Simultaneity
in Special Relativity  2
by
Ardeshir
Mehta
Ottawa,
Canada
Thursday,
October
25, 2001
Einstein's
Theory of Special Relativity claims
that if two events are simultaneous in an inertial
frame of reference,
then they cannot be simultaneous in another
inertial frame of reference
which is moving uniformly and rectilinearly at a
velocity v relative
to the first inertial frame of reference.
But
if the Lorentz
transformation equations are correct, this claim
results in a clear contradiction,
as follows:

Let
there be an inertial
frame of reference  which we shall designate as
I  in which
there are two clocks C_{1} and C_{2},
separated
from one another spatially, and synchronised:
so that whenever the
clock C_{1} indicates a moment in
time t_{1},
the other indicates a moment in time t_{2}
such that t_{1}
= t_{2} = t.

Let
there be another
inertial frame of reference  which we shall
designate as I' 
also moving rectilinearly and uniformly at a
velocity v relative
to I, in which there are two more
spatially separated clocks C'_{1}
and C'_{2}, which are also
synchronised: that is,
whenever the clock C'_{1}
indicates a moment t' coinciding
with the moment t indicated by the clock C_{1},
the
clock C'_{2} also indicates the same
moment t'
indicated by the clock C'_{1}.

Now
when the clock C_{1}
indicates any particular moment t_{1},
the moment t_{1}
must be related to the moment t'_{1}
indicated by the clock
C'_{1}
by the Lorentz transformation equation
t'_{1}
= (tvx/c^{2})/(1v^{2}/c^{2})^{0.5}
...
where x
is the distance, as measured by an observer in I,
between clock
C_{1}
and clock C'_{1}, and the
moment t = t_{1}
is that indicated by the clock C_{1}
... and c is
of course the speed of light.

So
when C_{1}
and C_{2} both indicate a
particular moment t = t_{1}
= t_{2}, C'_{1}
indicates a particular moment
t'_{1}.

And
when the clock C_{2}
indicates the same moment t_{2} = t =
t_{1} as is
indicated by Clock C_{1}, the
moment t_{2}
must be related to the moment t'_{2}
indicated by the clock
C'_{2}
by the Lorentz transformation equation
t'_{2}
= (tvy/c^{2})/(1v^{2}/c^{2})^{0.5}
...
where y
is the distance, as measured by an observer in I,
between clock
C_{2}
and clock C'_{2}, and the
moment t is, again,
that indicated by the clock C_{2}
... and c is again
the speed of light.

So
when the clocks C_{1}
and C_{2} both simultaneously
indicate a particular
moment t = t_{1} = t_{2},
the clock C'_{2}
indicates a particular moment t'_{2}.

Now
unless x = y
above  which is highly unlikely, though of
course not impossible  it
is clear that t'_{1} will not
be equal to t'_{2}
above: both of them being indicated, of course,
when both the clocks C_{1}
and C_{2} indicate the single
moment in time t
= t_{1} = t_{2}.

But
this contradicts
Point No. 2. above, according to which whenever C'_{1}
indicates
any particular moment t', C'_{2}
must also
indicate the same moment t', since
both C'_{1}
and C'_{2} are synchronised.
Or
expressed
in table form:
According
to Point
No. 2 above, when:
C_{1}
indicates

C_{2}
indicates

C'_{1}
indicates

C'_{2}
indicates

Such that

t_{1}
= t_{2} = t

t_{2}
= t_{1} = t

t'_{1}

t'_{2}

t'_{1}
= t'_{2}

According
to Points
Nos. 3, 5 and 7 above, when:
C_{1}
indicates

C_{2}
indicates

C'_{1}
indicates

C'_{2}
indicates

Such that

t_{1}
= t_{2} = t

t_{2}
= t_{1} = t

t'_{1}

t'_{2}

t'_{1}
=/= t'_{2} *

It
is of course abundantly
clear that the above two tables contradict one
another in their fifth columns.
Any
comments? email
me.
* At
least, not necessarily.
