Cause of time dilation when net gravity is zero

[DaveC426913]

I'm armchair knowledgeable about GR, but I've come up against something I can't eloquently explain (to someone who asserts otherwise and wants an unassailable answer).

What property results in time dilation even when net gravity is zero?The net gravity at the centre of the Earth is zero, but, as far as we expect, time dilation is at maximum.
(left side of green line in diagram).

Is it the gravitational potential?

 

[Ibix]

Is it the gravitational potential?

Yes.

Gravitational time dilation and gravitational redshift are the same phenomenon. Imagine watching a clock from orbit - if it ticked at the same rate as yours, light coming up from it couldn't be redshifted, since you could count oscillations of the emitter (the clock) and oscillations received and come up with a different number. And gravitational redshift is what photons do instead of slowing down as they climb out of a gravity well. If they didn't, you could drop a mass, convert it and its kinetic energy into radiation, let the radiation climb out of the gravity well without losing energy, and bingo! Free energy.

So redshift is required for conservation of energy and gravitational time dilation is gravitational redshift.

 

[1977ub]

Perhaps there is something analogous with charge? Someone in a chamber feeling no charges but being surrounded by a charged sphere puts potential energy between them and the observer away from masses & charges ?

 

[DaveC426913]

Yes.

Gravitational time dilation and gravitational redshift are the same phenomenon. Imagine watching a clock from orbit - if it ticked at the same rate as yours, light coming up from it couldn't be redshifted, since you could count oscillations of the emitter (the clock) and oscillations received and come up with a different number. And gravitational redshift is what photons do instead of slowing down as they climb out of a gravity well. If they didn't, you could drop a mass, convert it and its kinetic energy into radiation, let the radiation climb out of the gravity well without losing energy, and bingo! Free energy.

So redshift is required for conservation of energy and gravitational time dilation is gravitational redshift.

OK. Apparently I lied about being armchair knowledgeable about GR. All your words are words I know, I've just not strung them together quite that way before.

"Gravitational time dilation and gravitational redshift are the same phenomenon."

 

[Ibix]

"Gravitational time dilation and gravitational redshift are the same phenomenon."

So the textbooks say. And you can see it like this: attach a 1MHz clock to a 1MHz radio source. Lower it into a gravity well on the end of a string, leave it at the bottom for a while, and pull it back up. The number of wavecrests you receive must be the same as the number of ticks on the clock during the mission, and the twin-paradox-like setup gives an unambiguous start and stop to the counting. The number of received crests is low due to gravitational redshift and the number of ticks observed is low due to gravitational time dilation - but they must always be the same number however you modify the path. So you can't really describe the two things as separate phenomena.

 

[A.T.]

What property results in time dilation even when net gravity is zero?

You can turn this around, and say that the gradient of time dilation results in gravity.

 

[stevendaryl]

I don't know if this is helpful, or confusing, but gravitational time dilation is an artifact of your choice of coordinate system. The definition of gravitational time dilation is:
$$\frac{\delta \tau}{\delta t}$$
How much time $\delta \tau$ does a clock advance when coordinate time advances by $\delta t$? Obviously, the answer depends on how you set up your coordinate system. You can always choose a coordinate system so that the time dilation factor is 1 at any particular point in space-time---just use that clock as your standard for time for your coordinate system. So there is a sense in which time dilation isn't real/objective.

But what is real and objective is the comparison of different clocks. If you have two clocks, $A$ and $B$, and you pick two events (points in spacetime) $e_1$ and $e_2$, and you let $A$ take one path through spacetime to get from $e_1$ to $e_2$, and let $B$ take a different path, you can compare their two elapsed times: $\tau_A$ and $\tau_B$. For example, you can start $A$ and $B$ on the surface of the Earth, take $A$ to the center of the Earth, and wait a year, then bring it back up to rejoin clock $B$. Then it's objectively true, independent of your choice of coordinate systems, that $\tau_A$ will be different than $\tau_B$. So even though the quantity $\frac{\delta \tau_A}{\delta t}$ is coordinate dependent, if you integrate it:

$$\tau_A = \int_{t_1}^{t_2} \frac{d\tau_A}{dt} dt$$

you will get an answer that is independent of your choice of coordinate system.

