Inertial frames in changing gravitational fields

[Staticboson]

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.

If by gravitational acceleration you refer to the acceleration that an object experiences under the influence of a gravitational field, I understand, this is unrelated to gravitational time dilation.

What I believe to be related to gravitational time dilation is a clock inside an elevator that is being accelerated.

 

[Staticboson]

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.

Very nice... I'm going to spend a little time on the subject of congruence before I adventure into rotating frames. Considering that gravity disturbances propagate at a finite velocity (c?) I can see how a rotating gravity source
would form a spacetime vortex around the rotating object. I can imagine the mathematics describing the movement of a falling object into a rotating gravity field would be daunting, so I'm going to try and gain some familiarly with non-rotating fields first.

 

[Dale]

What I believe to be related to gravitational time dilation is a clock inside an elevator that is being accelerated.

Yes. That is another way that you can derive the dependence on the gravitational potential instead of the gravitational acceleration.

 

[PeroK]

Yes. That is another way that you can derive the dependence on the gravitational potential instead of the gravitational acceleration.

But, to be clear, an accelerating clock is not measured to have two components to time dilation: one for its velocity and one for its acceleration (equivalent to additionally being in a gravitational field).

Clocks inside an accelerating elevator are measured (from the accelerating reference frame of the elevator) to be time-dilated depending on their height inside the elevator). As measured from an inertial reference frame (outside the elevator if you like) the clocks have precisely the time dilation associated with their instantaenous velocity and no additonal "gravitational" component.

 

[Staticboson]

But, to be clear, an accelerating clock is not measured to have two components to time dilation: one for its velocity and one for its acceleration (equivalent to additionally being in a gravitational field).

Clocks inside an accelerating elevator are measured (from the accelerating reference frame of the elevator) to be time-dilated depending on their height inside the elevator). As measured from an inertial reference frame (outside the elevator if you like) the clocks have precisely the time dilation associated with their instantaenous velocity and no additonal "gravitational" component.

Based on this I have to deduce that an infinitesimally small clock would not experience gravitational time dilation when accelerating, as compared to any inertial observer.

 

[Ibix]

Based on this I have to deduce that an infinitesimally small clock would not experience gravitational time dilation when accelerating, as compared to any inertial observer.

No. The clock will have the same time dilation factor as any other clock moving at the same instantaneous speed. Otherwise you define an absolute state of rest by comparing time dilation factors between different clocks.

The point is that two nearby clocks who are accelerating won't typically agree that they are synchronised, nor that they are ticking at the same rate, even if they undergo the same proper acceleration. An inertial frame will conclude that they tick at the same rate if it sees the clocks as doing the same speed. The clocks' point of view can be understood in terms of the relativity of simultaneity from the inertial frame, or via the equivalence principle as gravitational redshift from the accelerating frame.

 

[PeroK]

Based on this I have to deduce that an infinitesimally small clock would not experience gravitational time dilation when accelerating, as compared to any inertial observer.

But, this is based on the misapprehension that a clock "experiences" time dilation. It does not.

Note that in the velocity-based case there is no experiment that could attribute an absolute "time dilation" to a clock. The time dilation is purely relative to the IRF in which measurements are made.

The gravitational case is similar. There is no local experiment that could attribute an absolute time dilation to a clock. The time dilation is relative to other clocks at different potential in the gravitational field.

To take a precise scenario that you may be alluding to. We have a clock at rest (in Schwarzschild coordinates) in the gravitational field of a star. If an infalling observer passes the clock, then in the local inertial frame of that observer the clock is moving and has a time-dilation associated with the (local) relative velocity of the two (based on the local application of SR).

There is no gravitational component to the time dilation in this scenario because there is no gravity in SR; if there were gravitational considerations then SR would not apply and there would be no local IRF(!).

Moreover, if you consider the scenario in Schwarzschild coordinates, then the clock and the observer are at the same radius when they are local to each other. So, any gravitational time dilation - relative to a distant clock - is the same for both when they pass each other.

 

[Staticboson]

This thread has been very, very helpful, much appreciated.

