# Failure of simultaneity at a distance

1. Jun 6, 2017

### Karol

1. The problem statement, all variables and given/known data

I get $~t_1-t_2=\frac{u}{c^2}(x_1-x_2)$

2. Relevant equations
Lorentz transformations:
$$x'=\frac{x-ut}{\sqrt{1-u^2/c^2}}$$
$$t'=\frac{t-ux/c^2}{\sqrt{1-u^2/c^2}}$$

3. The attempt at a solution
t' are times in the moving system S' and are equal, the times t in the fixed system S differ, so:
$$t'=\frac{t-ux/c^2}{\sqrt{1-u^2/c^2}}~\Rightarrow~t=t'\sqrt{1-u^2/c^2}+\frac{ux}{c^2}$$
$$\rightarrow~t_1-t_2=\frac{u}{c^2}(x_1-x_2)$$

2. Jun 6, 2017

### Staff: Mentor

The problem statement is a bit garbled. The events are simultaneous in S, not S'. So you are solving for Δt', not Δt.

3. Jun 6, 2017

### Karol

But it is clearly said:
"events that occur at two separate places, at the same time, as seen by Moe in S'..."
So the events are simultaneous in S'. and there is also the example given later, which i didn't include, is about a space ship with a clock made of a central source of light and two mirrors at the ends.

4. Jun 7, 2017

### Staff: Mentor

But then it immediately contradicts that by saying "If one event occurs at point x1 at time t0 and the other event at x2 at t0 (the same time)". So despite the first sentence they are talking about events simultaneous in S, not S'. Otherwise, why would they calculate t2' - t1'? It would just be zero!

What textbook is this from?

5. Jun 7, 2017

### Karol

You are right, it would be zero. my english isn't that good, maybe i don't understand.
I took it from a PDF i have of Feinman lectures, i attach it here, it's short

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6. Jun 7, 2017

### Staff: Mentor

Feynman's talking in general, saying that things simultaneous in one frame are not simultaneous in another. Doesn't matter whether its S' or S.

(I'll admit that it's confusing to follow up that first statement about events simultaneous in S' with a formula based on events simultaneous in S!)

7. Jun 7, 2017

### Karol

Thank you very much Doc Al