# Question on special relativity from "Basic Relativity"

PhysicsTruth
Summary:: Require confirmation regarding answers to a question posed by the book "Basic Relativity" by Richard A. Mould.

Here is a problem which I encountered while going through Basic Relativity by Richard A. Mould-

I'd like to receive a confirmation regarding the answers I've come up with to the question, since I'm an absolute beginner in Special Relativity.
a) The length of the train as measured by the train observer would be the length of the platform as measured by the train observer, and equal to L/gamma, as the platform is moving with respect to the train observer.

b) According to a ground observer, the reading on clock 2 at that instant would be 4 - L/v. Since the platform clocks are not synchronized with respect to the train observer, the reading on clock 2 at that instant for the train observer would be 4 - L/(gamma v), due to time dilation.

c) The time interval between the two events according to the train observer would be thus L/(gamma v).

Can someone confirm if the above answers are correct? If not, can someone tell me where am I possibly going wrong? Thanks!

Delta2

## Answers and Replies

a) The length of the train as measured by the train observer would be the length of the platform as measured by the train observer, and equal to L/gamma, as the platform is moving with respect to the train observer.
Remember that a thing in motion is length contracted. Someone who is at rest with respect to it measures it to be longer then someone who sees it in motion. How long does the platform observer say it is? How long, then, does the train observer say it is?
b) According to a ground observer, the reading on clock 2 at that instant would be 4 - L/v. Since the platform clocks are not synchronized with respect to the train observer, the reading on clock 2 at that instant for the train observer would be 4 - L/(gamma v), due to time dilation.
No - you need to use the full Lorentz transform for this one.
c) The time interval between the two events according to the train observer would be thus L/(gamma v).
No - again, you need to use the full Lorentz transform.

The general rule for doing problems of this nature is to write down the ##(x,t)## coordinates of each event of interest in one frame or another and then mindlessly apply the Lorentz transform to get the coordinates in the other frame. Then you can take time differences or whatever in the relevant frame. An even better approach is to draw a Minkowski diagram (and if you haven't come across those I'd recommend looking them up), but grinding through the maths will get you to the result.

PhysicsTruth
Ok, I'm trying those out using the Lorentz transformations. But, I didn't understand where did I go wrong in part (a).

Should the train length measured on the train be longer or shorter than that measured from the platform? How long is the train as seen by the platform? Is ##L/\gamma## longer or shorter than that?

PhysicsTruth
The platform observer should see the actual length of the train contracted by a factor of gamma. So, should the train observer just perceive the length of the train to be L? My way of thought was that the train observer would probably measure the length of the platform according to him in order to state the length of the train in his frame.

The platform observer should see the actual length of the train contracted by a factor of gamma.
Better to say "rest length" since there's nothing less actual about any frame's measurements than any other, but otherwise correct.
So, should the train observer just perceive the length of the train to be L?
No. Do you know how long the train is as measured in the platform frame?

PhysicsTruth
Better to say "rest length" since there's nothing less actual about any frame's measurements than any other, but otherwise correct.

No. Do you know how long the train is as measured in the platform frame?
In the platform frame, the length of the train would be the rest length contracted by a factor of gamma, which must be equal to the platform length. So, the rest length(as perceived by the train observer) should be gamma times the platform length, i.e. gamma*L. Is this reasoning correct?

Ibix
Is this reasoning correct?
Yes.

PhysicsTruth
Yes.
Thanks a lot for helping me out in figuring this.

ergospherical
for b, there are some different approaches

let the ##\tilde{K}## (train) system have velocity ##v## with respect to the ##K## (platform) system. the two clocks have worldlines ##\gamma_1## and ##\gamma_2## defined by ##x_1 = L## and ##x_2 = 0## in the ##K## coordinates. the spacetime point ##p## at which clock ##1## coincides with the front of the train has ##K## coordinates ##(4,L)## and ##\tilde{K}## coordinates ##(\tilde{t}_p, \tilde{x}_p)##

you are looking for the point ##q## on the worldline ##\gamma_2## which is ##\tilde{K}##-simultaneous with ##p##, i.e. ##\tilde{t}_q = \tilde{t}_p## so that ##q## has ##\tilde{K}## coordinates ##(\tilde{t}_p, \tilde{x}_q)## and ##K## coordinates ##(t_q, 0)##. and the number ##t_q## is what is displayed by clock ##2## at ##q##

it's faster to use the diagram rather than write out the coordinate transformations

PhysicsTruth
Remember that a thing in motion is length contracted. Someone who is at rest with respect to it measures it to be longer then someone who sees it in motion. How long does the platform observer say it is? How long, then, does the train observer say it is?

No - you need to use the full Lorentz transform for this one.