 

[jartsa]

Based on this I have to deduce that an infinitesimally small clock would not experience gravitational time dilation when accelerating, as compared to any inertial observer.

Well, it's really easy to find out if some clock is affected by gravitational time dilation: If there is a large mass nearby , then it is affected. Otherwise not.

If some guy in some windowless chamber can not figure out whether there is a large mass nearby, then figuring out whether his clock is affected by gravitational time dilation is impossible for him.So, is a clock that has been dropped from the tower of Pisa, and is now in free fall, affected by gravitational time dilation?

 

[jbriggs444]

Well, it's really easy to find out if some clock is affected by gravitational time dilation: If there is a large mass nearby , then it is affected.

Clocks are not affected by gravitational time dilation. Clock comparisons are affected by gravitational time dilation.

 

[Nugatory]

Well, it's really easy to find out if some clock is affected by gravitational time dilation: If there is a large mass nearby

That part is easy in principle: just look for tidal effects. Put two particles in freefall near you (hold one in each hand, drop them together); if the distance between them changes, you're seeing tidal effects and you know that you're in a gravitational field instead of accelerating through empty space.

then it is affected. Otherwise not.

The English language is not a precision instrument, so saying "affected by time dilation" is not exactly incorrect... it's more so imprecise that there's no unambiguous way of saying that it's wrong or right.

So, is a clock that has been dropped from the tower of Pisa, and is now in free fall, affected by gravitational time dilation?

Dunno... What do you mean by "affected by time dilation"? A clear answer to that question is also a pretty good hint that there's a better way of thinking about it:

Clocks are not affected by gravitational time dilation. Clock comparisons are affected by gravitational time dilation.

 

[bhobba]

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).

Exactly. The question is not really well posed . Generally inertial frames only exist locally when gravity is present. This is seen more easily by the definition of an inertial frame given by Landau where an inertial frame is one where all points are the same, all directions are the same, and all instants of time are the same as far as the laws of physics are concerned. Now imagine curved space-time then more or less by the definition of curvature that definition can't be true - except locally ie in an area so small the change in going to different points, directions, and instants of time is so small it can be neglected ie using the word pure mathematicians get concerned about - infinitesimal.

This actually leads to a sneaky way of deriving GR. You start with an inertial frame, then by transforming to general coordinates instead of ds^2 = nij dxi dxj you get ds^2 = gij dxi dxj where gij is called the metric and determines the motion of a free particle in those coordinates. In a sense it acts like a field. Usually fields are described by Lagrangian's so we want to see if we can find the Lagrangian of this field. There is a hard to prove, deep, but very useful theorem that helps here called Lovelocks Theorem:
https://en.wikipedia.org/wiki/Lovelock's_theorem

So the only equations consistent with this view is Einsteins Field Equations, which show that space-time is curved. But we assumed, by starting with an inertial frame - flat space-time. This shows that assuming the metric acts like a field with its own Lagrangian is inconsistent with the assumption of an inertial frame.

Sneaky hey.

Added later - fixed notation because of comments

Thanks
Bill

 

[Orodruin]

This actually leads to a sneaky way of deriving GR. You start with an inertial frame, then by transforming to general coordinates instead of ds^2 = Nuv dxi dxj you get ds^2 = Guv dxi dxj where Guv is called the metric and determines the motion of a free particle in those coordinates.

I am not sure I follow this part. In which sense does this derive GR. All you will get is Minkowski space in curvilinear coordinates. For GR you need curved spacetime with the metric satisfying the Einstein field equations.

Side note: I would abe cautious about using G for the metric so that it is not confused with the Einstein tensor.

 

[PeroK]

Now imagine curved space-time then more or less by the definition of curvature that definition can't be true - except locally ie in an area so small the change in going to different points, directions, and instants of time is so small it can be neglected ie using the word pure mathematicians get concerned about - infinitesimal.

The extent of a local inertial frame is determined by the accuracy to which you want to model something. There are no exact inertial reference frames - in the same way that there are no non-relativistic velocities. An inertial reference frame, therefore, covers a patch of spacetime that is small enough for the curvature to be neglected. This patch is finite rather than infinitesimal.

