stop watch (or stop watches, if there are

more than one) should register NO TIME AT ALL -- i.e.,

0.0000 ... 000 seconds, when measured with an accuracy of

fifty decimal places -- for the spaceship to pass the Sun's axis!(And this is EVEN if we ignore the spaceship's length

contraction; for if its length contraction IS also taken into

account, the time registered by the stop watches in the

spaceship's co-ordinate system would be even less ... if that

were at all possible!)Do you now see the VAST difference? Observe--

1. According to your calculation: the spaceship's stop watch

reads 1.0000 ... 000 second.2. According to Relativity: the spaceship's stop watch reads

0.0000 ... 000 seconds.And 1.0000 ... 000 second is NOT equal to 0.0000 ... 000

seconds!So either your calculation is wrong or the Theory of Relativity

is wrong!Are you NOW satisfied that you are ABSOLUTELY, COM-

PLETELY and CATEGORICALLY mistaken?

Best wishes,

Ardeshir.

_______________________________________________________

Subject: Come si deve fare la matematica

[How Mathematics Should be Done]

Date: Thu, 30 Aug 2001 01:26:21 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8

Caro Professore:

Le faccio vedere come si deve fare i calcoli matematici,

piano-piano, senza fare delle "balze" intellettuali, sicché

si ritiene tutti i legamenti fra le idee:[TRANSLATION]

I show you how mathematical calculations should be done,

slowly, without making intellectual "jumps", so as to retain

all the connections between the ideas:[END OF TRANSLATION]

[1] Let the rest length of the spaceship be L, and the velocity

between spaceship and buoy be v; and let all these

expressed, for the sake of convenience in making the

calculations, in units in which the speed of light, c, equals 1.[2] Let the event of the front end -- i.e., the "bow" -- of the

spaceship passing the buoy be denoted as E1, and the event

of the rear end of the spaceship -- its "stern" -- passing the

buoy be denoted as E2. (Note that E1 and E2 are events

common to the "lives" of both, the co-ordinate system

associated with the buoy AND the co-ordinate system

associated with the spaceship.)[3] Let the primed co-ordinates in the calculations below be

those pertaining to the spaceship, and let the un-primed co-

ordinates be those pertaining to the buoy.[4] The Lorentz time dilation formula is

<delta>t' = <delta>t*(sqrt.(1-v^2/c^2))^-1.[5] The Lorentz length contraction formula is

<delta>x' = <delta>x*(sqrt.(1-v^2/c^2))^-1.[6] The Lorentz transformation equations are as follows:

x' = (x - vt)/(sqrt.(1-v^2/c^2)),

y' = y,

z' = z,

t' = (t - vx/c^2)/(sqrt.(1-v^2/c^2)).[7] Let E1 occur in the co-ordinate system associated with the

buoy at t1= 0.[8] Then by the Lorentz transformation equation for time, E1

will also occur in the co-ordinate system associated with the

spaceship at t'1=0. (Note that at E1, x = 0, and thus

vx/c^2 also equals zero.)[9] According to the Lorentz length contraction formula, in the

co-ordinate system associated with the buoy, the length of

the spaceship will not be L but L', where

L' = L*(sqrt.(1-v^2/c^2))^-1.[10] Thus E2 will occur, in the co-ordinate system associated

with the buoy, at t2 = L'/v where L' = L*sqrt.(1-v^2/c^2).[11] Let 1/(sqrt.(1-v^2/c^2)) be denoted by the conventional

term [<gamma>].[12] Then E2 will occur in the co-ordinate system associated

with the buoy at t2 = L'/v, where L' = L*[<gamma>^-1].[13] Using the Lorentz time dilation formula, we calculate that

E2 will occur in the co-ordinate system associated with the

spaceship at t'2 = t2*[<gamma>^-1]. (Note: The same

result is attained if we use the Lorentz transformation

equation t' = (t - vx/c^2)/(sqrt.(1-v^2/c^2)), since at E2, x

is zero, and thus vx/c^2 is also zero.)[14] Since t2 = L'/v and L'=L*[<gamma>^-1],

t'2 = t2*[<gamma>^-1]

= (L'/v)* [<gamma>^-1]

= ((L*[<gamma>^-1])/v) )* [<gamma>^-1]

= ((L*[<gamma>^-1])*v^-1) )* [<gamma>^-1]

= L*v^-1*[<gamma>^-2].[15] E1 will occur at t1 = 0.00 and t'1 = 0.00 -- i.e., in both co-

ordinate systems E1 will occur at time 0.00.[16] Since both t1 and t'1 are zero, and E2 will occur at time

t2 = L'/v (according to [10] and [12] above), the time

registered by the stop watch aboard the buoy at event E2

will be T = (t2 - t1) = (t2 - 0) = t2; and the time registered by

the stop watch aboard the spaceship at the same event E2 will

be equal to T' = (t'2 - t'1) = (t'2 - 0) = L*v^-1*[<gamma>^-2]

(according to [14] above).[17] These will be the times registered by the two stop watches

when photographed -- that is, IF the primed co-ordinates

are those of the SPACESHIP.[18] But if the primed co-ordinates are those of the BUOY, then

the situation will be as follows:[19] E1 will again occur at t = 0.00 and t' = 0.00.

[20] E2 will occur in the un-primed co-ordinate system -- i.e.,

the one associated with the spaceship -- at t2 = L/v: or in

other words, at t2 = L*v^-1.[21] According to the Lorentz time dilation formula, E2 will

occur in the primed co-ordinate system -- the one

associated with the buoy -- at t'2 = t2*[<gamma>^-1].

(Again, the same result is obtained if we use instead the

Lorentz transformation t' = (t - vx/c^2) / (sqrt.(1-v^2/c^2)),

since at E2, x is zero, and thus vx/c^2 is also zero.)[22] Since t2 = L/v, by the Lorentz time dilation formula

t'2 = t2*[<gamma>^-1] = (L/v)*[<gamma>^-1].[23] And since L' = L*[<gamma>^-1],

(L/v)*[<gamma>^-1] = L'/v.[24] So the time recorded by the stop watch aboard the buoy

will be L'/v in EITHER case -- no matter which co-ordinate

system is represented by the primed letters.[25] BUT the time recorded by the stop watch aboard the

spaceship will be EITHER L*v^-1 OR

L*v^-1*[<gamma>^-2], depending on which co-ordinate

system is represented by the primed letters!So WHICH figure correctly represents the time recorded by the

stop watch of the co-ordinate system associated with the

spaceship: L*v^-1*[<gamma>^-2] or L*v^-1?According to the above calculations, which are done using the

Lorentz transformation equations, BOTH figures must be

correct!But this is mathematically IMPOSSIBLE.

Therefore the Lorentz transformation equations cannot be

MATHEMATICALLY valid.

Tanti saluti,

Ardeshir <http://homepage.mac.com/ardeshir/education.html>

_______________________________________________________

Subject: Re: Come si deve fare la matematica

[How Mathematics Should be Done]

Date: Fri, 31 Aug 2001 14:53:31 +0200

From: umberto bartocci <bartocci@dipmat.unipg.it>

To: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9

[I wish to reply to you immediately, I hope not to have made

mistakes...]Dear Mehta,

thanks even for your last mails, even if they confirm my

opinion: you make too many mistakes, not knowing

accurately enough relativity. So for friendship, and even for

"duty" as a teacher (which cannot last at "infinity"!), I send to

you a list of the new mistakes you have made in one of these

mails, but let me say once again very frankly that I have no

more time for this kind of discussion.As a matter of fact, there is a kind of "small" mistakes, which

can be corrected, and do not change the value of some

argument; there are instead "big" mistakes, which reduce the

value of some argument to zero. If you are not convinced that

you are making BIG mistakes, you can ask to some other

competent professor, and please let me know if there exists

even ONE which will appreciate these same objections against

relativity (of course in some point you are right, for instance

when you remark that the 1905 Einstein's paper is not so good

- but nowadays relativity is simply the Minkowski space-time

physics, and it is rather useless to express criticism which

would be as best accepted as possessing only some

HISTORICAL value)...Always best wishes,

from yours UB (I took the file of your work on Goedel, and

I shall try to study the whole matter in the following months,

I hope...)- - - - -

> [2] (Note that E1 and E2 are events common to the "lives" of

> both, the co-ordinate system associated with the buoy AND the

> co-ordinate system associated with the spaceship.)- This is not entirely correct. We have got three "observers", in

the language of relativity, X the buoy, B the bow, S the stern,

and E1 is an event which is common to X and B, while E2 is

an event which is common to X and S. So E1 and E2 are both

in the "life" ONLY of X, and not of B and S. This is in some

sense a source of error in what follows.> [4] The Lorentz time dilation formula is

> <delta>t' = <delta>t*(sqrt.(1-v^2/c^2))^-1.

> [5] The Lorentz length contraction formula is

> <delta>x' = <delta>x*(sqrt.(1-v^2/c^2))^-1.- This is the major source of misunderstanding. If this was true,

then you would simply get immediately a contradiction like the

following one:1 - <delta>t' = <delta>t*(sqrt.(1-v^2/c^2))^-1

2 - there is no difference whatsoever between t and t', Lorentz

transformations have a group structure, the situation is SYM-

METRIC,

so one must have even:<delta>t = <delta>t'*(sqrt.(1-v^2/c^2))^-1

3 - from 1 and 2 it follows for instance:

<delta>t' = <delta>t'*(sqrt.(1-v^2/c^2))^-2

[as a matter of fact, this is precisely what happens in your

point 16]whence:

(sqrt.(1-v^2/c^2))^-1 = 1 which implies v = 0 .

I HAVE SEEN THIS MISTAKE HUNDRED OF TIMES

IN MY LIFE...The true formulae, which one must "understand", are the ones

which I gave to you in my last mail, namely:A - (proper length of some object)*sqr(1-v^2/c^2) =

coordinate (or "apparent") length of this same object in an

(inertial) reference frame in which the object is moving with

speed vB - proper time of some phenomenon IN THE LIFE OF ONE

OBSERVER = sqr(1-v^2/c^2)*(coordinate time interval of the

same event) [meaning as before]> [9] According to the Lorentz length contraction formula, in the

> co-ordinate system associated with the buoy, the length of the

> spaceship will not be L but L', where L' = L*(sqrt.(1-v^2/c^2))^-1.- WRONG, the exponent ^-1 is wrong, the truth is:

L' = L*sqr(1-v^2/c^2) . As a matter of fact, you write that

correctly in your point N. 10, which is GOOD:[10] Thus E2 will occur, in the co-ordinate system associated

with the buoy, at t2 = L'/v where L' = L*sqrt.(1-v^2/c^2).> [13] Using the Lorentz time dilation formula, we calculate that E2

> will occur in the co-ordinate system associated with the spaceship at

> t'2 = t2*[<gamma>^-1]. (Note: The same result is attained if we use the

> Lorentz transformation equation t' = (t - vx/c^2)/(sqrt.(1-v^2/c^2)),

> since at E2, x is zero, and thus vx/c^2 is also zero.)- This is WRONG. True that x = 0, but if you make the com-

putation:t'2 = t2/sqr(1-v^2/c^2) ,

by using the formula which you write above in your point 10,

which

I said is correct, then you gett'2 = (L'/v)/sqr (1-v^2/c^2) = (L*sqr(1-v^2/c^2)/v)/sqr(1-v^2/c^2)

= L/v(as I already wrote to you in my previous mail!)

which implies:

t'2 = t2*gamma [and not gamma^-1!!]

REMARK 1 - It is not enough to give t'2, since in order to

give an event we must specify even the SPACE

COORDINATE, which in this case is -L , as it MUST BE

because, with the actual hypotheses, it happens in the stern of

the spaceship. As a matter of fact, Lorentz transformations

rightly give:-v*t2/sqr(1-v^2/c^2) ,

and since t2 is L'/v , we have:

-L'/sqr(1-v^2/c^2) = - L*sqr(1-v^2/c^2)/sqr(1-v^2/c^2) = -L !!

GOOD.

REMARK 2 - The formula t'2 = t2*gamma SEEMS at odd

with B, namely with:proper time of some phenomenon IN THE LIFE OF ONE

OBSERVER = sqr(1-v^2/c^2)*(coordinate time interval of

the same event) which we could indeed "tentatively" write as:Dt' = Dt*(gamma^-1)

but the point one has to understand is that UNTIL NOW

THERE IS NOT A PROPER TIME in the primed coordinates

of the spaceship, since E1 and E2 belong to DIFFERENT

OBSERVERS (or, if you prefer, they happen in DIFFERENT

SPACES), so you CANNOT make use of the previous

formula in this case.On the contrary, one has such a proper time in the life of X ,

so one could have indeed said immediately that, since L'/v is

the proper time for X , then this time is, in the primed

coordinate of the spaceship, equal to:coordinate time = (proper time)*(gamma) = (L'/v)/sqr(1-v^2/c^2)

= ((L*sqr(1-v^2/c^2))/v)/sqr(1-v^2/c^2)

= L/vAND EVERYTHING GOES WELL.

> [14] Since t2 = L'/v and L'=L*[<gamma>^-1],

> t'2 = t2*[<gamma>^-1]

> = (L'/v)* [<gamma>^-1]

> = ((L*[<gamma>^-1])/v) )* [<gamma>^-1]

> = ((L*[<gamma>^-1])*v^-1) )* [<gamma>^-1]

> = L*v^-1*[<gamma>^-2].- WRONG, WRONG, I hope you will start to get persuaded

of this! Everything which follows from now on is wrong.

FINAL REMARK 3 (for "students" - one can be "student" a

lot of time in his life, every time he studies something he

does not know yet, even if he is a "professor" in some other

matter: for instance, I shall now become a student about that

Goedel affair...)The event E1 is (0,0) (first space, second time) for the

coordinate system of the buoy, and then it is (0,0) even for the

coordinate system of the spaceship. The event E2 is (0,L'/v)

for the buoy, and it is (-L,L/v) for the spaceship. So the time

which is needed for the spaceship to pass the buoy, FROM

THE POINT OF VIEW OF THE BUOY, is L'/v , while this

"same" time, FROM THE POINT OF VIEW OF THE

SPACESHIP, is L/v . This means that S "knows" that at the

time 0 of the spaceship the buoy was in front of the bow, and

at L/v of its clock it SEES the buoy in front of it. He sees

that the buoy's clock marks in this event L'/v < L/v , while he

knows that it marked 0 when it was in front of the bow. S

then describes this phenomenon as a DILATION OF TIME

occurred to the buoy's clock. From the point of view of the

buoy, instead, the time measured L'/v was simply less of that

time L/v because this was due to the "contraction of the

length" L . [By the way, why do you say: "Parenthetically,

this is a mere ASSERTION on your part: there is no

DEMONSTRATION of it anywhere in your e-mail"? THIS IS

THE DEMONSTRATION, and it is exactly the same which I

have given before]. But we can ask: where is the TIME

DILATION that the buoy MUST see, according to relativity,

aboard of the spaceship? When the two clocks of X and of S

meet, the clock of X marks LESS than the clock of S , no

doubt about it. Is this then a real ASYMMETRY? Does this

mean that X DOES NOT see a time dilation occurring aboard

of the spaceship, but a TIME CONTRACTION? The answer

is simply NO, because in order to understand HOW THE

CLOCK OF S HAS GONE, slower or faster than its own, X

cannot assert that the time elapsed for S, from X's point of

view, is L/v. As a matter of fact, X should compute the

difference between L/v , the time marked in front of it, at the

unique meeting of the two clocks, AND THE TIME WHICH

THIS SAME CLOCK MARKED, IN THE REFERENCE

FRAME ASSOCIATED WITH THE BUOY, at t = 0 , when

X had in front of it the bow of the spaceship, with a clock

marking 0, exactly as its own: but it does not know anything

DIRECTLY about the clock of the stern, which was

SOMEWHERE ELSE. If this time, let us call T , was zero,

then we would have got problems, but this time is not zero.

