Special relativity (overtaking)

1. The problem statement, all variables and given/known data

This is an example problem although I don't really understand it.

Train A is moving at 4c/5 and train B moves at 3c/5 with respect to observer C.

Let E1 be the event "front of A passing back of B" and E2 be the event "back of A passing the front of B".

An traveller D (who is in train B) walks from the back of B to the front of B and coincides with both E1 and E2. What is the speed he walks at and how long does D measure the overtaking process to be?

2. ???

I just need a little hint here. This is actually an example problem from my book, so I do have the working and answer provided, but I don't actually understand what the significance of the observer D coinciding with both events is.

The book states immediately that as a result of the condition here, and looking at the point of view of D, A moves at speed v wrt D and B moves at speed -v wrt D. Then it uses the velocity addition again to get [tex]\frac{2v}{1+v^2/c^2} = \frac{5c}{13}[/tex]. The book concludes that v = c/5, upon solving the equation given earlier.

Once this is done, the rest is easy: using the velocity of either A or B relative to D, find the contracted length of each train and divide by c/5 to get the required answer, which is [tex]2\sqrt{6}L/c[/tex].

(From an earlier and easier half of the question, velocity addition finds that A moves at 5c/13 with respect to B.)
 
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I run into another problem once I go further along this question.

If I solve the equation given here, I actually get v = 2c/5 rather than v = c/5 quoted in the book.
 

Doc Al

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I just need a little hint here. This is actually an example problem from my book, so I do have the working and answer provided, but I don't actually understand what the significance of the observer D coinciding with both events is.
Look at things from D's perspective. D observes the length of train A pass him (front to back) in exactly the same amount of time that the length of train B passes him (back to front). That fact allows you to immediately conclude that the speed of A wrt D is the same as the speed of B wrt D (except for the sign, of course). (Assuming the trains have the same rest length L, of course.)

The book states immediately that as a result of the condition here, and looking at the point of view of D, A moves at speed v wrt D and B moves at speed -v wrt D. Then it uses the velocity addition again to get [tex]\frac{2v}{1+v^2/c^2} = \frac{5c}{13}[/tex]. The book concludes that v = c/5, upon solving the equation given earlier.
OK. Given my comment above, does this now make some sense?

If I solve the equation given here, I actually get v = 2c/5 rather than v = c/5 quoted in the book.
Double check your solution. (Using that same equation, I got the book's answer.)
 
Yes, I understand now, thanks Al!
 

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