 

[Staticboson]

The question is not really well posed

I learned from this thread how referring to the treatment of inertial frames in curved spacetime presents a contradiction of terms, not much different than asking "How does still air behave in a hurricane".

 

[Nugatory]

I learned from this thread how referring to the treatment of inertial frames in curved spacetime presents a contradiction of terms, not much different than asking "How does still air behave in a hurricane".

A big and important difference: the behavior of still air is never a good approximation for the behavior of hurricane air, but an inertial frame is an excellent approximation over any small region of spacetime. This is why in general relativity we can speak of local inertial frames, which still work the way we expect, instead of global inertial frames, which do not.
(Although natural language being what it is, people will omit the word “local” when they expect that their audience will understand from the context that it is intended. This leads to great confusion when non-specialists are listening to specialists without understanding this important bit of context).

You could think of general relativity as the way that we relate the local inertial frames that work in one place to the ones that work in another place.

 

[Orodruin]

the behavior of still air is never a good approximation for the behavior of hurricane air,

I would say this depends on the frame and volume extent of the region being described. I would actually say it is a pretty good analogy.

 

[Nugatory]

I would say this depends on the frame and volume extent of the region being described. I would actually say it is a pretty good analogy.

Ah, yes, I wasn’t considering a dust mote moving with the wind... for which the still air approximation is excellent.

 

[jartsa]

Dunno... What do you mean by "affected by time dilation"? A clear answer to that question is also a pretty good hint that there's a better way of thinking about it:

I meant that clocks are slowed down by gravitational time dilation. I forgot that that is wrong, at least according to some people.

Let me try to say it more correctly now. So, large masses are surrounded by a field of curved space-time, which causes observers inside that field to see clocks running extra fast, when they look out through the field.

So when I said that absence of large masses means the absence of gravitational time dilation, that was quite correct. No mass, no field, no gravitational time dilation. Well, except that some distant observer may be looking at a clock that we have placed far from all masses, and that observer may be surrounded by a gravity field. So maybe that part was wrong too.

When I said that next to a large mass there has to be gravitational time dilation, that was wrong. Because observers inside the field do not have to look through the field at a clock, if said clock is right next to them.

I just wonder why an astronaut visiting a planet orbiting a gigantic black hole observes that his buddies have aged a lot, when he gets back up from the planet. He is not looking at those buddies through the field when he is standing next to them.

 

[jbriggs444]

when he gets back up from the planet

Look at @Orodruin's signature. Relativity of simultaneity.

 

[PeterDonis]

I just wonder why an astronaut visiting a planet orbiting a gigantic black hole observes that his buddies have aged a lot, when he gets back up from the planet.

That is not time dilation; that is differential aging. It is accounted for by the simple fact that the astronaut and is buddies took different paths through spacetime, and the former's path was much shorter than the latter's. In other words, it's just geometry. No "slowing down of clocks" is involved--in fact the calculation of differential aging requires that the astronaut's clock ticks at the same "rate" as his buddies', one second per second. The only difference is in the length in seconds of his path through spacetime as compared to theirs.

It's no different from two cars taking two different routes to get from point A to point B, where one route is much shorter than the other. You don't say one car's odometer "slowed down" compared to the other's; both odometers measure one mile per mile. There are just fewer miles in one route than in the other.

 

[bhobba]

I am not sure I follow this part. In which sense does this derive GR. All you will get is Minkowski space in curvilinear coordinates.

That's the point - you are led to a contradiction. You start with flat space-time and end up with the the Lagrangian of GR. Add a matter term to the Lagrangian, vary it and you get the EFE's with the stress energy tensor on one side. Many of the solutions of which are curved space-time. The flat space-time you started with can only be true locally.

Thanks
Bill

 

[Orodruin]

That's the point - you are led to a contradiction.

