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I read that at the point of gravitational cancellation between two masses there was still time dilation, because this dilation depends on the gravitational potential and not on the gravitational force itself. Yet The equivalence principle explains gravitational time dilation by acceleration.

The equivalence principle explains that the time of an object will be more dilated in a gravitationnal field when located in a zone of greater acceleration than located in a zone of lesser acceleration. It is in this sense only that the potential plays any part.

But when we are between two masses, the situation is a very different one, as there is no acceleration. In my opinion, the clocks at the cancelling point are ticking as fast as at infinity. I deduce it from the equivalence principle : If one end of a rocket is at the cancelling point and the other is not at the cancelling point but a little in the gravitational acceleration of one of the two masses, this end will undergo a greater acceleration than the other end, and consequently from the equivalence principle its time will be slowed down comparative to the other.

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  • $\begingroup$ “The equivalence principle explains gravitational time dilation by acceleration” this is not true. “The equivalence principle explains that the time of an object will be more dilated in a gravitationnal field when located in a zone of greater acceleration than located in a zone of lesser acceleration” this is also not true. Either you are learning from a bad source or you are misunderstanding a good source. Where did you learn these $\endgroup$
    – Dale
    Sep 27 at 20:07
  • $\begingroup$ Length contraction of an accelerating object explains the gravitational time dilation. The back of the object accelerates more than the front and moves faster, so its time is more dilated. $\endgroup$
    – externo
    Sep 27 at 20:11
  • $\begingroup$ No, it doesn’t, using the equivalence it is the gravitational potential that explains gravitational time dilation, even without considering length contraction or differences in the acceleration. Again, where are you getting this information? You are either learning from a bad source or you are misunderstanding a good source $\endgroup$
    – Dale
    Sep 27 at 20:15
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    $\begingroup$ Vu du référentiel inertiel global, la distance entre deux horloges à l’intérieur de l’ascenseur diminue puisque l’ascenseur accélère : c’est la contraction des longueurs prédite par la relativité restreinte. Les horloges ne gardent donc pas une distance fixe dans le référentiel inertiel global : leurs vitesses diffèrent de même que leurs accélérations. Nous en concluons que leurs temps propres seront différents : c’est la désynchronisation cinématique des horloges parfaites prédite par la relativité restreinte. $\endgroup$
    – externo
    Sep 27 at 20:23
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    $\begingroup$ wikipedia : Gravitational redshift can be interpreted as a consequence of the equivalence principle (that gravity and acceleration are equivalent and the redshift is caused by the Doppler effect) $\endgroup$
    – externo
    Sep 27 at 20:38

4 Answers 4

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Time dilation is not caused by acceleration, even in the case of an accelerating rocket ship; it is caused by potential differences. Consider for example two clocks at the front and back of an accelerating spaceship. The front one will be found by an observer inside the ship to be ticking faster than the back one, even though both are accelerating at the same rate. There is a (pseudo) gravitational potential between the clocks, though: a test particle released at the front of the accelerating rocket will fall to the back.

Similarly, the time dilation of a particle moving at constant speed in a circular particle accelerator is measured to be the same as the time dilation of a particle moving in a straight line at the same speed, even though the circular path involves tremendous acceleration. This experiment has actually been done.

See https://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html for further details.

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  • $\begingroup$ You are wrong, the two sides of the rocket do not accelerate at the same rate, because there is length contraction. It is this length contraction which is responsible for the difference in the passage of time between the two sides, since they do not go at the same speed. If you don't believe me, draw an accelerating and contracting rocket in a Minkowski diagram. en.wikipedia.org/wiki/Bell%27s_spaceship_paradox#/media/… $\endgroup$
    – externo
    Sep 28 at 8:59
  • $\begingroup$ Bell's spaceship paradox assumes the string has negligible rigidity, which is not in general the case for the walls of a spaceship; we more often model that as being Born rigid. In any case my general point stands: the clock postulate (confirmed by experiment) states that time dilation depends only on velocity, and not on acceleration. $\endgroup$
    – Eric Smith
    Sep 28 at 11:49
  • $\begingroup$ Time dilation depends only on velocity, but if the velocity of the two ends of the rocket is not the same, time dilation will not be the same : due to the length contraction the acceleration is stronger at the rear than at the front, so the velocity is also greater at the rear than at the front. During acceleration, time dilation is therefore stronger at the rear than at the front. Gravitational dilation comes from this consideration. $\endgroup$
    – externo
    Sep 28 at 12:08
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the standard description of "time dilation from the equivalence principle" involves comparing an object "on the surface of earth" to an accelerating rocket far from earth.

