Inertial frames in changing gravitational fields

[Staticboson]

TL;DR Summary

Looking for clarification on whether and how Special Relativity would be applicable in a changing gravitational field.

The man floating inside the elevator travels through space at constant velocity, and soon reaches proximity to a planet.
To an outside observer, the elevator appears to change course and accelerate towards the planet, so he reasons there is a force acting on the elevator, changing its course and increasing its velocity.
To the man inside there is not such force, no perception of direction change or acceleration. The elevator is simply following a geodesic along curved spacetime.

Sidenote: I'm understanding the concept of inertial frames would be meaningless here, as for in free fall every infinitesimal increment in time would also change the frame of reference.

Here is the question:
As the elevator travels past the observer on his way to the planet his clock will be ticking slightly slower than the observer's (from the observer's perspective) due to the relative velocity. From the observer's perspective, will the clock on the elevator slow down even further as it approaches the planet?
SR indicates that to an outside observer the clock on an accelerating object will slow down, and since to the observer the elevator appears to be accelerating, he predicts its clock will slow down. However, you can argue the elevator is not accelerating but free-falling without any forces being exerted on it, and therefore you can predict its clock rate, from the observer's perspective, will not change.

Which one is it?

 

[Orodruin]

Summary:: Looking for clarification on whether and how Special Relativity would be applicable in a changing gravitational field.

To an outside observer, the elevator appears to change course and accelerate towards the planet, so he reasons there is a force acting on the elevator, changing its course and increasing its velocity.

This is not correct. You are trying to apply Newtonian gravity here. Regardless of the observer, the elevator is in free fall and following a geodesic.

Summary:: Looking for clarification on whether and how Special Relativity would be applicable in a changing gravitational field.

SR indicates that to an outside observer the clock on an accelerating object will slow down

No, SR says there is time dilation of an observer relative to the simultaneity of an inertial frame where the observer is moving.

Summary:: Looking for clarification on whether and how Special Relativity would be applicable in a changing gravitational field.

Sidenote: I'm understanding the concept of inertial frames would be meaningless here, as for in free fall every infinitesimal increment in time would also change the frame of reference.

The concept of global inertial frames is not generally applicanle in GR. This has more to do with spacetime being curved than any change in velocity (however you would define that).

 

[PeroK]

Here is the question:
As the elevator travels past the observer on his way to the planet his clock will be ticking slightly slower than the observer's (from the observer's perspective) due to the relative velocity. From the observer's perspective, will the clock on the elevator slow down even further as it approaches the planet?
SR indicates that to an outside observer the clock on an accelerating object will slow down, and since to the observer the elevator appears to be accelerating, he predicts its clock will slow down. However, you can argue the elevator is not accelerating but free-falling without any forces being exerted on it, and therefore you can predict its clock rate, from the observer's perspective, will not change.

Which one is it?

In SR acceleration does not "cause" time dilation; time dilation is explicitly a function of relative velocity only. In SR no two objects can change their state of relative motion without at least one of them being subject to proper acceleration. I.e. there is no ambiguity about which one is accelerating.

In curved spacetime this is not the case, so you cannot apply SR thinking and rules to a situation (curved spacetime) which explicitly SR does not cover.

However:

If you take the case of astronauts in the ISS, we have:

1) The ISS is at a higher gravitational potential, hence clocks would run slightly faster there, as measured from the Earth, due to this gravitational time dilation.

2) The ISS is moving relative to the surface of the Earth, hence clocks there would run more slowly there, as measured from the Earth.

You have to put these two calculations together to get the net differential ageing when an astronaut returns to Earth.

Note: These two effects are separate in this analysis from the reference frame of the surface of the Earth. But, of course, this separation is coordinate dependent. The invariant quantity is the eventual differential ageing.

Note also that in the Hafele-Keating experiment, they had to account for these two effects: gravitational and velocity-based differential ageing.

Finally, note that if you fly an aircraft west (against the Earth's spin), then you get a net positive differential ageing. The clock on the plane will have recorded more time than one on the surface when they are reunited. You may like to work out why.

