How Do Relativity and Speed Affect Radio Signals and Shuttles?

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The discussion revolves around a physics homework problem involving a spaceship traveling at 0.60c and the effects of relativity on radio signals and another shuttle. In the Earth frame, the radio signal travels at the speed of light (c), while the shuttle moves at 0.80c, with calculations indicating the signal reaches the spaceship in 3 hours and the shuttle in 6 hours. Transitioning to the spaceship's frame complicates matters, requiring the use of Lorentz transformations and the Lorentz factor to account for time dilation and length contraction. Participants express confusion over the calculations and the correct application of relativistic principles, particularly regarding the timing of events and the perceived speeds of the signal and shuttle from the spaceship's perspective. The conversation highlights the complexities of understanding relativity in different reference frames and the need for careful analysis of time and distance in such scenarios.
  • #31
I don't think my pointing out the error is a good idea because I'm not getting a sense that you understand the calculations. You just seem to be following a recipe; you're not demonstrating an understanding of why you're doing certain calculations or what the results mean. The reason I'm saying this is because this error, like some of your previous errors, is a very elementary one. It's not a subtle mistake; it's a blatant one, which you should easily be able to find on your own.
 
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  • #32
(Ignore my previous post if you read it, this is the better replacement)

Right I've gone back through my calculations to try and make sense of it all and spot where I've gone wrong.

So in the Earth reference frame, the signal reaches the ship at t=5.0~\textrm{h} and at that time the Earth and the ship are a distance of x=3.0~\textrm{lh} apart.

Then using Lorentz transforms:

ct' = \gamma(ct - \beta x) = 1.25 [5~\textrm{lh} - 0.6\cdot 3~\textrm{lh}] = 4~\textrm{lh}

x' = \gamma(x-\beta ct)=1.25[3~\textrm{lh}-1\cdot 5~\textrm{lh}]=-2.5~\textrm{lh}

So in the ship's reference frame, the signal reaches the ship at t=4.0~\textrm{h} and at that time the Earth and the ship are a distance of x=2.5~\textrm{lh} apart.

This seems to make sense since to someone on the ship in the ship's reference frame, the clock on Earth appears to be running slow, and there is also length contraction making the preceived distance to Earth shorter.

Ok, right so hopefully that's all good.

Then if I know, in the ship's reference frame, what time the signal left the Earth and what distance away the Earth was at that time, I can find the distance traveled by the signal and the time taken. Hence I can find the speed of the signal.

In the Earth frame, the event of the signal being sent is at the coordinates (ct,x)=(2 lh, 0 lh). Using the Lorentz transformations:

ct' = \gamma(ct - \beta x) = 1.25 [2~\textrm{h} - 0.6\cdot 0~\textrm{h}] = 2.5~\textrm{h}

x' = \gamma(x - \beta ct) = 1.25 [0~\textrm{lh} - 0.6\cdot 2~\textrm{lh})] = -1.5~\textrm{lh}

So from the ship's reference frame, the signal left Earth when the ship's clock read 2:30 and the Earth was 1.5~\textrm{lh} away.

We've already established that everything above is correct, at least I believe so, through previous posts.

Then I somehow seem to be going wrong from this step onwards.

I've then calculated that:

\Delta t=4.0-2.5=1.5~\textrm{h}

\Delta d=2.5-1.5=1.0~\textrm{lh}

I can do this right.

So then simply did:

v=\frac{1.0~\textrm{lh}}{1.5~\textrm{h}}=\frac{2}{3}c

That's just a calculation, don't see how that could be wrong.

.. and then I've run out of ideas.I've checked through everything. I can't spot where I've gone wrong, it's obviously more of a subtle mistake to me. Perhaps you could hint at which part in that method I've detailed I've gone wrong please? Then I should hopefully be able to see it :smile:
 
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  • #33
I've just looked through my notes again and thought I could use this to find the speed of the signal:

u_{x}'=\frac{u_{x}-v}{1-\frac{u_{x}v}{c^{2}}}=\frac{c-0.6c}{1-\frac{0.6c^{2}}{c^{2}}}=\frac{(1-0.6)c}{1-0.6}=\frac{c}{1}=c

where u_{x}' is the speed of the signal in the ship's reference frame and the speed of the signal in the Earth's reference frame is u_{x}

yes?
 
  • #34
Lissajoux said:
(Ignore my previous post if you read it, this is the better replacement)

Right I've gone back through my calculations to try and make sense of it all and spot where I've gone wrong.

So in the Earth reference frame, the signal reaches the ship at t=5.0~\textrm{h} and at that time the Earth and the ship are a distance of x=3.0~\textrm{lh} apart.

Then using Lorentz transforms:

ct' = \gamma(ct - \beta x) = 1.25 [5~\textrm{lh} - 0.6\cdot 3~\textrm{lh}] = 4~\textrm{lh}

x' = \gamma(x-\beta ct)=1.25[3~\textrm{lh}-1\cdot 5~\textrm{lh}]=-2.5~\textrm{lh}

So in the ship's reference frame, the signal reaches the ship at t=4.0~\textrm{h} and at that time the Earth and the ship are a distance of x=2.5~\textrm{lh} apart.

