Homework Help: Relative motion of train and platform observer

1. Oct 15, 2012

CAF123

1. The problem statement, all variables and given/known data
A train is passing a platform at 0.75c, and an observer, stationary and at the middle of the platform, sees two bolts of lightning simultaneously hit the front and back of the train when the train's middle was adjacent to the middle of the platform. A passenger on the train (and sitting in the middle) sees the two bolts of lightning at different times.

Given that the two bolts of lightning were 50m apart(as viewed by observer on platform), calculate the time difference between the two bolts of lightning as viewed by the passenger on the train.

3. The attempt at a solution

Before I delve into the computation, I just want to check something. From the platform observer's frame of reference, he sees the two bolts at the same time. If we define a coordinate system with x axis parallel to motion of train, then the two bolts strike the train at different spatial coordinates, but at the same time, relative to platform observer. Does this imply that the 50m is a proper length as viewed by the observer? Since the bolts are at either end of the train then does imply that the actual length of the train is 50m?
If so, I don't see how this makes sense given that we know length contraction of the train occurs relative to the platform observer. On the other hand, I believe the def of proper length conforms with what I said above.

2. Oct 15, 2012

tiny-tim

Hi CAF123!
No, the question says that the lightning bolts are 50m apart.

3. Oct 15, 2012

CAF123

Does that mean we can then work out the proper length of the train using $l = \frac{l_{o}}{\gamma}$, where l is 50m. Would the 50m be a proper length since the time coordinate is the same (I.e both bolts strike simultaneously)?

4. Oct 15, 2012

tiny-tim

I don't understand what you're asking, nor why.

Forget about proper length.

The question gives you the distance and time (= 0) between two events in one frame, and asks you for the time in another frame.

5. Oct 15, 2012

CAF123

I am asking about this because I want to make sure I understand what is meant by a proper length. I know the question may not require this, but I feel I should understand what is going on conceptually with the correct terminology. My understanding is that proper length is where the time coordinate is unchanged between two events as observed in some frame of reference. So would the 50m be a proper length since we have the two events (bolt 1 striking rear of train and bolt 2 striking front of train), each of which have the same t coordinate.

Many thanks

6. Oct 15, 2012

tiny-tim

Not really.

"Proper length" really only applies to a rigid object in the frame in which it is stationary (its "rest-frame").

The train only has a proper length as measured in its own rest-frame
It's the proper length between the two points on the platform next to where the bolts land (because it's measured at the same time in the platform's rest-frame).

7. Oct 15, 2012

CAF123

Ok, thanks for clarifying that. In terms of doing the question, I used the result of the 'moving clocks synchronisation' to get two values for t. For the front bolt, I computed it takes a time $$\frac{l_o}{2\gamma(c+v)}$$ and $$\frac{l_o}{2\gamma(c-v)}$$ for the rear bolt where $l_o = 50m?$ Will the difference of these times give me my answer?

I tried initially using the relativistic lorentz transformations, then realised they are only valid if the initial condition is that the frames are in the standard configuration.

8. Oct 16, 2012

CAF123

I think the above calculation is what the platform observer would think that the train passenger would observe. Since the platform observer sees length contraction in his frame, then in the train's frame, time must be dilated. So I believe the required time for the question is the difference of the two times I calculated in the previous post divided by gamma. Correct?

9. Oct 16, 2012

tiny-tim

i'm not following this logic

for the platform observer, the two events are at the same time, and at distance 50m

if there was a pole fixed to the platform where the lightning hits, then 50m is the distance between the poles at all times

for the train observer, the poles are moving (together), and the distance between them as measured at the same time by the train observer is contracted

but the distance between the events is not measured at the same time (asdnd anyway the question doesn't ask for it)

so far as time dilation is concerned, that only applies to one stationary clock (and you can't find a frame in which a stationary clock could join those two events … it would have to go faster than light, or infinitely fast in the platform frame)

don't take short-cuts! …

use the Lorentz transformation

10. Oct 16, 2012

CAF123

Thanks. Using the lorentz transformation I.e $$Δ t' = \gamma(Δt - \frac{Δ x\,v}{c^2}),$$ with $Δt = 0, Δx = 50m, v = 0.75c,$.

In my notes, this is derived assuming a standard configuration set up. Why is this applicable here? Also, I note that the above relation gives a negative value for $Δt'$. What is the physical significance?

11. Oct 16, 2012

tiny-tim

i don't understand your question

any two inertial frames will be in standard configuration provided they have a common origin, and provided they give the same name to their relative line of motion
if there are lots of bolts of lightning all hitting the ground at the same time as measured by someone stationary on the ground,

so that the line joining bottoms of all the lightning bolts is horizontal, and moving downward,

then the train passenger sees that line as sloping towards the front of the train, so that the further forward the lightning is, the earlier it reaches the ground

12. Oct 29, 2012

CAF123

There has been a fundamental disagreement between me (and some of my friends) and my professor over one of the issues rasied in this question.
According to the platform observer, he would 'see' the bolts 50m apart. SInce the bolts are applied at either end of the train, the effective contracted length of the train is also 50m, relative to the platform.

But according to many of the TA's who mark our work, we have to do a length contraction computation to work out the contracted length of the train (i.e l = 50/γ). I don't see why this makes sense. The train is moving at a relativistic speed relative to the platform, and so I believe the 50m would already be the length contraction.

The method I use yields an answer of 1.89E-7s for the answer to the question (which goes through a derivation based on leading clocks lag) and agrees with the method offered by the lorentz transformations.

I am going to talk to my professor on Wednesday - i just want to know what people here think. Many thanks.

13. Oct 29, 2012

tiny-tim

correct
that can't be right, the question is completely clear …

the 50m is the platform observer's distance between the bolts at the same time

it is therefore also the platform observer's distance between the front and back of the train at the same time

in other words, as you say, 50m is the contracted length of the train (and the rest length is longer)

enlarging slightly on what i said before, in the train's frame, the bottom of the front bolt is lower than the bottom of the back bolt

in the train's frame, the two bolts are less than 50m apart (length contraction! ), and the train is 50m long …

so how can the train "fit" between two bolts that are closer than the length of the train?

because the back of the train has had a little extra time to squeeze in between them!

14. Oct 29, 2012

CAF123

Thanks. So just to clarify you agree that the TA's have made a mistake somewhere? I didn't think I was wrong because the answer I had agrees with the lorentz transformation formula and an extra factor of gamma (which in the TA's soln gives a gamma squared somewhere) does not make much physical sense to me (length contraction twice???)

15. Oct 29, 2012

tiny-tim

yes: if the original question is correctly stated …
… then the 50m is the contracted length of the train, and the rest-length of the train is longer