 

[Staticboson]

This is not correct. You are trying to apply Newtonian gravity here. Regardless of the observer, the elevator is in free fall and following a geodesic.

Absolutely. I only stated that to the observer the free-falling object appears to change direction and accelerate towards the planet.

No, SR says there is time dilation of an observer relative to the simultaneity of an inertial frame where the observer is moving.

Yes, but it also states through the principle of equivalence that an accelerating object will experience clock slowdown equivalent to the clock slowdown within a gravitational field.

The concept of global inertial frames is not generally applicanle in GR. This has more to do with spacetime being curved than any change in velocity (however you would define that).

Understood, just looking for and answer and explanation to my question at this point.
Thanks!

 

[PeroK]

Yes, but it also states through the principle of equivalence that an accelerating object will experience clock slowdown equivalent to the clock slowdown within a gravitational field.

That is NOT the equivalence principle. The e.p. states roughly that being in a gravitational field is equivalent to being in an accelerating reference frame. It does not state that an accelerating object is measured to have time dilation by an inertial observer. It has no more or less than it does from its relative velocity.

 

[Staticboson]

In SR acceleration does not "cause" time dilation; time dilation is explicitly a function of relative velocity only. In SR no two objects can change their state of relative motion without at least one of them being subject to proper acceleration. I.e. there is no ambiguity about which one is accelerating.

In curved spacetime this is not the case, so you cannot apply SR thinking and rules to a situation (curved spacetime) which explicitly SR does not cover.

I understand, and I did not say that there is a lorentz dilation due to the acceleration, I was referring to the clock slowdown due to the equivalence principle in an accelerate frame of reference. Although there is not accelerated frame of reference here, to the observer the falling object appears to be accelerating.

I'm trying to understand what happens to the clock under the simplest of scenarios: An observer sufficiently far away from the planet to where the gravitational effect is negligible, and the guy inside the elevator free falling right down smack into the planet. There will be a lorentz time dilation due to the velocity of the elevator in relation to the observer, however does the clock rate change, as the elevator falls in.

I suspect that it doesn't because there is no force exerted on the elevator guy and therefore the equivalence principle does not apply, but I want to ask if I'm missing something, given that it is not a simple SR problem.

Thanks!

 

[Staticboson]

That is NOT the equivalence principle. The e.p. states roughly that being in a gravitational field is equivalent to being in an accelerating reference frame. It does not state that an accelerating object is measured to have time dilation by an inertial observer. It has no more or less than it does from its relative velocity.

A clock sitting on the surface of the planet will experience gravitational time dilation. Due to the EP, the clock will experience the same time dilation if subject to acceleration away from the gravitational field. How is this wrong?

 

[PeroK]

A clock sitting on the surface of the planet will experience gravitational time dilation. Due to the EP, the clock will experience the same time dilation if subject to acceleration away from the gravitational field. How is this wrong?

It's just wrong!

Gravitational time dilation relates to the asymmetric time dilation for points at different potentials in a gravitational field. A clock at a higher potential runs slower compared to a clock at a lower potential.

If you have two clocks in an accelerating reference frame (e.g. in an accelerating rocket) then the clock at the front runs slower compared to a clock at the rear. As measured by each other. But not as measured by an external inertial observer.

The equivalence principle states that these two scenarios are locally experimentally indistiguishable.

Furthermore: if two rockets pass you, one traveling at a steady speed and the other traveling at the same speed but accelerating then the time dilation is the same (it depends only on the speed). There is no element of the gamma factor for the acceleration. The gamma factor for an accelerating particle is $$\gamma = \frac{1}{\sqrt{1- v(t)^2/c^2}}$$
This is a function of the instantaneous speed, $v(t)$, only. There is no dependence on its instantaneous acceleration.

 

[Staticboson]

It's just wrong!

Gravitational time dilation relates to the asymmetric time dilation for points at different potentials in a gravitational field. A clock at a higher potential runs slower compared to a clock at a lower potential.

If you have two clocks in an accelerating reference frame (e.g. in an accelerating rocket) then the clock at the front runs slower compared to a clock at the rear. As measured by each other. But not as measured by an external inertial observer.

