Special Relativity, Train Question.

[mmmboh]

[PLAIN]http://img573.imageshack.us/img573/9443/1132y.jpg

So I did what I thought was right, but I'm a little confused on whose frame they are talking about when they say she arrived at the back of the train at the same time the back of the train leaves the tunnel..I think it makes a difference..
but anyway

a) I said the length of the train in the ground frame is L'=L/γ= (1-(0.6c/c)²)^1/2*L=4L/5.
Now the time in the ground's frame for the back of the train to pass the tunnel is t=(L+L')/v=(9L/5)/(3c/5)=3L/c, and since the person reaches the back of the train at the same time the train leaves the tunnel, it takes t=3L/c.

b) Now the person travels a distance L in this time, so v_{w.r.t ground}=L/(3L/c)=c/3.

c)t_train=t_ground/γ=(4/5)*(3L/c)=12L/5c...either that or t=3L/c...I am a little confused as to in whose frame she arrives at the back of the train at the same time as it leaves the tunnel.

Those are my answers, but I am not sure if I used the frames correctly.

Thanks for any help :).

 

[mmmboh]

Can anyone help?

 

[sylas]

So I did what I thought was right, but I'm a little confused on whose frame they are talking about when they say she arrived at the back of the train at the same time the back of the train leaves the tunnel..I think it makes a difference..
but anyway

I don't think it makes a difference. The event of "train leaving the tunnel" is when the back of the train is at the end of the tunnel, and that is when she arrives at the back of the train. So she is at the same location as the end of the tunnel and the back of the train. This holds for all frames.

a) I said the length of the train in the ground frame is L'=L/γ= (1-(0.6c/c)²)^1/2*L=4L/5.

Correct.

Now the time in the ground's frame for the back of the train to pass the tunnel is t=(L+L')/v=(9L/5)/(3c/5)=3L/c, and since the person reaches the back of the train at the same time the train leaves the tunnel, it takes t=3L/c.

Yes, that is the time taken for the walk in the ground frame.

b) Now the person travels a distance L in this time, so v_{w.r.t ground}=L/(3L/c)=c/3.

Yes, that is the velocity of the walker in the ground frame.

c)t_train=t_ground/γ=(4/5)*(3L/c)=12L/5c...either that or t=3L/c...I am a little confused as to in whose frame she arrives at the back of the train at the same time as it leaves the tunnel.

Here's the thing. The walker's watch has time dilation given by the walker's velocity; not the train's velocity.

Arriving at the back of the train is an event. You can consider that event in any frame you like. All the frames will give the same result for calculating the time passing on the person's watch; but they will do so with different numbers for her velocity and how for she has moved and how long it took in the frame's time.

You have been using the ground frame. Stick with it; that will give the answer.

For extra credit you could try calculating it in the frame of the train. Should get the same answer for time shown passing on her watch.

Cheers -- sylas

 

[mmmboh]

Ok, well I would say t_person=t_ground/γ = (1-(c/3)²/c²)^1/2*(3L/c)=(2*2^1/2)L/c.
Is that right?

 

[sylas]

Ok, well I would say t_person=t_ground/γ = (1-(c/3)²/c²)^1/2*(3L/c)=(2*2^1/2)L/c.
Is that right?

Yes, I think so. It is what I got, anyhow... :-)

Cheers -- sylas

 

[mmmboh]

thanks!