Given the equivalence principle, I would expect any object in freefall to have the same frame of reference, which would mean the same time dilation. I'd like to verify my understanding of this.

For example, let's say there are two space ships in orbit around a massive body, both at 5000 km. They then synchronize their clocks and both use equal and opposite thrust (such that they both experienced identical acceleration) to cruise at different heights relative to the massive body 5000 + X km and 5000 - X km. They then reverse the maneuver to end up back at 5000 km again traveling at the same speed. In this scenario, I would expect their clocks to still be synchronized. Is this the case? Or does the differing gravitational fields affect their clocks differently?

If so, this would lead me to the conclusions that time for someone at the center of earth would move at the same rate as someone in orbit / freefall. It would also lead me to the conclusion that clocks close to a black hole would be fundamentally moving at the same rate as clocks in orbit around earth.

For the benefit of readers (not necessarily answerers), this situation is materially different from the time dilation difference between someone at sea level on earth vs someone on a mountain top, because the acceleration needed to resist gravity at sea level is higher than on a mountain top (further away from the earth's center of gravity). Its also materially different from the difference in time dilation between someone on the surface of the earth vs someone in freefall in orbit, since according to general relativity only the person on earth is accelerating, while the person in orbit is not.

I want to be careful to distinguish fundamental time dilation from apparent time dilation. Eg, acceleration causes fundamental time dilation, where the object that undergoes more acceleration experiences less time. Whereas velocity differences between two objects cause apparent time dilation, where the objects each see each other's clocks running slower, but they both see each other's clocks slowed by the same rate. A secondary question is: does strength of the gravitational field affect apparent time dilation? I would assume not. Its harder for me to think up a scenario where two ships could sync up clocks, travel to two different gravitational fields using the same acceleration, and be also traveling at the same speed (so as to eliminate the apparent time dilation due to different speeds), such that any apparent time dilation would only be due to existing in different gravitational fields.

FYI, I've searched for the answer here already, but haven't found an answer that has been precise enough (or possibly accessible enough - since I'm not skillful enough with the math):

  • $\begingroup$ Clock gains potential energy - clock ticks faster. Clock gains kinetic energy - clock ticks slower. Energy is not well defined in general relativity - ticking rate is not well defined in general relativity. Falling guy sees a gravity field around earth. $\endgroup$
    – stuffu
    Nov 18, 2020 at 1:58

3 Answers 3


Given the equivalence principle, I would expect any object in freefall to have the same frame of reference, which would mean the same time dilation

EP is very useful, but we must always remember that it is valid locally, what means small $\Delta x$, $\Delta y$, $\Delta z$, and also $\Delta t$.

So, when orbiting / freefall bodies record their relative velocities, they will approach uniform linear velocities if they are close, and if the period of recording is small. That uniform velocities includes $\frac{dt}{d\tau} $.

For example, if 2 bodies orbit at the same height but opposing direction, when they come close and compare their clocks, each one will see the other clock ticking slower. The same result for special relativity when 2 inertial frames (far away of any gravitational fields) of different velocities compare their clocks.

But different orbits and long periods of time are out of the scope of EP. It is necessary to know the metric, and use it to calculate the proper time for each desired path.

In your example, due to the simplicity of the situation, we can say that the ship that spend some time at lower orbit, when rejoin the other (which went to higher one) at the original orbit, will have its clock recording less time. Provided that the period of orbits are much bigger then the transient period of changing them. The difference is the effect of the bigger gravitational well on the first one.

  • 1
    $\begingroup$ Interesting. Does time dilation change depending on whether you're in freefall or not, at all? $\endgroup$
    – B T
    Nov 18, 2020 at 6:51
  • $\begingroup$ If you are in free fall at a given height, time ticks different compared to a clock fixed in a building or mountain at the same height. You can explore the differences with the Schwarzschild equation. $\endgroup$ Nov 18, 2020 at 15:33
  • $\begingroup$ Hmm interesting, thanks $\endgroup$
    – B T
    Nov 21, 2020 at 23:56

You're not giving enough information to describe what you have in mind, but I will hazard a guess.

I'm guessing you are thinking about a scenario where the time spent in no-thrust orbital motion is long compared to transient periods of producing thrust to shift orbital height.

I'm taking terrestrial time as arbitrary reference point. (UTC is a globally maintained time, all over the world there are multiple centers for timekeeping, and they have procedures to maintain a worldwide coordinated time.)

For a satellite in low Earth orbit a smaller amount of proper time elapses than on Earth. This is because that satellite has a large velocity with respect to the Earth, and the height difference with Earth surface is small

For a satellite orbiting at a very large distance to the Earth a larger amount of proper time elapses than on Earth. Orbiting velocity at a large distance to the primary is very slow (as gravitational influence falls off with the square of the distance). The Earth's surface is deep in the Earth's gravitational well, which gives that on the Earth's surface a smaller amount of proper time elapses than far away from Earth.

