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vela
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Did you draw the spacetime diagram?
No. What are the endpoints of the world line of the signal?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 )
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. Did you bother to read what I wrote earlier?Lissajoux said:
vela said:In the Earth frame, the event of the signal being sent is at the coordinates (ct,x)=(2 lh, 0 lh).
..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.
Better. It's actually the mirror image of the actual diagram because x'=-1.5 lh, not +1.5 lh.Lissajoux said:
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.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
Lissajoux said:The signal reaches the ship when [itex]t=5~\textrm{h}[/itex] and the ship is [itex]x=3~\textrm{lh}[/itex] away, as observed in the Earth frame.
Using Lorentz transformations:
[tex]ct' = \gamma(ct - \beta x) = 1.25 [5~\textrm{lh} - 0.6\cdot 3~\textrm{lh}] = 4~\textrm{lh}[/tex]
[tex]x' = \gamma(x-\beta ct)=1.25[3~\textrm{lh}-1\cdot 5~\textrm{lh}]=-2.5~\textrm{lh}[/tex]
So to the observer on the ship, the signal arrives at 4h and at that point the Earth is 2.5lh away.