Relativity radio signal Question

[vela]

Did you draw the spacetime diagram?

 

[Lissajoux]

Yep

http://yfrog.com/5orelativ1j

(sorry it's not very good, was best I could do in paint )

 

[vela]

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.

 

[Lissajoux]

I've done so many different calculations it's a bit hard to keep track of the right ones (the few there have been )

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?

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

Now to the ST diagram:

umm, well wouldn't that give this though:

http://yfrog.com/5crelativ2j

??

 

[vela]

I've done so many different calculations it's a bit hard to keep track of the right ones (the few there have been )

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?

No. What are the endpoints of the world line of the signal?

 

[vela]

umm, well wouldn't that give this though:

http://yfrog.com/5crelativ2j

??

No. Did you bother to read what I wrote earlier?

 

[Lissajoux]

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

?

 

[Lissajoux]

In regards to the diagram:

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

2.

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?

 

[Lissajoux]

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

 

[vela]

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

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.

 

[Lissajoux]

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?

 

[Lissajoux]

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?

 

[vela]

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?

 

[Lissajoux]

.. the signal has to be received at x'=0 lh right?

does the x=2.5lh apply to the location of the Earth instead? or is that just wrong?

 

[Lissajoux]

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.