Two trains leave a station, and an observer misses them both what does he see

1. Oct 6, 2004

stunner5000pt

Using Speical relativity, of course

If two trains leave a station on the same track. An observer missed both these trains and is standing close to the track sees the westbound train recede at 0.6c and sees the eastbound train recede at 0.8c. There is a ticket collector on the westbound train going from the back of the train to the front at 0.4c, with respect to a passenger on the westbound train.

What would the speed of the eastbound train wit hrespect to the westbound train (call it Ur) according to:

a) Observer on the station: it's just as if the trains were approaching each other... right?

in taht case using Ux' = UW - UE / (1 - UW UE / C^2) yields 0.38c

b)Passenger seated on the westbound train?

The same framework as the previous question (which leads me to doubt part A) and i get 0.38c

c) Ticket collector on the westbound train?
The ticket collector would see hte same as the passenger, no?

Similarly what is the speed of the ticket collector :

d) According to the observer on the station?
I would think calculating the speed of the ticket collector with respect tothe train first, and then the observer to the train and then adding the velocities up

e) According to the passenger on the east train?
First find the relative velocities of the two trains and then add the velocity of the ticket collector with respect to the west train as done in the previous one

F) Relative to a passenger seated on the east train according tothe observer on the station?

STumped even more...

2. Oct 14, 2004

Staff: Mentor

Does that even make sense? After all, the speed of the eastbound train with respect to the platform is 0.8c--how can its speed with respect to the westbound train be less? :yuck:

Learn the relativistic velocity addition formula:
$$V_{c/a} = \frac{V_{b/a} + V_{c/b}}{1 + V_{b/a}V_{c/b}/c^2}$$
(Hint: signs--which signify direction--matter!)