Relativity of simultaneity -- an example in D.Morin's book

[Oz123]

Ok, so here's an example from David Morin's book that I seriously don't understand:
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 attached image). By how much?
Ok, so his solution is quite vague, he says that the time delay for the light going to the left relative to the light going to the right with respect to the train's frame is lv/c^2, that is clear. But why is it equal to the time delay of the clocks with respect to the outside observer?

Why can't it just be like this:
Supposed we do place the light in the middle of the train, so that the time it takes to travel for the left and right lights be t1=l/2(c+v) and t2=l/2(c-v) respectively. Now we suppose that the clock hands will move if the light reaches at each end, so in the train's frame, both light reaches at the same time, therefore the hands will tick at the same time. And for the outside observer, there will be a time difference of t1-t2, so that's the same time difference between the movements of the hands. So we get t1-t2=vl/(c^2-v^2) which is not the same as lv/c^2, since it has an extra factor of the square of the Lorentz factor...Why is this other thought experiment wrong (well, it gives a different answer) when it seems to be right?

 

[PeroK]

You need to take into account length contraction and time dilation. Imagine a third clock in the centre of the train in order to see the effect of time dilation.

 

[Oz123]

Does time dilation differ in different parts of the train even though they move at the same velocities relative to the outside frame? Also, this example didn't even consider time dilation, just simultaneity.

 

[BvU]

they move at the same velocities relative to the outside frame

One is moving towards, the other away from !

 

[Oz123]

oh ya, I'm going to think about that.. Thanks

 

[Oz123]

So suppose we do use the other thought experiment though, how would I change it by using time dilation so that it arrives to the same answer as his though experiment? I'm sorry for a lot of questions, but I am just very confused.

 

[PeroK]

Does time dilation differ in different parts of the train even though they move at the same velocities relative to the outside frame? Also, this example didn't even consider time dilation, just simultaneity.

Let me show you a derivation using the platform frame. The light source is in the middle of the train and we'll have clock A at the rear and clock B at the front of the train. The length of the train is ##l## in the platform frame.

In the platform frame, the light reaches clocks A and B at:

##t_A = \frac{l}{2(c+v)}, \ t_B = \frac{l}{2(c-v)}##

And ##t_B - t_A = \frac{lv}{c^2 - v^2} = \frac{\gamma^2 lv}{c^2}##

As these events are simultaneous in the moving frame, we know that the clocks at A and B must read the same time at these two events. So, at ##t_A## clock A will read some time ##t'## and at ##t_B## clock B will read the same time ##t'##.

Now, what you've missed is that in the time ##\Delta t = t_B - t_A##, the clock at A has moved on (and is subject to time dilation). At ##t_B## the clock at A will read:

##t' + \frac{1}{\gamma}\Delta t = t' + \frac{\gamma lv}{c^2}##

Clock A, therefore, is ahead of clock B by ##\frac{\gamma lv}{c^2}##

Finally, the length of the train in the moving frame is ##L = \gamma l## so we have that clock A is ahead by ##\frac{Lv}{c^2}##

 

[Oz123]

Let me show you a derivation using the platform frame. The light source is in the middle of the train and we'll have clock A at the rear and clock B at the front of the train. The length of the train is ##l## in the platform frame.

In the platform frame, the light reaches clocks A and B at:

##t_A = \frac{l}{2(c+v)}, \ t_B = \frac{l}{2(c-v)}##

And ##t_B - t_A = \frac{lv}{c^2 - v^2} = \frac{\gamma^2 lv}{c^2}##

As these events are simultaneous in the moving frame, we know that the clocks at A and B must read the same time at these two events. So, at ##t_A## clock A will read some time ##t'## and at ##t_B## clock B will read the same time ##t'##.

Now, what you've missed is that in the time ##\Delta t = t_B - t_A##, the clock at A has moved on (and is subject to time dilation). At ##t_B## the clock at A will read:

##t' + \frac{1}{\gamma}\Delta t = \frac{\gamma lv}{c^2}##

Clock A, therefore, is ahead of clock B by ##\frac{\gamma lv}{c^2}##

Finally, the length of the train in the moving frame is ##L = \gamma l## so we have that clock A is ahead by:

##\frac{Lv}{c^2}##

OMG! Thank you very much! You just made me understand it, that's so awesome! Thanks! Morin should've left this later on after time dilation and length contraction. Since it is an example of simultaneity, the way I thought of it in his derivation is that if the two lights traveling inside the train has a time difference of lv/c^2 when one reaches the other end and the clock are simultaneous which are in the train's frame, then in the outside frame, it is the reverse, that is the clocks now have time difference and the light beams reach each end simultaneously, then therefore the time difference on each clock must also be lv/c^2...This thought doesn't even make sense but it seems to give the correct answer if we only think about simultaneity.

 

[PeroK]

OMG! Thank you very much! You just made me understand it, that's so awesome! Thanks! Morin should've left this later on after time dilation and length contraction. Since it is an example of simultaneity, the way I thought of it in his derivation is that if the two lights traveling inside the train has a time difference of lv/c^2 when one reaches the other end and the clock are simultaneous which are in the train's frame, then in the outside frame, it is the reverse, that is the clocks now have time difference and the light beams reach each end simultaneously, then therefore the time difference on each clock must also be lv/c^2...This thought doesn't even make sense but it seems to give the correct answer if we only think about simultaneity.

