Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: saspinski on 04/05/2018 23:23:14
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A spaceship passes close enough to the earth to check the clocks here. Its speed is 0.8c, but just when the distance to here is minimum, the engine is turned on in order to decelerate it. It is regulated to keep an constant artificial gravity = g in the ship, not only until it stops, but until it is back to earth at 0.8c to the opposite direction.
According to the equations for relativistic uniform acceleration, and setting c = 1 light second/second, and g = 9.8/(3*108)ls/s2:
τ = 2*atan(v)/g = 389,2 days according to the ship clock.
t = 2*sinh(gτ/2)/g = 472,4 days according to the earth clock.
The situation is pretty symmetric. Both earth and ship can measure an acceleration = g at its own frame of reference all the time. Both have the same right to say: I am at rest.
What breaks the symmetry, if the principle of equivalence doesn't allow to differentiate between acceleration and gravity?
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The fact that both an Earth observer and the ship observer locally measure 1g is a red herring. Gravitational time dilation is due to potential difference. Think of potential difference as the energy per unit mass needed to lift something from a lower height to a higher height. For the Earth observer, the difference in gravitational potential and thus the time dilation between the ship and himself is due to a gravitational field that decreases with strength as you move away from the center of the Earth.
For the ship observer to claim that he is "at rest" he has to assume that the 1 g acceleration he feels is also due to a gravitational field, In his case, however it is a uniform field that does not diminish in magnitude with height ( if his ship was really long and he traveled from tail to nose, he would not measure and weakening in " gravity"). This means that "his" gravity field extends at a constant 1g all the way from him to the Earth. Now it would take an lot more energy to lift a mass that distance against 1g the whole way, than it does against a gravity field that weakens as you get higher. Thus for the ship observer there is a much greater potential difference and thus time dilation between the Earth and himself than the Earth observer would measure between the ship and himself. Therein is the break in symmetry between the two observers.
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It is common to say that the equivalence principle holds locally, for example inside a spaceship, because the tidal forces are too small. But I understand that locally also means small time interval, otherwise time dilation can show the difference as in the example.
Another point is: the ship has a more intense gravitational field compared to the earth’s one, in the meaning of same “g” but greater escape velocity and time dilation. And the associated spacetime is flat, while it is curved for the earth’s one. So, spacetime curvature is not a reliable indicator for the intensity of a gravitational field.