In order to compute this value T , we must realize that by

definition of the length of a moving object, at t = 0

(remember that the unprimed coordinates are relative to the

buoy), the observer associated with the buoy which had the

stern S in front of it, was exactly -L' (event (-L',0) )

[ATTENTION, one should be careful not to make confusion:

with our assumptions, L would "belong" in some sense to the

primed coordinates, and L' to the unprimed!]. Well, which is

the event corresponding to (-L',0) in the primed coordinates? It

is:x' = (-L')/sqr(1-v^2/c^2) = -L

(AS WE COULD HAVE SAID IMMEDIATELY, it must be

the stern S!), plus:t' = vL'/c^2*sqr(1-v^2/c^2) = vL/c^2 = T .

Now that we have T , we must compute, as we have said, the

difference:L/v - T = L/v - v*L/c^2 = (L/v)*(1 - v^2/c^2) .

THIS IS THE TIME WHICH, FROM THE POINT OF VIEW

OF THE BUOY (of the system associated to the buoy),

OCCURRED TO S's CLOCK IN ORDER TO ARRIVE IN

FRONT OF IT, when it marked the time L/v .The previous formula implies that the buoy sees exactly a

time dilation aboard of the spaceship, as it must be according

to relativity, exactly in the same quantitative form which is

foreseen by the theory.As a matter of fact this time interval (L/v - T) , WHICH IS

NOW A PROPER TIME OF ONE SINGLE OBSERVER,

namely S , is equal to:the time elapsed for the buoy, L'/v, namely a "coordinate

time" (which is even a proper time), multiplied by the factor

sqr(1-v^2/c^2) , and so at last:(L'/v)*sqr(1-v^2/c^2) = ((L*sqr(1-v^2/c^2))/v)*sqr(1-v^2/c^2)

= (L/v)*(1 - v^2/c^2) .EVERYTHING IS ALRIGHT ONCE MORE, AS IT OB-

VIOUSLY HAD TO BE.- - - - -

This is the promised FULL solution of this rather simple

exercise in relativity. I am not playing the part of the

ADVOCATUS DIABOLI in a very diabolically way, but the

truth is the truth. Pretending to have shown that "the Lorentz

transformation equations cannot be MATHEMATICALLY

valid", as you claim, with such wrong arguments, is counter-

productive with respect to all serious attempts to criticize

relativity...

Umberto Bartocci

Dipartimento di Matematica

Universita' di Perugia

06100 - Italy

http://www.dipmat.unipg.it/~bartocci

________________________________________________________

Subject: Re: Come si deve fare la matematica - Addendum

[How Mathematics Should be Done - Addendum]

Date: Sat, 01 Sep 2001 09:19:45 +0200

From: umberto bartocci <bartocci@dipmat.unipg.it>

To: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9

Short addendum to my previous "final" comment...

Dear Mehta,

since I have understood from the correspondence between you

and Del Larson that our discussion is going to be made public

in Internet, it is perhaps better to add some "public remark" to

my last mail, in order to be even more clear.Remark I - I could have explained better to "students" the

situation we were discussing, by using more simple words as

conclusion. But first let me sum up what I have said,

answering to your question. A buoy X is passing in front of a

spaceship, long L (proper length), from bow B to stern S, and

with some speed v (spaceship and buoy are equipped with

Lorentz coordinates etc.). Here it is the "picture" in the

reference frame associated with the ship:______________ <------- °

S B vector velocity X

the spaceship the buoy is "moving"

"is still"When B and X are one in front of the other, the clock of B marks

0 as the clock of X. When X is in front of S, the clock of X marks:(1) L'/v = [L*sqr(1-v^2/c^2)]/v

while the clock of S (synchronized with the clock of B) marks

(quite OBVIOUSLY!):(2) L/v .

Since (2) > (1) , one has indeed something "strange" from the

point of view of common sense. From aboard of the

spaceship, one must acknowledge that the clock of X is going

SLOWER, since it shows LESS time as it should have to.

This is called "time dilation". From the point of view of X, it

has needed less time in order to pass the spaceship, because

the spaceship was not long L , but less, namely L' = L*sqr(1-

v^2/c^2) , and to pass a shorter spaceship obviously requires

less time. This is called "length contraction". In the previous

formulae (1) and (2) , and in their comparison (2) > (1) , there

are simultaneously at work both time dilation and length

contraction.Remark II - Since the situation is quite symmetric, according

to the theory of relativity, X must even "see" a time dilation

in the clocks aboard of the spaceship, exactly in the same

amount as before, that is to say it must see two different rate

of times connected by the ratio sqr(1-v^2/c^2) . This is more

difficult to be understood, but it can be done (see the Remark

3 in my previous message). Of course, there is no possibility

to notice an "inverse" symmetric length contraction, since one

has supposed the buoy to be "pointlike". If one introduces

even a (proper) length for the buoy, let us call it M , then the

situation is FULLY SYMMETRIC. With the necessary

specifications (see one of my previous mails), the formulae (1)

and (2) become:(1a) L'/v + M/v

(2a) L/v + M'/v .

Remark III - There is no doubt that the relativistic description

of the "phenomenon" which Mr Mehta challenged to discuss

is exactly the previous one, and that THERE IS NO "logical

contradiction" at all in the relativistic treatment of spaces and

times. On the contrary, one has a great and still alive

discussion about the "physical interpretation" of the formulae

above. These contractions and dilations are in some sense

REAL, or they just depend on the assumptions introduced by

Einstein, and followers, in order to synchronize clocks (using

the II postulate of special relativity), to measure lengths of

moving objects, and so on? This is why some people talk of

"apparent phenomena", but I rather believe that in this case

relativity IS MAKING SOME PRECISE PHYSICAL

ASSERTION, which could be put to test (of course, from an

"ideal", if today not yet "practical", point of view). One way

of doing it, it is to use the famous "twin paradox": when an

inertial clock and another accelerated clock (in this case, NO

SYMMETRY AT ALL, Herbert Dingle famous "syllogism"

notwithstanding!) meet again, then, according to relativity, the

accelerated clock MUST show less time then the inertial one,

in the prescribed quantitative amount. Physicists claim that

this is so, after experiments made with particles in their

accelerators, but I think: first, that one could provide for

different interpretations (see for instance in Bartocci's website

the "cup-model" explained in point 4ter in the page

Foundations of Physics) of these observations, which I "must"

assume to be correct; second, that one could try to check the

whole matter MORE DIRECTLY. One famous trial is the

celebrated Hafele-Keating experiment, but it seems to meet the

strong criticism of Dr Kelly (see again in Bartocci's website

the point 18bis in the page Foundations of Physics).

Moreover, this experiment has to deal even with rather

complicated gravitational effects. For these reasons, I believe

that it would be convenient to perform this kind of

experimental tests, and it does not seem impossible to

imagine a Gedanken-Experiment with rotating clocks in some

Earth's laboratory. Supposing that the Earth's laboratory is

(with good approximation) an inertial one, it is rather funny to

observe that even some ether theory would foresee a

SLOWING DOWN of spinning "light-clocks", but with a

different ratio than the relativistic one, namely by (1 -

v^2/c^2) , with no square root in front of this term (this

would even imply that the well known modification in the

mean-life of particles WOULD NOT obey the same

quantitative law than the slowing down of light clocks!). So,

at least "in principle", all what we have said above is physics,

and not only mathematics (which of course cannot be

contradictory!).[Remark IV (less important) - When I have said at the end of

my last mail that:> the time elapsed for the buoy, L'/v, namely a "coordinate

> time" (which is even a proper time)...I was not quite correct, since the words between parentheses

could give rise to some misunderstanding. As a matter of fact,

we were using in that case the formula connecting a proper

time (in the life of S), which was L/v - T , with a coordinate

time, which was L'/v (computed as the difference of the two

"corresponding" times L'/v and 0 , which were observed by

TWO DIFFERENT observers in the reference frame associated

with X ). So the fact that L'/v is even a proper time for X is

(obviously) true, but it is a useless, if not worst, remark, in

the context I was discussing...]

Umberto Bartocci

Dipartimento di Matematica

Universita' di Perugia

06100 - Italy

http://www.dipmat.unipg.it/~bartocci

_______________________________________________________

Subject: Re: Come si deve fare la matematica (reply in English)

[How Mathematics Should be Done (reply in English)]Date: Thu, 13 Sep 2001 23:36:10 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10

Caro Professore:

(I am separately sending you a copy of this e-mail translated

into Italian, because I think you have not understood properly

some important points in my preceding e-mails. I hope I

haven't made too many serious linguistic mistakes! Here is the

original e-mail in English.)(Le spedisco in un altro e-mail una copia di questo e-mail

tradotto in italiano, perché penso che Lei non ha capito bene

dei punti importanti nei miei e-mails precedenti. Spero di non

aver fatto grandi sbagli linguistici! Le spedisco qui l'originale

di questo e-mail in inglese.)Thank you very much for your two last replies, but I am sorry

to say that although I did indeed make what you call SMALL

mis- takes in my previous e-mail to you (with the subject line

"Come si deve fare la matematica"), I did not make any BIG

mistakes which can reduce my argument to zero.On the contrary, it is YOU who did that! And I shall show

you hereunder just how your calculations result in

contradictions. (I have pointed out seven contradictions

hereunder, but essentially an infinite number of contradictions

can result from the fact that there is a fundamental

contradiction between the Lorentz transformation equations,

on the one hand, and on the other, the axioms of mathe-

matics and the propositions and postulates of Euclidean

geometry.)And I am certainly willing to show what I have written to

other competent people well versed in logic, mathematics and

physics, and I am sure that if they actually read -- and

consequently, understand -- my arguments, they WILL

appreciate and agree with what I am saying. (Try Dr Al Kelly

of Ireland or Clarence Dulaney of Texas, or Mrs Gertrude

Walton of England -- an excellent mathematician if ever there

was one! -- or for that matter try my fellow- citizen of Ottawa,

Prof. Paul Marmet of the National Research Council of

Canada, who does not even agree with my philosophical

views on the nature of Reality! Indeed try ANYONE with an

open and clear mind. Maybe there are even people among your

own students or assistants who would be interested in our

discussion -- perhaps Dr. Macri'?)And I am also exceedingly sorry to see that you say you "have

no more time for this kind of discussion." It smacks of what

the kind of remark the Church Cardinals might have made to

Galileo. People who think they already KNOW the truth, and

think they do not have to listen to any reason why they might

NOT know it, often say such things. But SEEKERS after the

truth -- as all scientists should be -- cannot say such things

and still claim to be seekers and scientists.There is not -- and never has been, and probably never will be

-- ANY scientific theory so completely true that it can

NEVER be refuted by an intelligence clever enough to find a

flaw in it! And I claim to have found just that in the Theory

of Relativity.But to grasp my arguments you must first *read carefully and

digest thoroughly what I am saying*: which I am sorry to see

you continue to neglect doing. THIS IS THE MOST

IMPORTANT POINT IN ALL OUR DISCUSSIONS. Unless

it is resolved there can be no agreement between us, because

you will not GRASP what I am saying!BUT FIRST OF ALL --

Let me see if *I* have understood YOU correctly. (It would be

wrong of me to point the finger while making the same

mistake you do!)A-i.

You say that if the spaceship is supposed to be MOVING,

while the buoy is NOT supposed to be moving, the stop

watch in the spaceship will record the time interval [E2 - E1]

as L/v -- right?A-ii.

You say that if the buoy is NOT supposed to be moving,

while the spaceship IS supposed to be moving, the stop watch

in the buoy will record the time interval [E2 - E1] as L'/v --

right?A-iii.

You say that if the spaceship is NOT supposed to be moving,

while the buoy IS supposed to be moving, the stop watch in

the spaceship will record the time interval [E2 -= E1] as L/v

-- right?A-iv.

You say that if the buoy is supposed to be MOVING, while

the spaceship is NOT supposed to be moving, the stop watch

in the buoy will record the time interval [E2 - E1] as L'/v --

right?Have I understood your answers correctly?

A-v.

If so, then you would have to be saying, when comparing A-i

to A-iii, that the stop watch in the spaceship will show the

SAME reading whether it is supposed to be moving or not!(Note that the interval [E2 - E1] is the SAME interval in both

A-i and A-iii. If they were not the same, it WOULD be

possible to tell when the spaceship was moving and when it

wasn't!)But A-v contradicts the Theory of Relativity, according to

which a stop watch which is MOVING must run slower than

the SAME stop watch when it is STATIONARY!

(Remember, we are speaking here about the very SAME stop

watch measuring the SAME time interval.)This proves that your answer, if I have understood it correctly,

contradicts the Theory of Relativity.

[I] -- THIS IS THE FIRST CONTRADICTION WHICH

RESULTS FROM YOUR CALCULATIONS.

Or else, if you say I do not understand the Theory of

Relativity, please explain how, according to the Theory of

Relativity, a stop watch which is carried in a MOVING object

can run at the SAME rate as the SAME stop watch carried in

the SAME object when the latter is STATIONARY?

Specifically, if that is so, where would be the length

contraction and the time dilation that should take place for a

moving object, and for any clock or stop watch in such an

object, in comparison for the SAME object -- and for its clock

or stop watch -- when said object is NOT in motion?For example, let's say the spaceship and the buoy are both

equipped with powerful rocket engines, capable of accelerating

and decelerating each of them to speeds of +v or - v. Let's say

the reading of the stop watch in the spaceship which given in

A-i is measured when the buoy passes the spaceship, the buoy

during this pass travelling along the x axis in the positive

direction at velocity v. Then let's say the buoy is caused to

decelerate, using its rocket engine, to the same speed as the

spaceship, so that they are now motionless relative to one

another: and at this point the engine on the buoy is shut off.

Now let's say the spaceship fires ITS rocket engine,

accelerating to speed v along the x axis in the positive

direction, at which speed ITS engine is shut off. So now it is

the spaceship which passes the buoy at speed v: and during

this pass, we obtain the reading of the stop watch in the space-

ship given in A-iii. Both the readings, according to your

calculations, must be the same: namely, L/v. SO WHERE IS

THE TIME DILATION WHICH OUGHT TO BE PRESENT,

ACCORDING TO THE THEORY OF RELATIVITY FOR

THE SPACE- SHIP'S STOP WATCH IN A-iii?If you say that the length contraction and the time dilation of

the spaceship and of its stop watch can only be observed by a

hypo- thetical observer on the buoy, but never by a

hypothetical obser- ver on board the spaceship itself, then the

length contraction and the time dilation must exist merely *in

appearance* and not in *reality*.Right?

But if that is the case, the Theory of Relativity must be

dealing only with *appearances*, and not with *reality*!If THAT is what you are saying, then I am in complete

AGREEMENT with you.But in that case the Theory of Relativity cannot be a truly

*physi- cal* theory, which must deal with *reality* and not

mere *appear- ances*! (Perhaps it would be more correct to

say, in that case, that the Theory of Relativity is a

PSYCHOLOGICAL theory :-) ... !(See also below, wherein I show that this is indeed the case.)

NOW NEXT --

Let me get MY few small mistakes in my previous e-mail out

of the way.B.

I admit that I made a careless mistake in [4], in which I wrote:

> [4] The Lorentz time dilation formula is

> <delta>t' = <delta>t*(sqrt.(1-v^2/c^2))^-1.I should have written instead:

> [4] The Lorentz time dilation formula is

> <delta>t' = <delta>t*(sqrt.(1-v^2/c^2)).(See equations 11.5 and 11.6 given at the following Web

Page: <http://theory.uwinnipeg.ca/mod_tech/node135.html>.

