# Free fall orbit time dilation

• B

## Main Question or Discussion Point

I had been informed that for objects in free fall orbit no proper acceleration would be measured.

But I also understood it that for objects in free fall orbit gravity time dilation would be observed.
https://www.physicsforums.com/threa...vitational-time-dilation.563173/#post-3685830 .

So I am not sure how proper acceleration and gravity are equivalent, as in the example given in the last thread referenced, each time B comes back to A the difference between the clocks would show an affect of gravity, and the affect could be dismissed as being an affect of proper acceleration, as no proper acceleration would have been measured by A or B. So can someone explain why proper acceleration and gravity are equivalent in this example?

With two satellites, going at the same velocity, in free fall orbit around a massive body at different altitudes for a few thousand years, as I understand it the clock on the one much closer to the massive body would run slower than the one in a higher free fall orbit so that if they were both quickly brought together and compared the clock that was in lower orbit would have measured less time passing.

Would the comparison be significantly different if one of the satellites was in free fall orbit at a much greater velocity than the other? If so, what type of difference would it make compared to if they were going at the same velocity? If the bringing the together (presumably a constant affect) were to influence the outcome, perhaps it could be explained how the difference (if there would be one) would vary dependent on how long they had been orbiting for, once the gravitational affect had been taken into account.

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mfb
Mentor
I had been informed that for objects in free fall orbit no proper acceleration would be measured.
By those objects.
But I also understood it that for objects in free fall orbit gravity time dilation would be observed.
As seen by others.

Different reference frames. An outside observer can tell if you fall towards a black hole or sit in a rocket, e.g. by detecting the black hole or the rocket.

For orbits driven by gravity, orbital velocity and gravitational potential energy (averaged over an orbit) are linked, you cannot change them individually.

Dale
Mentor
I am not sure how proper acceleration and gravity are equivalent
They are only equivalent locally. The scenarios you describe are non local.

If you restrict yourself to local scenarios then the equivalence principle applies. In fact, this local equivalence can be used to derive the gravitational time dilation e.g. for the Pound Rebka experiment.

I had been informed that for objects in free fall orbit no proper acceleration would be measured.
By those objects.
Yes neither of the objects involved would measure proper acceleration. And from what I understood from what was written in the link I supplied ( https://www.physicsforums.com/threads/stationary-frames-of-reference.899195/page-2#post-5658771 ):

"Proper acceleration, the kind that an accelerometer measures and that is the same no matter what coordinate system I choose, happens or not according to whether the object is in free fall or not. Freefall paths through flat spacetime appear in space like the straight lines of Euclidean geometry; freefall paths through curved spacetime do not. eiether way they are freefall paths and there is no proper acceleration."

The measurement of proper acceleration is coordinate system independent. Were you in agreement that in no coordinate system would those object be considered to be undergoing proper acceleration?

But I also understood it that for objects in free fall orbit gravity time dilation would be observed.
As seen by others.

Different reference frames. An outside observer can tell if you fall towards a black hole or sit in a rocket, e.g. by detecting the black hole or the rocket.

For orbits driven by gravity, orbital velocity and gravitational potential energy (averaged over an orbit) are linked, you cannot change them individually.
I am not sure what you meant by "as seen by others". In the link I supplied ( https://www.physicsforums.com/threa...vitational-time-dilation.563173/#post-3685830 ) I had assumed B performs the orbit, and comes back into the same rest frame as A to compare clocks, similar to the Hafele-Keating experiment. The result of that I had assumed was reference frame independent. Were you in disagreement that for objects in free fall orbit the affects of gravity time would be observed even though all coordinate systems could agree that there was no proper acceleration measured on board?

