Flat spacetime in a gravity well

[DaveC426913]

I am trying to explain to someone why there is still gravitational time dilation at the centre of the Earth.

(Not to over-tax the rubber sheet analogy, but...)
When we sketch a gravity well, we show a curve that starts out nearly flat, then curves away from the flat plane, until it reaches the centre of the mass. At that point, it is parallel with flat space, but not in the flat space plane. i.e. Y has a non-zero, negative value, yet a slope of zero.

So, the y-axis in this diagram does not represent gravitational force experienced , since otherwise it should a y=0.

So, how do we describe the gravitational state of the small area at the bottom of this well?

More to the point, how do I convince my colleague that, at that point - though gravitational forces are cancelled, they are not ... zero? And that GR time dilation applies here?

What is the term? Gravitational potential?

 

[PeterDonis]

Gravitational potential is the usual term for what you're describing. "Gravitational force" is the (negative of the) gradient of the potential. This terminology only works for a special class of spacetimes (stationary spacetimes), but spacetimes describing an object like the Earth and its vicinity meet this requirement (at least to a good enough approximation for most purposes).

 

[mfb]

Another point of view: If you are at the center of Earth, there is no local measurement you can do that would tell you that you are in a gravitational potential. To observe the time dilation, you have to compare your clock to a clock somewhere far away from Earth. You have to send a signal out of the gravity well, and get an answer that goes into the gravity well. This is where the time difference appears.

 

[Ibix]

There's no absolute time, so clocks don't run fast or slow - they run fast or slow compared to another clock at another location. Time dilation between two locations is closely related to the gravitational redshift between the two locations (they're basically the same phenomenon), which you can easily show depends on the difference in energy.

 

[DaveC426913]

Right. So, I can't exactly beg the question here.

I'm trying to show someone that the largest time dilation will exist at the centre of the Earth; I can't very well use time dilation existing at the centre of the Earth as the explanation. (And no one has actually done the experiment)

And 'gravitational potential' is just a phrase if I can't explain how it shows what's happening at the centre.

 

[Ibix]

Drop two rocks, one made of matter and one of anti-matter, from the top of a tower. Stop them at the bottom and use the kinetic energy for something. Combine the rocks to convert them to photons and send the photons to the top of the tower and reconstitute the rocks. Rinse and repeat for free energy! ...unless the photons lose energy as they climb. Hence redshift. Replace the tower with a mine to show that this applies at the centre of the Earth.

Redshift is related to time dilation fairly easily - you just imagine sending pulses of light from a lower clock to an upper one. Redshift means that the gap between the received pulses is longer than the gap between the emitted pulses. This is time dilation.

 

[DaveC426913]

Redshift is related to time dilation fairly easily - you just imagine sending pulses of light from a lower clock to an upper one. Redshift means that the gap between the received pulses is longer than the gap between the emitted pulses. This is time dilation.

This is missing the point.

I am not looking for a description of how to check of for time dilation. That doesn't help my colleague accept that it is strongest at the centre of a mass (since he can't actually prove it to himself by doing the experiment himself).

 

[Ibix]

This is missing the point.

No it's not - you can have a stack of clocks from the centre to the surface of the Earth. The upper clock of any pair must always run slower by the redshift argument. So a clock at the centre must run the slowest.

 

[DaveC426913]

No it's not - you can have a stack of clocks from the centre to the surface of the Earth. The upper clock of any pair must always run slower by the redshift argument. So a clock at the centre must run the slowest.

I think I see your point. I'm trying to wrangle it into an argument that shows, logically, why time dilation must be maximum at the centre, even though gravitational force cancels out.

My colleague seems to imagine that time dilation is is proportional to net force, not to absolute gravitational potential. (since grav pot doesn't seem to have an intuitive manifestation we can experience.)

 

[PeterDonis]

grav pot doesn't seem to have an intuitive manifestation we can experience

Sure it does. It's just height. The center of the Earth is lower than any other point in the Earth's gravity well, so it has the lowest potential (and thus the largest time dilation).

 

[DaveC426913]

Sure it does. It's just height. The center of the Earth is lower than any other point in the Earth's gravity well, so it has the lowest potential (and thus the largest time dilation).

