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An observer in a rocket moves toward a mirror at speed $v$ relative to the reference frame in which the mirror is stationary - call this frame $S$. A light pulse emitted by the rocket travels toward the mirror and is reflected back to the rocket. As measured by an observer in $S$, the front of the rocket is a distance $d$ from the mirror at the moment the light pulse leaves the rocket. What is the total travel time of the pulse as measured by an observer in (a) the $S$ frame and (b) the front of the rocket?

For (a), I'm thinking we can just use the usual formula: $\triangle t =2\frac{d}{c}$

For (b), let $\triangle t'$ be the time as measured by someone in the front of the rocket. By Lorentz:

$\triangle t'=\gamma (\triangle t-\frac{v \triangle x}{c^2})=\frac{1}{3} \triangle t$, after some algebra. I can show this if you want.

Thus it seems that the observer in the rocket measures a shorter time [would this be the proper time?] I am finding special relativity a bit confusing and just want to see what you guys think of this. This seems right to me, since the moving observer should measure the smallest time.

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closed as off-topic by Brandon Enright, BMS, Kyle Kanos, Ali, John Rennie Sep 24 '14 at 5:24

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  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/43591 $\endgroup$ – dmckee Sep 23 '14 at 18:58
  • $\begingroup$ Careful, the rocket is moving too! It will have traveled a distance when the light gets back to the rocket $\endgroup$ – ClassicStyle Sep 23 '14 at 19:06
  • $\begingroup$ @dmckee: I don't think this is a duplicate of that one. That one just was looking for a faster/slower than light answer. This one wants an actual calculation. $\endgroup$ – Ross Millikan Sep 23 '14 at 19:22
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Be careful on (a), the rocket is moving too and will have moved a distance of $v \Delta t$ by the time the light comes back. So we have:

$$2d-v \Delta t=c \Delta t$$

Solving for $\Delta t$:

$$\Delta t = \frac{2d}{c+v}$$

For (b), your answer certainly cannot be correct because it is independent of velocity. The easiest way to solve most problems in Special Relativity is by using the invariant interval. I'll use units where c=1 to make the algebra simpler, and we can put the c's back in using dimensional analysis later.

We have:

$$(\Delta t')^2 - (\Delta x')^2=(\Delta t)^2 - (\Delta x)^2$$

and:

$$\Delta t=\frac{2d}{1+v}$$ $$\Delta x=v\Delta t= \frac{2vd}{1+v}$$

In the rocket frame, both events occur at the same location (the front of the rocket) so $\Delta x' =0$. So we have:

$$(\Delta t')^2= (\Delta t)^2 - (\Delta x)^2= \left ( \frac{2d}{1+v} \right )^2 - \left ( \frac{2vd}{1+v} \right )^2$$

After a bit of algebra (and dimensional analysis to put the c's back in):

$$\Delta t' = \frac{2d}{c} \sqrt{\frac{1-v/c}{1+v/c}}$$

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