 

[A.T.]

But what is real and objective is the comparison of different clocks.

Similarly, the potential value at some point is arbitrary, only the difference in potential between two points is objective.

 

[PeroK]

OK. Apparently I lied about being armchair knowledgeable about GR.

I thought to be armchair knowledgeable, you only have to be the most knowledgeable person in your armchair. So, you're probably all right there!

 

[stevendaryl]

Similarly, the potential value at some point is arbitrary, only the difference in potential between two points is objective.

I would say that even that is not objective in General Relativity. It is in Newtonian physics, but not GR. In GR, you have to compare two paths with the same endpoints to get something objective.

 

[Janus]

OK. Apparently I lied about being armchair knowledgeable about GR. All your words are words I know, I've just not strung them together quite that way before.

"Gravitational time dilation and gravitational redshift are the same phenomenon."

Consider two clocks in at different heights in a gravity well. radio signals are traveling between them so that they can read each other's ticks. We will use a scheme where each clock shifts frequency on alternating second1 . One second at one frequency and then one second at the other. If we make frequency 1at 500 khz and frequency 2 at1 mhz, then each second also amounts to the transmission of 500,000 cycles or 1,000,000 cycles transmitted.

If you are at the lower clock, the signals you receive from the upper clock are shifted to a higher frequency, For instance, 500.001 khz and 1.000002 mhz . This means that you will receive the 500,000 or 1,000,000 cycles representing 1 sec for the upper clock in less than one sec. Since this is a continuous signal, this means that you will register the upper clock as accumulating seconds faster than your.
If you are at the upper clock the signals are shifted to to a lower frequency, it take more than one second to get one seconds worth of cycles from the lower clock, and it accumulates seconds slower than yours.
If you stop both clocks and bring them together, Each will still show the time that was observed on them before they were stopped.

Unlike The situation wihen there's a difference in velocity, when you have to factor out that component due to changing distance and propagation times from the observed frequency shift in order to get the time dilation, there is no change in propagation time with gravitational red shift, so what you see is what you get in terms of time dilation.

 

[PeterDonis]

gravitational time dilation is an artifact of your choice of coordinate system

It depends on what you mean by "gravitational time dilation". Many people (including me) use it to mean this:

what is real and objective is the comparison of different clocks.

However, this comparison is not quite as limited as you say here:

In GR, you have to compare two paths with the same endpoints to get something objective.

There is another way to compare paths: if both paths are integral curves of a timelike Killing vector field, and that KVF is hypersurface orthogonal. Then you can compare segments of the paths that both have endpoints on the same pair of hypersurfaces orthogonal to the KVF. This is the kind of comparison @Janus is making in post #11, and the kind of comparison that is usually implied when "gravitational time dilation" is being discussed.

 

[stevendaryl]

There is another way to compare paths: if both paths are integral curves of a timelike Killing vector field, and that KVF is hypersurface orthogonal. Then you can compare segments of the paths that both have endpoints on the same pair of hypersurfaces orthogonal to the KVF. This is the kind of comparison @Janus is making in post #11, and the kind of comparison that is usually implied when "gravitational time dilation" is being discussed.

Doesn't that just mean that in the case where there is a timelike Killing vector field, there is a most natural choice for a time coordinate (up to scaling, I guess), so you can compare two clocks using $\frac{d\tau}{dt}$ for that coordinate?

 

[PAllen]

I would say that even that is not objective in General Relativity. It is in Newtonian physics, but not GR. In GR, you have to compare two paths with the same endpoints to get something objective.

Two clocks sending periodic signals to each other produces a result that must be independent of coordinates used to compute it.

 

[PeterDonis]

Doesn't that just mean that in the case where there is a timelike Killing vector field, there is a most natural choice for a time coordinate (up to scaling, I guess), so you can compare two clocks using $\frac{d\tau}{dt}$ for that coordinate?

