# "Rear clock ahead" effect in special relativity

I've read (pretty much) all the questions in the site regarding the topic, however I wasn't able to comprehend the author's solution to the problem in its totality. The problem is from David Morin's book, Chapter 11, example 11.3.1:

Two clocks are positioned at the ends of a train of length $$L$$ (as measured in its own frame). They are synchronized in the train frame. The train travels past you at speed $$v$$. It turns out that if you observe the clocks at simultaneous times in your frame, you will see the rear clock showing a higher reading than the front clock (see Fig. 11.6). By how much?

My attempt:

Let $$S_t$$ be the train frame and $$S_g$$ the ground frame.
Since the clocks are synchronized in $$S_t$$, we may say that there is no time difference in the ticks of the clocks', namely $$\Delta t_t=0$$.
Moreover, $$\Delta x_t=L$$.
The Lorentz transformation give us the time elapsed between ticks in $$S_g$$ : $$\Delta t_g=\gamma(0+\frac{vL}{c^2}) = \frac{\gamma v L}{c^2}$$ To me, in my first attempt, this would be the final answer, but I've read here that time dilation must be taken into account when returning to the ground frame, which makes sense, since "counting time" inevitably changes in the frames (a second in $$S_t$$ clock passes differently in $$S_g$$). Ultimately, it yields:$$\Delta t_g = \frac{Lv}{c^2}$$ Morin's Solution

Now, what I fail to understand is Morin's solution using the Lorentz transformation (which can be found in the Appendix G):

The second of Eqs. (11.17) is $$\Delta t_g=\gamma(\Delta t_t+\frac{v\Delta x_t}{c^2})$$, where the subscripts refer to the ground and train frames. If two events (for example, two clocks flashing their times) located at the ends of the train are simultaneous in the ground frame, then we have $$\Delta t_g = 0$$. And $$\Delta x_t = L$$., of course. The above Lorentz transformation therefore gives $$\Delta t_t = \frac{-Lv}{c^2}$$. The minus sign here means that the event with the larger $$x_t$$ value has the smaller $$t_t$$ value. In other words, the front clock reads $$\frac{Lv}{c^2}$$ less time than the rear clock, at a given instant in the ground frame.

My problem lies in the following topics:

1. I get why he considered $$\Delta t_g = 0$$, but, because the clocks are synchronized in the train frame, shouldn't the event "clocks flashing their time" also have $$\Delta t_t = 0$$? Then the problem would have no solution.
2. He proceeds to calculate $$\Delta t_t = \frac{Lv}{c^2}$$ and state that this is the difference in time in the ground frame. To me, the Lorentz transformation clearly states that this result is with respect to the train frame, which, again, doesn't make sense, because we assumed the clocks to be synchronized.
• The point is that the delta t’s don’t represent “the difference between the readings on the clocks”. They represent “the time difference between two events in a given reference frame”. So if I want to consider what’s happening in different places but at the same time in the ground frame, then $\Delta t_g = 0$. It has nothing to do with how the clocks are synchronized, that simply is the definition of “at the same time”. Sep 19, 2023 at 13:32
• Similarly, if you consider the two clocks at the same time in the train frame, then $\Delta t_t = 0$. And because the clocks are synchronized in that frame, the difference in the two clock readings at these times is $\Delta \tau = 0$. More generally, since the clocks tick regularly, $\Delta t_t = \Delta \tau$ and the answer to the question is $\Delta \tau$. Sep 19, 2023 at 13:34
• So $\Delta t_t = 0$ does not represent directly the synchronization of the clocks, but that we (in the train frame) are sort of "taking a picture" of the train? Then, as the coordinates of the clocks are different, their readings will change as time passes? That makes more sense, however I am not 100% sure what I wrote is correct. Could you please elaborate further that last part of your last comment?
– Huye
Sep 19, 2023 at 14:07
• That’s almost it. Don’t think about time passing, just think about applying the Lorentz transformation to a single pair of events. Those events are at the same time in the ground frame. Now we want to compute the difference in clock readings corresponding to those events. So we transform to the train frame because we know the clocks are synchronized there. And in the train frame the two events are at different times, so they correspond to different clock readings (because the same clock reading corresponds to the same time in the train frame). Sep 19, 2023 at 14:14
• Ok, I think I get it now. The source of my confusion was that I was mistaking the synchronization with the definition of events. So, if we define E1 - see front clock; E2- see rear clock as events in the ground frame and set $\Delta t_g = 0$, these events happen at different times in the train frame. This means that, because they are synchronized, their readings must account for this difference for someone in the train. Therefore, when the ground frame sees the clock simultaneously, he confirms this.
– Huye
Sep 19, 2023 at 15:22

It really useful to consider events, and the difference between them, when considering relativistic effects. So start with a decent space-time diagram from the internet (or draw your own):

Then: define your frames. The train (station) is $$S$$ ($$S'$$).

Pick a good origin: $$(t=0,x=0)$$ and $$(t'=0,x'=0)'$$ is the front, when both observers measure the time on the clock there.

The world line of that clock in $$S$$ is:

$$f(t)=(t, 0)$$

the clock in the back has a world line:

$$b(t)=(t, L)$$

So 2 vertical lines (not shown) define the ends of the train in all frames.

To find out when the station observer measures the end of the train, figure out when $$b(t)$$ crosses the $$x'$$ axis (so $$t'=0$$).

That's high school geometry, with:

$$\tan\alpha=\beta$$

Remember that the tics on $$x'$$ are stretched by:

$$U' = U\sqrt{\frac{1+\beta^2}{1-\beta^2}}$$

From there you can write down events (points on the Minkowski diagram) and Lorentz transform them as needed.

Also, the fact that for any two events:

$$\Delta s^2 = \Delta t^2 - \Delta x^2$$

is the same in all reference frames, can come in handy.

Yes, the clocks are synchronised in S'. However, if you look at the two of them at the same instant in S, then you are looking at one of them before the other in S', hence they will show two different times.