# Light Signal Travels Between Earth and a Spaceship

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Hi All,

An observer, F, stands on Earth. A spaceship, F', is also on Earth. Their clocks are set at 0. The spaceship then leaves earth at .5c. After 10 seconds, F sends a light signal to F'. As soon as F' receives the signal, F' sends a light signal back. When does F receive this signal from F'?

When F sends a signal to F', the distance light has to travel is different for F and F'. Could I calculate this distance using .5c x 10 seconds, if I want to say in F's time frame? Then I could divide it by c to get the time light needs to travel to F'.

If so, I can just double this time to get the time F gets the signal from F', assuming F' responds instantly? Does this mean I can calculate all time it takes for all these events just staying in F's frame?

## Answers and Replies

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Orodruin
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When F sends a signal to F', the distance light has to travel is different for F and F'. Could I calculate this distance using .5c x 10 seconds, if I want to say in F's time frame?
No. This would give you the distance that the light has to travel in F in order to reach the point where F' was when the light was emitted. By that time, F' will have moved further away.

Ohh, right F' does not stop moving. So we could solve for it algebraically by having c*t=.5c(t+10), where t is the time light will reach F'.

Nugatory
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An observer, F, stands on Earth. A spaceship, F', is also on Earth. Their clocks are set at 0. The spaceship then leaves earth at .5c. After 10 seconds, F sends a light signal to F'.
That speed is .5c relative to earth, right? That's the natural interpretation of your question, but in relativity problems it is important to be precise, always. A much more important "be precise" question is: is that ten seconds according to a clock that is at rest relative to the earth, or a clock that is at rest relative to the spaceship?

There are several tricks for making these problems easier to calculate with. First, choose to measure all your distances in light-seconds so that $c$ is equal to one instead of some annoying number like 186,000 miles/sec. Second, choose the speed of the spaceship - in light-seconds per second, of course - to be either $3/5=.6c$ or $4/5=.8c$. This makes the factor $\gamma=1/\sqrt{1-v^2/c^2}$ that appears in all the equations come out to a round number.
Does this mean I can calculate all time it takes for all these events just staying in F's frame?
Yes, but not the way you did it because the spaceship is moving while the light is in transit. In all problems of this sort, you should look for and then use whatever frame makes the problem easy, and in this case that will be the frame in which the earth is at rest - assuming that the ten seconds is measured on a clock at rest in that frame and the .5c is the speed of the spaceship in that frame.
[Edit: By "F's frame" do you mean "the frame of F" the way that "F's father" means "the father of of F"? Or are you talking about the frame F'?]

It would, however, be a good exercise to try calculating the time on the spaceship clock when it receives the light signal and when the return signal gets back to earth using a frame in which the ship is at rest.... just to see that you get consistent results. The out and back times will be different in this case.

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Orodruin
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Indeed.

And .5c is the speed in the time frame of the Earth.

If I keep all of my events, signal send and signal received in F's frame and F is inert, then I do not need to use Lorentz' transformation?

If I am thinking about this correctly, the first event is F sending a signal at 10 seconds (his timeframe). The second event is F receiving a signal. If the event was F' sent the signal, we would be shifted to his frame, correct?

If the spaceship is at rest, then the earth is moving a .5c?

Thank you again.

By "F's frame" do you mean "the frame of F" the way that "F's father" means "the father of of F"? Or are you talking about the frame F'?]
Apologies: I meant frame of F.

and it is 10 seconds from the perspective of a clock on Earth.

No. This would give you the distance that the light has to travel in F in order to reach the point where F' was when the light was emitted. By that time, F' will have moved further away.
Thank you. I think I can calculate this with c*t=.5c(t+10).

Does it matter for the final answer that I am calculating all of these distances and times in frame of F even when F' is the one that sends a signal back?

Nugatory
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If I keep all of my events, signal send and signal received in F's frame and F is inert, then I do not need to use Lorentz' transformation?
That's right. The Lorentz transformation is used to calculate the $x$ and $t$ coordinates of an event in some frame when you know them in some other frame. So if you of you do all your calculations using the same frame, you don't need them.

Be aware that the "F is inert" bit above is redundant. The phrase "F's frame" is just a convenient shorthand for "a frame in which F is at rest" so when you said "F's frame" you also said that F is not moving.
If I am thinking about this correctly, the first event is F sending a signal at 10 seconds (his timeframe). The second event is F receiving a signal. If the event was F' sent the signal, we would be shifted to his frame, correct?
There are three events in this problem:
1) Observer F emits a flash of light. Using the frame in which the earth is at rest, this event has coordinates $(x=0,t=10)$
2) The spaceship receives the flash of light and sends the return signal.
3) The return signal is received on earth.

If the spaceship is at rest, then the earth is moving a .5c?
Actually, moving at $-.5c$ because it's going in other direction.

Nugatory
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Does it matter for the final answer that I am calculating all of these distances and times in frame of F even when F' is the one that sends a signal back?
It does not.

It does not.
This question, in the end, did not require the use of gamma at all? As long as I stayed in frame of F, it was largely an algebraic problem?

Orodruin
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This question, in the end, did not require the use of gamma at all? As long as I stayed in frame of F, it was largely an algebraic problem?
Yes.

Yes.
Assuming F' sends the reply signal instantly, I do not have to account for if F' moved between receiving and sending?

I can just double the time that I calculated with c*t=.5c(t+10).

Orodruin
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Assuming F' sends the reply signal instantly, I do not have to account for if F' moved between receiving and sending?
That is what "instantly" means. No time to move.

I can just double the time that I calculated with c*t=.5c(t+10).
Yes.

Thanks to Orodruin and Nugatory!

Erland
An observer, F, stands on Earth. A spaceship, F', is also on Earth. Their clocks are set at 0. The spaceship then leaves earth at .5c. After 10 seconds, F sends a light signal to F'. As soon as F' receives the signal, F' sends a light signal back. When does F receive this signal from F'?
It wíll take the light signal from F 10 seconds in the Earth frame to reach the ship, which left Earth 10 seconds before the light signal was sent, but with only half the velocity of the light. So F' receives the signal at t = 20 s in the Earth frame, and the signal back to F will take the same time, 10 seconds, in the Earth frame. Thus, F receives the signal from F' when F:s clock reads 30 s.
When F sends a signal to F', the distance light has to travel is different for F and F'. Could I calculate this distance using .5c x 10 seconds, if I want to say in F's time frame? Then I could divide it by c to get the time light needs to travel to F'.
This distance, 5 light seconds, should be multiplied by $\beta =2/\sqrt3$, which gives ca 5.77 light seconds, in F's frame. This can be divided by c, so that the time the light signal from F to F' travels is ca 5.77 seconds. But the time in F's frame when the signal from F to F' is sent is $t'=10*2/\sqrt 3\approx 11.55$ s. So the time in F's frame when this signal from F arrives to F' is the sum of these two, or $3*10/\sqrt 3 \approx 17.32$ s.
If so, I can just double this time to get the time F gets the signal from F', assuming F' responds instantly? Does this mean I can calculate all time it takes for all these events just staying in F's frame?
No, the distance between F' and F increases by time, so the signal from F' to F takes longer time than the signal from F to F', measured in F's frame.
The time in F's frame when the signal from F' arrives to F is $30*2/\sqrt 3\approx 34.64$ s, so the time for this latter signal in F's frame is 17.32 s, which three times the time of the signal from F to F' (in F's frame).

Hi, when you write F's, is that referring to F' or F?

Erland