I assume your comment about orbits driven by gravity, orbital velocity and gravitational potential energy (averaged over an orbit) are linked, you cannot change them individually. Had to do with where I wrote:

Would the comparison be significantly different if one of the satellites was in free fall orbit at a much greater velocity than the other? If so, what type of difference would it make compared to if they were going at the same velocity? If the bringing the together (presumably a constant affect) were to influence the outcome, perhaps it could be explained how the difference (if there would be one) would vary dependent on how long they had been orbiting for, once the gravitational affect had been taken into account.
I am not sure what need to be able to change any one of them individually. The change of the velocity could imply a change in orbit, but presumably that does not prohibit the calculation of what amount of time dilation would have been expected due to the gravitation effect (which I think is known as gravitational time dilation) and how much was due to the relative velocity differences (which I think is known as kinematic time dilation), and what as I understand it they did in the Hafele–Keating experiment. And therefore the difference in the proportion of the time dilation due to gravitational time dilation and the proportion due to kinematic time dilation could be calculated for example.

They are only equivalent locally. The scenarios you describe are non local.

If you restrict yourself to local scenarios then the equivalence principle applies. In fact, this local equivalence can be used to derive the gravitational time dilation e.g. for the Pound Rebka experiment.
Sorry I am not quite clear on what you mean by "locally". I assume it is not connected to the principle of locality vs "spooky action at a distance". Is it the same as being in the same frame of reference?

Dale
Mentor
Sorry I am not quite clear on what you mean by "locally". I assume it is not connected to the principle of locality vs "spooky action at a distance". Is it the same as being in the same frame of reference?
It means close enough in space and time so that tidal effects are negligible. I.e. a uniform gravitational field.

A.T.
So I am not sure how proper acceleration and gravity are equivalent,
They aren't. Gravity doesn't cause proper acceleration. The equivalence is between local frames of reference, which undergo the same proper acceleration. For example:
- Frame in free fall near Earth is locally equivalent to frame floating in deep space (0g proper acceleration).
- Frame at rest to Earth's surface is locally equivalent to frame of an accelerating rocket in deep space (both 1g proper acceleration).

clocks would show an affect of gravity, and the affect could be dismissed as being an affect of proper acceleration,
Proper acceleration doesn't affect proper time:
https://en.wikipedia.org/wiki/Clock_hypothesis

It means close enough in space and time so that tidal effects are negligible. I.e. a uniform gravitational field.
Thanks I think I get it now (undergoing the same g proper acceleration).

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They aren't. Gravity doesn't cause proper acceleration. The equivalence is between local frames of reference, which undergo the same proper acceleration. For example:
- Frame in free fall near Earth is locally equivalent to frame floating in deep space (0g proper acceleration).
- Frame at rest to Earth's surface is locally equivalent to frame of an accelerating rocket in deep space (both 1g proper acceleration).
Ok thanks.

With regards to:

With two satellites, going at the same velocity, in free fall orbit around a massive body at different altitudes for a few thousand years, as I understand it the clock on the one much closer to the massive body would run slower than the one in a higher free fall orbit so that if they were both quickly brought together and compared the clock that was in lower orbit would have measured less time passing.

Would the comparison be significantly different if one of the satellites was in free fall orbit at a much greater velocity than the other? If so, what type of difference would it make compared to if they were going at the same velocity? If the bringing the together (presumably a constant affect) were to influence the outcome, perhaps it could be explained how the difference (if there would be one) would vary dependent on how long they had been orbiting for, once the gravitational affect had been taken into account.
Have you got any answer, taking into account:

The change of the velocity could imply a change in orbit, but presumably that does not prohibit the calculation of what amount of time dilation would have been expected due to the gravitation effect (which I think is known as gravitational time dilation) and how much was due to the relative velocity differences (which I think is known as kinematic time dilation), and what as I understand it they did in the Hafele–Keating experiment. And therefore the difference in the proportion of the time dilation due to gravitational time dilation and the proportion due to kinematic time dilation could be calculated for example.
The reason I am asking is that if the change is in a certain direction, then how were the velocities relative, could you not tell from the end clock measurement which had been travelling faster?

pervect
Staff Emeritus
I had been informed that for objects in free fall orbit no proper acceleration would be measured.