Note that this Web page is part of a Physics Course written by

Dr. Randy Kobes and Prof. Gabor Kunstatter of the Physics

Department, University of Winnipeg, Canada, in September,

1999 -- in other words, a course given by a professor much

like your good self, and his assistant!)But as you see from my e-mail, I CORRECTED this mistake

in the first part of my [13] *et seq.*: so my conclusion in [14]

and [16] IS correct!Note also that THIS is the formula that is actually

RELEVANT to my argument, since if you read the text of the

Web article above, you will see that this formula refers to the

rate at which moving clocks tick at different rates. In my

Challenge I have only TWO stop watches, so if Relativity is

correct, the stationary stop watch must tick <gamma> times

faster than the moving stop watch!Now I anticipate that you may claim that THIS formula is just

as wrong as the PREVIOUS one, because we can always

obtain a contradiction like the one you had already noted

earlier, in the following manner (I quote your own words,

with a small change):> If this was true, then you would simply get

> immediately a contradiction like the following one:

>

> 1 - <delta>t' = <delta>t*(sqrt.(1-v^2/c^2))

>

> 2 - there is no difference whatsoever between t and t', Lorentz

> transformations have a group structure, the situation is SYM-

> METRIC, so one must have even:

>

> <delta>t = <delta>t'*(sqrt.(1-v^2/c^2))

>

> 3 - from 1 and 2 it follows for instance:

>

> <delta>t' = <delta>t'*(sqrt.(1-v^2/c^2))^2

>

> ...

>

> whence:

>

> (sqrt.(1-v^2/c^2)) = 1 which implies v = 0 .

[II] -- THIS IS THE SECOND CONTRADICTION WHICH

RESULTS FROM YOUR CALCULATIONS.

IT IS PRECISELY BECAUSE SUCH CONTRADICTIONS

ARE DERIVED that I claim that the Lorentz transformation

equations cannot be mathematically valid! (Note that the

Lorentz transformation equations can only be constructed

using such equations as Eq. 11.5 and 11.6 of the above-

mentioned Web page, which as you rightly point out above,

result in mathematical contradictions.)C.

I also made another careless mistake in saying at [5]:

> [5] The Lorentz length contraction formula is

<delta>x' = <delta>x*(sqrt.(1-v^2/c^2))^-1.I should instead have written:

> [5] The Lorentz length contraction formula is

<delta>x' = <delta>x*(sqrt.(1-v^2/c^2)).(See equation 11.7 at

<http://theory.uwinnipeg.ca/mod_tech/node136.html>.)

And I made a similar careless mistake in [9] -- which you

have already pointed out -- wherein I mistakenly wrote:> [9] According to the Lorentz length contraction formula, in the

> co-ordinate system associated with the buoy, the length of the

> spaceship will not be L but L', where L' = L*(sqrt.(1-v^2/c^2))^-1.But as you yourself noted, I CORRECTED both these

mistakes in [10] *et seq.* -- which AGAIN makes my

conclusion in [14] and [16] quite correct.Note also that with THIS formula ALSO one can obtain a

contradiction such as:> If this was true, then you would simply get

> immediately a contradiction like the following one:

>

> 1 - <delta>x' = <delta>x*(sqrt.(1-v^2/c^2))

>

> 2 - there is no difference whatsoever between x and x', Lorentz

> transformations have a group structure, the situation is SYM-

> METRIC, so one must have even:

>

> <delta>x = <delta>x'*(sqrt.(1-v^2/c^2))

>

> 3 - from 1 and 2 it follows for instance:

>

> <delta>x' = <delta>x'*(sqrt.(1-v^2/c^2))^2

>

> ...

>

> whence:

>

> (sqrt.(1-v^2/c^2)) = 1 which implies v = 0 .In any case, you surely cannot claim the equation <delta>x' =

<delta>x*(sqrt.(1-v^2/c^2)) to be wrong -- when written as "L'

= L*sqr(1-v^2/c^2)", as expressed in your e-mails -- because

you YOURSELF use this equation in calculating the length of

the spaceship as when it is in movement L' -- as opposed to

L !So if you YOURSELF have used the above equation to

calculate the length of the spaceship when in movement, how

is it that this very equation can give a contradiction such as

the one described above?Either you yourself must have used an equation, namely "L' =

L*sqr(1-v^2/c^2)", in your calculation of the contracted length

of the spaceship, which you must claim is wrong; or else if

you claim it is RIGHT, then your own words (as slightly

changed above for the sake of the context) prove that such an

equation results in a mathematical contradiction!

[III] -- THIS IS THE THIRD CONTRADICTION WHICH

RESULTS FROM YOUR CALCULATIONS.

SURELY NOW YOU MUST ADMIT THAT YOUR OWN

EQUATION CONTRADICTS MATHEMATICS AS WE

HUMAN BEINGS KNOW IT!And if again you say that the length contraction of the

spaceship can only be observed by a hypothetical observer on

the buoy, but never by a hypothetical observer on board the

spaceship itself, then the length contraction must be merely

*apparent* and not *real*.And if that is the case, then again, as I said, the Theory of

Relativity must be dealing only with *appearances*, and not

*reality*!And in that case, as I said earlier, the Theory of Relativity

cannot be a truly *physical* theory, which must deal with

*reality* and not mere *appearances*!D.

I did NOT make a mistake -- careless or otherwise! -- in the

FIRST part of [13], as you imply in the following words of

yours (in which you also quote my [13] for ease of reference):

On the contrary, it is YOU who made the mistake!Note that I had written:

> > [13] Using the Lorentz time dilation formula, we calculate

> > that E2will occur in the co-ordinate system associated with

> > the spaceship at t'2 = t2*[<gamma>^-1]. (Note: The same

> > result is attained if we use the Lorentz transformation

> > equation t' = (t - vx/c^2)/(sqrt.(1-v^2/c^2)),since at E2,

> > x is zero, and thus vx/c^2 is also zero.)To which you had replied:

> - This is WRONG. True that x = 0, but if you make the

> computation:

>

> t'2 = t2/sqr(1-v^2/c^2) ,

>

> by using the formula which you write above in your point

> 10, which I said is correct, then you get

>

> t'2 = (L'/v)/sqr (1-v^2/c^2) = (L*sqr(1-v^2/c^2)/v)/sqr(1-v^2/c^2)

> = L/v

>

> (as I already wrote to you in my previous mail!)

>

> which implies:

>

> t'2 = t2*gamma [and not gamma^-1!!]Here -- look carefully, now! -- it is YOU who made a mistake!

For on the one hand you say that (i) t'2 = t2/sqr(1-v^2/c^2) --

which is ESSENTIALLY THE SAME as my ORIGINAL [4] --

is WRONG (and which I ADMIT is wrong), and yet (ii) you

make USE of my WRONG original [4] to arrive at YOUR

answer "t'2 = t2*gamma [and not gamma^-1!!]"!How then can your answer NOT be wrong?

YOUR OWN WORDS prove that YOU MUST BE WRONG

in what you have written above, and that I was actually RIGHT

in the FIRST part of my [13] !

[IV] -- THIS IS THE FOURTH CONTRADICTION WHICH

RESULTS FROM YOUR CALCULATIONS.

E.

I DID make a somewhat more serious, but still not fatal,

mistake in the SECOND part of [13], which I quote once

again in full for ease of reference:> [13] Using the Lorentz time dilation formula, we calculate that

> E2 will occur in the co-ordinate system associated with the

> spaceship at t'2 = t2*[<gamma>^-1]. (Note: The same result is

> attained if we use the Lorentz transformation equation

> t' = (t - vx/c^2)/(sqrt.(1-v^2/c^2)),since at E2, x is zero, and

> thus vx/c^2 is also zero.)The SECOND part, between the parentheses after the word

"(Note: ...)", does not agree with the FIRST part. I made a

mistake in saying that it does.But the important question is, WHY does the second part of

[13] not agree with the first part of [13], which is based on my

COR- RECTED [4], and which in turn is based on Equations

11.5 and 11.6 of the above-noted Web site, and which in *its*

turn is based on a simple *Gedankenexperimente* conducted

on the postulate of the constancy of the speed of light?For it is to be noted that if the result obtained by ME using

my corrected [4] -- which is based on equations 11.5 and 11.6

of the above-noted Web site -- is *different* from the result

obtained by YOU using the Lorentz transformation equation

for time, then there must be a CONTRADICTION between

the Lorentz trans- formation equation for time on the one

hand, and the postulate of the constancy of the speed of light,

on the other!But how can that be, if the Lorentz transformation equations

are THEMSELVES obtained by using the postulate of the

constancy of the speed of light?The answer is to be found in YOUR mistake -- and this is the

BIG one, which reduces the value of your ENTIRE argument

to zero! -- namely, to consider the so-called "proper" time as

being the time RECORDED by the two stop watches -- the

one on the buoy and the one at the mid-point of the spaceship.

THIS IS JUST NOT THE CASE.(Remember I had written earlier that I had NOT asked for any

such thing as "proper time" in my Challenge? I think it is fair

to say -- and I am sure you would agree! -- that anyone who

does not actu- ally READ my Challenge before ANSWERING

it will not have TRULY answered it at all!!)I NEVER asked for "proper time", and your introducing this

notion into your answer is the source of your BIG mistake.But if you INSIST on bringing the notion of "proper time"

into the argument, then please note that the DEFINITION of

"proper time" is as follows:"... the time which is measured by an observer who is

present at the same location as the events which mark the

beginning and the ending of the event. Another way to

say this is that the proper time is measured in a reference

frame in which the events occur at the same spatial point.

... Time intervals as measured by any other inertial

observers are always greater than the proper time."(See < http://aci.mta.ca/Courses/Physics/4701/EText/Proper.html>.

This is from Mount Allison University of New Brunswick, Canada .)Now this term "proper time" is really the source of one HUGE

problem -- among several -- in the Special Theory of

Relativity. Even the TERM "proper time" is a misnomer: it

should instead be called "IMproper time". Because according

to this definition -- which I would call a very stupid

definition! -- a clock observed from a great distance would

not be ticking at a rate that indicates the "proper time" at all!A clock observed from a great distance -- say, a gigantic clock

on the moon observed through a very powerful telescope on

earth -- does not *really* show a different time from an

identical clock on earth which is synchronised with the

former: it only *appears* to the earthly observer to show a

different time (ticking about two seconds slow compared to

the identical and synchronised clock on earth.) That is simply

because it takes light from the clock on the moon about two

seconds to each the earth, whereas it takes light almost no

time at all to reach the observer from the clock on earth!But to grasp this, one has to truly understand what the term

"synchronised clocks" means. When two clocks are *really*

(or *actually*) synchronised, it means they are *actually* or

*really* displaying the same reading simultaneously. They do

not, however, always have to *appear* to be displaying the

same reading simultaneously! If the clocks are at different

distances from the observer, then the clock which is farther

away from the observer will *appear* to be displaying its time

with a delay compared to the display observed on the other

clock. And the magnitude of this delay will depend on the

difference in the distances of each of the clocks from the

observerBut that will only be an *apparent* difference in the two

displays, not a *real* one! It will be due to the TIME

DIFFERENCE THE SIGNAL TAKES TO REACH THE

OBSERVER FROM EACH CLOCK -- much like in the case

of lightning and thunder, which *appear* not to be

simultaneous, though everyone knows that they *are*, and

indeed MUST BE and cannot be otherwise.Now the Lorentz transformation equations I had given at [6],

which I mistakenly applied in the second part of [13], are

obtained not ONLY by applying a *Gedankenexperimente* to

the postulate of the constancy of the speed of light, but ALSO

a second *Gedankenexperimente*: the one on which the

argument that "proper time" REALLY exists is based! This,

as I said, is Einstein's well-known "Train *Gedanken-

experimente*" regarding the lack of simultaneity of lightning

flashes for different inertial observers. The logic of this

*Gedankenexperimente*, when examined carefully, can only

prove that the famous (or maybe we should say, infamous!)

"lack of simultaneity for different inertial observers" can only

*apparently* exist, and not *really*.And why is this? It is because the two observers in this

*Gedankenexperimente* do not observe the lightning flashes

at the precise LOCATIONS at which they occur, but at a

DISTANCE from those locations!And to arrive at the conclusion that the two observers do not

observe the lightning flashes simultaneously, Einstein forgets

to take into account the ACTUAL and CALCULABLE

difference in time it takes for the light from each flash to reach

each of the observers!He pretends that since one of the two observers *perceives*

the lightning flashes simultaneously while the other does not,

that the flashes *are* in fact simultaneous for one observer and

not for the other. But that is absolute nonsense. It is almost as

bad as arguing that since we do not *perceive* the thunder at

the same time as we perceive the lightning, the lightning and

thunder are in *reality* not simultaneous. Just because light

travels much faster than sound doesn't mean that light takes

absolutely NO TIME AT ALL to travel from one location to

the other!Indeed in the same *Gedankenexperimente* Einstein could

also argue -- and validly -- that the fellow on the platform

would hear the two THUNDERCLAPS simultaneously, while

the fellow on the train would hear them non-simultaneously!

And yet the difference in time between the two thunderclaps

for the fellow on the train would be GREATER than the

difference in time for the two lightning flashes! Would

Einstein be so stupid as to claim, as a result, that the

thunderclap from one side occurred *before* the lightning

flash that gave rise to it, and/or that the thunderclap from the

other end occurred *after* the lightning flash that gave rise to

IT?No, of course, not. Any reasonable and realistic person would

argue that it was absolutely necessary, in order to know at pre-

cisely what instant in time the flashes and the thunderclaps

ACTUALLY occurred, to take into account NOT ONLY the

time at which each of them was *observed* to have occurred,

but ALSO the speeds of the signals -- whether light or sound -

- which conveyed the relevant information to the observer, as

well as the distances over which these signals had to travel!(But of course Einstein was neither reasonable nor realistic, so

he actually *might* have argued that the thunderclaps occurred

at a time different from the lightning flashes that gave rise to

them ... ! ;-) And then we could have all had a huge laugh at

his expense, and at Relativity as well.)The fact is that Einstein developed this half-baked idea in

order to remove what he called "the apparent" incompatibility

of the law of propagation of light with the principle of

relativity. It is not an *apparent* incompatibility at all, but a

very *real* one!And as for Einstein's even more stupid remarks in Chapter

VIII of his well-known book "Relativity: The Special and

General Theory", viz.:After thinking the matter over for some time you then offer

the following suggestion with which to test simultaneity.

By measuring along the rails, the connecting line AB should

be measured up and an observer placed at the mid-point M

of the distance AB. This observer should be supplied with

an arrangement (e.g. two mirrors inclined at 90 deg.) which

allows him visually to observe both places A and B at the

same time. If the observer perceives the two flashes of

lightning at the same time, then they are simultaneous.I am very pleased with this suggestion, but for all that I

cannot regard the matter as quite settled, because I feel

constrained to raise the following objection: "Your defi-

nition would certainly be right, if I only knew that the

light by means of which the observer at M perceives the

lightning flashes travels along the length A --> M with the

same velocity as along the length B --> M. But an exami-

nation of this supposition would only be possible if we

already had at our disposal the means of measuring time.