Choosing the natural time coordinate certainly makes it easier to compute the comparison, yes. But the KVF is a coordinate-independent feature of the spacetime, as are the set of spacelike hypersurfaces orthogonal to it (if the KVF is hypersurface orthogonal). So I would argue that the comparison done in this way is not entirely dependent on your choice of coordinates; it makes use of coordinate-independent features of the spacetime. The remaining element of choice is that you have to choose to interpret events lying on the same orthogonal spacelike hypersurface as happening at the same time, i.e., you have to adopt the simultaneity convention that uses those hypersurfaces. For observers who are static, i.e., whose worldlines are integral curves of the timelike KVF, this is the "natural" simultaneity convention.

 

[DaveC426913]

I will go through each of the replies slowly to absorb them, but the impression I'm getting is akin to this (and the more I think about it, the more obvious it becomes)

Time dilation is not A Thing That Exists at the centre of the Earth. Time dilation is - like other relativistic phenomena - a relative effect dependent on two reference points separated by a distance.

One does not experience time dilation - one only ever observes it occurring in a different reference frame.

Time dilation is an effect of the difference in states between two points.

And, for GR, what stands between them is a curvature of space. GR time dilation occurs when an observer on one side (say, Earth orbit) of a curvature of space (Earth's well) observes something on the other side (the centre of Earth). And the direction - whether they're looking down into it or up out of it - is what determines the "bias" of time dilation.

(And GR time dilation is biased - it's not subjective, like SR dilation is. The two observers are not equivalent - and they know it. Simply by observing each other's time dilation, they will each always know which one of them is "above" the other.)

So that's how an observer at the centre of the Earth e̶x̶p̶e̶r̶i̶e̶n̶c̶e̶s̶ observes time dilation - only by comparison to events outside the well - and to look outside the well for him means looking up that curvature. Likewise time dilation at the centre of Earth is only observed because we are outside looking in.

And that's why time dilation cannot be zero at the centre. No matter what he does during his normal day, an observer at the centre of the Earth ultimately must look up-well to his reference point - through the Earth's spacetime curvature - out to "flatter" space. Likewise, the only way to observe events at the centre of the Earth is to look down-well - down that curvature of space.TL;DR: it's not "that" you're at the centre of the Earth that is the source of time dilation - it's where you are looking - what (curvature) you are looking through.

Right?

 

[A.T.]

So that's how an observer at the centre of the Earth e̶x̶p̶e̶r̶i̶e̶n̶c̶e̶s̶ observes time dilation - only by comparison to events outside the well - and to look outside the well for him means looking up that curvature. Likewise time dilation at the centre of Earth is only observed because we are outside looking in.

And that's why time dilation cannot be zero at the centre. No matter what he does during his normal day, an observer at the centre of the Earth ultimately must look up-well to his reference point - through the Earth's spacetime curvature - out to "flatter" space. Likewise, the only way to observe events at the centre of the Earth is to look down-well - down that curvature of space.TL;DR: it's not "that" you're at the centre of the Earth that is the source of time dilation - it's where you are looking - what (curvature) you are looking through.

Right?

Just keep in mind, that It's not about the curvature at the two points. The curvature would be zero in a cavity at the center, just like in outer space. But there still would be gravitational time dilation between these points.

Curvature is the tidal effect, and related to the second derivative of time dilation. Gravity in the Newtonian sense is related to the first derivative of time dilation.

 

[jbriggs444]

TL;DR: it's not "that" you're at the centre of the Earth that is the source of time dilation - it's where you are looking - what (curvature) you are looking through.

It's not about curvature.

It is not about local curvature at the center. But it seems clear that @DaveC426913 is seeing past that now. He is no longer talking about curvature at the center, but curvature on the path(s) between an observer and a clock at the center -- the curvature through which one is making observations. That means that we are talking about the global effects of local curvature along those paths.

 

[stevendaryl]

(And GR time dilation is biased - it's not subjective, like SR dilation is. The two observers are not equivalent - and they know it. Simply by observing each other's time dilation, they will each always know which one of them is "above" the other.)