I do not see where the contradiction is. There is no a priori requirement that the metric is a dynamic field. You just have it expressed in a different set of coordinates. This is no stranger than polar coordinates in the two-dimensional Euclidean plane. Lovelock’s theorem apply only if you assume that the metric is dynamic, the action at most contains second derivatives of the metric, and is linear in those second derivatives. In fact, there are studies of so-called f(R)-theories, where the action is a (not necessarily linear) function of the Ricci scalar.

The flat space-time you started with can only be true locally.

This is only true if you assume all of the above and that the stress-energy tensor is non-zero. Minkowski space is a perfectly fine vacuum solution to the EFEs. However, there is no a priori requirement to make the metric dynamic just because it has a different form in curvilinear coordinates.

 

[bhobba]

This is only true if you assume all of the above and that the stress-energy tensor is non-zero.

You are of course correct. If I gave the impression the assumptions I made are a-priori correct then I explained it badly. Everything you mention is an assumption I made - but a suggestive one to try and get across the idea if you start with an inertial frame and follow a reasonable - but of course assumption laden - line of thought things to not work out correctly - you are led to assuming what you started with was only locally inertial.

I think since it is causing confusion I would have been better giving a different argument based on linearised gravity in an inertial frame as found in Gravitation and Spacetime by O'hanian. Briefly using EM as a guide one can construct linearised gravity in an inertial frame. Then you can show in this theory space-time acts as though it has an infinitesimal curvature. The strange thing is if you assume space-time can have more than just an infinitesimal curvature, which seems quite treasonable, you immediately end up with the full theory of GR. The details can be found in chapter 7 of O'hanians book.

Thanks
Bill

 

[Orodruin]

If I gave the impression the assumptions I made are a-priori correct then I explained it badly.

I think the key word that gave me this impression was ”deriving”. I agree of course that it is a way of making GR plausible. When I teach GR I usually go by the EH action being the simplest non-trivial Lagrangian you can write down and ”oh look! you reproduce Newtonian gravity in the weak field non-relativistic limit and make a ton of other predictions that have been verified to high precision”.

 

[bhobba]

When I teach GR I usually go by the EH action being the simplest non-trivial Lagrangian you can write down and ”oh look! you reproduce Newtonian gravity in the weak field non-relativistic limit and make a ton of other predictions that have been verified to high precision”.

That is an excellent way of doing it.

Thanks
Bill

 

[Android Neox]

The gravitational time dilation at any point in space is determined by the gravitational potential so, as a clock (or observer) moves downward, gravitational time dilation will increase and time will slow. The freely falling observer will have increasing time dilation (slower and slower time) both due to increasing speed (from the perspective of an observer stationary with respect to the massive body) and due to moving to space at lower gravitational potential.

 

[bhobba]

The freely falling observer will have increasing time dilation (slower and slower time) both due to increasing speed (from the perspective of an observer stationary with respect to the massive body) and due to moving to space at lower gravitational potential.

Is there a gravitational field for a freely falling observer?

I think I see what you are saying, but maybe it could be expressed a bit clearer.

Thanks
Bill

 

[Android Neox]

Is there a gravitational field for a freely falling observer?

I think I see what you are saying, but maybe it could be expressed a bit clearer.

Thanks
Bill

I should have been more clear.

No, all observations within the falling observer's frame and local to the observer (e.g. within a falling elevator) would not show any acceleration, gravitational or otherwise. Equivalence Principle holds. However, that's not the determining factor.

Since all correct models & thought experiments are correct (or, at least, not wrong) under all correct interpretations... a single correct thought experiment will yield observations that are logically consistent with those of all observers in all frames.

Thought experiment: Consider a platform, stationary with respect to a Schwarzschild black hole, at some finite coordinate distance above the Schwarzschild radius, $R_s$. On the platform is a winch that can lower a mirror. There is a laser that shines a beam onto the mirror. There is also a device that detects reflected laser light.

At any mirror position along the light beam, it will have some blueshift which follows, exactly, the difference in gravitational potential between the platform and the mirror. The change in time rate exactly follows this, too. If the frequency is doubled, the time rate is halved. That value of time dilation is an attribute of that frame. If the mirror is held stationary with respect to the black hole then that is the frame in which time passes at the maximum rate, for that point in space.