The equivalence principle states that these two scenarios are locally experimentally indistiguishable.

Furthermore: if two rockets pass you, one traveling at a steady speed and the other traveling at the same speed but accelerating then the time dilation is the same (it depends only on the speed). There is no element of the gamma factor for the acceleration. The gamma factor for an accelerating particle is $$\gamma = \frac{1}{\sqrt{1- v(t)^2/c^2}}$$
This is a function of the instantaneous speed, $v(t)$, only. There is no dependence on its instantaneous acceleration.

I understand about gamma being only velocity dependent.
However if I'm accelerating away from an observer, isn't that the equivalent of him being at a lower gravitational potential (actually zero gravitational potential) and me being in a higher gravitational potential?

 

[Ibix]

A clock sitting on the surface of the planet will experience gravitational time dilation.

Ask yourself how you know if a clock is time dilated. The only possible answer is that you compare it to another clock. So "a clock experiences time dilation" is an incomplete sentence (at best) - actually it's time dilated compared to some other clock.

So then when you want to say that an accelerating clock is time dilated, the question is: compared to what clock? Not answering that question makes your attempt at a statement of the equivalence principle meaningless, I'm afraid.

So what clock should you care about? The point about the equivalence principle is that in a small region there's no difference between "at rest in a gravitational field" and "at rest in an accelerating reference frame". So you need to be comparing nearby clocks. For example, if you are in a small room with two accurate clocks, you can nail one to the ceiling and one to the floor. In an accelerating lift in deep space or sat on Earth, the higher clock will tick faster.

 

[Staticboson]

Furthermore: if two rockets pass you, one traveling at a steady speed and the other traveling at the same speed but accelerating then the time dilation is the same

The time dilation would be the same only for the infinitesimally short moment that the two rocket velocities are equal, as the accelerating rocket speed increases, so will the time dilation.

 

[Ibix]

However if I'm accelerating away from an observer, isn't that the equivalent of him being at a lower gravitational potential (actually zero gravitational potential) and me being in a

No, because you don't share a notion of gravitational potential. If you have a gravitational potential then you have a notion of "up", a direction in which you have to do work in order to move. Your friend is floating in space - he has no up or down. You do.

 

[Staticboson]

Ask yourself how you know if a clock is time dilated. The only possible answer is that you compare it to another clock. So "a clock experiences time dilation" is an incomplete sentence (at best) - actually it's time dilated compared to some other clock.

So then when you want to say that an accelerating clock is time dilated, the question is: compared to what clock? Not answering that question makes your attempt at a statement of the equivalence principle meaningless, I'm afraid.

So what clock should you care about? The point about the equivalence principle is that in a small region there's no difference between "at rest in a gravitational field" and "at rest in an accelerating reference frame". So you need to be comparing nearby clocks. For example, if you are in a small room with two accurate clocks, you can nail one to the ceiling and one to the floor. In an accelerating lift in deep space or sat on Earth, the higher clock will tick faster.

I believe I stated this, the clock on the surface will run slower compared to a clock located far away from the gravitational field.

 

[Staticboson]

No, because you don't share a notion of gravitational potential. If you have a gravitational potential then you have a notion of "up", a direction in which you have to do work in order to move. Your friend is floating in space - he has no up or down. You do.

If my friend moves away from the gravitational field, at what point does he loose the notion of up or down? In this thought experiment down is always down, in the direction of the planet.

 

[PeroK]

I understand about gamma being only velocity dependent.
However if I'm accelerating away from an observer, isn't that the equivalent of him being at a lower gravitational potential (actually zero gravitational potential) and me being in a higher gravitational potential?

No, it's not. There's a lot of confusion of ideas here. If he is at a higher potential then you must be deeper in the field. If you are on the surface of the Earth and someone is far away, then you are at the lower potential. You would have to either accelerating or in a strong gravitational field. In any case, your local circumstances cannot depend on what someone else is doing.

Also, "zero" gravitational potential is not what you think. This is why it's a good idea to study Newtonian gravity before dealing with GR.