I don't go along with this distinction between SR and GR. In both cases, the time dilation factor, $\frac{\Delta \tau}{\Delta t}$ depends on the coordinate system used for computing $t$. I think that some of the responses while correct are misleading on this point. There are certain special cases where there is a most natural way to compare two clocks, but that's true for SR, as well. For a complicated situation in which gravity is not constant in time, there is no unique way to talk about one clock having more time dilation than another clock except relative to a coordinate system. That's exactly the same situation as in SR.

What is objective, and not coordinate-dependent is the relative elapsed times for two different clocks that start together, go along different paths, and then reunite. That's true for both SR and GR.

I really think that it's very misleading/false to say that time dilation is more "absolute" in GR than in SR. The time dilation for clocks at different heights in a gravitational field is no more "absolute" than the time dilation for clocks at different locations within an accelerated spaceship. The latter can be computed using SR. So I don't agree at all with the claim that GR makes time dilation more real or absolute or anything than SR.

 

[jartsa]

The time dilation for clocks at different heights in a gravitational field is no more "absolute" than the time dilation for clocks at different locations within an accelerated spaceship. The latter can be computed using SR.

Well some observers say the spaceship accelerates and contracts, and the clocks at the rear are the faster moving and more time-dilated ones, while other observers say that same spaceship decelerates and gets longer, and the clocks closer to the rocket motors are the slower moving and less time-dilated ones.

But when said spaceship stands on a planet, all observers agree what clock are the slower ones.
("Height in a gravity field", is that an absolute thing?)

 

[stevendaryl]

Well some observers say the spaceship accelerates and contracts, and the clocks at the rear are the faster moving and more time-dilated ones, while other observers say that same spaceship decelerates and gets longer, and the clocks closer to the rocket motors are the slower moving and less time-dilated ones.

But when said spaceship stands on a planet, all observers agree what clock are the slower ones.

("Height in a gravity field", is that an absolute thing?)

There is no difference in principle between the spaceship case and the gravity case. One is not more absolute than another. And no, all observers do not agree which clocks are time-dilated. If you operationalize the comparison--say exactly what you are doing to compare the two clocks---then it becomes observer-independent and coordinate-independent. That's true for both GR and SR.

In GR:

1. Take two clocks.
2. Bring them together on the surface of the Earth.
3. Synchronize them.
4. Bring one clock to the top of a mountain. (Or put the other clock down a deep hole.)
5. Wait a year.
6. Bring the clocks back together.

Then you can say absolutely that one clock (the one that was highest) had more elapsed time than the other (the one that was lowest).

In SR:

1. Take two clocks.
2. Bring them together in the rear of an accelerating spaceship.
3. Synchronize them.
4. Bring one clock to the front of the spaceship.
5. Wait a year.
6. Bring the clocks back together.

Then you can say absolutely that one clock (the one that was in the front of the spaceship) had more elapsed time than the other (the one that was in the rear).

In both cases, there is an absolute difference in the elapsed times on the two clocks. But in neither case can it be attributed to an absolute difference in clock rates.

What you can say in both cases is that for the noninertial clocks, there is a natural noninertial coordinate system in which the two clocks are at rest (except for the beginning and the end of the experiment). Relative to this coordinate system, you can say that one clock experiences more time dilation than the other. But that's a coordinate-dependent fact. And---it's as true for the accelerated rocket as it is for clocks at different heights on a planet.

So this notion that time dilation is more "absolute" or "observer-independent" in GR than in SR is just not correct. There are situations in both GR and SR where there is a natural way to compare clocks, and if you use the natural way, you'll get a sort-of objective answer. But GR is no different from SR in this respect.

 

[stevendaryl]

But when said spaceship stands on a planet, all observers agree what clock are the slower ones.

That's not true. If the two clocks are close enough together, then a freefalling observer can use a local inertial coordinate system (approximately) to describe the two clocks, and if the observer starts falling from rest, then according to that coordinate system, the two clocks will (instantaneously) be at rest, and will tick at the same rate as the free-falling clock.

 

[DaveC426913]

Caveat: this is essentially a sidebar. I was the one who made the claim in post 16 that (to paraphrase) the GR scenario is absolute whereas SR is relative.