Still not much use in helping my colleague, as it's still begging the question.

 

[Ibix]

I think I see your point. I'm trying to wrangle it into an argument that shows, logically, why time dilation must be maximum at the centre, even though gravitational force cancels out.

The drop-a-rock argument works for any two points where you have to do work to get from A to B - the clock at B must tick faster. Although the force is zero at the centre, that just means you have to do work to move away in any direction.

My colleague seems to imagine that time dilation is is proportional to net force, not to absolute gravitational potential. (since grav pot doesn't seem to have an intuitive manifestation we can experience.)

He's not the only one...

 

[PeterDonis]

Still not much use in helping my colleague, as it's still begging the question

No, it isn't, because we have direct experimental evidence that clocks higher in a gravity well run faster than clocks lower in a gravity well. Tell your colleague to look up the Pound-Rebka experiment.

 

[DaveC426913]

No, it isn't, because we have direct experimental evidence that clocks higher in a gravity well run faster than clocks lower in a gravity well. Tell your colleague to look up the Pound-Rebka experiment.

Right. But my colleague is essentially associating time dilation with change in y-displacement, rather than absolute y-displacement.

(I realize at this point that I am playing Devil's Advocate.)

So, in his mental model, time dilation will vary directly with the net gravitational force - increasing as we go deeper into the gravity well, until we reach the surface, and then decreasing back to zero by the time it reaches the centre.

 

[PeterDonis]

my colleague is essentially associating time dilation with change in y-displacement

That's fine; it makes no difference where you set the "zero point" of potential (height). The difference in height, and which observer is lower, is all that matters, and both of those things have been directly tested. The difference in height in the Pound-Rebka experiment was, IIRC, 22.5 meters; that's true regardless of whether you set the" zero" of height at the Earth's surface, the top of the tower they used to do the experiment, or spatial infinity.

in his mental model, time dilation will vary directly with the net gravitational force

No, it won't, it will vary with difference in height, i.e., change in vertical displacement. That's not the same as net gravitational force, unless I'm misinterpreting what you mean by "change in displacement". A change in displacement is usually interpreted to mean a distance, not a force.

 

[PeterDonis]

A change in displacement is usually interpreted to mean a distance, not a force.

Here's another way of looking at this, which might help to elucidate the error your colleague is making. It's true that it requires a force to raise an object against gravity; but even so, the "change in displacement" from one height to another is not the same as the force, and it's not even proportional to the force; it's independent of the force. The quantity that combines the two is the force times the distance (height change), i.e., the work done. And the change in gravitational potential is just the work done per unit mass. That is the thing that is the same as the time dilation factor/redshift factor (modulo a change of units if you're not using "natural" GR units where ##G = c = 1##). (The redshift is there, heuristically, because the photons have to do work and thus lose energy to climb up to a higher point in the gravity well, and since they're photons, that reduces their frequency.)

 

[DaveC426913]

That's fine; it makes no difference where you set the "zero point" of potential (height). The difference in height, and which observer is lower, is all that matters, and both of those things have been directly tested.

My colleague thinks the gravity well illustration in post 1 is incorrect.
He thinks that the centre of the well - to be accurate - should be at the same y-value as it would be a distance infinity. i.e. as long as they the same outside Earth and at its centre.

The difference in height in the Pound-Rebka experiment was, IIRC, 22.5 meters; that's true regardless of whether you set the" zero" of height at the Earth's surface, the top of the tower they used to do the experiment, or spatial infinity.

Correct me if I'm wrong, but both test points were above the Earth's surface. The curvature does not begin to inflect (shallow out) until below the Earths surface.

... in his mental model, time dilation will vary directly with the net gravitational force

No, it won't,

It does in his mental model.
Other than 'trust me, you're wrong', I can't think of a way to show him.

 

[mfb]

So, in his mental model, time dilation will vary directly with the net gravitational force - increasing as we go deeper into the gravity well, until we reach the surface, and then decreasing back to zero by the time it reaches the centre.

That is ruled out by experiments done at various places on the surface and above it. In particular, clocks at sea level run at the same speed everywhere, despite a ~1% difference in gravitational force for different latitudes.