It would thus appear as though we were moving here in a

logical circle."... does Einstein forget that we have plenty of ways of

measuring time that have nothing to do with the speed of

light? What then are pendulums, tuning forks or caesium

atoms? The "examination of the proposal" he speaks of is

eminently possible! There is absolutely NO logical circle here

at all.Besides, his entire objection in his second paragraph quoted

above deals with the speed of a LIGHT signal. But is it not

necessary to use a LIGHT signal in order to judge whether two

events occur simultaneously! One could use sound, or a

rolling ball-bearing ball, or any other signal whose speed -- or

even rate of change of speed over time or over distance -- is

unchanging and actually known with precision.For instance, if the two lightning bolts were arranged to each

set a ball-bearing ball rolling down an inclined and polished

steel groove towards the observer, then by measuring the time

difference between the two times when the two balls arrive at

the observer, and if the rate at which they roll down their

respective inclined groove were also known with precision, the

observer could STILL make the necessary calculations to

judge whether the two lightning flashes *actually* occurred

simultaneously or not!(Note also that in the above case the balls would not even be

rolling at the same speed throughout their trips! And neither

would they be required to both reach the observer at the same

instant in time, even if the two lightning flashes WERE in

fact simultaneous! All that would be needed would be to

know with precision how MUCH time it takes any GIVEN

ball to roll down any GIVEN length of any GIVEN groove.

And of course one should remember that this method would

not be too accurate, because the margin of error might be too

great to judge simultaneity any better than with optical means.

But it just illustrates that ANY signal may be used for the

purpose. A quite adequately accurate signal, as already

described in my previous e-mails, could be an electrical one in

a wire of known length. Of course NO signal would ever be

100% accurate, but that's beside the point: absolute accuracy

in judging simultaneity impossible even when events occur

right in front of the observer's nose!)The whole problem lies in EINSTEIN'S definition of

simultaneity, namely "If the observer PERCEIVES the two

flashes of lightning at the same time, then they ARE

simultaneous." (His own words, no less!) This essentially

implies that "Perception IS Reality" -- which idea, when thus

expressed, does not even pass the laugh test, at least in

physics.(Of course in ADVERTISING he is right: and Einstein,

though a lousy physicist and mathematician, was indeed a

great PR man and an advertising genius: witness his hairstyle,

which I have myself adopted, and which has made him

famous even among government bureaucrats, who haven't the

foggiest idea what he actually wrote!)You see now: it is not *I* who do not understand

RELATIVITY well enough: rather it is *you* -- and Einstein

-- who do not under- stand REALITY well enough! (As I said,

you should have asked any competent engineer about this.)And when Relativity contradicts Reality, it is Relativity that

must give way and not Reality! For Reality mocks, not just at

Relativity, but at ALL so-called "theories" which do not

conform to it!So the difference between my [4] and the Lorentz

transformation equation for time is that the former is derived

ONLY from the postulate of the constancy of the speed of

light, while the latter is derived from BOTH, (i) the postulate

of the constancy of the speed of light AND (ii) the argument

used by Einstein in his above- mentioned "Train *Gedanken-

experimente*" to try and show that simultaneity at a distance

is impossible. (To see how the Lorentz transformation

equations is derived using not just one, but BOTH, of these

hypotheses, see the Web article entitled "Construction of the

Lorentz Transformation" which can be found at the URL

<http://origins.Colorado.EDU/~ajsh/sr/construction.html> .)In my Challenge I have got RID of this nonsense of "proper

time" which is derived from the notion of "lack of

simultaneity-at-a- distance" -- and which in turn implies that

an *apparent* lack of simultaneity is in fact a *real* one. I

have done this by insisting that there be only ONE stop watch

in the spaceship (note again, this is not a "clock" but a *stop

watch*), placed exactly at the MID-POINT between the ends

of the spaceship. And I have insisted that the signals to start

and to stop the stop watch both travel from the relevant sensor

to the stop watch at the SAME speed (whatever that speed

may be) over the SAME distance (whatever that distance may

be) in the SAME Inertial Frame of Reference or IFR (once

again, whatever that IFR may be)!So THIS stop watch does not measure the "proper time" of

ANYTHING! Instead it measures the *actual* time interval

between the events E1 and E2. The actual time interval

recorded by this stop watch is NOT the so-called "proper

time" at all.As I already wrote in previous e-mails, this stop watch at the

mid- point of the spaceship measures the time interval with a

DELAY compared to the stop watch on board the buoy, for it

will take the signal indicating that E1 has occurred a FINITE

AMOUNT OF TIME (as measured in the Inertial Frame of

Reference -- or IFR -- of the spaceship) to reach the stop watch

from the bow. But since the signal that E2 has occurred will

also take the exact SAME amount of time (as measured in the

same IFR) to reach the stop watch from the stern, these two

delays will CANCEL EACH OTHER OUT!So if -- to illustrate -- the speed of the signal (in the IFR of

the spaceship) is u, and the length it has to travel (again in the

IFR of the spaceship) is l, then the stop watch on board the

spaceship will START ticking with a delay of l/u compared

with the stop watch at X on the buoy. But the stop watch on

the spaceship will also STOP ticking with a delay of l/u

compared with the stop watch at X on board the buoy! So the

time INTERVAL which the two stop watches will record will

be the *real* or *actual* time interval which the spaceship

takes to pass the X marked on the buoy.You STILL haven't grasped this ELEMENTARY point,

which I have repeated many times. You do not even

MENTION it in your replies! But THAT is where one of your

major mistakes lies. You apparently do not read, or give any

indication that you understand, my arguments.Of course, if we use this method of measuring the time

interval [E2 - E1] in the system of co-ordinates associated

with the spaceship, my calculation given in the second part of

[13] is NOT correct, because the single stop watch at the mid-

point of the spaceship cannot be at E1 when it takes place, nor

can it be at E2 when IT takes place. Thus it cannot measure

the "proper time" of E1 *or* of E2. The second part of [13] is

a formula based on the notion of "proper time", and since

there IS no proper time in this case, this notion and this

formula is not applicable at all here.(I DID make a mistake in thinking initially that it IS

applicable: I admit that. And it was a serious mistake. But

even so, it does NOT vitiate my conclusions in [14] and [16]!)But as an alternative -- as you also note -- I could also have

per- formed the same calculation by putting two

SYNCHRONISED clocks (N.B.: this time, it is *not* stop

watches), one at the bow and the second at the stern. I would

arrange to have them photo- graphed automatically on two

different occasions: first, when the bow of the spaceship

passes the X marked on the buoy, a photo- graph would be

taken of the clock at the bow; and the second time, when the

stern of the spaceship passes the X marked on the buoy, a

photograph would be taken of the clock at the stern.

Subtracting the reading of the photograph of the stern clock

from that of the photograph of the bow clock will also give

the time interval exactly as calculated by me.But before you even START to object, please, please, P L E

A S E note that to conduct such an experiment correctly, these

two clocks have got to be SYNCHRONISED. Since in your

e-mails YOU do not give a way to synchronise clocks which

are not moving relative to one another, I shall do so myself.

In order to synchronise such clocks, one sends a SIGNAL

whose speed is KNOWN from one clock to the other along a

distance whose length is KNOWN, and by dividing the latter

by the former, calculates the time interval it takes for the

signal to go from one clock to the other: and one ALLOWS

FOR THAT TIME INTERVAL WHEN SETTING THE TWO

CLOCKS!For example, if from the stern of the spaceship one observes,

say through a telescope, the clock at the bow reading 0.00

seconds, the clock at the stern should be set to read at that

moment -(L/c) seconds! Then although a hypothetical observer

at the stern will not *observe* the two clocks ticking

simultaneously, they will in fact *actually* and *really* be

ticking simultaneously.It is THAT simple. No need for Special Relativity at all!

(I have already written much of this in previous e-mails and

given several details from which the above can easily be

deduced, and so I won't repeat myself here any further: I have

already said more than enough.)So you are COMPLETELY WRONG when you write in your

"FINAL REMARK 3":> ...

> it [i.e., the clock at X in the buoy] does not know anything

> DIRECTLY about the clock of the stern, which was SOME-

> WHERE ELSE. If this time, let us call T , was zero, then we

> would have got problems, but this time is not zero. ...As a matter of fact the time which you have called "T" IS zero,

and there are N O P R O B L E M S ! ! !None WHATSOEVER. (If you can think of any, let me know!)

[V] -- SO THIS IS THE FIFTH CONTRADICTION. It results

more from Einstein's thinking than from yours, but since you

are supporting Einstein in the nonsensical notion of "proper

time", it is yours also.

But again, it is of course true that to a hypothetical observer at

X, for example, or at the bow of the spaceship, it will not

*appear* that the two clocks in the spaceship -- the one at the

bow and the one at the stern -- are synchronised with each

other at time t = 0.00. But that will only be an *apparent*

lack of synchronicity, because of the time it takes for the light

to reach the observer at the bow from the stern -- not a *real*

lack of synchronicity, any more than there is indeed a *real*

lack of synchronicity between thunder and lightning, even

though we see the lightning a short time before we hear the

thunder.(Heck, I figured this out from the very first time I read

Einstein's argument, over forty years ago, when I was still in

my teens! I just didn't make an "waves" about it until now

because I always thought that Einstein, being a certified

genius, must surely have had some additional argument up his

sleeve to counter what I had figured out. Only this year, when

I made a serious effort to study the Theory of Relativity in

every single detail, did I find out that the genius did NOT

have any such argument!)But note that a hypothetical observer at the MID-POINT of

the SPACESHIP will, if he uses a telescope and a simple

system of mirrors to see both clocks together, OBSERVE

them both as being synchronised -- though of course this

observation will take place with some DELAY from the

ACTUAL or REAL time that the ticks do take place.F.

Indeed you have hit the nail on the head with your remark in

your "addendum":> ... one has a great and still alive discussion about the "physical

> of interpretation"the formulae above. These contractions and dilations

> are in some sense REAL, or they just depend on the assumptions

> introduced by Einstein, and followers, in order to synchronize

> clocks (using the II postulate of special relativity), to measure

> lengths of moving objects ...... but you are mistaken when you continue on by saying:

> but I rather believe that in this case relativity IS MAKING SOME

> PRECISE PHYSICAL ASSERTION ...... which is wrong.

But I see that you are open-minded enough about it to say:

> ... which could be put to test (of course, from an "ideal", if today

> not yet "practical", point of view).However, I would like to point out that there is ALREADY

today a physical test and a PROOF that Relativity is wrong

about this: namely the possibility -- which astronomers have

known since before Einstein was even born -- that the EXACT

POSITIONS of ALL the planets can ALL be calculated with

GREAT PRECISION: so great a precision indeed that we can

now even land spacecraft on a precise predetermined location

on Mars! This would be impossible if we could not calculate

just WHERE Mars would be at any given time as measured

on earth, EVEN THOUGH WE CANNOT OBSERVE

EITHER MARS OR OUR SPACECRAFT AT THAT

PRECISE TIME, namely the moment when they are

ACTUALLY at the location where they meet!Therefore there must be the true or real simultaneity possible

ALL THROUGHOUT THE SOLAR SYSTEM, even though

all the planets are separated by many light-minutes and even

light-hours.So the fact that the Theory of Relativity is contradicted by

this very elementary fact is ALREADY a physical test

disproving it, which should satisfy your longing for a

PHYSICAL proof of the falsity of the Theory of Relativity.

[VI] -- SO THIS SIXTH CONTRADICTION CONTRADICTS

THE NOTION OF "PROPER TIME", according to which

the time which is measured by an observer who is NOT

present at the same location as the events which mark the

beginning and the ending of the event does NOT measure the

"proper time".

And the fact that Special Relativity is MATHEMATICALLY

self-contradictory is proven by my arguments in my previous

e- mails, and also above.G.

And if anyone is interested enough to discuss my PHILOSO-

PHICAL argument against the postulate of the constancy of

the speed of light regardless of the speed of the source of the

light or of the observer thereof, I can also do that: namely that

such a postulate cannot be part of mathematics and geometry

as we know them, since it contradicts the Galilean

transformations which *are* a part of mathematics and

geometry as we know them, and since mathematics -- and

geometry -- cannot be self-contradictory. (I see that you have

avoided mentioning anything against THIS argument of mine

in any of your e-mails, though I consider it to be even

stronger than my "Challenge" as an argument against Special

Relativity.)

[VII] -- AND THIS IS THE SEVENTH CONTRADICTION:

BETWEEN THE AXIOMS AND THEOREMS OF

MATHEMATICS AND THE LORENTZ

TRANSFORMATIONS.

Now it is up to you, of course, whether you wish to reply to

this e-mail or not: but I for one am so confident that I am

right and you are wrong, that I am willing to put all of our e-

mail correspondence regarding Relativity on the Web, so that

any persons with competence in the fields of physics, logic

and mathematics -- in ALL of these fields, of course, not in

just one or two of them -- who might find themselves reading

our correspondence can judge who is right and who is wrong!But in closing let me assure you that I remain always your

friend in our search for a strong argument against Relativity.

The only difference between us is that I am pretty sure (not

100% sure, of course, but *almost* sure!) that I have FOUND

it, and anyone who actually reads and understands my

arguments will see that this is so; while you STILL do not

understand just HOW I have found it, and so you THINK I

have NOT found it!So, in friendship, as always, I remain,

Yours

Ardeshir <http://homepage.mac.com/ardeshir/education.html>.

______________________________________________________

Subject: Re: Come si deve fare la matematica (Your reply in

Italian, including errata, translated by me into English)

Date: Sat, 29 Sep 2001 16:33:40 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12

Dear Mehta,

Here quickly is my reply to some of your claims.

> There is not -- and never has been, and probably never will be --

> ANY scientific theory so completely true that it can NEVER be

> refuted by an intelligence clever enough to find a flaw in it! And I

> claim to have found just that in the Theory of Relativity.NO!

> But to grasp my arguments you must first *read carefully and

> digest thoroughly what I am saying*: which I am sorry to see you

> continue to neglect doing.You surely know very well that there are thousands of people

who write thousands of pages, claiming that they contain

marvellous things capable of advancing knowledge, and claim

to be read by everyone, accusing others of treating them like

Galileo in case they are not! But before dedicating one's time

to the study of something one should have a fair hope that the

time will be well spent, OR ELSE ACCORDING TO YOU,

NO, and one must read EVERYTHING that arrives?

(extravagant arguments regarding perpetual motion,

demonstration that the Greek "pi" is a rational number, etc.).

There are in fact some of my colleagues who after having read

some lines such as these throw them into the trash-can right

away (one must also admit that there are competent persons

who write nonsense, but a competent person can tell that

quickly), or do not even take them up in hand. I instead read

you in general with great attention, but NOT INFINITE. I

have already told you frankly that I think that you, as a

student, should be "failed" for what you have written, but

naturally you may refute my judgement in this case, even if it

does harm, and it would be better to study further certain

questions before asserting that scientists commit the errors

you accuse them of committing. Incidentally, I have read at

your site the material concerning Cantor's Theorem on the

denumerability of real numbers, but it is possible that no one

has ever told you that what you have written in this case is

without any sense, even worse than what you have sent me

concerning Relativity? You can try sending these reflections to

ANY mathematician in the world and see what he replies. You

write in fact a natural number with an INFINITE number of

digits!> Now consider the natural number X = x1x2x3x4x5 ... xk ... ,

> where x1 is any digit other than d11; x2 is different from d22;

> x3is not equal to d33; x4 is not d44; and so on. Now, X is

> a natural number, so it should be in our list. But where is it?- Obviously your X is NOT a natural number - it is not even

any number!, and you should address to yourself your

conclusion that: "One wonders where their authors' and

editors' heads are at!"Returning to the preceding discussion, though it is true that

one risks in some cases ignoring, or persecuting in certain

cases, a person such as Galileo, it is also necessary to defend

oneself from those who make so many errors and do not wish

to admit them, do not wish to recognise them with an

incredible and presumptuous obstinacy. Or are you not in

agreement with my words? What other attitude would it

suggest to a professor in front of students who SHOULD be

failed?Now I come without further delay to the subject of Relativity,

saying once again that you obviously are mistaken in all your

considerations, despite having some merit in having

understood that there is something that does not seem right.