Importantly, this derivation has an analysis based on the time-of-fight of a light ray in the accelerated frame, which has an integral of the acceleration built into it.

For the general case, consider the following (very, very handwavey) argument:

If we are going to arbitrarily compare the "time" in two stationary frames at two different points in spacetime, we are going to have to find a way of comparing them. This will involve takign the "time" vector of frame a, parallel translating it along a geodesic that connects to frame b, and then comparing the two vectors. Since the "force" term arises from the geodesic equation, this amounts to taking a vector product of the time vector and the geodesic, and then integrating that along the path, which will look like $\int F\cdot dx$, which will depend on a potential.

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  • $\begingroup$ So he is correct to say that a weightless person in the center of the earth or between two equal masses experiences time dilation? $\endgroup$
    – jelly ears
    Sep 27 at 21:49
  • $\begingroup$ @jellyears: relative to what? It's a comparative statement, not an absolute one. $\endgroup$ Sep 28 at 1:26
  • $\begingroup$ relative to the same position as the masses are moved further away? that seems to be what he is implying. $\endgroup$
    – jelly ears
    Sep 28 at 2:26
  • $\begingroup$ @jellyears a local clock will always read 1s has ticked every second $\endgroup$ Sep 28 at 18:15
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If two observers are in the same region where the gravitational field is zero, e.g. between those two masses, then there will be no time dilation between them, exactly as you argue. If you enclose the observers in a rocket, that rocket does not accelerate.

However, they could still experience time dilation with respect to an third observer somewhere else. To see this, imagine a whole series of rockets joining the first two observers to the third. This sequence does pass through gravitational fields, so some of the rockets have to accelerate.

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  • $\begingroup$ What you are saying seems logical to me, but it is contrary to what is being said almost everywhere. We read everywhere that time dilation does not depend on the gravitational force but on the potential, and as the potential does not cancel between two masses there is time dilation. This seems to me to contradict the principle of equivalence. $\endgroup$
    – externo
    Sep 27 at 20:51
  • $\begingroup$ Time dilation does depend on the potential. That is, when comparing elapsed time for two observers, the difference depends on the difference in the potentials that they reside at. In the above example, our first two observers lie at the same potential (since the gravitational field is zero there), while the third lies at a different potential. $\endgroup$
    – Sten
    Sep 27 at 20:57
  • $\begingroup$ Very good. I agree with you. But look at this : physics.stackexchange.com/questions/520350/… $\endgroup$
    – externo
    Sep 27 at 21:11
  • $\begingroup$ I re-read the answer and there is a misunderstanding. My question was is there time dilation for the occupants of the rocket relative to infinity ? In fact, it sounds like you think like everyone else and say that there is time dilation. $\endgroup$
    – externo
    Sep 27 at 21:26
  • $\begingroup$ Not sure if I'm understanding you correctly, but to consider time dilation with respect to infinity, just imagine that the third observer in my example is at infinity. That is to say, there is time dilation relative to infinity, because if you were to line up a bunch of rockets out to infinity, most of them would be accelerating. $\endgroup$
    – Sten
    Sep 27 at 21:29
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The equivalence principle explains gravitational time dilation by acceleration.

I deduce it from the equivalence principle : If one end of a rocket is at the cancelling point and the other is not at the cancelling point but a little in the gravitational acceleration of one of the two masses, this end will undergo a greater acceleration than the other end, and consequently from the equivalence principle its time will be slowed down comparative to the other.

