Time Dilation: Clock A & B Confusion

[Haorong Wu]

TL;DR Summary

time dilation

Suppose there are two clocks. Clock B is moving with a speed of ##v## relative to clock A. Then clock A is moving with a speed of ##-v## relative to clock B.

Let ##t_0## be the time interval for two events in the framework of clock A, while ##t## be the time interval for the same two events in the framwork of clock B.

Then ##t=\gamma t_0##. Since ##\gamma## is greater than one, so for clock B time runs slower than for clock A.

But, also, ##t_0=\gamma t##, then time in clock A runs slower than in clock B.

This seems wrong. But where?

Thanks!

 

[Dale]

Then ##t=\gamma t_0##.

This is not true in general. It is only true in the special case that the two events occur at the same location in frame A.

 

[Mister T]

This seems wrong. But where?

They can't possibly be the same two events except in the trivial case where ##v=0##.

 

[Ibix]

As others have pointed out, ##t=\gamma t_0## is a special case that only applies when the events happen at the same spatial location. That will only apply to at most one frame (unless ##v=0##, as @Mister T notes).

The general expression is the Lorentz transforms:$$\begin{eqnarray*}
t'&=&\gamma\left(t-\frac v{c^2}x\right)\\
x'&=&\gamma\left(x-vt\right)
\end{eqnarray*}$$and their inverse:$$\begin{eqnarray*} t&=&\gamma\left(t'+\frac v{c^2}x'\right)\\ x&=&\gamma\left(x'+vt'\right) \end{eqnarray*}$$Note that you need to specify the time and position of each event and transform each one separately to get its transformed time and position. Then you can work out the time difference in either frame. (A useful trick is to declare that the origin is where and when one of the events is - ##(x,t)=(0,0)## transforms to ##(0,0)##.)

We generally recommend that you forget you ever heard of the time dilation formula and always use the Lorentz transforms until you are comfortable enough with them that you can see instantly whether you are allowed to use the time dilation formula in a given situation or not.

 

[Haorong Wu]

Thanks, guys. I got it now. I forgot that the clock A remain still relative to the two events. Thanks again!

 

[PeroK]

I forgot that the clock A remain still relative to the two events.

Events are points in spacetime. To say that a clock is at rest relative to two events does not make sense. Instead, what you might mean is what @Dale said: that the two events occur at the same place in the rest frame of clock A.

 

[Erland]

This is because events at different locations which are simultaneous in one intertial frame are not simultaneous in a frame which moves with constant velocity relative to the first frame.

Let me recommend an old post I wrote here, where this is explained.
https://www.physicsforums.com/threa...multaneity-easier-to-see-with-a-train.468826/