There is in fact a cross-over point. That is, there is an orbital altitude such that for a satellite orbting at that altitude the same amount of proper time will elapse as on the Earth's surface. This is at a height of about 3000 km above the Earth surface. In a recent video Matt Parker proposed to call this orbital altitude 'time-dilatopause'.

In all the above cases the total time dilation effect is cumulative, which I suppose qualifies as what you refer to as 'fundamental time dilation' (as opposed to what you refer to as 'apparent time dilation')

I have to say, I recommend against introducting nomenclature of your own. It makes your writing less accessible, and in this case there is no need for it.

When I write 'cumulative' I mean the following: an orbiting satellite can over a long stretch of time compare the accumulated internal time with accumulated time on Earth.

On wikimedia Commons there is a diagram that plots time dilation as a function of orbital height

Note that the diagram is in terms of accumulation: the time plotted is the amount of microseconds per day.

More specific to the scenario you are asking about.

For simplicity let the starting orbital altitude be the time-dilatopause (about 3000 km above the Earth surface).
Let one satelite remain at that orbital altitude and let the other one migrate to another altitude. Then a long time is allowed to pass, so that the time spent in transfer from altitude to altitude is small compared to the total time.

Then when the two satellites rejoin the two onboard clocks will not have counted the same amount of proper time.

But time dilation effects associated with orbital motion is a very tricky case, because there is that cross-over altitude. With this kind of scenario the risk of wrongfooting yourself is high. I recommend against using it as a starting point for thinking about time dilation effects

  • $\begingroup$ You're introducing time as viewed from someone standing on earth - I described a situation with orbiting satellites from their point of view precisely to eliminate the potential complication of time dilation when on the surface of a planet vs in freefall in space - I wanted to ask about the freefall situation (at least first). I also want to eliminate the time dilation an observer sees in a moving object. So yes, reviewing the exact meaning of proper time, it looks like that is what I meant by "fundamental time dilation" - I think.. $\endgroup$
    – B T
    Nov 18, 2020 at 0:50

I noticed a wrong statement, I'm addressing it separately.

Eg, acceleration causes fundamental time dilation, where the object that undergoes more acceleration experiences less time.

The factor that is decisive for difference in elapsed proper time is difference in pathlength travelled.

If two travellers travel to the same point, but along different routes, then on rendez-vous for the traveller who has taken the longest journey a smaller amount of proper time has elapsed.

Comparison of amount of proper time elapsed is the best way, because it is unambiguous. You compare at departure and again on rejoining. You don't even try to specify where or when during the journey the effect happened, that only leads to unnecessary complexity.

The shape of the path of the traveller is immaterial, in the end the only thing that counts is the difference in pathlength travelled.

Specifically: the traveller who is travelling a longer pathlenght can do so in multiple ways:
The traveller can do one long haul away, a single U-turn, and a long haul back.
The traveller can follow a zigzag route, thus making multple U-turns.

If the single U-turn journey and the zigzag journey are the same pathlength then on rejoining for both the same amount of proper time has elapsed.

That is: the amount of acceleration that a traveller accumulates is immaterial. Only one factor counts: the difference in pathlength travelled.

  • $\begingroup$ Hmm, this goes against what I think I know about time dilation. I compare at departure and again on rejoining to be precise with when the clocks are compared (ie what specifically I mean by "time dilation"). I could specify that what I think happens is that both travelers experience time dilation (slower time) during acceleration and proportional to acceleration only, and any time drifting (not accelerating) they do not. $\endgroup$
    – B T
    Nov 17, 2020 at 23:37
  • $\begingroup$ Oops, submitted early. What goes against what I think I know about time dilation is that I have never heard that distance traveled is what matters. Take this example: let's say we have two spaceships far away from any gravitational field. One accelerates at 10 Gs for 5 seconds, then immediately decelerates at 10 Gs for 5 more seconds. That's 10 seconds of 10 G acceleration going a distance of 2450 m in 10 seconds. The other travels that distance by accelerating for 1 G for 5 seconds, drifting for 40 seconds, and then decelerating at 1 G for 5 seconds, going a distance of 2450 m in 50 seconds. $\endgroup$
    – B T
    Nov 17, 2020 at 23:46
  • $\begingroup$ In this scenario, they both went the same distance, so you're telling me when they meet up again, their clocks will be synchronized as they were at the beginning of their trip? $\endgroup$
    – B T
    Nov 17, 2020 at 23:46
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    $\begingroup$ @BT In the question, you state that acceleration causes time dilation. That's not correct. Acceleration per se doesn't affect time dilation, although speed changes due to acceleration (of course) affect time dilation. That is, time dilation is a function of the slope of the worldline, not the rate of change of that slope. So gravitational acceleration $GM/r^2$ doesn't directly affect time dilation, but gravitational potential $-GM/r$ does. $\endgroup$
    – PM 2Ring
    Nov 18, 2020 at 3:21
  • 1
    $\begingroup$ Hmm, interesting. So is there a difference in time dilation between a rocket accelerating up exactly countering gravity hovering at 500 km above a planet vs something in orbit at 500 km? $\endgroup$
    – B T
    Nov 18, 2020 at 6:48

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