But I also understood it that for objects in free fall orbit gravity time dilation would be observed.
https://www.physicsforums.com/threa...vitational-time-dilation.563173/#post-3685830 .

So I am not sure how proper acceleration and gravity are equivalent, as in the example given in the last thread referenced, each time B comes back to A the difference between the clocks would show an affect of gravity, and the affect could be dismissed as being an affect of proper acceleration, as no proper acceleration would have been measured by A or B. So can someone explain why proper acceleration and gravity are equivalent in this example?
Proper acceleration and gravity are equivalent locally, in a small enough region of space-time. So if you consider a small orbiting object, say the space-station MIR, and we assume the space station isn't rotating, to the first order you won't have any "gravity" inside the space station. We will say the astronauts inside the space station are in "free fall", you've probably seen the videos.

Note that you'll still see second order effects (which would be called tidal forces) - the larger the space station, the more important these tidal accelerations will become. Also, the longer you observe an object, the more important these effects become.

The tidal forces are small, but over time they will have noticeable effects, so ignoring them completely is simply wrong. Suppose, for instance, that we had a (non-rotating) space station in Earth orbit, as follows

......................................................................A
E
......................................................................B

E is the Earth, we are looking at a side view so that A and B have a velocity that's out of the plane of the paper, i.e. if the paper is in the x-y plane, the velocity of A and B is in the z direction.

In such a situation, if there are no external forces acting on A and B, they will both follow perfectly circular orbits. These circular orbits will intersect twice every orbital period. This is noticably different from how A and B would act if they were in an inertial frame.

This does not show that the principle of equivalence is fundamentally wrong, it does show that it's purely a local approximation, valid to the first order over a short enough time period. And that if we attempt to apply it to an entire orbital period (or some large fraction thereof), we have misused it. While there is no "gravity" inside the space station, in Newtonian theory we say there are tidal forces. In General Relativity we might use a slightly different language (but tidal forces will do, and it will be less distracting than using the exact description that GR uses).

On a related note, it's possible you've fallen into the trap thinking that it's "acceleration" that explains time dilation, and have a fundamentally non-relativistic view of the structure of time based on the absolute time idea we've inhereted from Newtonian physics.

The simple way of describing the facts is this. Two objects, each of which caries their own clock, will in general have different clock readings if they take different paths through space-time. There is no logical reason to assume they won't - the only reason to think they would is due to the fact that it works that way in classical, Newtonian physics. For a worthwhile analogy, if we have two cars with odometers, and they drive along different routes and rejoin each other, we don't even expect the odometer readings to be the same when they re-unite. IT's logically consistent for objects to have clocks that read differently when they move differtrently, just as it's logically consistent that odometers won't read the same for cars driving different routes. It's just not something that we are used to. If we use the block universe interpretation (not required, but helpful), we can even view clocks as being a kind of space-time "odometer".

Over long time periods, in the full GR case one can draw no conclusion over which clock will read more and which clock will read less when they re-unite based solely on their acceleration profiles. To use the orbital example, suppose that one clock is orbiting the Earth in a circular orbit. They other clock moves only in a radial direction, it moves outwards in such a manner that it rejoins the orbiting clock after one orbit. Neither clock experiences any proper acceleration, but they will have different time readings when the re-unite. The clock moving outwards winds up having the longest elapsed proper time.

At the risk of confusion, I'll add that there is a principle that says that if all possible clocks take different paths through space-time and re-unite after some finite time, there will be at least one clock that has the maxium time reading, an upper bound. We can say that this clock will be a clock that experiences no proper acceleration, that this is a necessary condition.. Having no proper acceleration is a necessary condition for having the maximum proper time, but it's not sufficient. The previous example shows an case where we have two clocks, neither one of which experiences any proper acceleration, but yet both read different times when they re-unite, which serves as a direct counter-example.

