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Re: Variable speed of light (was: Relativity conundrum)



Pentcho Valev wrote:

I still believe that the setting of the problem is correct and consistent with
traditional relativity analysis. Two events ("light enters train" and "light leaves
train") happen that are identifiable in either frame (otherwise any such analysis
should be abandoned) and we are to compare the distances between them, x and x',
and the time intervals between them, t and t'. I think the only reasonable
criticism of my solution should consist in offering a solution proving that x/t =
x'/t'. Einstein's theory is expected to be able to resolve problems as simple as
this one.


I agree that this is a useful example - basically because it uses the width, W, of the
moving train, for which both observers measure the same value. I think that when this
discussion line ends - if we come to an agreement on the analysis - we will have a
useful bit of pedagogy.

The person on the ground sees the light travel perpendicular to the tracks, crossing
the width of the train in a time t = W/c. The person in the train, who can consider
themselves at rest if they wish, sees the light going through the train at a diagonal.
The diagonal distance is D = sqrt( W^2 + (v t')^2 ) as measured from inside the train
(using vt' as the distance the light moves sideways while traversing the width of the
train.) The time t' is simply D/c (c is used because the person on the train cannot
detect the motion by a physical experiment.) Obviously c = W/t = D/t' because it was
assumed in the analysis, but that's not the point. The real underlying concept is the
assumption of compete equivalence of inertial frames - that one cannot make any
physical measurement, including the speed of light, that would give a clue that an
inertial frame is not at rest.

Bob at PC