This seems to make sense since to someone on the ship in the ship's reference frame, the clock on Earth appears to be running slow, and there is also length contraction making the preceived distance to Earth shorter.
This is a perfect example of what I'm talking about. First, the numbers are wrong. This part should be just plug-and-chug. Now it could be a careless mistake, in which case, you should have easily found it, or it could be you're just plugging numbers in somewhat randomly because you don't understand what the variables stand for. If this is the case, you need to learn what the variables mean.

Second, the interpretation is wrong. In part A, you found that the ship and signal meet at x=3 lh. Note that you were solving for where in space this event occurred; you weren't solving for how far away from Earth it occurred. The distance between the Earth and the ship happens to also be 3 lh because the Earth is at the origin in the Earth frame. In the ship's frame, however, the Earth isn't at the origin, so x' isn't the distance between the ship and Earth when it receives the signal.

The Lorentz transformations convert the coordinates of an event in one frame to the coordinates of the same event in a different frame. In this particular case, you found that the ship and signal meet at the spacetime coordinates (ct, x) = (5 lh, 3 lh) as observed in the Earth's frame. The transformed coordinates (ct', x') tell you where in spacetime the ship and signal meet as observed in the ship's frame. To repeat, x' is not the distance between the ship and Earth.
Then if I know, in the ship's reference frame, what time the signal left the Earth and what distance away the Earth was at that time, I can find the distance traveled by the signal and the time taken. Hence I can find the speed of the signal.

In the Earth frame, the event of the signal being sent is at the coordinates (ct,x)=(2 lh, 0 lh). Using the Lorentz transformations:

ct' = \gamma(ct - \beta x) = 1.25 [2~\textrm{h} - 0.6\cdot 0~\textrm{h}] = 2.5~\textrm{h}

x' = \gamma(x - \beta ct) = 1.25 [0~\textrm{lh} - 0.6\cdot 2~\textrm{lh})] = -1.5~\textrm{lh}

So from the ship's reference frame, the signal left Earth when the ship's clock read 2:30 and the Earth was 1.5~\textrm{lh} away.

We've already established that everything above is correct, at least I believe so, through previous posts.

Then I somehow seem to be going wrong from this step onwards.

I've then calculated that:

\Delta t=4.0-2.5=1.5~\textrm{h}

\Delta d=2.5-1.5=1.0~\textrm{lh}

I can do this right.

So then simply did:

v=\frac{1.0~\textrm{lh}}{1.5~\textrm{h}}=\frac{2}{3}c

That's just a calculation, don't see how that could be wrong.

.. and then I've run out of ideas.


I've checked through everything. I can't spot where I've gone wrong, it's obviously more of a subtle mistake to me. Perhaps you could hint at which part in that method I've detailed I've gone wrong please? Then I should hopefully be able to see it :smile:
As I suggested a few posts ago, draw spacetime diagrams for this problem. I think it'll help you clarify what's going on in the problem.
 
  • #35
Back in post 18 I believe we established that according to the ship, the signal left Earth when the ships clock read 2.5h and the ship at that time was 1.5lh away from the earth. Then in post 19, that according to the ship the signal was received when the ships clock read 4.0h and at that point the ship was 2.5lh from earth. So that'a a time difference of 1.5h, and a distance difference of 1.0lh. So why is any of that now incorrect? (or have I just plugged the wrong numbers into those lorentz transforms? ;))
 
  • #36
Did you draw the spacetime diagram?
 
  • #37
  • #38
Bad news. That's completely wrong. For one thing, you didn't bother to fix your earlier, erroneous calculations. Nevertheless, answer these questions from your diagram. How much time elapsed between the emission and reception of the signal? What distance did the light travel between emission and reception? You'll find you get different answers than from your previous calculations. They'll still be wrong, but at least, you should see the mistake you were making with your earlier method.

Now back to your diagram. In the ship's reference frame, the ship will be at rest, so its world line, x'=0, should extend vertically from the origin. Back in post 18, I calculated the coordinates of the signal's emission event in this frame for you. Locate that point on the diagram. The world line of the Earth is the line passing through that point and the origin.
 
  • #39
I've done so many different calculations it's a bit hard to keep track of the right ones (the few there have been :wink:)

At least I should be able to answer the questions with the help of the diagram.

Time between emission and reception of signal: 4.0-2.5=1.5h

Distance traveled by signal between emission and reception: 2.5-1.5=1lh

OK? :smile:

(.. but I did get these results earlier?)

Now to the ST diagram:

umm, well wouldn't that give this though:

http://yfrog.com/5crelativ2j

??
 
  • #40
Lissajoux said:
I've done so many different calculations it's a bit hard to keep track of the right ones (the few there have been :wink:)

At least I should be able to answer the questions with the help of the diagram.

