# Free fall orbit time dilation

• B
Only until the radiation from the destroyed shell has passed it.
Why would the passing of the radiation slow it?

Have you even read the post #6 which I referred you to?
Yes Dale stated that local meant in a unified gravitational field. Why would the whole region once the radiation had passed not be a unified 0g gravitational field?

jbriggs444
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Yes Dale stated that local meant in a unified gravitational field. Why would the whole region once the radiation had passed not be a unified 0g gravitational field?
What does "unified" mean?

Dale
Mentor
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 ?
Please re read the definition I gave previously: "Local means close enough in space and time so that tidal effects are negligible. I.e. a uniform gravitational field". Is the gravitational field uniform in the region containing both clocks? Are tidal effects negligible?

A.T.
Why would the passing of the radiation slow it?
The energy distribution affects the gravitational potential and thus the clock rates between distant points.

What does "unified" mean?
Sorry that should be "uniform"

Please re read the definition I gave previously: "Local means close enough in space and time so that tidal effects are negligible. I.e. a uniform gravitational field". Is the gravitational field uniform in the region containing both clocks? Are tidal effects negligible?
Would they not be once the radiation had passed the distant clock a million years before at c?

Dale
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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?
Yes, but note that the metric will have a different form in these different charts. With a correct expression for the metric, all invariants will be correctly calculated.

The energy distribution affects the gravitational potential and thus the clock rates between distant points.
So considering just the distant clock

1) Before the shell had radiated (imagine it was a light year away)

What energy distribution difference between 1 and 2 were you thinking would explain the affect?

A.T.
So considering just the distant clock
Gravitational time dilation is always between two separated clocks. With just one clock there is nothing to explain.

Dale
Mentor
These two clocks are not local:
(the clock in flat spacetime in distant space outside the shell and the clock in flat spacetime in flat spacetime inside the shell)

These two clocks are local:

Dale
Mentor
What energy distribution difference between 1 and 2 were you thinking would explain the affect?
The distribution of the energy in the shell

Gravitational time dilation is always between two separated clocks. With just one clock there is nothing to explain.
Then consider both the clocks.

1) Clock 1 in the shell in flat spacetime before the shell had radiated, and Clock 2 a light year outside the shell
2) Clock 1 two years after the shell has radiated away, the space time flat as it was in scenario 1, Clock 2 a year after the radiation from the shell had passed it.

The distribution of the energy in the shell
So the gravitational energy causes the time dilation?

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?
Yes, but note that the metric will have a different form in these different charts. With a correct expression for the metric, all invariants will be correctly calculated.
If each had a ruler and a clock on board, when they were at rest together could they not all go by their own clock and ruler, and agree that their clocks and rulers were metrically equivalent?

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?

Dale
Mentor
So the gravitational energy causes the time dilation?
Also energy due to "stuff" like matter or light or whatever.

Dale
Mentor
If each had a ruler and a clock on board, when they were at rest together could they not all go by their own clock and ruler, and agree that their clocks and rulers were metrically equivalent?
Yes, completely.

The question is ambiguous. I cannot tell if you are asking about proper rotation or coordinate rotation.

The question is ambiguous. I cannot tell if you are asking about proper rotation or coordinate rotation.
Proper rotation assuming that it what causes the proper acceleration that can be measured.

Dale
Mentor
Proper rotation assuming that it what causes the proper acceleration that can be measured.
Whether or not a given object is undergoing proper rotation is an invariant. It does not depend on the coordinates chosen. You can choose coordinates where a proper-rotating object is at coordinate-rest. In such coordinates there will be "fictitious forces" which will lead to the correct amount of proper rotation.

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?
Yes, but note that the metric will have a different form in these different charts. With a correct expression for the metric, all invariants will be correctly calculated.

If each had a ruler and a clock on board, when they were at rest together could they not all go by their own clock and ruler, and agree that their clocks and rulers were metrically equivalent?
Yes, completely.
So A would agree with B about the expected differences between their clocks due to gravitational time dilation. And the affect due to bringing the clocks together could be considered insignificant if the orbiting had gone on for long enough.

Would not A think, using the metric of A's clock, that B's clock had ticked less than would have been expected (taking gravitational time dilation into account) if it had been in A's rest frame, and B, using the metric of B's clock, think that A's clock had ticked less than would have been expected (taking gravitational time dilation into account) if A had been in B's rest frame?

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Whether or not a given object is undergoing proper rotation is an invariant. It does not depend on the coordinates chosen. You can choose coordinates where a proper-rotating object is at coordinate-rest. In such coordinates there will be "fictitious forces" which will lead to the correct amount of proper rotation.
So with the A, B, and C coordinates it seems arguable that while the maths can model the relative motion regardless of which is thought to be at rest, only one rest frame would give the coordinate rotation which corresponds to the correct measurable proper rotation, without the addition of fictitious forces? Also I have read that fictitious forces are used with non-intertial frames of reference, but would not both A and B at rest both be intertial frames of reference (what intertial frame are they undergoing accceleration relative to)?

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Dale
Mentor
So A would agree with B about the expected differences between their clocks due to gravitational time dilation. And the affect due to bringing the clocks together could be considered insignificant if the orbiting had gone on for long enough.
Yes, as edited.

I have no idea how to parse the other question. In GR, the proper time on any clock undergoing any motion in any frame in any spacetime is given by ##\int \sqrt{ g_{\mu \nu} dx^{\mu} dx^{\nu}}##

• vanhees71
I have no idea how to parse the other question. In GR, the proper time on any clock undergoing any motion in any frame in any spacetime is given by ##\int \sqrt{ g_{\mu \nu} dx^{\mu} dx^{\nu}}##
Could you make up some figures for the scenario, and use them in the equation to demonstrate please (just so I can see how it works)?

P.S. Regarding where I had written: "So A would agree with B about the expected differences between their clocks due to gravitational time dilation." I had thought it was invariant how much difference would be expected due to gravitational time dilation. I thought in the Hafele-Keating experiment they had split the affects expected due to gravitational time dilation and kinematic time dilation to come up with their estimates, and so was thinking that you might be able to explain whether the expected gravitational time dilation affects would be same regardless of frame of reference, and if they were, and the coming together was insignificant (given long enough orbit), then simply explain why the kinematic time dilation expectations would be the same from each perspective, or how else they came to the same answer when added to the gravitational time dilation effects.

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Dale
Mentor
Why don't you work through the first few chapters of Sean Carroll's lecture notes on general relativity first.