Over short enough time periods in GR, or in the case of flat and simply connected pace-time of special relativity, it turns out that the clock that does not accelerate has the largest reading, that in these cases the zero acceleration is sufficient as well as necessary for zero proper acceleration to imply maximal aging. Unfortunately, if we consider the more general case of GR over long time periods, we simply can't say this.

• Mister T
Proper acceleration and gravity are equivalent locally, in a small enough region of space-time. So if you consider a small orbiting object, say the space-station MIR, and we assume the space station isn't rotating, to the first order you won't have any "gravity" inside the space station. We will say the astronauts inside the space station are in "free fall", you've probably seen the videos.

Note that you'll still see second order effects (which would be called tidal forces) - the larger the space station, the more important these tidal accelerations will become. Also, the longer you observe an object, the more important these effects become.

The tidal forces are small, but over time they will have noticeable effects, so ignoring them completely is simply wrong. Suppose, for instance, that we had a (non-rotating) space station in Earth orbit, as follows

......................................................................A
E
......................................................................B

E is the Earth, we are looking at a side view so that A and B have a velocity that's out of the plane of the paper, i.e. if the paper is in the x-y plane, the velocity of A and B is in the z direction.

In such a situation, if there are no external forces acting on A and B, they will both follow perfectly circular orbits. These circular orbits will intersect twice every orbital period. This is noticably different from how A and B would act if they were in an inertial frame.

This does not show that the principle of equivalence is fundamentally wrong, it does show that it's purely a local approximation, valid to the first order over a short enough time period. And that if we attempt to apply it to an entire orbital period (or some large fraction thereof), we have misused it. While there is no "gravity" inside the space station, in Newtonian theory we say there are tidal forces. In General Relativity we might use a slightly different language (but tidal forces will do, and it will be less distracting than using the exact description that GR uses).

On a related note, it's possible you've fallen into the trap thinking that it's "acceleration" that explains time dilation, and have a fundamentally non-relativistic view of the structure of time based on the absolute time idea we've inhereted from Newtonian physics.

The simple way of describing the facts is this. Two objects, each of which caries their own clock, will in general have different clock readings if they take different paths through space-time. There is no logical reason to assume they won't - the only reason to think they would is due to the fact that it works that way in classical, Newtonian physics. For a worthwhile analogy, if we have two cars with odometers, and they drive along different routes and rejoin each other, we don't even expect the odometer readings to be the same when they re-unite. IT's logically consistent for objects to have clocks that read differently when they move differtrently, just as it's logically consistent that odometers won't read the same for cars driving different routes. It's just not something that we are used to. If we use the block universe interpretation (not required, but helpful), we can even view clocks as being a kind of space-time "odometer".

Over long time periods, in the full GR case one can draw no conclusion over which clock will read more and which clock will read less when they re-unite based solely on their acceleration profiles. To use the orbital example, suppose that one clock is orbiting the Earth in a circular orbit. They other clock moves only in a radial direction, it moves outwards in such a manner that it rejoins the orbiting clock after one orbit. Neither clock experiences any proper acceleration, but they will have different time readings when the re-unite. The clock moving outwards winds up having the longest elapsed proper time.

At the risk of confusion, I'll add that there is a principle that says that if all possible clocks take different paths through space-time and re-unite after some finite time, there will be at least one clock that has the maxium time reading, an upper bound. We can say that this clock will be a clock that experiences no proper acceleration, that this is a necessary condition.. Having no proper acceleration is a necessary condition for having the maximum proper time, but it's not sufficient. The previous example shows an case where we have two clocks, neither one of which experiences any proper acceleration, but yet both read different times when they re-unite, which serves as a direct counter-example.

Over short enough time periods in GR, or in the case of flat and simply connected pace-time of special relativity, it turns out that the clock that does not accelerate has the largest reading, that in these cases the zero acceleration is sufficient as well as necessary for zero proper acceleration to imply maximal aging. Unfortunately, if we consider the more general case of GR over long time periods, we simply can't say this.
Thanks for the response. As I now understand it, acceleration does not affect time dilation, although the amount of curvature does, the greater the curvature the slower the clock ticks. So in the satellite example I gave where from the perspective of the mass there are two satellites, in free fall, orbiting the mass in circular orbits, one closer to the mass and one further away, and at different velocities (from the mass's perspective). Could you explain how the clock time difference would vary when they were compared after bringing them together, if they had been observed orbiting (post clock synchronisation) for millions of years?

