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Okay, I still don't get the solution (which I will lay out) to the following problem:

Suppose that A', B', and C' are at rest in frame S', which moves with respect to S at speed v in the positive x-direction. Let B' be located exactly midway between A' and C'. At t'= 0, a light flash occurs at B' and expands outward as a spherical wave.

I know that according to an observer in S', the wave fronts arrive simultaneously at A' and C'. I also know that they are not simultaneous in S frame. Now I find the time difference between the events as recorded by an observer in S.

Let distance from B' to C' and B' to A' be L. Then the time difference between the events as seen by frame S is:

$$\delta T = T(B' \to C') - T(B' \to A') = L/(c-v) - L/(c+v).$$

My question is, does this not contradict Einstein's Second Postulate? I thought the speed of light to any observer is always $c$? So why in the deltaT equation can we write $c-v$ and $c+v$? Shouldn't the speed of light be $c$ to any observer?

cheers

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    $\begingroup$ The $c\pm v$ term is a relative speed between two things, not a speed of a single object. Nothing's violated. $\endgroup$ – BMS Jan 26 '14 at 1:25
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    $\begingroup$ DWade64: It's great that you gave explicit distinct names (A', B', C') to the relevant members of "frame" (inertial system) S'; therefore +1. However, it would be even more helpful to also name relevant members of system S; such as (1) the member of S whom B' observed passing in coincidence with stating the "flash"; (2) the member of S whom A' observed passing in coincidence with observing the "flash" of B', and (3) the member of S whom C' observed passing in coincidence with observing the "flash" of B'; and, once those names have been decided, even additional relevant members of S. $\endgroup$ – user12262 Feb 25 '14 at 6:12
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The speed of light c is the same for every observer.

While the two observers disagree on the lenght of time it took to get from one place to the other, they also disagree on the distance bwtween those places. So the v-c term corrosponds to a disagreement about the length L not the speed c.

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  • $\begingroup$ But wouldn't the disagreement apply equally to the two legs of the trip, from B to A, and from B to C? $\endgroup$ – DJohnM Jan 25 '14 at 22:48
  • $\begingroup$ Yes, no the disagreement applies differently in each leg depending of your speed and direction just the right ammount to make the speed of light c always. $\endgroup$ – user288447 Feb 27 '14 at 18:32
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The calculation of the time difference in based on the velocity of light being the same for all observers.

The difference is caused by the difference in time to catch up, travelling at $c$, with a target initially at some distance, but fleeing at $v$, and to "ram" another target, initially at the same distance, but approaching at $v$

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