Failure of simultaneity at a distance

[Karol]

Homework Statement

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

Homework Equations

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

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:
$$t'=\frac{t-ux/c^2}{\sqrt{1-u^2/c^2}}~\Rightarrow~t=t'\sqrt{1-u^2/c^2}+\frac{ux}{c^2}$$
$$\rightarrow~t_1-t_2=\frac{u}{c^2}(x_1-x_2)$$

 

[Doc Al]

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

 

[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 spaceship with a clock made of a central source of light and two mirrors at the ends.

 

[Doc Al]

But it is clearly said:
"events that occur at two separate places, at the same time, as seen by Moe in S'..."

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?

 

[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

 

[Doc Al]

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!)

 

[Karol]

Thank you very much Doc Al