Simultaneity between events in reference frames

[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?

 

[yuiop]

... 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?

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 textbook, 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

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

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 $\Delta t'$.

 

[PeroK]

Thanks, I've definitely got it now.