If I had left that (contentious) statement out completely, I'd still have my answer.

Now that I'm satisfied I have an answer, I don't see a problem with the thread proceeding along this sidebar.

There is no difference in principle between the spaceship case and the gravity case. One is not more absolute than another.

OK, your scenarios in post 21 show that both SR and GR can both be absolute - but what about the other side of the coin - scenarios where they are both relative?

i.e. In what circumstances could the guy at the bottom of a gravity well claim that he is the one observing events moving slower on the other side of this gravity slope?

 

[jartsa]

There is no difference in principle between the spaceship case and the gravity case. One is not more absolute than another. And no, all observers do not agree which clocks are time-dilated. If you operationalize the comparison--say exactly what you are doing to compare the two clocks---then it becomes observer-independent and coordinate-independent. That's true for both GR and SR.

Well now I think that if we used sensible names then there would be:
1: SR time dilation
2: SR differential aging
and:
3: GR gravitational differential aging

And this name would not be used:
4: GR gravitational time dilation
SR differential aging is caused by things taking different paths in space-time. Like your clocks did.
GR differential aging is caused by things being in different gravitational potentials.

 

[stevendaryl]

OK, your scenarios in post 21 show that both SR and GR can both be absolute - but what about the other side of the coin - scenarios where they are both relative?

i.e. In what circumstances could the guy at the bottom of a gravity well claim that he is the one observing events moving slower on the other side of this gravity slope?

It's not a matter of circumstances, it's a matter of how you set up your coordinate system. The circumstances come into play only in that for certain situations, there may be an obvious coordinate system that is more convenient than others. In Special Relativity, for inertial observers, the most convenient coordinate system is one where the observer is at rest, and the plot of an unaccelerated object is a straight line. In SR, for an observer undergoing constant proper acceleration, the most convenient coordinate system is the noninertial coordinate system where the observer is at rest and where the metric tensor components are independent of time. In GR, if you are standing on the surface of a planet, the most convenient coordinate system is one where the ground is at "rest".

But in GR, you can use any coordinate system you like, because the equations of GR work for any coordinate system whatsoever. So if you have a clock at the top of a mountain and another clock at sea level, you can certainly choose a coordinate system such that the coordinate time between "ticks" of the higher clock is just about anything you like, and the coordinate time between "ticks" of the lower clock is also anything you like. Those numbers don't mean anything in isolation, they are just labels on events. The only constraint on those labels is that, in order to use calculus, the labels must change continuously as you move around in spacetime.

A difference in attitude between Einstein's development of Special Relativity and his development of General Relativity was that for SR, he tried to operationalize the meaning of coordinates, by specifying how you would set up a coordinate system using clocks and rulers and light signals, etc. In GR, there is no operational way (except in certain special situations) to give a unique coordinate system. So instead, GR allows you the freedom to choose any coordinate system you like. But if the time coordinate can be just about anything, then the ratio $\Delta \tau/\Delta t$ is not particularly meaningful, physically.

 

[stevendaryl]

Well now I think that if we used sensible names then there would be:
1: SR time dilation
2: SR differential aging
and:
3: GR gravitational differential aging

And this name would not be used:
4: GR gravitational time dilation
SR differential aging is caused by things taking different paths in space-time. Like your clocks did.
GR differential aging is caused by things being in different different gravitational potentials.

No, I would not agree to that characterization. In both cases, differential aging is caused by things taking different paths through spacetime. In both cases, you can define "time dilation" as a coordinate effect: Time dilation factor = (Change in elapsed time on clock)/(Change in coordinate time). In neither case is "time dilation" an absolute property of clocks.

There really is no difference between GR and SR in these matters. SR is a special case of GR in which the metric tensor has a particularly simple form.

GR doesn't actually have a notion of a "gravitational potential". All it has is a metric tensor, which can be used for computing elapsed time along paths. But for situations such as the spacetime near a planet, you can choose a convenient coordinate system (Schwarzschild coordinates) such that the metric tensor is related to the gravitational potential that you would calculate using Newtonian gravity. But that's only for the purpose of showing the correspondence between Newtonian gravity and GR. The "gravitational potential" does not play a role in GR.