Morin's derivation is quite clever. By using the off-centre source and making the events simultaneous in the plaftorm frame, he avoids needing time dilation and length contraction. This shows that the relativitity of simultaneity is just as fundamental as time dilation.

The derivation I gave, I think, is more standard. But, I like Morin's as well.

 

[Oz123]

Ok so I am confused again. The length contraction I get, so the square of the lorentz factor is gone, but the time dilation I don't. So I used Lorentz transformation and got this:

Since the times for each events in s' frame (0,t'r) and (L, t'f) are synchronized so t'r=t'f, and we want the time difference in the s frame which is tr-tf. It still gives a gamma term.

 

[Oz123]

In Morin's derivation, he used t'r-t'f...Which doesn't make sense, because it means he is taking the difference in the train's frame, where the clocks are synchronized, so it should be zero.

 

[Mister T]

It still gives a gamma term.

That's the time dilation factor.

 

[Mister T]

In Morin's derivation, he used t'r-t'f...Which doesn't make sense, because it means he is taking the difference in the train's frame, where the clocks are synchronized, so it should be zero.

I don't have Morin's book, so I don't know what he's deriving. My guess is that he's considering the locations of the ends of the rod at some time ##t## in the unprimed frame, so that ##t=t_L=t_R##. The difference in those two locations will then be the rod's length in the unprimed frame.

By the way, if you're trying to derive the time dilation relation you need two events that occur at the same location in one of the frames.

 

[Oz123]

But we want the time delay in the S frame, because the question is asking what the time delay is as observed from the platform. The one Morin derived using Lorentz transformation is the same thing in the example, the time delay between the clocks. What he did though is to subtract the time for the events on each ends of the train in the train's frame which is the primed frame...But the question states that the clocks are synchronized on the train's frame so if we subtract the times in the primed frame, they should be zero..But he completely did the reverse

 

[Oz123]

Here's Morin's derivation for the result. 1st one is the one above, 2nd he used Lorentz transform, but he used the primed frame to get the time delay, which should be zero in that frame but not zero in the unprimed frame...For the unprimed frame, what he should get is the one I did above

 

[Oz123]

In his 2nd derivation he states that two events are simultaneous in the ground frame...But in his question, the clocks are simultaneous on the train frame, and the ground frame they are not simultaneous...Is he wrong then?

 

[PeroK]

In Morin's derivation, he used t'r-t'f...Which doesn't make sense, because it means he is taking the difference in the train's frame, where the clocks are synchronized, so it should be zero.

Morin is talking about two different events that take place at different times in the S' frame.

Synchronising clocks won't ensure that all events take place at the same time!

 

[Oz123]

But why did he used the time difference in the S' frame when it is supposed to be observed in the S frame? Why can't it be Tr-Tf since it is the time difference in the train as observed in the platform?

 

[PeroK]

But why did he used the time difference in the S' frame when it is supposed to be observed in the S frame? Why can't it be Tr-Tf since it is the time difference in the train as observed in the platform?

The time difference in the S frame is 0. He's arranged the experiment that way.

 

[Oz123]

hmm, ok the experiment he did is for the light going to each sides at the same time in the S frame. But what if we don't have those lights? Suppose we just use the clocks and use Lorentz transformations only. So using the clocks only, the time difference in S' is 0, but not in S...So we do this experiment and calculate the time difference in S, which I did using the Lorentz transformations above. Then why is the solution different? The one I wrote on paper I didn't arrange the experiment that way, I arranged it so that the there are no time differences in the S' frame as originally said in the question.

 

[vanhees71]

I'm not sure what Morin is discussing either (but better say the "light signals" hit the end of the train rather than "photons" :-(). Perhaps he's discussing Einstein's train thought-experiment? Two versions where discussed in this thread, perhaps it helps (I even have drawn Minkowski diagrams, although I think the formulae explain it much more clearly, but some people seem to like Minkowski diagrams more the formulae :-)):

https://www.physicsforums.com/threads/in-c-v-what-is-v-the-trains-speed-relative-to.828679/

 

[Oz123]

Ya, I seriously don't know. Like if we forget about the lights and just focus on the clocks, or if I just gave the question without Morin's solution. The answers might arrive differently.

 

[PeroK]

Ya, I seriously don't know. Like if we forget about the lights and just focus on the clocks, or if I just gave the question without Morin's solution. The answers might arrive differently.

What Morin is doing is using the Lorentz Transformation to show the relativity of simultaneity. The two clocks on the train are being illuminated by light pulses at the same time in the ground frame, which would follow on from his previous example of an off-centre light source. In this case, each clock is recording the time in the moving frame of an event at its location.

The clocks on the train are in sync in the moving frame, but are being illuminated at different times. Whereas, in the ground frame they are being illuminated simultaneously.

 

[Oz123]

ya I think I am getting it, thank you. I still am a bit confused, but I guess if we forget about the clocks and just think about the lights, it would work. Thanks a lot :)