A clock at the tail end of an accelerating rocket runs slower than a clock at the nose end. The situation is similar if the rocket is standing upright on earth, so the clock at the tail end is slower. But what happens if a rocket is falling freely towards earth with the tail end towards the earth? This rocket is indistinguishable from a non accelerating rocket away from gravity, so shouldn't the time dilation disappear? My confusion is, the tail end is still closer to the earth but the time difference disappears if the rocket falls freely?
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2 Answers
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Actually, a rocket accelerating upwards will have the tail clock ticking more slowly in the rocket's frame, but not because of the Earth. Via the equivalence principle, the rocket's large acceleration is creating a strong local gravitational field which causes the difference between the nose and tail clocks. If the rocket were pointed down at the ground, the tail clock would still tick more slowly (assuming the rocket's acceleration was more than 1g, which is almost certain).
You are exactly correct, that in free fall there would be no difference between the nose and tail.
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4$\begingroup$ Your last sentence would be true only in a uniform gravitational field. For a rocket falling towards Earth the two clocks would experience difference gravitational fields due to variation of gravity with distance, so they could not be both falling freely. One or both of them would have a force applied to them by their attachments to the rocket so their proper acceleration(s) would not be zero. $\endgroup$ Commented Sep 28, 2021 at 4:22
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2$\begingroup$ Good point! I suppose I was implicitly assigning the rocket with opposing clocks to be an infinitesimal, so that locally in free fall it is a Mikowski space. $\endgroup$– RC_23Commented Sep 28, 2021 at 4:25
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$\begingroup$ Actually, now that I think some more, I wonder if that is correct. If two clocks are falling on identical geodesics, one slightly behind the other, wouldn't the distance between them remain constant, and no forces would need manifest? $\endgroup$– RC_23Commented Sep 28, 2021 at 4:32
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1$\begingroup$ Gravitational time dilation is caused by differences in the gravitational potential, which in Newtonian mechanics is $-GM/r$, it has units of Joules/kg. The gravitational field strength is $GM/r^2$, it has units of Newtons/kg. $\endgroup$– PM 2RingCommented Sep 28, 2021 at 6:38
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1$\begingroup$ Consider two 1 kg bodies in vacuum, freefalling towards the Earth. One body is at the initial distance $r$ from the centre of the Earth, the other is directly above the first, initially at $r+h$. The lower body is accelerating slightly faster, so the separation between them increases. If we do this experiment near a black hole, the differential acceleration can be quite large, as shown at physics.stackexchange.com/q/631414/123208 $\endgroup$– PM 2RingCommented Sep 28, 2021 at 6:46
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I think I found the answer today. A rocket is standing on a launchpad. The clock at the tail end is running slower than the clock at the nose end due to gravity.
But if this same rocket were falling freely towards the Earth with the tail towards the Earth, this time difference would disappear. There are two ways to understand this.
First way,a freely falling object is just like an inertial object. So no time dilation.
But the second way is this; the tail end is closer to the Earth and is subject to time dilation compared to the higher nose end. But then, the nose end is like the rear end of an accelerating rocket, accelerating downwards at g. So this clock also experiences exactly the same time dilation. Both these effects cancel out