1. Sep 3, 2008

### jianxu

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

Two Rockets A and B depart from Earth at constant speeds of 0.6c in opposite directions, having synchronized clocks with each other and with Earth at departure. After one year as measured in Earth's reference frame, rocket B emits a light signal(call this event E1). At what times, measured in the reference frames of the earth and of rockets A and B, does rocket A receive the signal(event E2)

2. Relevant equations
Lorentz Transformation
kinematics
Time Dilation

signal travels at speed of light

3. The attempt at a solution
So first I just did a simple conversion to get seconds, tearth = 3.15X107s.

Next, I solved for each of the rocket's respective distances at that point with simple kinematics and obtained (-)5.67X1015m.

After this, I drew myself a coordinate system saying the earth is the origin, rocket B(transmitting signal) in the positive direction, rocket A in the negative direction.

Then I argued that at a time t, the signal and rocket A will be at the same position for the signal to be received and therefore:

position of rocket A = -5.67X1015m- 0.6ct
position of signal = 5.67X1015m - ct

position of rocket A = position of signal

I then solved for t and got 9.45x107s after the signal is sent. I then added this to the original 1 years time and got 1.26x108s.

That was for the earth's frame.

For the frame of rocket B and A,
I first took the time dilations for them which should be the same since their speeds are the same just in different directions.

for rocket A/B time = tearth*$$\gamma$$
where $$\gamma$$= 1/($$\sqrt{1-(v^2/c^2)}$$

I got 3.94x10^7s for their dilated year

and for their velocities, I got (-)1.875x108m/s after applying velocity addition in relativity

with those information, I solved for the distance as observed by rocket a/b and obtained:
1.48x1016m from each other.

Now for observer b, I used the following kinematics:
position of signal = -ct
position of rocket A = -1.48x1016m - 1.875x108*t

position of signal = position of rocket A

the answer I got when I solved for t was 1.32x108s from the time the signal was sent.

I added that t to the dilated years time and got 1.71x108s total for observer rocket B

Lastly, for rocket A I used the same information. The only difference is thekinematics equation:

speed of signal = 1.48x1016- ct
speed of rocket A= 0

solving for t I got 4.93x107seconds and adding to the dilated year I got

8.87x107s total.

Please check if my thought process/math is right? Thanks!!!!! :)

2. Sep 4, 2008

### Goddar

Your answer about the time at which the signal reaches rocket A in earth's reference frame looks right to me; after that, i think it gets more complicated than it should:
The factor gamma reduces to 5/4, which is then in a sense the ratio of your values (time and space) between the 2 reference frames; since time is "dilatated" in the rocket, it simply tells you that 4/5 of your 1.26x10^8 seconds have passed in the rockets, which reduces to 1.00x10^8 seconds..

3. Sep 4, 2008

### jianxu

Just applying time dilation seems strange to me? It seems unintuitive that rocket B and rocket A will see the same time? Wouldn't the gamma be altered since one of the rocket is going in the positive direction and rocket A is going in the negative direction?

4. Sep 4, 2008

### Goddar

Well, rocket A and B surely won't see the same thing: the signal only reaches rocket A.. But their clocks run at the same pace relative to earth, since their speed is the same; now if you know at what time on Earth the signal catches rocket A (which is not the time at which you SEE it catch it from there, of course), you really only need time dilatation to convert into the rockets' reference.. Am I wrong?

5. Sep 4, 2008

### jianxu

I think I understand. Part of the inner me wants to reject this because it seems too weird :P So if I apply time dilation, both rockets will measure the 1.00x10^8 when rocket A receives the signal since they both move at the same speed relative to the earth?

6. Sep 4, 2008

### Goddar

This is the time that will definitely be seen in rocket A, which is what you are to determine here; for rocket B, you can say that it is the time at which the signal would be seen if it were sent from rocket A in the same conditions.. The thing is: we can't help trying to think in terms of an absolute frame of reference, which doesn't make sense in special relativity!
When the signal reaches A, B is not measuring anything because it will take a few light-years to get any "news" from A..

7. Sep 4, 2008

### jianxu

is that why in my first solution the time is so much larger for observer B? the fact that it has to take so much more time to receive the news from A?

8. Sep 5, 2008

### Goddar

Yes: if you need to know at what time the signal is SEEN reaching A from Earth and B's reference, you just need to repeat the kinematic calculation you did (and then convert for B the same way as above); the point is: with this kind of problem, from the "fixed" frame of reference (Earth), you can do all the basic calculations you need, and then you really only need time-dilatation to convert; with this, you should be able to give all the answers you need with minimal maths...

9. Sep 5, 2008

### jianxu

Ah it's making a bit more sense now :P thanks