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bigevil
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Homework Statement
Two planets A and B are at rest with respect to each other and are L apart in this frame. They have synchronised clocks. A spaceship flies at speed v with respect to planet A and synchronises its clock with A-B.
We know that when the spaceship reaches B, B's clock reads L/v and the ship's clock reads L/γv. How would someone account for the fact that B's clock reads L/v, which is more than its own L/γv, considering that the spaceship sees B's clock as running slow.
Homework Equations
I'm not sure if I got the correct solution, please help me to check if the reasoning is correct.
Let the frame A-B measure (x,y,z,t).
Let the ship measure coordinates (x',y',z',t').
Using the Lorentz transformation for time,
[tex]t' = \gamma (t - vL/c^2)[/tex]
When A-B resets its clock at zero (t=0), t' registers [tex]- vL\gamma / c^2[/tex]. Relative to the time of A-B, the ship's clock runs at vLγ/(c*c) slower.
When the ship passes B, time at AB is t = L/v. Then,
[tex]\gamma (\frac{L}{v} - \frac{Lv}{c^2}) = \frac{L}{\gamma v}[/tex].
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My solution:
When I reset my clock at A, this event is not simultaneous with the time reset at the two planets. In fact, the planets have a "head start" of vLγ/c^2. In the planet's time it takes L/v for me to reach there. In my time, to adjust for the head start, the time to reach B is L/γv.
The resolution of the paradox (that my clock is slower on my ship than the planet's, although due to time dilation, I should read the planet's clock as slower) lies in the assumption that the time reset of the A-B system and the reset of the ship are simultaneous, but they are not.
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