 

[Staticboson]

No, it's not. There's a lot of confusion of ideas here. If he is at a higher potential then you must be deeper in the field. If you are on the surface of the Earth and someone is far away, then you are at the lower potential. You would have to either accelerating or in a strong gravitational field. In any case, your local circumstances cannot depend on what someone else is doing.
Also, "zero" gravitational potential is not what you think. This is why it's a good idea to study Newtonian gravity before dealing with GR.

I apologize for stating it backwards, but isn't a clock on the surface of the planet running slower than a clock far away from the surface of the planet?

 

[Staticboson]

Getting back to the question, and based on the comments, is the answer that since the falling is following a geodesic and not changing its speed (not accelerating) that the time dilation is constant?
Thanks

 

[PeroK]

Getting back to the question, and based on the comments, is the answer that since the falling is following a geodesic and not changing its speed (not accelerating) that the time dilation is constant?
Thanks

You're still seriously confusing SR (flat spacetime) with GR (curved spacetime). Changing velocity has no real meaning in curved spacetime. For example, we are on the surface of the Earth and subject to continuous "proper" acceleration "upwards". But, we are not changing velocity (in the reference frame of the Earth). A falling ball, however, has zero proper acceleration but is accelerating in our reference frame.

Which one is changing velocity? The one that with proper acceleration or the one with zero proper acceleration? There is no such ambiguity in SR.

 

[Ibix]

Getting back to the question, and based on the comments, is the answer that since the falling is following a geodesic and not changing its speed (not accelerating) that the time dilation is constant?
Thanks

Time dilation relative to what?

In general, the tick rate of a falling clock as observed by a hovering clock will be changing. This can be seen trivially by considering the infalling clock crossing the event horizon of a black hole, where the tick rate measured is asymptotic to zero.

 

[Staticboson]

You're still seriously confusing SR (flat spacetime) with GR (curved spacetime). Changing velocity has no real meaning in curved spacetime. For example, we are on the surface of the Earth and subject to continuous "proper" acceleration "upwards". But, we are not changing velocity (in the reference frame of the Earth). A falling ball, however, has zero proper acceleration but is accelerating in our reference frame.

Which one is changing velocity? The one that with proper acceleration or the one with zero proper acceleration? There is no such ambiguity in SR.

That was well put, thank you.

 

[pervect]

Getting back to the question, and based on the comments, is the answer that since the falling is following a geodesic and not changing its speed (not accelerating) that the time dilation is constant?
Thanks

Observers are tricky in GR. Some authors suggest not even using them. (Misner, "Precis of General Relativity", for instance.).

One first banishes the idea of an “observer”. This idea aided Einsteinin building special relativity but it is confusing and ambiguous in generalrelativity.

As far as inertial frames in General Relativity goes, they are valid in a small enough region. For instance, if we consider a pair of test particles, each free-falling towards the planet, they'll tend to follow different geodesics and separate. This can be attributed to "tidal forces" in the Newtonian paradigm, in the GR paradigm it's called "geodesic deviation" and it can be mathematically modeled by the Riemann curvature tensor. In fact, one can view the components of the Riemann tensor as just being the tidal forces in many cases.

In some circumstances, such as a sufficiently small region of space-time, one can ignore the effects of curvature - in such circumstances, the problem is one of SR, and the problem becomes one of reciprocal time dilation. A thinks B's clock is slow, B thinks A's clock is slow - who is right? The answer is they both are, because they use different notions of simultaneity. In GR, the notion of the relativity of simultaneity takes on a different form, one of how to express the laws of physics in a matter that is independent of the choice of coordinates.

If we follow Misner's approach, rather than focusing on the observer, and "time dilation", we focus instead on "proper time". In fact, time dilation can be thought of as the ratio of coordinate time to proper time. Coordinates are purely a matter of convention in GR, there are a large number of coordinate systems one might use., and one can get good results with any of them with the right techniques, though these "right techniques" may not be familiar or reconizable without some background in GR, they are not necessarily similar to the Newtonian techniques.