But we are dealing with a problem much more complicated

than you think, which can be confronted only from an

experimental point of view, not logical and mathematical, a

problem which can be quickly solved with a very simple

argument, but all the same INVALID, and which as a result

does nothing but increase the strength of the Relativists, who

have good reason to say: See, those who are not in agreement

with us are not only not those who do not understand science,

and make huge errors ...> Let me see if *I* have understood YOU correctly. (It would

be wrong > of me to point the finger while making the same

mistake you do!)In all that follows you report correctly enough what I have

indicated should be the right result, but you seem not to

understand that all the expressions you use, such as " if the

spaceship is supposed to be MOVING", have no exact

meaning. They have meaning only to introduce special co-

ordinate systems, in which certain objects can be in motion or

not, but these co-ordinate systems are only AUXILIARY, for

the purposes of being able to perform the calculations better. If

the calculations are performed in ONE system, one knows that

the result obtained must be the same in ALL systems, I could

for example use a SINGLE system in which BOTH bodies,

the spaceship and the buoy, ARE MOVING, and all the same

I would get the same result. When the clock is in the rear (the

poop), of the spaceship, and the one on board the buoy are

next to one another, they indicate different times, to be exact,

the one in the spaceship L/v , the one on board the buoy L'/v ,

period, there cannot be any doubt, and it is quite obvious that

it is so.The fact is that to you this conclusion still seems

contradictory to other things such as time dilation, but which

you know only imprecisely (a contradiction in a claim can be

revealed only along with other elements, the claim by itself

not enough), despite my preceding efforts to explain. In fact

you write:> But A-v contradicts the Theory of Relativity, according to which

> a stop watch which is MOVING must run slower than the SAME

> stop watch when it is STATIONARY! (Remember, we are speak-

> ing here about the very SAME stop watch measuring the SAME

> time interval.)And with these words you demonstrate once again that you

have not understood the Theory of Relativity, like thousands

of people like you. In Minkowski space-time a clock is neither

in movement nor is it not in movement: it "is", and that's all:

a clock is a geometrical CURVE of a particular type, which

can be a (geodesic) straight line or not, and the proper time is

nothing more than a certain "length" of this curve between two

of its points (events). The clock is in movement or not only

in respect to other clocks, in other curves, (it would be

different if one were to think for example of the ether, but that

would be quite another discussion.) But what can one do? I

cannot explain better than I have already done. I have already

told you that the preceding result demonstrates in an obvious

manner that time in the spaceship seems to go more slowly

relative to a clock situated in the buoy, because clearly L/v >

L'/v , and it is a little less easy to understand that, according

to the Theory of Relativity, without any contradiction, TIME

IN THE BUOY ALSO SEEMS TO GO MORE SLOWLY,

and by the same factor, relative to a clock in the spaceship!

The fact is that the preceding inequality is undoubtedly valid -

it is ONE only, and cannot be reversed - and does NOT mean

anything, because it is not necessary to see in ONE SINGLE

INSTANT that a clock indicates less than another to say that

this clock runs more quickly (I speak from the point of view

of the "poop" of the spaceship, which sees the clock in the

buoy indicating a lesser time than its own.) That clock

perhaps goes more slowly! The fact is that it is necessary to

calculate the DIFFERENCE between TWO different

INSTANTS , as I have demonstrated in my preceding e-mail

(which I can re- send to enable you to study it better), and one

finds that the spaceship also finds the clock in the buoy going

more slowly, as it must according to Einstein's Theory! (and

note that we are not speaking of EXPERIMENTS, but only of

a purely mathematical theory of space-time, whose fault, if

any, is to claim to describe "reality".) I can anticipate a less-

perspicacious student's doubt: How is it that in the first case,

that is to say in the conclusion that there is a time dilation in

the spaceship as seen from the buoy, we haven't spoken of the

these TWO instants which are necessary? But THE FACT IS

THAT we have spoken about it implicitly when we have

supposed that the clock in the buoy and the clock in the

FRONT (that is, the prow) of the spaceship indicated the same

time, for example zero, when they are next to one another, and

there you are, the whole "mystery" is explained to him who

succeeds in understanding it, or has the necessary humility to

exert himself to learn it ...I wish to emphasise at this point in our correspondence (this

file dedicated to you has come to 156 pages) that you will

NOT remain persuaded what I have said, even though perhaps

it is owing to some problems due to the language, and due to

the fact that writing continuously I could have made some

mistake or confused something. But on one point I would be

curious to get a non-scientific comment from you: WOULD

YOU ADMIT, OR WOULD YOU NOT, THAT IT IS RIGHT

FOR A PROFESSOR SOMETIMES TO FAIL SOMEONE?

(obviously one can always be mistaken.) Or do you think that

all should be promoted?I hope you won't find my frankness offensive if I add sincerely

that I do not consider our discussion to be between peers, just

as I don't feel when I explain things to students (who can be

of any age, I am myself a student in many things) that the

discussion is between peers. In certain cases it is I who know

and the other who must learn, from me or from books, at least

up to a certain point, until it is demonstrated that they have

arrived at a peer level, or superior, and then can request

attention to suggest something of possible interest. This

discussion instead is NOT INTERESTING, except as an

didactic exercise for me, and as an eventual help for you to

learn and correct your evident errors, but it does not seem to

me that you are disposed to do that ...Always in friendship, yours, UB

(If you wish, I can give you addresses of professors who are

certainly more expert in the matter than myself, to whom you

can address yourself to verify if they do not say the same

things I do: but you will have to write to them in extreme

synthesis, otherwise they will not read it ...)Short Errata Corrige:

I had written:

> I have already told you that the preceding result demonstrates

> in obvious manner that time in the spaceship seems to go more

> slowly relative to a clock situated in the buoy, because clearly

> easy L/v > L'/v , and it is a little less to understand that,

> according to the Theory of Relativity, without any contradiction,

> TIME IN THE BUOY ALSO SEEMS TO GO MORE SLOWLY,

> and by the same factor, relative to a clock in the spaceship!I should have written the opposite instead! L/v > L'/v

demonstrates in fact that the clock in the buoy seems to be

SLOWER with respect to THOSE in the spaceship, precisely

because L'/v is smaller, and that is to say less time has passed

in the buoy than in the spaceship, which is to say, for the

slower clock. Vice versa, it is more difficult to understand that

the clocks in the spaceship also seem to be slower than the

one in the buoy, DESPITE the fact that when the clock in the

poop of the spaceship encounters the one on board the buoy,

the one in the spaceship indicates MORE time than the one on

board the buoy (which would make one believe, as in fact you

say, that the clocks in the spaceship are FASTER than the one

on board the buoy) ...I wish to emphasise the word "THOSE " (plural)! There have

to be in fact TWO distinct clocks in two distinct positions to

verify the phenomenon of which we speak, and your entire

"paradox" is in fact based on a sort of ASYMMETRY: in the

spaceship there are TWO clocks, while in the buoy there is

ONE only. If we leave things thus, then certainly the buoy

cannot experience the time dilation which takes place on the

spaceship, and the fact that it does not experience it does not

mean it doesn't take place!Cordially as always, and excuse me for the mistake,

UB

_______________________________________________________

Subject: Re: Come si deve fare la matematica (reply in English)

[How Mathematics Should be Done (reply in English)]

Date: Sat, 29 Sep 2001 16:10:09 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12

Dear Professor:

I am making an effort to write concisely.

I.

As regards Galileo, etc.: as Bernard Shaw says, it is true that a

lot of people speak nonsense, but that doesn't mean that ALL

people speak nonsense. But to know who does and who

doesn't, one has to listen to them all!(And besides, as you yourself say, one should always suspect

that even professors and so-called "experts" sometimes do

speak and write nonsense.)I have already specified several contradictions in your

preceding words, to which you do not seem disposed to reply

to me. Well, okay -- but I cannot see how, in that case, you

can claim to know more than I do in this matter. (Perhaps you

do not know that in mathematics, not even ONE single

contradiction is permitted?)Besides, I am not your STUDENT but your INTELLEC-

TUAL ADVERSARY -- a FRIENDLY adversary, of course,

but an adversary nevertheless. I have CHALLENGED you

intellectually: and in such a challenge, I have to exercise rather

forcefully my intellectual ability, and I must NOT accept

everything you say -- at least not unless you can PROVE what

you say.(You do not seem to understand the difference between an

intellectual challenge and an instruction. In a challenge,

especially one which involves mathematics, if I contradict

myself, I'd have "failed" and you'd have won, while if you

contradict yourself, you would have "failed" and I'd have won!

This is what is called the "failure criterion" of the challenge.

And I have already pointed out several contradictions in your

words, so it is obvious that you have failed, at least

mathematically.)Moreover, although you are certainly at liberty to IGNORE

my challenge, ignoring an intellectual challenge seems to me

to demonstrate a clear lack of intellectual courage!In any case, if you think I have NOT won my challenge

already, we can ask the world community of rational people to

judge between us: as indicated by me before, we should

publish our entire correspondence on the Web! I am sure that

ANY rational person who actually reads and understands all

that I have written will agree that I am in the main right, and

that you have contradicted yourself. (I might have made one or

two small mistakes here and there, but they would not nullify

everything I have said.)II.

Incidentally, as regards the idea of perpetual motion, which

you ridicule: your words made me smile. Every physicist

knows full well that there are MANY systems in existence

which are clearly in perpetual motion! In the microcosm some

of them are called "atoms". Aside from the obvious fact that

EVERY atom possesses electrons in perpetual motion, even

the atoms THEMSELVES cannot CEASE to be in perpetual

motion -- it is precisely because of this perpetual motion of

atoms that we cannot solidify Helium, even at a temperature

of absolute zero (i.e., at 0 degrees Kelvin)!And if we are to speak of the macrocosm, the Universe itself

is a system in perpetual motion. Ever since the Universe

began (whether with a "Big Bang" or otherwise), everything in

it has been in motion, and will remain in motion till the very

end of time.And this means that all these things will be moving for ALL

time -- i.e., in PERPETUAL motion!It is in fact perpetual REST that is impossible, not perpetual

MOTION.In any case, whoever finds arguments about perpetual motion

"extravagant" cannot be living in the real world!III.

Also incidentally, as regards my critique of Cantor's theorem

at:(<http://homepage.mac.com/ardeshir/ArgumentAgainstCantor.html>)

... you are mistaken when you say that I write "a natural

number with an INFINITE number of digits". Or to be more

accurate, you are mistaken in not specifying WHAT KIND of

infinity you are talking about: potential or completed.You being a mathematician must surely know that according

to Aristotle's and Kant's arguments, there cannot be a

COMPLETED infinity, only a POTENTIAL one. It is

certainly true that a natural number cannot have a

COMPLETED infinity of digits, for the simple reason that a

completed infinity cannot exist! But for the SAME reason,

neither can a real number possess a COMPLETED infinity of

digits. Both natural numbers AS WELL AS real numbers can

have only a POTENTIALLY infinite number of digits.If natural numbers can have only a POTENTIALLY infinite

number of digits, it is certainly possible to construct a table

using natural numbers, exactly in the same manner as Cantor

has constructed a table using real numbers!As I wrote recently to Allan Gillis, who teaches mathematics

at the Faculty of Mathematics, Ryerson University, Toronto,

Canada (and who incidentally is in complete AGREEMENT

with me!):Essentially, Cantor tries to make his argument by confusing

a POTENTIAL infinity -- that is, an infinity according to which

there can always be one digit more than the digits actually

specified -- with a COMPLETED infinity, that is to say one

according to which all the infinite number of digits are actually

SPECIFIED. As Aristotle and Kant pointed out, while the former

is possible, the latter is impossible.As you can see, there ARE mathematicians who are in

agreement with me in this regard: in fact I am not even the

first person in the world to have described this argument

against Cantor's theorem, and I have already discussed my

article with several other people who know mathematics very

well -- and who know also that in mathematics it is not

permissible to have even one single contradiction, which you

don't seem to have understood. (Remember that you had

written: "You can try sending these reflections to ANY

mathematician in the world and see what he replies.")IV.

Now regarding the Theory of Relativity: as I said, I see that

you do not wish to discuss some of the contradictions in your

previous e-mail, the ones which I have indicated in MY

previous e-mail by the numbers [II] to [VII].Now in science as in mathematics, even a SINGLE

contradiction suffices to invalidate a mathematical theorem or

a scientific theory.So if you do not reply to the contradictions which I have

pointed out, I must assume that you do not contest them.

(Certainly it is your right to NOT respond to my arguments --

but NOT TO RESPOND to an argument is certainly not the

same thing as REFUTING it!)On the other hand if you do contest my above-mentioned

points, how can you deny all the contradictions I have pointed

out in them? I see no RATIONAL way for you to do that.V.

As regards the ONLY point to which you HAVE responded,

you are mistaken once again. Note that you had written:> A clock is not in movement or at rest, in Minkowski space-time: it

> "is", and that's all: a clock is a geometrical CURVE of a particular

> type, which can be a (geodesic) straight line or not, and the proper

> time is nothing other than a certain "length" of this curve between

> two of its points (events). The clock is in movement or not only

> in respect to other clocks ...This is certainly FALSE, and in fact I believe you know so

too!ANY Minkowski space-time can be represented by a

Minkowski diagram.And in ANY Minkowski diagram one can ALWAYS

represent at least ONE object which is NOT in motion

according to that diagram. In such a diagram this object would

be represented by a VERTICAL STRAIGHT LINE -- i.e., a

line parallel to the time axis.If after a period of time this SAME object were to be set in

motion, the line in the Minkowski diagram would not remain

parallel to the time axis any more!A clock which is NOT in motion, and THE SAME CLOCK

when it is moving, can thus be represented in the SAME

Minkowski diagram: TWO diagrams are not needed.Which proves without any doubt that it is NOT true, as you

claim, that "A clock is not in movement or at rest, in

Minkowski space-time" or a clock "is in movement or not

only in respect to other clocks" -- since in such a SINGLE

Minkowski diagram, there is ONLY one clock, which can be

EITHER in movement OR at rest ... and there ARE no other

clocks!Nor can one argue that acceleration is not part of Special

Relativity, but only of General Relativity -- because

acceleration can be represented perfectly well in such a

Minkowski diagram in FLAT space-time. There is no NEED

for General Relativity here, which deals with CURVED space-

time.You remember I had written:

For example, let's say the spaceship and the buoy are both

equipped with powerful rocket engines, capable of accelerating

and decelerating each of them to speeds of +v or - v. Let's say the

reading of the stop watch in the spaceship which is given in A-i is

measured when the buoy passes the spaceship, the buoy during

this pass travelling along the x axis in the positive direction at

velocity v. Then let's say the buoy is caused to decelerate, using

its rocket engine, to the same speed as the spaceship, so that they

are now motionless relative to one another: and at this point the

engine on the buoy is shut off. Now let's say the spaceship fires

ITS rocket engine, accelerating to speed v along the x axis in the

positive direction, at which speed ITS engine is shut off. So now

it is the spaceship which passes the buoy at speed v: and during

this pass, we obtain the reading of the stop watch in the space-

ship given in A-iii. Both the readings, according to your calcula-

tions, must be the same: namely, L/v.Surely one cannot claim that according to the Theory of

Relativity, after having reached velocity v from its initial

velocity -- which was zero -- the stop watch in the spaceship

does NOT experience a time dilation!Remember that Einstein himself writes in his book

"Relativity -- The Special and General Theory" (1920):As a consequence of its motion the clock

goes more slowly than when at rest.(Clearer and more unambiguous -- and briefer -- words cannot

be found! See paragraph 6 of Chapter XII of the above-

mentioned book).Nor can you say that EINSTEIN himself, in the year 1920,

did not adequately understand the Theory of Relativity! (So

let's hear no more of this claim of yours made several times,

saying to me: "You do not understand Relativity well

enough", okay? If I don't, neither did Einstein.)According to you, the stop watch in the spaceship would

indicate the interval [E2 - E1] as L/v equally when the

spaceship is stationary as when it is in motion at speed v ;

and according to Einstein's above-mentioned words, the

moving stop watch must go slower than when the stop watch

is stationary. (And what possible meaning can there be to the

phrase "goes slower" than that the clock indicates a lesser

reading for the same time interval, or for an equivalent time

interval?)This means that:

(a) According to you, the stop watch in the spaceship must

indicate the interval [E2 - E1] when the spaceship is motion as

L/v -- and equally, it must indicate L/v also when the

spaceship is at rest. (We observe clearly that L/v = L/v .)(b) But according to Einstein, in measuring the interval [E2 -

E1], the stop watch in the spaceship must indicate a time

period -- let us call it T' -- which is SMALLER than the time

period, which we may call T , indicated by it for an

equivalent interval when the spaceship is at rest. (We clearly

see that T > T' .)But (a) and (b) CONTRADICT one another, even if either of

the time periods indicated, T or T' , equals L/v ! For if T =

L/v, then T' cannot equal L/v ; while if T' = L/v, then T

cannot equal L/v -- for T is not equal to T' .In that case, it is so very obvious that from a mathematical

point of view, one of you two -- yourself or Einstein -- must

be mistaken! (I can't understand how you cannot see this

utterly clear contradiction, which even a high school student

should be able to see.)Moreover, it is unimportant in our discussion which of the

two of you is mistaken -- yourself or Einstein. If you are

mistaken there must be an error in the Lorentz

transformations, while if Einstein is mistaken there must be

an error in the Theory of Relativity!So the rest of your argument cannot be valid, since we have

already found a contradiction in your above-quoted words.VI.