Presumably the bringing them together would become less significant the longer they orbited, and the amount of time dilation due to curvature would be frame of reference independent, but what about the observed velocities from the mass being orbited's perspective? If the satellites were labelled A and B and the mass C then would the clock comparison figure be correctly calculated no matter which you considered at rest?

Also I am not quite clear how A or B could be considered at rest and not rotating, as while C (the mass) could be considered to orbit A or B, if it was so considered, would it not also need to be considered that it was rotating, for an explanation of why the same section of C was not always facing A (or B), and could it not be objectively measured that it was not rotating?

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A.T.
As I now understand it, acceleration does not affect time dilation, although the amount of curvature does, the greater the curvature the slower the clock ticks.
No. Curvature just means that simple SR rules about who ages more don't apply.

No. Curvature just means that simple SR rules about who ages more don't apply.
Are you suggesting that if the curvature in a scenario was increased it would not make a clock within the curvature that was increased slow?

Dale
Mentor
Are you suggesting that if the curvature in a scenario was increased it would not make a clock within the curvature that was increased slow?
No, in fact it can be the opposite. Consider a spherical shell of mass. The spacetime inside the shell is flat with no curvature, while the spacetime outside the shell is curved. A clock outside the shell (in the curved portion of spacetime) will be faster than one inside the shell.

• No, in fact it can be the opposite. Consider a spherical shell of mass. The spacetime inside the shell is flat with no curvature, while the spacetime outside the shell is curved. A clock outside the shell (in the curved portion of spacetime) will be faster than one inside the shell.
I did not realise it would be flat inside the shell. Does it show some distortion relative to totally undistorted spacetime (distant from any mass), such that if the spherical shell of mass disappeared there would be a change to the spacetime within the shell, else why would clocks go slower inside the shell if spacetime was undistorted compared to the outside in distant space far from the shell where the clocks would be going faster?

Also regarding the questions about the two satellites A and B orbiting mass C, could you help there:

1) Presumably the bringing them together would become less significant the longer they orbited, and the amount of time dilation due to curvature would be frame of reference independent, but what about the observed velocities from the mass being orbited's perspective? If the satellites were labelled A and B and the mass C then would the clock comparison figure be correctly calculated no matter which you considered at rest?

2) I am not quite clear how A or B could be considered at rest and not rotating, as while C (the mass) could be considered to orbit A or B, if it was so considered, would it not also need to be considered that it was rotating, for an explanation of why the same section of C was not always facing A (or B), and could it not be objectively measured that it was not rotating?

Dale
Mentor
Does it show some distortion relative to totally undistorted spacetime (distant from any mass),
No, it is just flat spacetime. All local experiments will give the same result as the same local experiment performed infinitely far away.

such that if the spherical shell of mass disappeared there would be a change to the spacetime within the shell,
The continuity equations forbid the shell of mass disappearing. However, you could posit that the shell of mass suddenly turns into a spherical shell of radiation moving outward at c. In that case there would be no change to the spacetime within the shell. As the radiation passed a region, it would transition from curved to flat, but regions that were already inside would be unaffected.

why would clocks go slower inside the shell if spacetime was undistorted compared to the outside in distant space far from the shell where the clocks would be going faster?
That is a nonlocal comparison. Nothing is happening to either clock. They are both functioning normally locally, but between them spacetime is curved.

Regarding A and B, I don't really want to read a whole other thread to answer a question. I would rather you ask a self contained question.