Proper time is physical and independent of the coordinate system. So if one can reformulate one's question in terms of proper time, rather than "time dilation", we can give a better and more physically motivated answser without having to request detailed information about what coordinates are being used. To be clear, proper time is something we can directly calculate with the machinery of GR. And it's also something that we can measure via experiment, use of it sharpens the definition of the problem so that it has a clear-cut answer with a minimum of extraneous information.

 

[Staticboson]

Time dilation relative to what?

In general, the tick rate of a falling clock as observed by a hovering clock will be changing. This can be seen trivially by considering the infalling clock crossing the event horizon of a black hole, where the tick rate measured is asymptotic to zero.

The tick rate as measured by whom? If by the observer, would the tick rate be asymptotic to infinity?

 

[Ibix]

The tick rate as measured by whom? If by the observer, would the tick rate be asymptotic to infinity?

As measured by a hovering clock, the tick rate of an infalling clock is asymptotic to zero as it approaches the horizon. I'm not sure what you are expecting to be asymptotic to infinity. The infalling clock does not see the hovering clock ticking infinitely fast, if that's what you mean.

 

[PeterDonis]

As measured by a hovering clock, the tick rate of an infalling clock is asymptotic to zero as it approaches the horizon.

In view of the OP's confusion over these concepts, I think this needs to be stated much more carefully. The hovering observer cannot directly measure the tick rate of the infalling clock, because they are not co-located. The hovering observer can only directly measure the frequencies and arrival times of light signals coming from the infalling clock. From those signals, the hovering observer can calculate a "tick rate" for the infalling clock which goes to zero asymptotically as the infalling clock approaches the horizon. However, this "tick rate" is not something any observer will directly observe. Certainly the infalling clock itself observes no such thing; it continues to tick away at a rate of one second per second. The "tick rate" the hovering observer calculates is only a coordinate "tick rate" (corresponding to $d\tau / dt$ for the infalling clock in Schwarzschild coordinates). The fact that this "tick rate" goes to zero asymptotically as the infalling clock approaches the horizon is telling you about a problem with the coordinates, not with the infalling clock.

 

[Staticboson]

Observers are tricky in GR. Some authors suggest not even using them. (Misner, "Precis of General Relativity", for instance.).
As far as inertial frames in General Relativity goes, they are valid in a small enough region. For instance, if we consider a pair of test particles, each free-falling towards the planet, they'll tend to follow different geodesics and separate. This can be attributed to "tidal forces" in the Newtonian paradigm, in the GR paradigm it's called "geodesic deviation" and it can be mathematically modeled by the Riemann curvature tensor. In fact, one can view the components of the Riemann tensor as just being the tidal forces in many cases.

In some circumstances, such as a sufficiently small region of space-time, one can ignore the effects of curvature - in such circumstances, the problem is one of SR, and the problem becomes one of reciprocal time dilation. A thinks B's clock is slow, B thinks A's clock is slow - who is right? The answer is they both are, because they use different notions of simultaneity. In GR, the notion of the relativity of simultaneity takes on a different form, one of how to express the laws of physics in a matter that is independent of the choice of coordinates.

If we follow Misner's approach, rather than focusing on the observer, and "time dilation", we focus instead on "proper time". In fact, time dilation can be thought of as the ratio of coordinate time to proper time. Coordinates are purely a matter of convention in GR, there are a large number of coordinate systems one might use., and one can get good results with any of them with the right techniques, though these "right techniques" may not be familiar or reconizable without some background in GR, they are not necessarily similar to the Newtonian techniques.

Proper time is physical and independent of the coordinate system. So if one can reformulate one's question in terms of proper time, rather than "time dilation", we can give a better and more physically motivated answser without having to request detailed information about what coordinates are being used. To be clear, proper time is something we can directly calculate with the machinery of GR. And it's also something that we can measure via experiment, use of it sharpens the definition of the problem so that it has a clear-cut answer with a minimum of extraneous information.

I appreciate that information.

I'm not sure how to formulate the question in terms of proper time, is there a way to calculate what the clock of the falling object is doing in comparison to the clock of the hovering observer? And if so, how would the proper time tick intervals of the falling object compare to the time tick intervals of the hovering observer? Would they slow down as the falling observer approaches the planet?

