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I am trying to understand the relativity of simultaneity in different frames, and I am trying to work out an example.

Suppose along the x-axis there are two points 2000m apart. Event A happens at t=0 and event B happens at $t=2\cdot10^{-6}s$ after event A happens in the rest frame. If there is an observer moving along the positive x-axis, how fast must he move in order to see the two events happen simultaneously?

So a diagram might look like this:

A------------------------------B
Observer ->

My understanding is that the observer should "see" the two events simultaneously if the information reaches him at the same time. Thus, if he stands somewhere between A and B such that after A happens, the information of A and B happening reach him at the same time, then he would think that A and B are simultaneous events. If my logic is correct, then he should stand somewhere to the right of the middle point between A and B.

But the example is asking for the velocity of the observer. I am not sure how to do this since I think the initial position of the observer should matter as well (if it does matter, can we assume he starts at the left point?).

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1 Answer 1

up vote 3 down vote accepted

Your logic is correct but for one misunderstanding right at the beginning. The observer is smart enough to know that just because the photons are received simultaneously doesn't mean they were emitted at the same time. The emitting of a photon and the receiving at a different time and different place are two distinct events, as explored further in this post, among others.

Basically, you need to find a position and a velocity such that the observer infers that the photons were emitted simultaneously. The observer makes this inference by saying the travel time for a photon is the distance between his current position and the source, as measured in his moving frame, divided by $c$.

In fact, you don't even need to solve for the position to get the velocity, but I'll let you crank through the appropriate Lorentz transformations.

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I see what you mean! Thanks for the clarification. –  Enzo Apr 19 '13 at 3:42

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