As regards your TWO clocks in the spaceship, I have

ALREADY told you several times that it is not necessary to

have TWO clocks: ONE stop watch, situated at the mid-point

of the spaceship, is quite enough to precisely measure the

interval [E2 - E1] !(And as I had already said, if you don't believe me you can

consult an engineer.)VII.

As regards the 156 pages of our correspondence: I do not

know how much effort YOU wish to dedicate to the pursuit of

the truth, but as for myself, I am willing to dedicate my

ENTIRE LIFE to it. The number of pages is altogether

irrelevant.VIII.

As regards your following question:

> WOULD YOU ADMIT, OR NOT, THAT IT IS RIGHT FOR A

> PROFESSOR SOMETIMES TO FAIL SOMEONE? (obviously

> one can always be mistaken.) Or do you think that all should

> be promoted?... I would say that sometimes a professor himself can have failed

a test!(What determines who fails or passes depends only on who

says what's true and who doesn't -- and not on who is a

professor or isn't. And what you say can't POSSIBLY be true,

because you keep on contradicting yourself -- or else you

contradict Einstein, as above!)IX.

The rest of your e-mail doesn't seem to me to be relevant to

the present discussion. Even if you BELIEVE yourself to be

right it doesn't mean that you ARE. On the contrary, when

you make statements that result in clear contradictions, which

I point out, and you do not respond to the contradictions, it is

abundantly clear that you must be WRONG.Rather than send my arguments in "extreme synthesis" to

those who are -- according to you -- "even more expert" than

you are, I'd rather publish our ENTIRE correspondence on the

Web, so that EVERYONE who is interested will be able to

read it, and moreover will be able to judge personally and

independently who between the two of us is right and who is

wrong! I think that even if students or colleagues of yours read

our correspondence they will be able to see that it is you who

are in error and not I.For such publication purposes I have translated this e-mail, as

well as your preceding e-mail, into English (since most

scientists in the world read English.) I am sending you a copy

of the translation of your preceding e- mail for approval, so

that you can personally judge whether I have translated your

words correctly.If anyone sends me comments on our e-mail I can forward

them to you, along with any reply I might send them, so that

you can see for yourself what the world is saying about your

intellectual position regarding the Theory of Relativity.But let me assure you that our correspondence should be taken

-- and IS taken by me -- only as a LOGICAL and

INTELLECTUAL challenge, and not at all personal.So I remain yours always in friendship,

Ardeshir <http://homepage.mac.com/ardeshir/education.html>.

_______________________________________________________

Subject: Re: Come si deve fare la matematica (reply in English)

[How Mathematics Should be Done (reply in English)]

Date: Sun, 30 Sep 2001 10:49:33 +0200

From: umberto bartocci <bartocci@dipmat.unipg.it>

To: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13

Dear Mehta,

Your knowledge of Italian is indeed excellent, and this

enables me to reply to you immediately, even if not as

exhaustively as you would have liked, but now I do not have

much time for correspondence, whether because tomorrow the

classes at the university begin, or whether because, as I already

explained to you, I sincerely consider the question completely

resolved.I do not wish to hurt anyone's susceptibility, but I am

convinced that it is necessary always to note that "amicus

Plato sed magis amica veritas" ("Plato is a friend, but truth is

a greater friend"). When you write:> Besides, I am not your STUDENT but your

> INTELLECTUAL ADVERSARY -- a FRIENDLY

> adversary, of course, but an adversary nevertheless. ...I cannot be in agreement. It is obvious that you aren't one of

my students (otherwise I would have other duties with respect

to your challenges, seeking to better explain things, as much

as is possible*), but your level of knowledge of physics and

mathematics seems to me that of a student (or if you prefer, an

"amateur"), and as a result I cannot comport myself towards

you other than as a professor. It is not a question of age, when

I go to study music or Arabic I am at my age of 60 a student

in respect of the youngsters who teach me. I cannot in

frankness consider your challenge as one BETWEEN PEERS.Other comments on your words:

> As regards Galileo, etc.: as Bernard Shaw says, it is true

> that a lot of people speak nonsense, but that doesn't mean that

> ALL people speak nonsense. But to know who does and who

> doesn't, one has to listen to them all!- And I do listen to all (or almost), but one should also agree

that often/sometimes it is possible to become aware right

away when someone is speaking inappropriately of things that

he does not know at all, having not understood them well (I

maintain that even in respect to your observations on Cantor's

Theorem.) I always remember a colleague who teaches music,

who used to say that one could tell AFTER A FEW

SECONDS whether a person could play the piano well or not:

I find that this is true also in the field of science (taking into

consideration that we are talking of a high level, a university

level, of scientific knowledge, and not of simple chatting

between friends, even though it is true that at times even

professors can be in error.)> (And besides, as you yourself say, one should always

> suspect that even professors and so-called "experts"

> sometimes do speak and write nonsense.)Very true, but that does not mean that anyone who writes

nonsense is a professor!> I have already specified several contradictions in your

> preceding words, to which you do not seem disposed to

> reply to me.- Because there are no contradictions in my words, or at least I

am not convinced that you have revealed them, on a question

so simple, and completely understood, on which it is not

possible to add anything (I have never claimed to be

"original": I have only explained things as they stand, as a

"good" professor should). Your comments are without

foundation and demonstrate that you do not understand certain

questions as you should (before desiring to speak of certain

levels): you have still to study a bit more, as Dr Larson has

invited you to do, and this is friendly advice, not an insult.> (What determines who fails or passes depends only on who

> says what's true and who doesn't -- and not on who is a

> professor or isn't. And what you say can't POSSIBLY be

> true, because you keep on contradicting yourself -- or

> else you contradict Einstein, as above!)- Here I am obviously completely in agreement with you! But

how can one get out of the dilemma of knowing who is right

and who is wrong? If the one who is wrong continues not to

admit it, and to exclaim stridently that he is being persecuted

like Galileo was, what can one do? I do not have a

prescription in that regard. As regards Einstein, I admit that I

do not know his writings well, nor have I ever claimed the

contrary (besides, as you well know, I do not like Einstein nor

his theory). I can SUPPOSE many things, that perhaps the

"great" scientist had not understood the consequences of his

theory, as undoubtedly have understood mathematicians such

as Hilbert, Minkowski, Weyl, etc. (remember that Hilbert said

that physics was too important to be left to the physicists, and

perhaps he was at least a little right!), or in certain writings

(certainly the "popular" ones, which could even be the ones he

himself wrote) his words can be interpreted in more than one

way, etc. Even if this is an interesting question, I personally

don't find it essential to dedicate my time to it (nowadays

even less so!) I think Relativity can be combated only in the

experimental field, and to this task I dedicate much (though

indeed not all, there are many other important things, I don't

know if you have had a chance to see my review "Episteme"

...) of my now scarce intellectual resources. You ought to

study Relativity from a modern well-written book, as for

example the book written by O'Neill which I cite several times

in my Site.> In any case, if you think I have NOT won my challenge

> already, we can ask the world community of rational people to

> judge between us: as indicated by me before, we should

> publish our entire correspondence on the Web! I am sure that

> ANY rational person who actually reads and understands all

> that I have written will agree that I am in the main right, and

> that you have contradicted yourself. (I might have made one or

> two small mistakes here and there, but they would not nullify

> everything I have said.)If you want to lose time in this manner I certainly cannot

prevent you. As far as I am concerned these are mere worthless

people who do not know things well: why don't you follow

my advice and turn to true experts in Relativity? Certainly

there may be a consensus on the part of those who do not

understand Relativity, I know people who have studied the

question for years together and yet do not succeed in

understanding the essence of the theory, and claim to refute it

with banal/simple pseudo-logical arguments. Unfortunately

such people are numerous, even among my own anti-

Relativistic friends -- but as usual, not everything they say is

in error, and from some of them one can understand, all the

same, many interesting things. Do therefore as you wish, I

shall be grateful if you would keep me informed of any

comments received as a result.Always and in friendship,

yours UB,

* For example, I could add to what I have already said this

regard, that one should really be aware of a certain

ASYMMETRY in the example which you propose for

analysis, the one which according to you would put Relativity

into a crisis. When the spaceship and the buoy cross each

other, or shall we say also, as is more "usual", a train and a

passenger who is on the station platform, people who are on

the train (let's say two people, on in the front end of the train

and the other in the rear end) observe ONE SINGLE CLOCK -

the one of the passenger on the platform - during two different

events (spatially and temporally) and can thereupon decide

whether this clock goes slower or not comparing their TWO

DISTINCT observations. The gentleman who sees the train

passing in front of him, sees TWO DISTINCT CLOCKS, and

not one SINGLE one, during events which are temporally

distinct though not spatially. From this knowledge one

CANNOT deduce at all whether these clocks tick more slowly

or more rapidly: there is a need for him too to have a

"companion", who can furnish some further information on

these clocks!-- Prof. Umberto Bartocci Dipartimento di Matematica -

Universita' Via Vanvitelli 06100 PERUGIA (ITALY)http://www.dipmat.unipg.it/~bartocci

_________________________________________________

Subject: Re: Come si deve fare la matematica (reply in

English)

[How Mathematics Should be Done (reply in English)]

Date: Sat, 06 Oct 2001 12:16:08 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 ,

14

Dear Professor:

I was indeed surprised to learn that, as you put it,

> there are no contradictions in my words, or at least I am

> not convinced that you have revealed them, on a question

> so simple, and completely understood, on which it is not

> possible to add anything ...Is it really possible that you do not see your extraordinarily

simple contradictions -- which, as I had said, even a high

school student can see? (And I describe below only two of the

SEVERAL contra- dictions in your e-mails -- and note that

these are purely MATHEMATICAL contradictions, and have

nothing to do with ANY understanding of Relativity, whether

good or otherwise):

FIRST CONTRADICTION:

You said my formula [4] (in my e-mail dated August 30,

2001), namely:<delta>t' = <delta>t*(sqrt.(1-v^2/c^2))^-1

... is WRONG, and then you yourself USED the formula

t'2 = t2/sqrt.(1-v^2/c^2) -- which is essentially the SAME as

<delta>t*(sqrt.(1-v^2/c^2))^-1 -- to obtain your answer, as

follows:t2 = L'/v (from my Point No. [10], which you said is correct)

...

t'2 = (L'/v)/sqrt.(1-v^2/c^2)

= (L*sqrt.(1-v^2/c^2)/v)/sqrt.(1-v^2/c^2)

= L/vSo you must have used a formula which according to you

yourself is wrong, in order to obtain your conclusion! How

then can your conclusion NOT be wrong?(In this case you CLEARLY contradict yourself -- and that too

not just once but TWICE: see your e-mail of 31 August 2001,

and Section B as well as Section D in my e-mail of

September 13, 2001. In Section B in fact you contradict

yourself in a manner similar to what is written below):

SECOND CONTRADICTION.

You yourself have USED the formula L' = L*sqrt.(1-v^2/c^2)

-- see your e-mail dated August 19, 2001 -- but then you

imply (in your e-mail dated August 31) that if this formula

were correct, one could get a contradiction of the following

type (and I quote your own words, which were written in

English):1 - L' = L*sqrt.(1-v^2/c^2)

2 - there is no difference whatsoever between L and L', Lorentz

transformations have a group structure, the situation is

SYMMETRIC, so one must have even:

L = L'*sqrt.(1-v^2/c^2)

3 - from 1 and 2 it follows for instance:

L' = L'*sqrt.(1-v^2/c^2)^2

whence:

sqrt.(1-v^2/c^2) = 1 which implies v = 0 .(See Section C in my e-mail of 13 September 2001. In

Section B thereof you contradict yourself in a similar manner,

using t and t' instead of L and L'.)This proves that the formula you have used in your e-mail of

August 19, namelyL' = L*sqrt.(1-v^2/c^2)

... CANNOT be mathematically valid, because from it one can

arrive at a contradiction, as you yourself have pointed out

above!

You doubtless know that even ONE mathematical

contradiction suffices to completely negate the validity of the

entire mathematical theorem in which it appears.The rest of your previous e-mail is irrelevant, having nothing

to do with contradictions, and therefore with the truth or

falsehood of your claims.The above-mentioned contradictions in your e-mails -- even if

we do not speak of the others which I have pointed out in my

e-mail of September 13 -- suffice to prove to any independent

and impartial mathematician that you must be MATHE-

MATICALLY mistaken, and indeed quite REGARDLESS of

any understanding the mathematician may or may not have of

the Theory of Relativity.How can you possibly NOT see these contradictions? I am

truly surprised that you do not see them.In friendship, as always,

Yours,

Ardeshir <http://homepage.mac.com/ardeshir/education.html>.

________________________________________________

Subject: Re: Come si deve fare la matematica (Reply in Italian)

[How Mathematics Should be Done (reply in English)]

Date: Thu, 11 Oct 2001 19:51:01 +0200

From: umberto bartocci <bartocci@dipmat.unipg.it>

To: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15

Dear Mehta,

I thank you as always for your kind consideration, even if, as I

have told you, the lessons having started again, I do not have

much time to dedicate to correspondence. However I reply at

once, because it seems to me that it is possible to quickly

clarify an ambiguity in which you seem to me to have fallen.

You write in fact things such as the following:>You have said that the formula ... is WRONG, and then you

> yourself have USED the formula ...but the fact is that all these formulae have full significance or

are radically wrong depending on the meaning of the terms

with which one wishes to describe them. I had said for

example that it is wrong to claim, as you had done:> [4] The Lorentz time dilation formula is

> <delta>t' = <delta>t*(sqrt.(1-v^2/c^2))^-1.