And if the answer is yes, if instead of a planet we have the EH of a BH, would the proper time intervals of the falling object, from the hovering observer's perspective, approach infinity as the object reaches the EH?

 

[Staticboson]

In view of the OP's confusion over these concepts, I think this needs to be stated much more carefully. The hovering observer cannot directly measure the tick rate of the infalling clock, because they are not co-located. The hovering observer can only directly measure the frequencies and arrival times of light signals coming from the infalling clock. From those signals, the hovering observer can calculate a "tick rate" for the infalling clock which goes to zero asymptotically as the infalling clock approaches the horizon. However, this "tick rate" is not something any observer will directly observe. Certainly the infalling clock itself observes no such thing; it continues to tick away at a rate of one second per second. The "tick rate" the hovering observer calculates is only a coordinate "tick rate" (corresponding to $d\tau / dt$ for the infalling clock in Schwarzschild coordinates). The fact that this "tick rate" goes to zero asymptotically as the infalling clock approaches the horizon is telling you about a problem with the coordinates, not with the infalling clock.

I read this after posting my last response, I believe it answers my question.

 

[PeterDonis]

is there a way to calculate what the clock of the falling object is doing in comparison to the clock of the hovering observer?

There is no unique way to do this because there is no unique way to make the comparison. Any such comparison requires a choice of coordinates, and different coordinate choices will lead to different comparisons.

To put this another way, the underlying concept that you are implicitly relying on, that it makes sense to ask what is happening "now" at some distant location, is no longer a well-defined concept in relativity.

 

[pervect]

I appreciate that information.

I'm not sure how to formulate the question in terms of proper time, is there a way to calculate what the clock of the falling object is doing in comparison to the clock of the hovering observer? And if so, how would the proper time tick intervals of the falling object compare to the time tick intervals of the hovering observer? Would they slow down as the falling observer approaches the planet?

And if the answer is yes, if instead of a planet we have the EH of a BH, would the proper time intervals of the falling object, from the hovering observer's perspective, approach infinity as the object reaches the EH?

Specifying a hovering observer still requires one to much about with coordinates, but it is almost enough to specify which coordinates one needs to use.

The reason it's only almost enough is that hovering observer's clocks, if they are not rate-adjusted, do not stay synchronized, they run at different rates depending on how far away they are from the mass. This is usually called "gravitational time dilation".

So, we're almost ready to answer your question. All we need is more background on the hovering observer. Are they using a "rate adjusted" clock, to keep all the different hovering observer in synch? If so, where is the "reference clock", to which all clocks are rate-adjusted.

Alternatively, you might be asking about a non-rate adjusted clock. Note that if you compare a falling clock to a hovering clock at the same loaction , the time dilation factors will depend only on the relative velocity as measured between the two clocks at their common, shared location.

 

[Staticboson]

The reason it's only almost enough is that hovering observer's clocks, if they are not rate-adjusted, do not stay synchronized, they run at different rates depending on how far away they are from the mass. This is usually called "gravitational time dilation".

It is this gravitational time dilation that I was expecting to affect the clock of the falling object in comparison to the hovering observer. As the objects falls deeper into the field its clock tick rate becomes slower in relation to the observer's.

So, we're almost ready to answer your question. All we need is more background on the hovering observer. Are they using a "rate adjusted" clock, to keep all the different hovering observer in synch? If so, where is the "reference clock", to which all clocks are rate-adjusted.

Understood, the field extends to infinity so there is some degree of gravitational time dilation effect at any distance from the gravity source. So I'm assuming no clock tick rate absolutes, just a comparison of what happens to the clock tick rate to the falling object in comparison to the hovering observer.

Alternatively, you might be asking about a non-rate adjusted clock. Note that if you compare a falling clock to a hovering clock at the same loaction , the time dilation factors will depend only on the relative velocity as measured between the two clocks at their common, shared location.

Understood, there is a lorentz gamma that differentiates the tick rate of the falling object's clock from the observer's at the moment they share the same location. Away from the effect of a gravitational field this difference in tick rate would be constant and depending only on the relative velocity between object and observer, which would also be constant.