 

[DaveC426913]

It's not a matter of circumstances, it's a matter of how you set up your coordinate system.

Yes.

But in GR, you can use any coordinate system you like, because the equations of GR work for any coordinate system whatsoever.

Right, so what coordinate system would have the guy at the bottom of Earth's g-well thinking the guy on the surface is moving slower?

That's what I'm trying to ask.

 

[stevendaryl]

Yes.Right, so what coordinate system would have the guy at the bottom of Earth's g-well thinking the guy on the surface is moving slower?

That's what I'm trying to ask.

I can give you an answer, but it would be pretty meaningless, because coordinates are not particularly meaningful. But if you insist:

Let's take a little region of spacetime that includes both clocks. The usual coordinate system would be $z, t$ where $z$ it the height above the Earth, and $t$ is the Schwarzschild time coordinate, scaled so that $\frac{d\tau}{dt} = 1$ on the surface of the Earth. According to this coordinate system, $\frac{d\tau}{dt} \approx 1 + \frac{gz}{c^2}$, where $g$ is the surface acceleration due to gravity. (For simplicity, I'm ignoring the rotation of the Earth).

So in this coordinate system, the higher clock runs faster. If you want the top clock to run slower, scale the time coordinate differently. Instead of using $t$, use a different coordinate $t' = t /(1 - \frac{2gz}{c^2})$. In this new coordinate system, $\frac{d\tau}{dt'}$ is greater for the lower clock.

That might seem like a cheat, but it doesn't matter. You can use any coordinate system you like.

 

[DaveC426913]

I can give you an answer, but it would be pretty meaningless, because coordinates are not particularly meaningful. But if you insist:

Let's take a little region of spacetime that includes both clocks. The usual coordinate system would be $z, t$ where $z$ it the height above the Earth, and $t$ is the Schwarzschild time coordinate, scaled so that $\frac{d\tau}{dt} = 1$ on the surface of the Earth. According to this coordinate system, $\frac{d\tau}{dt} \approx 1 + \frac{gz}{c^2}$, where $g$ is the surface acceleration due to gravity. (For simplicity, I'm ignoring the rotation of the Earth).

So in this coordinate system, the higher clock runs faster. If you want the top clock to run slower, scale the time coordinate differently. Instead of using $t$, use a different coordinate $t' = t /(1 - \frac{2gz}{c^2})$. In this new coordinate system, $\frac{d\tau}{dt'}$ is greater for the lower clock.

That might seem like a cheat, but it doesn't matter. You can use any coordinate system you like.

Yeah, I'm afraid I don't really follow.

Are you saying the guy at Earth's centre could play with his coordinate system in a way that would result in a clock at the surface moving slower?
What would happen when he took the elevator and compared clocks?My gut tells me you're right - even in SR, it is valid to say the spaceship is stationary while rest of the universe is time and length contracted, so I may have to take it on faith that GR is similarly symmetrical.

 

[stevendaryl]

Yeah, I'm afraid I don't really follow.

Are you saying the guy at Earth's centre could play with his coordinate system in a way that would result in a clock at the surface moving slower?
What would happen when he took the elevator and compared clocks?

As I have said before, the total elapsed time on a clock is coordinate-independent. But by choosing a different coordinate system, you can change the value of $\frac{d\tau}{dt}$ at any point along the path. That's exactly the same situation with Special Relativity and the Twin Paradox. One twin stays at home. One twin rockets away for 10 years, turns around, and comes home. They get back together to compare ages. It's an objective fact, independent of coordinates, that the stay-at-home twin ages more. But depending on the coordinate system, there are different ways to account for this differential aging.

If you take the inertial coordinate system in which the stay-at-home twin is at rest, then time dilation relative to that coordinate system tells us that the traveling twin always has a constant time dilation factor that makes his clock run slower than the stay-at-home twin.