> [5] The Lorentz length contraction formula is

> <delta>x' = delta>x*(sqrt.(1-v^2/c^2))^-1.where you had specified the meaning of the terms involved in

the following way:> Let the primed co-ordinates in the calculations below be those

> pertaining to the spaceship, and let the un-primed co-ordinates be

> those pertaining to the buoy.If these are the terms, the preceding formulae are WRONG, by

virtue of the absolute symmetry between the "primed" and

"un-primed" co-ordinates, to which in fact I had referred in the

following words:- This is the major source of misunderstanding. If this was true,

then you would simply get immediately a contradiction like the

following one:1 - <delta>t' = <delta>t*(sqrt.(1-v^2/c^2))^-1

2 - there is no difference whatsoever between t and t', Lorentz

transformations have a group structure, the situation is

SYMMETRIC, so one must have even:<delta>t = <delta>t'*(sqrt.(1-v^2/c^2))^-1

3 - from 1 and 2 it follows for instance:

<delta>t' = <delta>t'*(sqrt.(1-v^2/c^2))^-2

[as a matter of fact, this is precisely what happens in your point 16]

whence:

(sqrt.(1-v^2/c^2))^-1 = 1 which implies v = 0 .

I HAVE SEEN THIS MISTAKE HUNDRED OF TIMES IN MY

LIFE...I had done my best to explain to you what ought to be the

meaning given to the length contraction and time dilation

formulae:- The true formulae, which one must "understand", are the

ones which I gave to you in my last mail, namely:A - (proper length of some object)*sqr(1-v^2/c^2) =

coordinate (or "apparent") length of this same object in an

(inertial) reference frame in which the object is moving with

speed vB - proper time of some phenomenon IN THE LIFE OF ONE

OB- SERVER = sqr(1-v^2/c^2)*(coordinate time interval of

the same event) [meaning as before]You see, one can say that the formulae ARE THE SAME, but

it is necessary to pay close attention to how they are used:

there is NO connection between primed and un-primed co-

ordinates, which cannot be other than symmetric, but rather a

connection between proper and apparent measurements in the

above-specified sense. If one uses the formulae in this sense,

NO CONTRADICTIONS ARISE. Which is to say that, once

again (and I do not take any pleasure in this), I have to repeat

that you are wrong when you sustain that:> How can you NOT see these contradictions? I am truly

> surprised that you do not see them.I could say that it is I who am truly surprised that you do not

understand how one should treat space and time -- which are

"relative" -- in the Theory of Relativity: a manner which is

certainly not in consonance with common sense, but not

impossible to understand. Because instead of trying to find

contradictions, WHICH DO NOT EXIST, in my theory, why

don't you try to study and understand it well, so as to be able

to direct thereby your efforts to refute it which are effectively

refutable? (These aspects could be the EXPERIMENTAL

aspects: for example, if one could find an electromagnetic

phenomenon which does not satisfy Maxwell's equations, and

which are not Lorentz invariant, then Einstein's theory would

be in a crisis, but certainly not with INFANTILE mistakes -

and that is the correct word! - manipulations in your

fundamental conceptions ...)I exchange friendly greetings, hoping that you can soon

understand the root of your errors, and overcome them once

and for all,Yours, UB

-- Prof. Umberto Bartocci

Mathematics Department

University

Via Vanvitelli

06100 PERUGIA (ITALY)http://www.dipmat.unipg.it/~bartocci

________________________________________________

Subject: Your Own Words Refute the Theory of Relativity!

Date: Fri, 19 Oct 2001 13:17:30 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16

Dear Professor:

I just noted that YOUR OWN WORDS are sufficient to mathe-

matically refute the Theory of Relativity!Note that the formulae which you said were "true" were as

follows:> - The true formulae, which one must "understand", are the ones

> which I gave to you in my last mail, namely:

>

> A - (proper length of some object)*sqr(1-v^2/c^2) = coordinate

> (or "apparent") length of this same object in an (inertial) reference

> frame in which the object is moving with speed v

>

> B - proper time of some phenomenon IN THE LIFE OF ONE OB-

> SERVER = sqr(1-v^2/c^2)*(coordinate time interval of the same

> event) [meaning as before]I presume you mean by your sentence B above as follows (I

have replaced your words "meaning as before" by your EXACT

WORDS from your sentence A above):> B - proper time interval of some phenomenon IN THE LIFE OF

> ONE OBSERVER = sqrt.(1-v^2/c^2)*(coordinate time interval of

> the same phenomenon) in an (inertial) reference frame in which

> another observer is moving with speed vThis completely and absolutely DISPROVES the so-called

Relativistic "loss of simultaneity between clocks moving rela-

tive to one another". You see:[1] If two events -- let's call them F and G -- are SIMULTANE-

OUS in the life of ONE observer, that means that according to

THAT observer's clock, they occur at the SAME TIME.

(After all, that is the very DEFINITION of the word "simul-

taneity".)[2] Let us call the times at which they occur, as indicated by that

observer's clock, t(F) and t(G) : where t(F) = t(G) . Then obvi-

ously [t(G) - t(F)] must be absolutely ZERO. In other words,

the "proper" time interval between them in the life of THAT

observer must be absolutely ZERO.[3] And since in mathematics, zero multiplied by ANY number

whatsoever -- even by [1/sqrt.(1-v^2/c^2)] -- is STILL zero,

this means that the CO-ORDINATE time interval between

events F and G in ANY other (inertial) reference frame in which

ANY other observer is moving with ANY speed whatsoever

must ALSO be absolutely zero![4] And this in turn means that if two the events F and G are si-

multaneous in ONE (inertial) reference frame, they must be

simultaneous in ALL (inertial) reference frames.Thus if what you say is true, the famous Einsteinian argument

about the "loss of simultaneity" in reference frames moving

relative to one another must be MATHEMATICALLY IN-

CORRECT -- and in that case, ALL the other mathemati-

cal/logical arguments against the Special Theory of Relativity

become superfluous, be- cause as Einstein himself says in his

book *Relativity: The Special and General Theory* (1920):Now before the advent of the theory of relativity it had

always tacitly been assumed in physics that the statement

of time had an absolute significance, i.e. that it is inde-

pendent of the state of motion of the body of reference.

But we have just seen that this assumption is incompatible

with the most natural definition of simultaneity; if we

discard this assumption, then the conflict between the law

of the propagation of light in vacuo and the principle of

relativity (developed in Section VII) disappears.But as we have seen above, the so-called "loss of simultaneity

for different inertial frames" argued by Einstein cannot be

logically valid.Thus it is abundantly clear that time can no longer be

considered to be relative, but instead must be regarded as

absolute (for if simultaneity must be the same for ALL clocks,

wherever they may be and however they may be moving, then

one can always synchronise all the clocks in the universe with

a SINGLE clock situated, for example, at Greenwich in

England, using the method given by Einstein himself in his

above-quoted book.)As a result, what Einstein calls above "the conflict between

the law of the propagation of light in vacuo and the principle

of relativity" does NOT disappear -- and this logical conflict

cannot be overcome.So the rest of my present e-mail becomes, as I said,

superfluous; but if you want to see how one can still get

contradictions from your previous arguments, EVEN IF what

you said were true, you can find them below.I hope you are happy now ... ? It should be clear now to all

who read the above that Relativity cannot be a mathematically

and logically sound theory. (And as I have already explained,

we are not dealing here with an understanding -- good or bad -

- of the Theory of Relativity, but only of mathematics and of

logic.)I remain, as always, yours in friendship,

Ardeshir <http://homepage.mac.com/ardeshir/education.html>.

****************************************************

P.S.: Nor can one argue -- as do certain people, among them Dr

Tom Van Flandern -- that according to the Theory of Relativity

there can be NO simultaneity between two events if these events are

separated spatially. According to the above-mentioned book, the

Theory of Relativity claims no such thing. Quoting Einstein's own

words:Note 1. We suppose further that, when three events A, B

and C take place in different places in such a manner that,

if A is simultaneous with B, and B is simultaneous with C

(simultaneous in the sense of the above definition), then the

criterion for the simultaneity of the pair of events A, C is

also satisfied.Note the words "in different places".

****************************************************

To reveal the contradictions that arise in the Theory of Relativity

even if what you say is true, I repeat what you had written, for

ease of reference:> - The true formulae, which one must "understand", are the ones

> which I gave to you in my last mail, namely:

>

> A - (proper length of some object)*sqr(1-v^2/c^2) = coordinate

> (or "apparent") length of this same object in an (inertial) reference

> frame in which the object is moving with speed v

>

> B - proper time of some phenomenon IN THE LIFE OF ONE OB-

> SERVER = sqr(1-v^2/c^2)*(coordinate time interval of the same

> event) [meaning as before]If this is correct, then if the symbols L and T represent the

proper length and time, respectively, and L' and T' the co-

ordinate length and time, then with respect to the spaceship

mentioned in my "Challenge", we can say that:A - L*sqrt.(1-v^2/c^2) = L' < L ,

... and

B - T = T'*sqrt.(1-v^2/c^2) < T' .

Suppose now that in my "Challenge" there IS an observer on

board the spaceship and another on the buoy. The

"phenomenon" under consideration in my "Challenge" is, of

course, the spaceship passing the buoy: that is, as you had

written, the time interval (E2 - E1).Now there are ONLY TWO POSSIBILITIES here, as follows:

I.

The proper time interval between E1 and E2 is the time

interval observed by the observer on the BUOY -- in which

case the co- ordinate time interval corresponding to that proper

time interval is the one observed by the observer on board the

spaceship;... or

II.

The proper time interval between E1 and E2 is the time

interval observed by the observer on board the SPACESHIP --

in which case the co-ordinate time interval corresponding to

that proper time interval is the one observed by the observer

on the buoy.Let us consider each of these in turn. Both of them will result

in contradictions -- though different ones.I.

In the first case, if we claim that the proper time interval for

this phenomenon is the one observed by the hypothetical

observer on the buoy, and the co-ordinate time interval for the

same phenomenon is the one observed by the hypothetical

observer in the space- ship, then we get the following

contradiction.Your B above can be now expressed as:

> B - proper time interval of the phenomenon of the spaceship

> passing the buoy IN THE LIFE OF THE OBSERVER ON

THE > BUOY = sqrt.(1-v^2/c^2)*(coordinate time interval of

the same > phenomenon) in the (inertial) reference frame of the

spaceship, in > which the buoy is moving with speed v .This means that if the proper time interval of the phenomenon

of the spaceship passing the buoy in the life of the observer on

the buoy is T, then the co-ordinate time interval of the

phenomenon of the spaceship passing the buoy in the life of

the observer on the spaceship must be T', where T = sqrt.(1-

v^2/c^2)*T' .And since sqrt.(1-v^2/c^2) < 1 , we get T < T' .

However:

(1) In the IRF of the buoy, the spaceship -- and its stop watch -- is

moving with speed v.(2) According to Relativity, if in any (inertial) reference frame --

which we shall abbreviate as IRF -- a timepiece C[mov.] (of any

nature) is moving with speed v , time passing in the timepiece

C[mov.] must be dilated in comparison with time passing ac-

cording to another timepiece C[stat.] which is NOT moving in

that same IRF. (See also footnote 1.)(3) Therefore according to Relativity, in the IRF of the buoy, any

timepiece in the spaceship -- which spaceship, in the IRF of

the buoy, is moving with speed v -- ought to experience its time

dilated in comparison with any timepiece on board the buoy,

which is NOT moving in the IRF of the buoy. (See also foot-

note 2.)(4) This means that for every unit of time -- say, for every nano-

second -- indicated by a stop watch on the buoy, the stop

watch on the spaceship must indicate LESS than one unit of

time (so in the above example, the stop watch on the spaceship

will indicate LESS than one nanosecond for every nanosecond

indicated by the stop watch on the buoy.)(5) This in turn means that if the stop watch on the buoy indicates

a time interval equal to T nanoseconds for (E2 - E1), then the

stop watch on the spaceship, which is moving in the IRF of the

buoy, must indicate for (E2 - E1) a time interval of T' nanosec-

onds, where T' < T .(6) This contradicts your B above, according to which T < T' .

(Note that this is a purely mathematical contradiction, and not

a physical one.)II.

On the other hand, if the proper time interval is the one observed

by an observer on the spaceship and the co-ordinate time interval is

the one observed by the observer on the buoy, then we get another

(and even simpler) contradiction, as follows:1. The relative velocity between the spaceship and the buoy must

be the SAME, regardless of which of the two above-mentioned

observers calculates it.2. Thus v = L/T = L'/T' .

3. By A above, L > L' .

4. So L/v > L'/v .

5. From 2, 3 and 4 above, we get T > T' .(See footnote 3.)

6. But according to your B above, T < T' , which contradicts 5

above.Either way -- whether by argument I above or by argument II above

-- we get a MATHEMATICAL contradiction: which proves that

the theory of Relativity must be MATHEMATICALLY self-

contradictory.

****************************************************

(Footnote 1):

Otherwise the postulate of the constancy of the speed of light

would be violated. For a theoretical proof of this, see for example:<http://www.btinternet.com/~j.doyle/SR/sr7/sr7.htm>.

Many other Web sites on Relativity argue in much the same way.

(Of course the argument must be WRONG, because the postulate

of the constancy of the velocity of light -- regardless of the velocity

of the source of light or of the observer -- is mathematically inad-

missible, since it contradicts the theorem of the addition of veloci-

ties which, being a THEOREM (that is, a mathematical statement

already possessing proof) may not be contradicted; but it cannot be

denied that this IS what Relativity TRIES to argue.)

(Footnote 2):

It cannot mathematically be argued that BOTH the following are

correct (although Relativity tries it also):(i) In the IRF of the buoy, any clock in the spaceship -- which in

the IRF of the buoy, is moving with speed v -- ought to have

its time dilated in comparison with any clock on board the

buoy, which is NOT moving in the IRF of the buoy;... AND

(ii) In the IRF of the spaceship, any clock in the buoy -- which in

the IRF of the spaceship, is moving with speed v -- ought to

have ITS time dilated in comparison with any clock on board

the spaceship, which is NOT moving in the IRF of the space-

ship.But that is mathematically impossible: for if (i) and (ii) above

were BOTH true, then by (i) above, ANY time interval of,

say, T(b) nanoseconds indicated by the stop watch on board

the buoy would correspond to a time interval of T(s) nano-

seconds indicated by a stop watch on board the spaceship,

where T(s) < T(b), while by (ii) above, ANY interval of T(b)

nanoseconds indicated by the stop watch on board the buoy

would correspond to a time interval of T(s) nanoseconds

indicated by a stop watch on board the spaceship -- but in this

case T(s) > T(b). So (i) above contradicts (ii) above.

(Footnote 3):

The calculation is simple:

L/v = L/(L/T) > L'/v = L'/(L'/T')

So cancelling out the two L s on the left of the inequality sign and

the two L' s on its right, we get T > T' .

______________________________________________________

Subject: Re: Your Own Words are Sufficient to Refute the Theory of Relativity!

Date: Sun, 21 Oct 2001 09:42:28 +0200

From: umberto bartocci <bartocci@dipmat.unipg.it>

To: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17

Dear Mehta,

Despite the fact that I have told you several times that I

sincerely consider this discussion a MERE LOSS OF TIME,

I reply at once in friendship because it is very easy to do so,

and basically you al- ready know the answer! In fact VAN

FLANDERN IS RIGHT , according to your words:Nor can one argue -- as do certain people, among them Dr

Tom Van Flandern -- that according to the Theory of Relativity

there can be NO simultaneity between two events if these events are

separated spatially.In fact this is the argument which one CAN MAKE, and

MUST MAKE (properly explaining all the concepts), and

this resolves all the usual mistakes made by beginners-

amateurs. In my opinion it is not worth wasting time to see

if Einstein explained himself well or not in one or another

place (in reality it is much easier to explain oneself badly, IT

MUST HAVE HAPPENED TO ME TOO IN OUR

CORRESPONDENCE, because one speaks in a hurry, does

not lay out all the hypotheses, etc. as you say: "According to

...." I don't know, but strictly speaking Relativity is not

what Einstein has written, in different times in his life, and

which different "rigor" depending on the readers he was

addressing, but rather Minkowski's theory of space-time, and

every mathematician can create it himself without "copying"

it off anyone.Here is the complete explanation. If two events are NOT

separated from a spatial point of view with respect to an

observer, and are also not separated from a temporal point of

view with respect to the same observer, then it IS THE

SAME EVENT, whether for that observer or for any other, an

d is in effect one part, BUT ONLY ONE PART, of your

following reasoning is correct, but also obvious, and thus

useless.