No - again, you need to use the full Lorentz transform.

The general rule for doing problems of this nature is to write down the ##(x,t)## coordinates of each event of interest in one frame or another and then mindlessly apply the Lorentz transform to get the coordinates in the other frame. Then you can take time differences or whatever in the relevant frame. An even better approach is to draw a Minkowski diagram (and if you haven't come across those I'd recommend looking them up), but grinding through the maths will get you to the result.
I'm a bit confused about the wording of the question. The question says - "Clock 1 reads 4:00 when it coincides with the front of the train..." Does this mean that the clock reads 4:00 with respect to the platform, or the train observer? If it reads 4:00 with respect to the platform, then Lorentz transformations give me the reading on clock 2 with respect to the train observer to be gamma*(4 - L/v - vL/c^2). It would be really helpful if someone could confirm.

Homework Helper
Gold Member
2021 Award
I'm a bit confused about the wording of the question. The question says - "Clock 1 reads 4:00 when it coincides with the front of the train..." Does this mean that the clock reads 4:00 with respect to the platform, or the train observer? If it reads 4:00 with respect to the platform, then Lorentz transformations give me the reading on clock 2 with respect to the train observer to be gamma*(4 - L/v - vL/c^2). It would be really helpful if someone could confirm.
The time that a clock reads when it coincides with a point on a physical object is the same in all reference frames. How could it be otherwise?

PhysicsTruth
The time that a clock reads when it coincides with a point on a physical object is the same in all reference frames. How could it be otherwise?
Yeah, thanks for clarifying. I've been overthinking this whole time. Can you kindly confirm if the Lorentz transformations evaluate to the result that I've got? I've taken both the frames' origins to be coinciding at the beginning of the platform.

Homework Helper
Gold Member
2021 Award
Yeah, thanks for clarifying. I've been overthinking this whole time. Can you kindly confirm if the Lorentz transformations evaluate to the result that I've got? I've taken both the frames' origins to be coinciding at the beginning of the platform.
Where did the number ##4## come from. In your answer?

PhysicsTruth
Where did the number ##4## come from. In your answer?
Since , at 4:00, the front end of the train coincides with clock 1.

Homework Helper
Gold Member
2021 Award
Since , at 4:00, the front end of the train coincides with clock 1.
I don't agree with your answer in a number of respects. Not least that the gamma factor can't apply to the reading of ##4:00##. Think about it.

The question doesn't seem.to say what units are used there.

PhysicsTruth
PhysicsTruth
I don't agree with your answer in a number of respects. Not least that the gamma factor can't apply to the reading of ##4:00##. Think about it.

The question doesn't seem.to say what units are used there.
I've been using the basic transformations so far, it seems that I've got to change the equations now. If I set clock 1's position at the platform to zero, then the space coordinate for the train observer is also zero. If the coefficients in the Lorentz matrix are a1,a2,a3,a4 (the upper 2*2 matrix), then a3 turns out to be zero. Do I need to use (x')^2 = c^2(t')^2 as my next set of equations in this process?

Homework Helper
Gold Member
2021 Award
I've been using the basic transformations so far, it seems that I've got to change the equations now. If I set clock 1's position at the platform to zero, then the space coordinate for the train observer is also zero. If the coefficients in the Lorentz matrix are a1,a2,a3,a4 (the upper 2*2 matrix), then a3 turns out to be zero. Do I need to use (x')^2 = c^2(t')^2 as my next set of equations in this process?
I must say that for part b) I would not use the Lorentz transformation. The question is what precise event are you transforming?

There's a quicker way.

PhysicsTruth
Homework Helper
Gold Member
2021 Award
PS in any case, I would first restate the whole problem in train frame. What do you know about the sequence of events in that frame?

PhysicsTruth
I must say that for part b) I would not use the Lorentz transformation. The question is what precise event are you transforming?

There's a quicker way.
Actually, I haven't read about Minkowski diagrams yet, so as suggested by previous replies, I decided to use the Lorentz transformations. Also, the reading on clock 2 during the event of the front end of the train coinciding with clock 1 is precisely what I'm trying to transform to the train's frame.

PhysicsTruth
PS in any case, I would first restate the whole problem in train frame. What do you know about the sequence of events in that frame?
The observer on the train sees himself coincide with clock 1 on the platform (considering the platform and the train to be separated by minimal width). The platform is moving at velocity v with respect to him, in the opposite direction. He can see the clock (clock 2) at the far end of the platform as well, during this event. After that, he sees himself coincide with clock 2.