My question centered around the change on the tick rate as the object falls deeper into the gravitational field, in the following way: Imagine a second observer, hovering much deeper inside the field (closer to the planet). at the moment the falling object passes him, will the difference in the clock rate between the falling object and the second observer be due only to the Lorentz time dilation from to the object's velocity?

Basically I'm undestanding from the responses that as the falling object passes a chain of hovering clocks located between the first observer and the surface of the planet, the only difference in the tick rate between the falling object and each clock will be due to gamma at the moment the object passes each clock.

 

[PeroK]

Basically I'm undestanding from the responses that as the falling object passes a chain of hovering clocks located between the first observer and the surface of the planet, the only difference in the tick rate between the falling object and each clock will be due to gamma at the moment the object passes each clock.

Yes, because inertial reference frames and the laws of SR only apply locally in curved spacetime.

 

[Staticboson]

Yes, because inertial reference frames and the laws of SR only apply locally in curved spacetime.

Thank you, that's the answer I was looking for.

 

[Staticboson]

To put this another way, the underlying concept that you are implicitly relying on, that it makes sense to ask what is happening "now" at some distant location, is no longer a well-defined concept in relativity.

This is the very thing I'm trying to grasp as it applies to the falling object in a changing field. I have some understanding of simultaneity relationships between inertial frames of reference and also now understand why those relationships can't be used in curved spacetime, where the concept of an inertial frame of reference is undefinable. The thread has been very helpul.

 

[pervect]

There is no unique way to do this because there is no unique way to make the comparison. Any such comparison requires a choice of coordinates, and different coordinate choices will lead to different comparisons.

To put this another way, the underlying concept that you are implicitly relying on, that it makes sense to ask what is happening "now" at some distant location, is no longer a well-defined concept in relativity.

This is the very thing I'm trying to grasp as it applies to the falling object in a changing field. I have some understanding of simultaneity relationships between inertial frames of reference and also now understand why those relationships can't be used in curved spacetime, where the concept of an inertial frame of reference is undefinable. The thread has been very helpul.

Curved space-time isn't that much different than flat space-time as far as simultaneity conventions go. Simultaneity conventions don't arise from physics - they arise from one's viewpoint. They aren't really necessary to do physics.

You can think of a space-filling array of observers in GR as one possible way to define a curved space-time equivalent of a frame of reference. Picking out which worldline a space-time event lies on defines "where" the event occurs. If two events lie on the same worldline, they happen "at the same point in space". Picking out a worldline defines space and position. Picking out a particular event on a worldline specifies the time.

A full discussion of such time-like congruences gets rather technical, see for instance https://en.wikipedia.org/w/index.php?title=Congruence_(general_relativity)&oldid=925963622. This may be more adnvaced than you want.

The main thing you have to beware of is something you haven't asked about and might not be interested in. This is that some knotty issues regarding simultaneity in "rotating frames". With a non-rotating congruence of worldlines (formally we'd say the vorticity is zero), there is a unique hypersurface orthogonal to the congruence that defines a notion of simultaneity. If the worldlines are rotating (have a non-zero vorticity), things become much more complicated, and you're likely to run into issues you are not equipped to deal with. IT's a bit of a digression, I feel I have to warn you about the issue, but I don't want to spend a lot of time on it if it's not of interest.

This issue with rotation arises both in the flat time of special relativity as well as in the curved space-time of GR. So it's not really a GR issue.

 

[Dale]

Yes, but it also states through the principle of equivalence that an accelerating object will experience clock slowdown equivalent to the clock slowdown within a gravitational field.

Actually, you have this a little backwards. The special relativity experiments with accelerating clocks show conclusively that the only slowdown is related to velocity and that there is no additional time dilation related to acceleration.

Correctly applying the equivalence principle would therefore indicate that time dilation in a gravitational field would be unrelated to the gravitational acceleration. In fact, it is found to be related to gravitational potential.

 

[PeterDonis]

Specifying a hovering observer still requires one to much about with coordinates

Not in a stationary spacetime; in a stationary spacetime a "hovering" observer can be defined in a coordinate-independent way, as an observer whose worldline is an integral curve of the timelike Killing vector field.