A remote planet is flat like pancake, and its gravitation field is uniform g=9.83m/s²?

Question

Three identical pendulum clocks have pendulums of equal length L = 25cm.
Two clcoks A and B are located at two stations connected by straigh railroad track, at distance
D = speed of light x 1/2 hour, and are always syncronized (in motionless frame).

The third clock C is aboard a train moving at constant speed c/2. When the train is passing station A, engineer syncronizes clock C to clock A.

What is the differece between clocks B and C, when the train passes station B 1 hour later?

The most basic assumption of GR is that uniform gravitation field is equivalent to uniformely accelerating frame of reference. There are other more mathematical approaches, of course, but this one is the most bullet-proof.

Answer 1

I'm thinking, really am.

Is the gravitational field sufficient to seriously affect the special relativity calculation of a time dilation of 1/sqr 2? First impression was no.

But I'm also thinking that the train will perceive the planet as having more mass and gravity. You have a pendulum clock. Its Period = 2*pi*sqrt(l/g) -- at least for small perturbations. Since g would increase by sqr 2. And the period would decrease by the same amount. However, the electronic wrist watch would not increase in speed. Therefore the observer on C would see his pendulum clock run faster.

Therefore, even though the observers on the planet would say time is dilated on the train, the clocks would be synchronized.

Interesting and unexpected result. You are tricky.

Now I get the GR. Looking from the outside the observer will see the same acceleration affecting all objects the same way. The pendulum stays the same!

Source(s)

(Ω) The cat promised me a glass of aquatofana and vodka. Do you know anything about that?

Answer 2

Would it not be the equivalent of a return trip? (if it happened in space, not on a pancake, and the distance would be speed of light x1/4 hour, it would take 1 hour at C/2 to make the whole return trip)
Then the difference between B and C would be the same as A and C.

Answer 3

The Lorentz factor γ = 1 / sqrt ( 1 - v^2 / c^2 ). This is the factor by which time passes more slowly in the moving frame than in the reference frame. Compute γ for v = .5c and find how much time passes in the moving frame when 30 minutes pass in the reference frame.