Homework Helper
Gold Member
2021 Award
The observer on the train sees himself coincide with clock 1 on the platform (considering the platform and the train to be separated by minimal width). The platform is moving at velocity v with respect to him, in the opposite direction. He can see the clock (clock 2) at the far end of the platform as well, during this event. After that, he sees himself coincide with clock 2.
There is no observer on the train. Let's assume that in any case. There are only two clocks on a moving platform.

Homework Helper
Gold Member
2021 Award
PS you need a diagram in the train frame

ergospherical
no in the platform frame the two events are simultaneous, whilst in the train frame the front end hits the front clock before the back end hits the back clock

PhysicsTruth
PhysicsTruth
no in the platform frame the two events are simultaneous, whilst in the train frame the front end hits the front clock before the back end hits the back clock
Yeah, I can realize now. At this point, I am just confusing myself. I've been trying to set up the equations for a while now but haven't been able to succeed yet.

Homework Helper
Gold Member
2021 Award
Let me help you. I have platform moving right to left in the train frame, so its velocity is ##-v##.

The platform has length ##L/\gamma##. Which is shorter than the train.

Clock 2 is at the " front" of the platform, in the sense that it reaches the front of the train first. We have no information about this event.

Clock 1 then reaches the front of the train. Shall we take this to be the common origin?

Then clock 2 reaches the rear of the train. We have coordinates for this event in the platform frame.

We are not interested in subsequent events.

Does that seem to you to be a fair description of the problem in the train frame?

PhysicsTruth
Homework Helper
Gold Member
2021 Award
PS you need a diagram in the train frame
Or, get your train set out!

PhysicsTruth
Let me help you. I have platform moving right to left in the train frame, so its velocity is ##-v##.

The platform has length ##L/\gamma##. Which is shorter than the train.

Clock 2 is at the " front" of the platform, in the sense that it reaches the front of the train first. We have no information about this event.

Clock 1 then reaches the front of the train. Shall we take this to be the common origin?

Then clock 2 reaches the rear of the train. We have coordinates for this event in the platform frame.

We are not interested in subsequent events.

Does that seem to you to be a fair description of the problem in the train frame?
Ok, I'm trying to proceed from here.

PeroK
PhysicsTruth
Or, get your train set out!
If you could just confirm...

I've taken the coordinates for the front end coinciding with clock 1 to be-
a) (0,4) for both the frames
and the coordinates for the rear end coinciding with clock 2 to be-
a) (-l_o,t1') for the train's frame
b) (-L,4) for the platform frame

Using these coordinates and applying the Lorentz transformations, I arrive at this result, as shown in the pics. Is this correct?

Homework Helper
Gold Member
2021 Award
The origin must be ##(0,0)##, by definition. The Lorentz transformation won't work with ##t =4## at the origin.

You've drawn the scenario in the platform frame. What about the train frame?

The results are hard to read. They don't look right at all.

Homework Helper
Gold Member
2021 Award
Imagine there are two more clocks next to 1 and 2, but starting at 0, rather than 4. Run the experiment and think about these new clocks, for which you can use Lorentz.

The original clocks always read the same as the new clocks plus 4.

PhysicsTruth
Imagine there are two more clocks next to 1 and 2, but starting at 0, rather than 4. Run the experiment and think about these new clocks, for which you can use Lorentz.

The original clocks always read the same as the new clocks plus 4.
Can't I just use a shifting matrix which shifts the time by 4 from zero, as it's just a linear transformation?

PhysicsTruth
The origin must be ##(0,0)##, by definition. The Lorentz transformation won't work with ##t =4## at the origin.

You've drawn the scenario in the platform frame. What about the train frame?

The results are hard to read. They don't look right at all.
But, I've taken into account the change of frames, right? I've used the inverse of the Lorentz matrix in this case.

PhysicsTruth
Imagine there are two more clocks next to 1 and 2, but starting at 0, rather than 4. Run the experiment and think about these new clocks, for which you can use Lorentz.

The original clocks always read the same as the new clocks plus 4.
Ok, so I'm totally confused now. If I use the time transformation, I arrive at-

If I use the space transformation equation, I arrive at a different answer (which I posted just some time ago). Why is this happening? I don't seem to notice any mistake in any of the approaches, and I'm thoroughly confused.

PhysicsTruth
If you could just confirm...
View attachment 288549View attachment 288552
I've taken the coordinates for the front end coinciding with clock 1 to be-
a) (0,4) for both the frames
and the coordinates for the rear end coinciding with clock 2 to be-
a) (-l_o,t1') for the train's frame
b) (-L,4) for the platform frame

Using these coordinates and applying the Lorentz transformations, I arrive at this result, as shown in the pics. Is this correct?
Ok, I panicked here and made a calculation mistake, it does turn out to be ##\gamma vL/c^{2}## using the space transformation as well.