Relativity question

Homework Statement 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:

kuruman
Homework Helper
Gold Member
2021 Award
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?

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:
kuruman
Homework Helper
Gold Member
2021 Award
Right, I forgot the sign; changed it in the OP.
How do you get plus (or minus) 0.2C?

kuruman
Homework Helper
Gold Member
2021 Award
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.
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.

vela
Staff Emeritus
Homework Helper
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. In this frame, the distance between the two events is unknown (?), and the time is given by $$5x/c$$ where $$x$$ is the distance between the two events. Is this right so far? What am I overlooking?
Which two events are you referring to?

According to the postulates of special relativity radar pulses (which are electromagnetic pulses) travel at the speed of light, c.

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.

vela said:
Which two events are you referring to?
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:
vela
Staff Emeritus
Homework Helper
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?
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.

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:
vela
Staff Emeritus
Homework Helper
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 $\Delta x = v\Delta t$.

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 $\Delta x = v\Delta t$.

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:
Doc Al
Mentor
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
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?

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).

Doc Al
Mentor
The earth clock read 12:50 when the signal was sent.
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.)

vela
Staff Emeritus
Homework Helper
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?
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

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

Doc Al
Mentor
Right! So then if

$$\Delta t' = \gamma\Delta t - \gamma\frac{v}{c^2}\Delta x$$
$$\Delta t' = 82 mins$$

So the earth clock reads 1:22 when the signal is sent. In the earths frame, it takes (from the calculation above) 124 minutes for the signal to reach earth...
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.

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?

vela
Staff Emeritus
Homework Helper
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.

vela
Staff Emeritus