# Relativity question

1. Jul 1, 2010

### novop

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

2. Relevant equations

3. The attempt at a solution

First off this isn't really homework, just a problem I can't seem to wrap my head around. I worked through the questions in the frame of the Earth and got the right answers:

a) 12:50
b) 7.2 x 10^11 m
c) 1:30
d) doesn't matter

Solving (c) in the earth frame is easy. The rocket is moving away from the earth with v = 0.8c, emits a pulse at v=c at a distance 7.2 x 10^11 m away, so time is just d/v.

My question is in solving part (c) in the frame of the rocketship. In this frame, the earth is moving away from the rocketship with v=-0.8c, so now there is a relative speed (-0.2c) between the pulse emitted from the rocket (stationary in this frame) and the earth moving away with v=-0.8c.

Last edited: Jul 2, 2010
2. Jul 1, 2010

### kuruman

If the relative velocity of the rocket relative to the Earth is 0.8c, then the Earth's velocity relative to the rocket is -0.8c. Where does the 0.2c come from?

3. Jul 1, 2010

### novop

Right, I forgot the sign; changed it in the OP. The rocket emits a radar pulse with v=-c, hence the relative velocity between the pulse and the earth as measured by the rocket is -0.2c.

Last edited: Jul 1, 2010
4. Jul 1, 2010

### kuruman

How do you get plus (or minus) 0.2C?

5. Jul 1, 2010

### kuruman

According to the postulates of special relativity radar pulses (which are electromagnetic pulses) travel at the speed of light, c. You need to find the distance of the rocket from the Earth at the time the pulse is emitted, and then find the time it takes the signal to get to Earth. You can do this in either frame as long as you are consistent.

6. Jul 1, 2010

### vela

Staff Emeritus
Which two events are you referring to?

7. Jul 2, 2010

### novop

Yes, but in the rocket's frame, the earth is moving away from the pulse at speed v=-0.8c. And v=-c for the pulse, hence in this frame, the relative velocity between the pulse and the rocket is -0.2c.

The pulse emission on the tail of the rocket and the receiving of it at earth.

In the earth frame, the time is simply the distance (at the instant of the pulse) between the rocket and the earth divided by the speed of light.

But in the rocket's frame, it takes longer to reach the earth because it's moving away relative to the pulse (as measured by an observer in the rocket). Does this make sense?

Last edited: Jul 2, 2010
8. Jul 2, 2010

### vela

Staff Emeritus
Yeah, I know what you're trying to do, but, as I suspected and you have confirmed, you're looking at the wrong two events if you want to do it that way.

Think of it this way: The Earth and the signal are both racing away from the ship. The signal obviously moves faster than the Earth, but the Earth has a head start. The two will meet when the signal finally overcomes this head start, which it closes at a rate of 0.2c. You just have to calculate how big this head start is, which is the distance between the Earth and the ship at the time the signal is emitted. It has nothing to do with where the Earth is when the signal finally reaches it.

9. Jul 2, 2010

### novop

That's just what I did. If 30 minutes have passed in the rocket frame, the earth is at a distance:

d = t*v = (1800 s)(0.8)(3x10^8 m/s)

time = d/0.2c = (0.8)(1800 s)(3x10^8 m/s) / (0.2)(3x10^8 m/s) = 7200 s = 120 mins

Last edited: Jul 2, 2010
10. Jul 2, 2010

### vela

Staff Emeritus
It's because the two events, emission of the signal and reception of the signal, are separated in both space and time in both reference frames. What you should be able to show is

$$\Delta x^2 - c^2\Delta t^2 = \Delta x'^2 - c^2\Delta t'^2$$

Note you will recover the usual time dilation formula in the case where $\Delta x'=0$ and $\Delta x = v\Delta t$.

11. Jul 2, 2010

### novop

Okay... that makes sense.

Using the invariant above, with:

$$\Delta x = 4.32x10^{11} m$$
$$\Delta t = 7200 s$$
$$\Delta x' = 7.2x10^{11}m$$

I'm still not getting the right answer. What gives?

Last edited: Jul 2, 2010
12. Jul 2, 2010

### Staff: Mentor

This looks fine to me. According to the ship, it takes 7200 s for the signal to reach earth. During that time, how much time has passed on the earth clock? What time did the earth clock read when the signal was sent?

13. Jul 2, 2010

### novop

The earth clock read 12:50 when the signal was sent. In my previous post, I attempted to use the invariant to find the time passed in the earth frame during the "signal process", but the answer is much too large (it should be 40 minutes).

14. Jul 2, 2010

### Staff: Mentor

No, the clock on the space station read 12:50. According to the ship, the earth clock and the space station clock do not read the same. (Hint: The relativity of simultaneity.)

15. Jul 2, 2010

### vela

Staff Emeritus
You're using the wrong value for $\Delta x$. Remember, the two events are the emission of the signal and the receipt of the signal. In the rocketship's frame, these events are separated by two hours in time and by $\Delta x=c(7200~\textrm{s})=2.16\times 10^{12}~\textrm{m}$ in space. It's not $4.32\times 10^{11}~\textrm{m}$ because the Earth is moving away from the ship while the signal propagates.

In the rocketship's frame, you'll find the invariant is equal to 0, which makes sense because the two events are connected by a light ray. In the Earth frame, you therefore get

$$\Delta t' = \Delta x'/c = 7.20\times 10^{11}~\textrm{m}/c = 2400~\textrm{s} = 40~\textrm{min}$$

16. Jul 2, 2010

### Staff: Mentor

Not exactly. According to the ship, how far off are the two clock readings? You know it takes 120 minutes for the signal to travel according to the ship clock. How much time elapses on the earth clock during that time (according to the ship observers)? Add the time elapsed to the starting time (according to the ship) and you'll get the same answer that you calculated from the earth's frame.

17. Jul 2, 2010

### novop

Thanks that's perfect. One more thing though. How is this consistent with what Doc Al said about the relative times of the station and earth?

18. Jul 2, 2010

### vela

Staff Emeritus
The rocketship observers say 120 minutes elapsed on their clocks between the two events. How much time do they say elapsed on an Earth clock during this time? Note this is different than what an Earth observer would say elapsed on his clock on Earth between the two events.

19. Jul 2, 2010

### vela

Staff Emeritus
There are different ways to calculate these quantities, and as long as you keep everything straight, you'll find everything works out correctly. It's just really easy to get confused because you have four different pairs of times between two events: what A sees on A's clock, what A sees on B's clock, what B sees on B's clock, and what B sees on A's clock.

If A is Earth and B is the ship, you know that between emission and reception, A sees 40 minutes elapse on his clock and B sees 120 minutes elapse on his clock. What Doc Al wants you to find is how much time B says elapses on A's clock and what time A's clock read (according to B) when the signal was sent.

Last edited: Jul 2, 2010
20. Jul 2, 2010

### novop

All is clear, thanks much for your help.