Time between emission and reception of signal: 4.0-2.5=1.5h

Distance traveled by signal between emission and reception: 2.5-1.5=1lh

OK? :smile:
No. What are the endpoints of the world line of the signal?
 
  • #41
Lissajoux said:
umm, well wouldn't that give this though:

http://yfrog.com/5crelativ2j

??
No. Did you bother to read what I wrote earlier?
 
  • #42
So that second diagram was correct? And that is for the spaceship's reference frame isn't it? I should use this latest diagram now then and ignore the first I take it.

Those answers were based on my first diagram, which was wrong. But for the signal:

Start: t=2.5h, x=0lh

End: t=4.0h, x=2.5lh

Oh, so the time period = 1.5h and the distance = 2.5lh

?
 
  • #43
In regards to the diagram:

1. The world line for the ship extends vertically upwards from the origin. That's shown on there.

2.

vela said:
In the Earth frame, the event of the signal being sent is at the coordinates (ct,x)=(2 lh, 0 lh).

I took those values. But that's clearly wrong, that's in the Earth reference frame and we're in the spaceship's reference frame for this diagram.

..from the ship observer's point of view, the signal left Earth when the ship's clock read 2:30 and the Earth was 1.5 lh away.

So should be using t=2.5h and x=1.5lh?
 
  • #44
http://yfrog.com/10relativ2j

how's that?

Then should there be a line from the Earth line to the shuttle line, from t=2.5h to the point that the signal is received. I've noted that this would mean that whilst the Earth and the ship continue to move further apart, the signal would move towards the ship.

PS. I'll give you a moment to look through my last few little replies :smile:
 
  • #45
Lissajoux said:
http://yfrog.com/10relativ2j

how's that?
Better. It's actually the mirror image of the actual diagram because x'=-1.5 lh, not +1.5 lh.
Then should there be a line from the Earth line to the shuttle line, from t=2.5h to the point that the signal is received. I've noted that this would mean that whilst the Earth and the ship continue to move further apart, the signal would move towards the ship.

PS. I'll give you a moment to look through my last few little replies :smile:
Right. Can you see now that x' for the event where the ship receives the signal has to be 0? Take a look at your calculation of x' for this event in post 32, where you got x'=-2.5 lh. Figure out what went wrong there.
 
  • #46
Yes I think I do see that now. So the ship receives the signal at different times, but at the same location on the diagram as those on the ship don't think there moving. If that makes sense, maybe I'm not wording it in the best way. I get it though.

Here's a new spacetime diagram:

http://yfrog.com/63relativ2j

Need to put the signal line on though, once calculate when it is received.

So I'm looking at the Lorentz transformations, right?

Here they are from post #32:

ct' = \gamma(ct - \beta x) = 1.25 [5~\textrm{lh} - 0.6\cdot 3~\textrm{lh}] = 4~\textrm{lh}

x' = \gamma(x-\beta ct)=1.25[3~\textrm{lh}-1\cdot 5~\textrm{lh}]=-2.5~\textrm{lh}

In the Earth frame, the event of the signal being sent was at ct=2h, x=0lh. So the equations should actually read:

ct' = \gamma(ct - \beta x) = 1.25 [2.0~\textrm{lh} - 0.6\cdot 0~\textrm{lh}] = 2.5~\textrm{h}

x' = \gamma(x-\beta ct)=1.25[0~\textrm{lh}-0.6\cdot 2.0~\textrm{lh}]=-1.5~\textrm{lh}

.. that's correct now?
 
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  • #47
Lissajoux said:
The signal reaches the ship when t=5~\textrm{h} and the ship is x=3~\textrm{lh} away, as observed in the Earth frame.

Using Lorentz transformations:

ct' = \gamma(ct - \beta x) = 1.25 [5~\textrm{lh} - 0.6\cdot 3~\textrm{lh}] = 4~\textrm{lh}

x' = \gamma(x-\beta ct)=1.25[3~\textrm{lh}-1\cdot 5~\textrm{lh}]=-2.5~\textrm{lh}

So to the observer on the ship, the signal arrives at 4h and at that point the Earth is 2.5lh away.

I think this still applies?

So the ST diagram would be:

http://yfrog.com/4jrelativ2j

So the speed of signal:

v=\frac{\Delta d}{\Delta t}=\frac{1.5lh}{1.5h}=1l=c

(not sure about the correct notation for the last bit of that equation)

.. any of this any good?
 
  • #48
Don't you see a contradiction between your spacetime diagram and your calculation that says x'=-2.5 lh is where the signal is received?
 
  • #49
.. the signal has to be received at x'=0 lh right? :smile:

does the x=2.5lh apply to the location of the Earth instead? or is that just wrong?
 
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  • #50
I'm just looking at the spacetime diagram now.

So the signal was sent at t=2.5h, where x=1.5lh. And the signal was received at t=4h, where x=0lh. Is that interpretation correct?

And was the other parts of the last few posts correct i.e. the diagram then I assume so and also the speed calc.
 

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