[1] If two events -- let's call them F and G -- are SIMULTANE-

OUS in the life of ONE observer, that means that according to

THAT observer's clock, they occur at the SAME TIME.

(After all, that is the very DEFINITION of the word "simul-

taneity".)- CORRECT, but it is well to understand what it means to

say "in the life of an observer". In this case, the two events

which you speak of are MORE THAN SIMULTANEOUS,

they are in fact COINCIDENT, F = G . Thus everything that

follows is unimportant, superfluous! One should not confuse

simultaneity, which the observer registers in the Relativistic

sense between events which are NOT in his "life", and events

which are in his life, for which simultaneity is in fact

coincidence. To understand it better, would you say that the

event that you read this e-mail there where you are is an event

in MY life as an observer? NO, I can however try to say that

this event is simultaneous with some other event in my life,

or in the life of anyone else that I am "observing".

[2] Let us call the times at which they occur, as indicated by that

observer's clock, t(F) and t(G) : where t(F) = t(G) . Then obvi-

ously [t(G) - t(F)] must be absolutely ZERO. In other words,

the "proper" time interval between them in the life of THAT

observer must be absolutely ZERO.- OBVIOUS.

[3] And since in mathematics, zero multiplied by ANY number

whatsoever -- even by [1/sqrt.(1-v^2/c^2)] -- is STILL zero,

this means that the CO-ORDINATE time interval between

events F and G in ANY other (inertial) reference frame in which

ANY other observer is moving with ANY speed whatsoever

must ALSO be absolutely zero!- OBVIOUS.

[4] And this in turn means that if two the events F and G are si-

multaneous in ONE (inertial) reference frame, they must be

simultaneous in ALL (inertial) reference frames.- MISTAKEN. You forget that in the preceding point [1], the

two events were supposed to have belonged to one single

observer, were not any two events, which were "observed

from a distance"! This is a FUNDAMENTAL

HYPOTHESIS, without which con- clusion [4] (which in the

correct hypothesis only means F = G , and not in general "F

is simultaneous with G", a circumstance which is in general

quite "relative") mistaken.I repeat, if you make the hypothesis that two events are NOT

spatially separated (or in other words, are spatially coincident))

for a certain observer, they your conclusion would be correct,

because if two events have a difference of proper time equal to

zero, then they ARE THE SAME EVENT, and it is obvious

that if they are the same event in an IFR, they must be the

same event in any IFR.The strange aspect of Relativistic time is not this, but is

instead that two events SPATIALLY SEPARATED in an

IFR can have the same temporal co-ordinate in that same IFR

(being that is to say relatively "simultaneous") and not

having the same temporal co-ordinate in another IFR (being

that is to say not simultaneous.) That is to say, in

Minkowski space-time, simultaneity is relative to a given

observer, and not "absolute", just as I imagine Einstein has

already well said, who has forgotten to explain the simple and

particular case of two events in the life of one single observer,

but I repeat that I don't think it very interesting to test if it

truly is or not, or whether in one sense yes and in another not.I could add, for a "student", that it should not be forgotten

that an "observer" is a special CURVE (a straight line in an

inertial case), and that if two events are points on such a

special curve then they are not any two events. If one does

not exert oneself to understand these things one will never

understand Relativity, not even the most simple, that is the

restricted. Given two events in Minkowski space-time one

can calculate the ds^2 of the vector which joins them, and it

makes a great difference whether the value one obtains is

positive, negative or zero (for what follows I underline that I

am using the so-called "geometrical unities", that is, c=1). If

two events are separated and spatially separated in an IFR,

this ds^2 is positive, and vice versa, in the sense that if ds^2

is positive then there exists and IFR for which the two events

belong to a single space of simultaneity (the square root of

this ds^2 expresses their spatial distance, and is "relative").

On the other hand, if the two events have a negative ds^2,

this means that there exists an IFR in which they occur in the

same position, but in different times, calculated with respect

to an inertial observer who "lives" both of them (in purely

geometrical terms, this observer is simply the straight line

which joins them: if we take another observer who "lives"

both of them, this time will be less, and this is the essence of

the "twin paradox" which is asymmetric because in one case

we have a straight line, aa geodesic, and in the other, no.)Finally, I can ask once more, with a sincere spirit of

friendship: why, instead of seeking stupid contradictions that

do not exist, do you not seek to study/understand the theory

better? Basically it is not difficult ...Salutations, yours UB

-- Prof. Umberto Bartocci Dipartimento di Matematica - Universi-

ta' Via Vanvitelli 06100 PERUGIA (ITALY)http://www.dipmat.unipg.it/~bartocci

_________________________________________________

Subject: If you wish, let us end this correspondence here

Date: Sat, 27 Oct 2001 15:43:06 -0400

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

To: umberto bartocci <bartocci@dipmat.unipg.it>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 , 17 , 18

Dear Professor,

If you wish to terminate this correspondence, believing it to

be a "mere waste of time" (as you put it), then let us end it

here, and let us publish it, as it stands up to and including

this e-mail, on the Web, so that anyone -- among them even

your own students and colleagues -- may judge who is right:

you or I.And hereunder I explain your errors -- in a way which any

rational and well-educated person should be able to understand

-- one last time.You claim that I do not understand Relativity, but I on the

contrary think that not only do I understand it, but I

understand it even better than you do yourself. What I do

understand, and what you do not, is that the "Minkowski

space-time" is ITSELF a concept which is in- valid from a

logical and mathematical point of view -- and the proof of that

is clearly laid out by me here in this e-mail.Of course WITH the concept of Minkowski space-time one can

"explain" the twin paradox, and refute Dingle; but one cannot

prove that the VERY CONCEPT of Minkowski space-time is

a part of mathematics, because in order to FORMULATE this

concept, one must ADD an postulate to those of mathematics:

the additional postulate being that of the constancy of the

velocity of light! So Minkowski space-time ITSELF

CANNOT FORM A PART OF MATHEMATICS (as we

know it).Try to formulate the concept of Minkowski space-time as a

series of mathematical theorems, WITHOUT the postulate of

the constancy of the velocity of light, and I am certain that

you will not be able to do so! But all the REST of

mathematics CAN be developed perfectly well using only the

axioms, propositions and postulates of mathematics itself:

namely, those of Peano, or those enunciated by Zermelo,

Fraenkel and John von Neumann, and the propositions and

postulates of Euclid (at least his first four postulates) -- that is

to say, WITHOUT the postulate of the constancy of the

velocity of light.So that what YOU do not understand is what mathematics IS.

Mathematics -- including geometry -- is a series of theorems

obtained from a limited number of given axioms, propositions

and postulates, to which are applied a limited number of rules

of inference. And one cannot add to the existing axioms,

propositions or postulates another postulate, nor another rule

to the existing rules of inference, without going outside the

bounds of mathematics itself.Therefore Minkowski space-time, and the Lorentz

transformation equations, formulated with the help of an

ADDITIONAL postulate to those of mathematics as we know

it, are only a kind of "pseudo- mathematics", and cannot be

part of the rest of mathematics known to humanity.What I do not accept, therefore, is the FUNDAMENTAL

BASIS of what you accept uncritically: namely, the very

concept of Minkowski space-time. Besides, it results -- and

MUST result, due to the postulate of the constancy of the

velocity of light, which is illogical -- in contradictions ...

despite your efforts at explaining them away: for your

explanations only cause other contradictions to arise,

endlessly, for every one you succeed in explaining away. And

this is once again clear from what I have written below.I remain always,

Yours in friendship,

Ardeshir <http://homepage.mac.com/ardeshir/education.html>.

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The proof that you are mistaken, and that I am right, is given

here below in the form of a *reductio ad absurdum* type of

proof.Note that I have NEVER spoken of events which are NOT

separated spatially. One should not forget in fact that there

must ALWAYS be a distance -- and for that very reason, a

time interval -- between the occurrence of an event and the

observation of the same event: because of the distance, which

must exist, even if it be a very tiny one, between any event

and its observer. So that strictly speaking, absolutely NO

event is EVER observed at the exact SAME moment and at

the exact SAME location at which it occurs.So that if you define "proper time" as the interval between two

events when observed at the EXACT LOCATION AT

WHICH THEY OCCUR, that itself is strictly speaking an

impossibility from a practical point of view. In fact strictly

speaking, EVERY time interval must be a "co-ordinate" one,

and can never be a "proper" one.But there can always be two events at a distance from one

another, which are nevertheless simultaneous in the life of an

observer. It doesn't matter whether we call the (zero) interval

between them the "proper time" or not; what matters is that

the interval between them, according to THAT particular

observer, must be ZERO.Then the interval between them must be zero ALSO according

to any OTHER observer. And the proof is as follows:Let us suppose that the Theory of Relativity, and the concept

of "Minkowski space-time", are NOT mistaken.Then it is obvious that:

1. If two events F and G were to occur, according to any observer

O , at a distance D from one another,2. and if the observer O , with a clock C in his hand, were to ob-

serve the two events, the event F being at a distance d from O

and the event G being at an equal distance d from O , so that

the points O , F and G form, in the IRF of the observer O , an

isosceles triangle, in such a manner that the line OF = d and the

line GO also equals d , while the line FG = D ,3. and if the observer O were to observe the event F at a moment

t(F) and the event G at a moment t(G) -- both moments indi-

cated, of course, by his SINGLE clock C ,4. and if the moments t(G) and t(F) were THE SAME -- which is

to say, according to the clock C in the hands of the observer O ,

t(F) = t(G) = t ,5. then the OBSERVATION of the two events F and G must have

been SIMULTANEOUS in the life of the observer O , because

the observation of BOTH events must have occurred in his life at

the SAME moment, i.e., at the precise moment t = t(F) = t(G) .6. But this is only the moment at which F and G are OBSERVED in

the life of the observer O -- not the moment at which the events

F and G OCCURRED in his life. The moment at which the event

F OCCURRED would be T(F) , where clearly T(F) = [t-(d/c)] --

and the moment at which the event G occurred must have been

the moment T(G) , where clearly T(G) = [t-(d/c)] also: so that

obviously T(F) = T(G) = T ... and so of course the moment

T = [t-(d/c)] (according to the clock C ).7. And if the moments when F and G OCCURRED *in the life of

the observer O* -- that is to say, according to his SINGLE

clock C -- are T(F) and T(G) , when clearly T(F) = T(G) = T ,

this means that the time interval between T(F) and T(G) ,

which we may call <theta>, must, in the life of O , be absolutely

ZERO.8. And this in turn means that *in the life of the observer O* --

that is according to his single clock C -- the events F and G

must certainly have been SIMULTANEOUS, but at the same

time certainly NOT COINCIDENT (because of the distance

D between them.)9. And additionally, according to the Lorentz transformation equa-

tions, and according to my point No. 7 above, the time interval

indicated by ANY other clock C' , which may even be in motion

relative to the clock C -- which interval we may call <tau> -- for

the occurrence of the events F and G must also be zero, because

of the mathematical fact that zero multiplied by any other num-

ber is still zero ... so that quite clearly <theta> = <tau> = zero.10. And this means that in the life of ANY other observer O' --

which is to say according to HIS clock C' -- who may even be

moving relative to O , the events F and G must ALSO have

occurred simultaneously.And this completely refutes your statement:

> One should not confuse simultaneity, which the observer regis-

> ters in the Relativistic sense between events which are NOT in his

> "life", and events which are in his life, for which simultaneity is in

> fact coincidence.As you can see above, the events F and G are

SIMULTANEOUS in the life of the observer O , but NOT

COINCIDENT (since they are separated by a distance D ,

which can be enormous: even many kilometres) -- and which

are also simultaneous in the life of any other observer O' ,

who may even be in motion relative to the observer O .(Of course another observer might not actually OBSERVE the

events F and G simultaneously, unless he too happens to be

equi- distant from them at the moment when the observation

takes place; but the events themselves must nevertheless

OCCUR simultaneously for the observer O' as for the observer

O , because of the fact that the moment of the

OBSERVATION of an event can never be the same as the

moment of its OCCURRENCE.)And the above-described conclusion, arrived at PURELY

LOGICALLY AND MATHEMATICALLY, completely

contradicts the notion of "Minkowski space-time" according to

which -- as you yourself have explained -- there can be

simultaneity between two (or more) events as indicated by a

clock in one particular IFR, and yet NOT according to a clock

which is in motion relative to the first-mentioned clock.Thus it is clear that the very notion of "Minkowski space-

time" is contrary to logic and to mathematics, since it

contradicts a conclusion arrived at under the assumption that it

is NOT mistaken.Q.E.D.

______________________________________________________

Subject: Re: Se vuole, terminiamo questa corrispondenza qui

[If you wish, let us end this correspondence here]

Date: Sun, 28 Oct 2001 09:46:49 +0100

From: umberto bartocci <bartocci@dipmat.unipg.it>

To: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 ,

15 , 16 , 17 , 18 , 19

[TRANSLATED FROM THE ITALIAN BY ARDESHIR MEHTA]

Dear Mehta,

I thank you once again for your constant kindness and con-

sideration, but it is obvious that, despite all the good inten-

tions and education, I cannot change my opinion on that

which I consider a fact, and that is, that your reasoning is

completely erroneous, and demonstrates a lack of knowledge

of the fundamentals of Relativity. I have tried to help you a

little to correct your numerous misunderstandings, but it is

clear that at such a distance it is difficult to do so, and seeing

as how I have not yet been able to convince you of anything,

it is preferable to end the discussion here. The "story" which

you would recount if you were to succeed in convincing the

world of claims such as "the concept of 'Minkowski space-

time' is itself contrary to mathematics" (we are looking at a

simple example of variety of Lorentzian geometry: to say that

it is "contrary to logic" is to say that Euclidean geometry too

is contrary to logic).The only thing that I can repeat, in all sincerity and

friendship, is the well-meaning suggestion Dr Larson has

already given you: "Please take the time to study and

understand relativity. I would suggest contacting your closest

quality university and seeking out a mentor. Once you

understand relativity I would hope that you could join the

battle against it. But fighting an intellectual battle without an

understanding of its underpinnings is simple folly".

Always very cordially,

from your UB

-- Prof. Umberto Bartocci

Dipartimento di Matematica

- Universita'

Via Vanvitelli 06100

PERUGIA (ITALY)http://www.dipmat.unipg.it/~bartocci

________________________________________________

Subject: Ho spiegato tutto nel mio ultimo articlo Web

[I have explained everything in my latest Web article]

Date: Sun, 11 Nov 2001 14:44:38 -0500

From: "Ardeshir Mehta, N.D." <ardeshirmehta@myself.com>

References: 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 ,

16 , 17 , 18 , 19 , 20

[TRANSLATED FROM THE ITALIAN BY ARDESHIR MEHTA]

Dear Professor:

I see once more that you do not carefully read what I write.

In any case I have explained everything in my latest Web article,

entitled 'The "Mathematics" of Relativity, which can be found at:<http://homepage.mac.com/ardeshir/RelativityMath.html>

And our entire correspondence has now been published on the

Web at:<http://homepage.mac.com/ardeshir/BartocciComments.html>

Best wishes,

Ardeshir <http://homepage.mac.com/ardeshir/education.html>

************************************************************