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A rod of rest length $L$ is moving away from a stationary observer at a constant velocity $v$. At time $t_1$, a light beam is fired from the front of the rod in the direction of the observer. At time $t_2$ in the rod's reference frame, the light beam reaches the back of the rod at which time another light beam is fired. The time between these events can be expressed as

$$\Delta t=\frac{L}{c}$$

In the observer's reference frame, the the light beam reaches the back of the rod at ${t_2}^{\prime}$. Due to the movement of the rod, the time between $t_1$ and ${t_2}^{\prime}$ is expressed as

$$\Delta t^\prime=\frac{L-v\Delta t^\prime}{c}$$

which can be rearranged in terms of $\Delta t^\prime$ as such:

$$\Delta t^\prime=\frac{L}{c+v}$$

In this situation, the observer sees the first light beam get to the back in less time than what the moving rod sees, or in other words $\Delta t^\prime<\Delta t$. This means a clock on the rod would seem to tick faster than a stationary clock from the perspective of the observer. Isn't this the opposite of what should happen in time dilation? What gives?


In fact, the opposite would happen if the rod was moving towards the observer, with time for the stationary observer now being

$$\Delta t^\prime=\frac{L}{c-v}$$

in which case $\Delta t^\prime$ is now more than $\Delta t$. This is now compatible with the time dilation formula (I think)

$$\Delta t^\prime=\frac{\Delta t}{\sqrt{1-v^2/c^2}}$$

but how does changing the direction of the rod's movement have this alternating time dilation/contraction effect? Am I missing something obvious?

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You made a few mistakes in your calculations. First of all, the length of the rod in the observer's frame isn't $L$, it's $\frac{L}{\gamma}$. Even taking this into account doesn't give you the right answer because you're still missing an important effect: space and time are interwoven.

If you have one spacetime event in one frame with coordinates $(t,x)$, the coordinates of the same event in a different frame moving with velocity $v$ relative to the first will be $\left(\gamma\left(t-\frac{vx}{c^2}\right),\gamma\left(x-vt\right)\right)$. Notice how in addition to time dilation and length expansion, there is also an effect on time from position and vice versa.

In the rod's coordinate frame, the two events in question happen at $(0,L)$ and $\left(\frac{L}{c},0\right)$. In the observer's frame (which is moving with velocity $-v$ relative to the rod), these events are at $\left(\frac{\gamma vL}{c^2},\gamma L\right)$ and $\left(\frac{\gamma L}{c},\frac{\gamma vL}{c}\right)$, respectively. In the rod's frame, the two events were separated by a time difference $\Delta t=\frac{L}{c}$ and a position difference $\Delta x=-L$, but in the observer's frame, the time difference is $\Delta t'=\gamma\left(1-\frac{v}{c}\right)\frac{L}{c}$ and the position difference is $\Delta x'=\gamma\left(\frac{v}{c}-1\right)L$.

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