No, it is just flat spacetime. All local experiments will give the same result as the same local experiment performed infinitely far away....That is a nonlocal comparison. Nothing is happening to either clock. They are both functioning normally locally, but between them spacetime is curved.
Why are they (the clock in flat spacetime in distant space outside the shell and the clock in flat spacetime in flat spacetime inside the shell) not local (which I understand means that they under the same g) to each other ?

Consider satellites A and B, going at different velocities, in free fall orbit around a massive body C at different altitudes for a million years, then being brought together and the clocks on them compared. Presumably the bringing them together would become less significant the longer they orbited, and the amount of time dilation due to curvature would be frame of reference independent, but what about the observed velocities from the mass being orbited's perspective? If the satellites were labelled A and B and the mass C then would the clock comparison figure be correctly calculated no matter which you considered at rest?

Also regarding A, B and C being at rest, I am not quite clear how A or B could be considered at rest and not rotating, as while C (the mass) could be considered to orbit A or B, if it was so considered, would it not also need to be considered that it was rotating, for an explanation of why the same section of C was not always facing A (or B), and could it not be objectively measured that it was not rotating?

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A.T.
That is a nonlocal comparison. Nothing is happening to either clock. They are both functioning normally locally, but between them spacetime is curved.
Why are they not local (which I understand means that they under the same g) to each other ?

They aren't. Gravity doesn't cause proper acceleration. The equivalence is between local frames of reference, which undergo the same proper acceleration. For example:
- Frame in free fall near Earth is locally equivalent to frame floating in deep space (0g proper acceleration).
- Frame at rest to Earth's surface is locally equivalent to frame of an accelerating rocket in deep space (both 1g proper acceleration).

jbriggs444
Homework Helper
2019 Award
Sorry I am not quite clear on what you mean by "locally".
It means close enough in space and time so that tidal effects are negligible. I.e. a uniform gravitational field.
Why are they (the clock in flat spacetime in distant space outside the shell and the clock in flat spacetime in flat spacetime inside the shell) not local (which I understand means that they under the same g) to each other ?
It's what Dale wrote in post #6 that is relevant.

As Dale explains, "local" means nearby. How nearby it has to be depends on how strong tidal effects are and how carefully you measure (i.e. how small is "negligible"). "Distant", on the other hand means far away. Something far away is unlikely to be local. If you have curved space-time between two areas of flat space-time then the two areas of flat space-time are not local to one another. Curved space-time and tidal effects are the same thing.

It's what Dale wrote in post #6 that is relevant.

As Dale explains, "local" means nearby. How nearby it has to be depends on how strong tidal effects are and how carefully you measure (i.e. how small is "negligible"). "Distant", on the other hand means far away. Something far away is unlikely to be local. If you have curved space-time between two areas of flat space-time then the two areas of flat space-time are not local to one another. Curved space-time and tidal effects are the same thing.
Dale wrote:

The continuity equations forbid the shell of mass disappearing. However, you could posit that the shell of mass suddenly turns into a spherical shell of radiation moving outward at c. In that case there would be no change to the spacetime within the shell. As the radiation passed a region, it would transition from curved to flat, but regions that were already inside would be unaffected.
So if the shell disappeared like that, then Dale said the space inside would be unaffected, and how much affect could the shell of had on the region of distant space? Both clocks would end up at 0g proper acceleration, but the one, that had been outside the shell, going faster than the other, and I wasn't clear why. Were you thinking the clocks would not be local once the shell had radiated away like Dale suggested?

jbriggs444
Homework Helper
2019 Award
Were you thinking the clocks would not be local once the shell had radiated away like Dale suggested?

You could apply the definition of "local" to try to decide for yourself whether two clocks separated by a large distance in flat space-time are "local" to one another.

You could apply the definition of "local" to try to decide for yourself whether two clocks separated by a large distance in flat space-time are "local" to one another.
Well you were suggesting that the that Dale gave was what was important. So given the situation described
Were you thinking the clocks would not be local once the shell had radiated away like Dale suggested?

A.T.
...but the one, that had been outside the shell, going faster than the other
Only until the radiation from the destroyed shell has passed it.

jbriggs444