by Sunday, October 7, 2001 Here is another very simple argument which proves that Special Relativity must be mathematically flawed, because it uses a set of equations — the Lorentz transformation equations — which cannot form part of mathematics as we know it. [1] Let’s say a train is moving at a velocity of magnitude v relative to the rails. Let’s assume for the purposes of this argument that the rails are absolutely straight and in the unprimed frame: that is, the rails will be (arbitrarily) assumed by us to be stationary. Let’s say that when the train is moving at speed v along the rails, the train’s length is L' metres. And of course we all know that the Lorentz <gamma> factor, by which the train must have contracted due to its movement, is {1/[1(v^{2}/c^{2})]^{0.5}} . [2] Now suppose the train comes to a stop at a terminus to allow passengers to embark and disembark; and during the stop, its length is carefully measured by the station master, who measures it to be exactly L metres in length. Then using the Lorentz <gamma> factor we calculate L' = L/{1/[1(v^{2}/c^{2})]^{0.5}} . Thus L' cannot be greater than L , but must be less (because v^{2}/c^{2} must be a positive number, and so [1(v^{2}/c^{2})] must be less than 1, so the square root of [1(v^{2}/c^{2})] must also be less than 1, which means that the absolute value of {1/[1(v^{2}/c^{2})]^{0.5}} must be greater than 1.) [3] Now after the passengers have
embarked and disembarked,
the train moves back towards the place it came
from — that
is, going in the opposite direction to its
original direction of
motion — till it again reaches a velocity of magnitude
v relative
to the rails. That is, the train is now moving in the
opposite direction
to the direction it was moving in [1] above. So now
its velocity, taking
into account not just the magnitude but also
the direction
of its movement, is v (i.e., minus
v)
relative to the rails.
[5] If it is, then the relative velocity of the train in [1] as compared to its velocity in [3] must have been zero! That’s because according to Relativity, the same object cannot have the same length in two different situations unless it was travelling at the same velocity in both situations — that is to say, unless the relative velocity of its movement in the two different situations, comparing the one situation with the other, was zero. Or in other words, we’d have to say that +v = v . That obviously can’t be the case mathematically speaking (whatever it might be in Zen!) [6] On the other hand, if L' isn’t equal to L" , then the Lorentz <gamma> factor — which is indispensable in Relativity — can’t have been applied in calculating L" , for (+v)^{2} is exactly equal to (v)^{2}, and as a result, {1/[1(v^{2}/c^{2})]^{0.5}} must be exactly equal to {1/[1((v)^{2}/c^{2})]^{0.5}} ! [7] So whichever way you squirm, the
answer’s bound
to be wrong — proving that the Lorentz <gamma>
factor
—
which is used in Special Relativity for the Lorentz
transformation equations
—
must be selfcontradictory. (And in mathematics, any
set of equations in
which a selfcontradictory formula is used must itself
be selfcontradictory,
and therefore cannot be a part of mathematics as we
know it!)
Any comments? email
me.
It has been brought to my attention that velocity, defined as distance divided by time interval, cannot be negative, since neither a distance nor a time interval can have a negative value. However, I would like to point out that by the weird "mathematics" of Relativity, distances and times can be negative. Remember that in Relativity, a distance L in one frame is related to the same distance L' in another frame moving relative to the first at a velocity v by the equation L = L' {1/[1(v^{2}/c^{2})]^{0.5}}, and of course the equation [1(v^{2}/c^{2})]^{0.5 }has two solutions: positive as well as negative. The same argument applies to time intervals. Thus Relativists cannot claim that L = L' {1/[1(v^{2}/c^{2})]^{0.5}} is a valid equation for distance, and then turn around and claim that a velocity cannot be negative !
