Re: relativity, speed of light postulate

From: Igor (thoovler_at_excite.com)
Date: 07/28/04 

Date: 28 Jul 2004 14:34:05 -0700

"Androcles" <androc1es@nospamblueyonder.co.uk> wrote in message news:<SFpNc.9079$DZ5.91648036@news-text.cableinet.net>...
> "Jon Bell" <jtbellj3p@presby.edu> wrote in message
> news:ce43ur$bfg$1@jtbell.presby.edu...
> | In article <19a422d8.0407261500.4ea9e16a@posting.google.com>,
> | abs <ciitv@yahoo.com> wrote:
> | >
> | > A->
> | > (Light)
> | > B
> | >
> | >
> | >consider A and B to be 2 inertial reference frames. there is a light
> | >source as shown. A is moving with constant velocity(therefore its a
> | >inertial frame)V towards light. according to B speed of light is C.
> | >but according to A, speed of light is C-V
> |
> | No, the speed of the light emitted by the source equals c in both
> | reference frames. It doesn't depend on which direction the light is
> | emitted, either.
> |
> Yes, he's right and you are wrong.
> Reference :
> http://www.fourmilab.ch/etexts/einstein/specrel/www/
> "But the ray moves relatively to the initial point of k, when measured in
> the stationary system, with the velocity c-v, so that x'/(c-v) = t."
> Since x' is defined as x'=x-vt, and v = x/t, it follows that x' = x-x = 0,
> and hence t = 0.
> Einstein has a divide by zero in
>
> 1/2 [1/(c-v) + 1/(c+v)]dtau/dt = dtau/dx' + 1/(c-v) dtau/dt
>
> at dtau/dx' and the Lorentz transforms cannot be derived.

And just where is there a division by zero here?

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