If you take the inertial coordinate system in which the traveling twin is at rest during his outward journey, then in that coordinate system, the stay-at-home twin initially has more time dilation (his clock runs slower), but then on the return trip, the traveling twin has more time dilation.

If you take the inertial coordinate system in which the traveling twin is at rest during his return journey, then in that coordinate system, the traveling twin initially has more time dilation (his clock runs slower), but on the return trip, the stay-at-home twin has more time dilation.

All three coordinate systems agree on the total elapsed time experienced by the two twins, but they disagree about whose clock is running faster at what time.

The same thing is true in GR. You have one clock that remains on the top of a tall mountain. You have another clock that starts on the top of the mountain, is carried down to the surface of the Earth, remains there for a year, and is carried back up to the top of the mountain. It's an objective, coordinate-independent fact that the clock that went down to the surface will have less elapsed time than the clock that remained on top of the mountain throughout. But how you account for that difference is coordinate-dependent. In some coordinate systems, the higher clock always runs slower. In another coordinate system, it might run slower for part of the time and faster for part of the time. There is no coordinate-independent answer to the question of "Which clock is running faster right NOW?" (actually, there is no coordinate-independent meaning to the phrase "right now", either).

 

[PAllen]

No, I would not agree to that characterization. In both cases, differential aging is caused by things taking different paths through spacetime. In both cases, you can define "time dilation" as a coordinate effect: Time dilation factor = (Change in elapsed time on clock)/(Change in coordinate time). In neither case is "time dilation" an absolute property of clocks.

There really is no difference between GR and SR in these matters. SR is a special case of GR in which the metric tensor has a particularly simple form.

GR doesn't actually have a notion of a "gravitational potential". All it has is a metric tensor, which can be used for computing elapsed time along paths. But for situations such as the spacetime near a planet, you can choose a convenient coordinate system (Schwarzschild coordinates) such that the metric tensor is related to the gravitational potential that you would calculate using Newtonian gravity. But that's only for the purpose of showing the correspondence between Newtonian gravity and GR. The "gravitational potential" does not play a role in GR.

A well defined gravitational potential can be introduced in an exact, invariant meaner in any spacetime with a timelike killing vector field. Unlike Newtonian gravity, this is a special case, but it is an important and very useful special case.

 

[stevendaryl]

A well defined gravitational potential can be introduced in an exact, invariant meaner in any spacetime with a timelike killing vector field. Unlike Newtonian gravity, this is a special case, but it is an important and very useful special case.

In those cases, I would say that the significance is only that the time-symmetry makes calculations a lot easier. It's exactly like the fact that spherical symmetry makes calculations easier (and makes quantities such as angular momentum into constants of the motion).

 

[PeterDonis]

It's not a matter of circumstances, it's a matter of how you set up your coordinate system.

I think your comments along these lines are causing more confusion than they are solving.

As @PAllen noted in post #14, if two observers exchange light signals, the results (for example, the observed frequency shift of each observer's signals as observed by the other, and the elapsed time on each observer's clock between successive signals) must be independent of any choice of coordinates.

When @DaveC talks about a clock at the center of the Earth running slower than a clock at the Earth's surface, he is talking about something that can be measured by the two clocks exchanging light signals. The surface clock will see the center clocks' light signals redshifted, and the center clock will see the surface clock's light signals blueshifted, and the center clock's elapsed time between successive signals will be less than the surface clock's elapsed time between successive signals. These are all coordinate-independent invariant facts about the scenario, and they are the facts that people are referring to when they talk about one clock "running slower" than the other.

Note, also, that these facts are different from the facts in the case of two spaceships in relative motion in flat spacetime. Say the ships are moving away from each other. Then each ship sees the other ship's light signals as redshifted (i.e., symmetric, not asymmetric shifts as in the gravity case above); and each ship's elapsed time between successive signals increases from signal to signal, but in the same way (i.e. ,symmetric, not asymmetric elapsed time behavior). When you say...

There is no difference in principle between the spaceship case and the gravity case. One is not more absolute than another.

...you give the strong impression that you are simply ignoring the above facts. I know you are well aware of those facts, and I don't think anyone in this thread actually disagrees about the physics; but the words you are using to describe the physics are making it very hard for other people in this thread to see that we're all talking about the same physics.