These statements are based on a misunderstanding of the equivalence principle. The equivalence principle does not state that all effects of gravitation are based on the gravitational acceleration. Instead, the equivalence principle states that locally being in a gravitational field is equivalent to being in a uniformly accelerating reference frame without gravity. So, to apply the equivalence principle to this scenario means that we need to determine the time dilation in a uniformly accelerating reference frame, then this will be the local equation for calculating time dilation in a gravitational field.

Gravitational time dilation is equivalent to the redshift between two observers that are at rest in the gravitational field. So the equivalence principle says that we can determine that by calculating the redshift between two observers that are at rest in a uniformly accelerating reference frame. That, in turn, can be calculated by calculating the redshift between two observers that are accelerating with the same acceleration in an inertial reference frame.

Suppose that we have a top observer at a distance $h$ away from a bottom observer in the $z$ direction. Both are at rest momentarily and both are accelerating with an acceleration $g$ in the $z$ direction. So the equations for the top and bottom observers are $$z_{T}(t)=\frac{1}{2}g t^2 +h$$$$z_{B}(t)=\frac{1}{2} g t^2$$

The bottom observer emits light pulses at $t_{B0}=0$ and $t_{B1}=\Delta t$. So the equation for the two light pulses are $$z_{0}(t)=(t-t_{B0}) c + z_B(t_{B0})= ct$$$$z_{1}(t)=(t-t_{B1}) c + z_B(t_{B1})= ct-c \Delta t + \frac{1}{2}g \Delta t^2$$ Now, to determine the redshift we can solve for the times when the light pulses arrive at the top observer $$z_T(t_{T0})=z_0(t_{T0})$$$$z_T(t_{T1})=z_1(t_{T1})$$ then we can take those times and calculate to first order in $\Delta t$ that $$\Delta t_T=t_{T1}-t_{T0} \approx \frac{c}{\sqrt{c^2 - 2 g h}}\Delta t$$

Notice that the acceleration $g$ does not show up alone anywhere, but only as $gh$ which is the gravitational potential. So the equivalence principle indeed shows that gravitational time dilation is locally a function of the gravitational potential $gh$ and not of the gravitational acceleration $g$.

To calculate a non-local gravitational time dilation then we integrate the local time dilations, which winds up just being a function of the total change in gravitational potential between the two observers. Therefore, between a gravitational observer at the midpoint and one at infinity, since there is a potential difference there is also time dilation.

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  • $\begingroup$ Thanks for the answer. First, I want to make one remark. The French paper I quoted explains that many physicists even today believe that SR redshift explains gravitational time dilation. But this is not the case because this redshift is deduced from Newtonian mechanics only and a redshift in itself don't explain time dilation. The real reason for the dilation is due to length contraction. Because of it the speed of the back part is greater than that of the front part and time dilation in the back is stronger than in the front. You can investigate this phenomenon to ensure the veracity of it. $\endgroup$
    – externo
    Sep 28 at 19:37
  • $\begingroup$ On the question of the potential, I would say that you have demonstrated that the gravitational time dilation depends on the potential in the case where there is only one attractive mass. The situation of an accelerated rocket is similar to that of a house on the surface of the earth, but on the condition that there is no gravitational influence other than that of the earth. But at the point of cancellation between two masses there is no acceleration, therefore the deduction drawn from the equivalence principle is no longer valid and neither is the formula obtained from it. $\endgroup$
    – externo
    Sep 28 at 19:48
  • $\begingroup$ If you are not convinced about the reason for gravitational time dilation there is a simple way to convince yourself. This phenomenon, if real, must be seen in a Minkowski diagram. Make the drawing and you will see that the difference in the passage of time between the two extremities comes from lengths contraction. Without length contraction this effect does not exist. $\endgroup$
    – externo
    Sep 28 at 20:41
  • $\begingroup$ If you would like to have a discussion then I recommend a discussion forum, like physicsforums.com, rather than a Q&A forum like here. Comments are not meant for discussion, particularly such a long one as you seem to want $\endgroup$
    – Dale
    Sep 28 at 20:53
  • $\begingroup$ I will do it now $\endgroup$
    – externo
    Sep 28 at 21:16

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