# Simultaneity between events in reference frames

1. Nov 9, 2013

### PeroK

Hi,

I've been trying to understand the (lack of) simultaneity between events in reference frames moving wrt each other. I'd be grateful if someone could confirm that I've got things right:

If two events are simultaneous in one reference frame (S'), then they will not be simultaneous in a reference frame (S) moving with velocity v relative to S' if the events are separated by any distance (D' or D) in the direction of motion. In which case, an observer in S will observe a time difference of:

$\Delta t = \frac{Dv}{(c^2 - v^2)} (*)$

And, if two events are separated by distance D' and time t' in S', then, as observed in S, they are separated by time:

$\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(t' \pm \frac{D'v}{c^2})$

(*) The book I'm reading expressed this equation in terms of an observer in S reading the times off clocks synchronised in S' and getting:

$\Delta t = \frac{D'v}{c^2}$

Which seems to me an odd way to express things and I think this confused me somewhat. Hopefully, I've now understood this simultaneity issue?

2. Nov 9, 2013

### yuiop

It is easier to start in reference frame S in which the rod of proper length $D'$ is moving with relative velocity v. We know that due to Length contraction the length of the rod in S is $D =D'/\gamma$. Now we boost to S' (the frame in which the rod is at rest) and from the Lorentz transform the time interval is:

$\Delta t' = \gamma(\Delta t - D v /c^2) = \gamma(\Delta t - (D'/\gamma) v /c^2)$

When measuring the length in S the measurements of the ends of the rod are made simultaneously so $\Delta t = 0$ and the above expression reduces to:

$\Delta t' = \gamma( - (D'/\gamma) v /c^2) = -D' v/c^2$

which is basically the expression in your text book, other than they have $\Delta t$ rather than $\Delta t'$. This is the amount the clock at the front of the rod is behind the clock at the rear of rod when viewed simultaneously in S. Note that I am using $\Delta t'$which is the proper time shown by the clocks at rest with the rod.

It is a little confusing because we are talking about the difference between two clock readings as seen in S when those clocks are not at rest in S. Note that you said
If an observer in S reads the times off clocks at rest in his reference frame, then the time interval is $\Delta t$ but if he is reading the times off clocks synchronised in S' and at rest in S' then the time interval is $\Delta t'$.

Last edited: Nov 9, 2013
3. Nov 10, 2013

### PeroK

Thanks, I've definitely got it now.