# I Regarding relativity of simultaneity

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1. Mar 26, 2016

### wildebeest44

I recently started studying Special Relativity an my book discusses the following:

Say I have synchronized two separated clocks in a reference frame S, if then an observer in another reference frame S' for whom the clocks are moving sees the clocks he would say those clocks are out of synchronization for him, as for him the light that was used to sync the clocks reaches one of them before the other.

Up to this point, I think I understand what is happening, but if then the observer in S' knowing about the relativity of simultaneity applies a Lorentz transformation to try and "see" the clocks as an observer in S would, it will, nevertheless, still find a non-zero difference between the times "shown" by the clocks, proportional to the distance separating them.

What does this mean?
Isn't the Lorentz transformation an attempt to find the space and time coordinates as they would be seen from other reference frames, and if so, why is it unsuccessful here?

What's the physical interpretation of the difference in time the observer in S' just found, considering is neither the difference he sees from his reference frame or the one seen by the original observer in S?

Thank you, and excuse me for not showing the math I don't how to write it here in a readable way.

2. Mar 26, 2016

### Ibix

Welcome to PF.

This is difficult to answer without the maths. Just below the reply box there's a link to how to use LaTeX for maths. An example: $$x'=\gamma (x-vt)$$ Reply to my post to see how it's done.

An obvious guess is that if you transformed from S to S' using v=u, did you transform from S' back to S using v=u (wrong) or v=-u (right)? If that's not it have a read of the LaTeX link and post what you did. Then we can help without guessing.

3. Mar 26, 2016

### wildebeest44

I just saw the guide, thanks, i'll give it a try.

So, suppose that clocks are placed in $x_1$ and $x_2$, in S they are in sync so,
$\Delta t=0$

However in S', which is moving with speed u relative to S, we have for each clock
$t_1'=\gamma(t_1-\frac{u}{c^2}x_1)$
$t_2'=\gamma(t_2-\frac{u}{c^2}x_2)$
then for this observer the time interval between them is
$\Delta t' = t_2'-t_1'=\gamma(\Delta t - \frac{u}{c^2}\Delta x)$
with $\Delta t=0$ because the clocks are in sync in S, then
$\Delta t' =- \gamma \frac{u}{c^2}\Delta x$
Therefore, the clocks are out of sync for the observer in S'. But if I now tried to apply
$\Delta t'=\gamma \Delta t$
I would get
$\Delta t=-\frac{u}{c^2}\Delta x$
Which is different to 0. This is the result I find troubling.

Try the Lorentz transform again, but reversing the sign of $u$. Don't just use the time dilation formula.