A freely falling observer will also have two other effects that will change the appearance of the light beam: Doppler and relativistic effects of motion with respect to the source.

If it's not clear that the change in laser beam frequency must be the exact inverse of the change in time rate, consider that every light wave cycle that reaches the mirror was first generated at the laser. The light beam is a causal sequence just as falling dominoes are, and so this sequence is the same for all observers in all frames. So, if a lower observer is seeing these wave cycles arriving at twice the rate, they must be measuring time half as fast.

Because the blueshift down to an event horizon, from any point above the EH, is infinite, by the time the front of the beam reaches the event horizon, the beam will be infinitely blueshifted. The beam will contain infinitely many wave cycles. This means that, before the front of the beam can reach the event horizon, infinite time must pass for the light source on the platform. And, since the rope supporting the mirror has been payed out at a constant rate, an infinite amount of rope will be payed out before the front of the light beam could reach the event horizon. Since the mirror is lowered slower than c, it will take event longer than infinite time for the mirror to arrive.

Idealized: All equipment is of infinite strength, negligible mass, and otherwise idealized.

 

[PeterDonis]

at some finite coordinate distance above the Schwarzschild radius

In what coordinates? You appear to be implicitly using Schwarzschild coordinates in the patch exterior to the horizon, but you should not leave that implicit. You should state it explicitly. That way you will be forced to consider the limitations of the coordinates you are using; see further comments below.

At any mirror position along the light beam, it will have some blueshift

Measured by what?

If the mirror is held stationary with respect to the black hole then that is the frame in which time passes at the maximum rate, for that point in space.

"For that point in space" is ambiguous. If you mean "for objects held at that point in space" (and remember, to properly define "point in space" in a coordinate-independent fashion is not trivial, though in this case it can be done), then your statement is vacuous, since only one worldline--path through spacetime--is possible for any such object, and so only one possible rate of time passage is possible.

If, OTOH, you mean "for any objects whose worldlines pass through two selected events, chosen to be at that point in space", then your statement is false. If an observer held stationary along with the mirror throws a clock up in the air vertically, then catches it at some later time, the elapsed time on the clock he threw up will be greater than the elapsed time on the clock he carries with him.

If it's not clear that the change in laser beam frequency must be the exact inverse of the change in time rate

If you mean that the observed redshift/blueshift is exactly proportional to the ratio of gravitational potentials, of course it is. I don't see why you're making such a big deal out of it, since this is a well-known property of stationary observers in Schwarzschild spacetime and has been for decades.

Because the blueshift down to an event horizon, from any point above the EH, is infinite,

Wrong. There are no stationary observers at the EH, nor is it possible to hold the mirror stationary there. So there is no such thing as "the blueshift down to an event horizon", since it's impossible for an observer to exist who could measure it.

before the front of the beam can reach the event horizon, infinite time must pass for the light source on the platform

Wrong. The coordinates you appear to be implicitly using are singular at the horizon, so "infinite time must pass" is not correct because it is attempting to apply coordinates at a coordinate singularity.

since the rope supporting the mirror has been payed out at a constant rate, an infinite amount of rope will be payed out before the front of the light beam could reach the event horizon

Wrong. The proper distance from your laser platform to the mirror, as the mirror gets lowered, approaches a finite limit as the mirror approaches the horizon. (We cannot directly measure such a distance with the mirror at the horizon, since the mirror can't be stationary there, but we can realize the limiting process I have just described by letting the mirror get closer and closer and measuring how the distance varies.)

It looks like you need to read the Insights series on the Schwarzschild geometry:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

You are making a number of elementary errors that are common with people who are not sufficiently familiar with the actual properties of this spacetime geometry.

 

[Android Neox]

It looks like you need to read the Insights series on the Schwarzschild geometry:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

You are making a number of elementary errors that are common with people who are not sufficiently familiar with the actual properties of this spacetime geometry.

Thanks for the response. I'll do that.
I'll probably be back, later, with some new misinterpretations.

 

[bhobba]

I'll probably be back, later, with some new misinterpretations.

Don't worry - that's how you learn.

Thanks
Bill