Let's take a little region of spacetime that includes both clocks.

You can't. A clock at the center of the Earth and a clock at the Earth's surface cannot be covered by a single local inertial frame. Tidal gravity is highly non-negligible between the two. The most obvious manifestation of this is the fact that, as has been noted, there is zero "gravity" at the center of the Earth: if you release a rock there, it hangs motionless next to you (supposing you and the rock are in a tiny cavity at the center), whereas a rock released at the Earth's surface will behave quite differently.

I actually don't think it's possible to find any coordinate chart in which the coordinate time dilation of the two clocks in question (one at Earth's center and one at the surface) would be reversed. But if there is one, it certainly can't be a local inertial coordinate system. Your analogy here with the standard twin paradox in flat spacetime simply doesn't work because of this.

In both cases, differential aging is caused by things taking different paths through spacetime.

This statement is correct, but the "differential aging" you are talking about here is somewhat different from the comparison involving exchanging light signals that I described above.

GR doesn't actually have a notion of a "gravitational potential".

As @PAllen pointed out in post #31, it does for stationary spacetimes, which are the kind we are talking about in this discussion.

In those cases, I would say that the significance is only that the time-symmetry makes calculations a lot easier.

The timelike KVF is not just a calculational convenience. It's an invariant geometric feature of the spacetime, and its presence makes a number of intuitions carried over from Newtonian gravity applicable which are not applicable in non-stationary spacetimes. I agree that if one's goal is to learn GR in full generality, one should learn not to rely on such intuitions; but if one's goal is simply to understand particular scenarios like the ones being discussed in this thread, I don't see why relying on those intuitions is a problem.

 

[PeterDonis]

My gut tells me you're right - even in SR, it is valid to say the spaceship is stationary while rest of the universe is time and length contracted, so I may have to take it on faith that GR is similarly symmetrical.

GR is "similarly symmetrical" in that you can always decide to adopt coordinates in which any particular object or observer is at rest. But it is not "similarly symmetrical" when you start looking at the particular predictions for curved spacetimes vs. the flat spacetime of SR. It certainly won't be as simple as "the rest of the universe is time and length contracted". I discuss some of the differences in my post in response to @stevendaryl just now.

 

[stevendaryl]

I think your comments along these lines are causing more confusion than they are solving.

Well, in my opinion, it's just the opposite. Focusing on gravitational potential will lead to the mistaken impression that there always is such a thing. It will lead to the conclusion that "time dilation" is always objective and coordinate-independent in a way that it isn't in SR. That's a completely mistaken impression.

In contrast, focusing on comparison of elapsed times for different paths is always applicable.

As @PAllen noted in post #14, if two observers exchange light signals, the results (for example, the observed frequency shift of each observer's signals as observed by the other, and the elapsed time on each observer's clock between successive signals) must be independent of any choice of coordinates.

The same is true for clocks at different heights on board an accelerated rocket. So this is not a difference between SR and GR.

When @DaveC talks about a clock at the center of the Earth running slower than a clock at the Earth's surface, he is talking about something that can be measured by the two clocks exchanging light signals.

The same is true of clocks at different heights on board an accelerated rocket. But in the case of the accelerated rocket, it is not interpreted as implying that there is an absolute time dilation for the rear rocket that is greater than for the front rocket.

Note, also, that these facts are different from the facts in the case of two spaceships in relative motion in flat spacetime.

The case I was comparing it to was clocks at different heights on board an accelerated rocket. In that case, everything said about GR time dilation applies equally well to clocks aboard the rocket.

The timelike KVF is not just a calculational convenience. It's an invariant geometric feature of the spacetime, and its presence makes a number of intuitions carried over from Newtonian gravity applicable which are not applicable in non-stationary spacetimes.

How is that different from a calculational convenience? The existence of a timelike KVF is (to my mind) exactly the same sort of thing as the existence of spherical symmetry. It makes calculations simpler, and it leads to conservation laws that don't apply in the more complicated case.