 

[DaveC426913]

That is ruled out by experiments done at various places on the surface and above it. In particular, clocks at sea level run at the same speed everywhere, despite a ~1% difference in gravitational force for different latitudes.

Yes. But not below sea level.

His mental image is, essentially:
1] As net gravity increases (eg. above the Earth's surface), so does time dilation,
2] As net gravity decreases (eg. below the surface), so does time dilation, so that
3] At the centre, where net gravity is zero, so too should time dilation be zero.
4] "Gross" gravity has no real meaning. i.e. if gravity cancels out at the centre then it cancels out.

I haven't been able to show him how to break out of that mental model. And I'm beginning to see why. I know his model is wrong, yet, even with your help, I cannot find the words to explain how.

 

[mfb]

But not below sea level.

I'm sure there have been experiments in some mines below sea level. But even if you don't find any: So what? Sea level has a ~40 km difference in distance to the center of Earth across different latitudes, and we have measurements that go up by ~20,000 km. We can rule out "time dilation is proportional to gravitational acceleration" or even "is monotonous with gravitational acceleration" from thousands of experiments done everywhere over the surface.

By the way: Gravitational attraction increases slightly as you go down until you reach the (very dense) core, and only decreases afterwards.

 

[DaveC426913]

I'm sure there have been experiments in some mines below sea level. But even if you don't find any: So what? Sea level has a ~40 km difference in distance to the center of Earth across different latitudes, and we have measurements that go up by ~20,000 km.

None of which intuitively contradict the notion of what would start happening as gravity starts to cancel out. They have all been in areas where a movement toward the centre also results in an increase in gravity.

Another way of saying it is: all the experiments have been in zones where the rate of change of the slope is increasing (negative). None have been in zones where the change in slope is decreasing (leveling off).

 

[A.T.]

My colleague thinks the gravity well illustration in post 1 is incorrect.
He thinks that the centre of the well - to be accurate - should be at the same y-value as it would be a distance infinity. i.e. as long as they the same outside Earth and at its centre.

Then his misconception has nothing to do with General Relativity. His ideas about the potential well would be wrong in Newtonian Gravity as well.

The GR view in simple terms: Local gravity is the gradient of time dilation. At the center the time dilation has a local maximum, so the gradient is zero.

 

[PeterDonis]

He thinks that the centre of the well - to be accurate - should be at the same y-value as it would be a distance infinity.

The fact that the center of the Earth is below the surface should be sufficiently obvious to make this claim absurd on its face. If your colleague does not see why this is so, then he needs to recalibrate his intuitions.

Correct me if I'm wrong, but both test points were above the Earth's surface.

Actually, no. One was at the top of a tower at Harvard; the other was in the subbasement of the same building, so it was actually below the surface.

However, as @mfb has pointed out, the "inflection" of the slope of the potential actually doesn't start until well below the Earth's surface, because of the way density varies inside the Earth. So to run a Pound-Rebka type experiment across an inflection point, you would have to do it, IIRC, a few hundred miles down or more.

Other than 'trust me, you're wrong', I can't think of a way to show him.

Tell him that his claim is equivalent to the claim that the center of the Earth is above the surface, which is obviously absurd.

 

[DaveC426913]

The fact that the center of the Earth is below the surface should be sufficiently obvious to make this claim absurd on its face. If your colleague does not see why this is so, then he needs to recalibrate his intuitions.

Tell him that his claim is equivalent to the claim that the center of the Earth is above the surface, which is obviously absurd.

He is simply of the mind that zero gravitational force should result in zero time dilation (just as it would be in the empty outer reaches of space).

Other than simply saying 'you're wrong', I have, as yet, no way to tell him otherwise.

 

[DaveC426913]

I've asked him to draw his idea of the well, and label it, with particular attention to any inflections and cusps he must apply.
I am certain he will find that some inflections and cusps needed to realize his idea just won't make sense once seen on paper.

 

[mfb]

He is simply of the mind that zero gravitational force should result in zero time dilation (just as it would be in the empty outer reaches of space).

As there is no point with zero gravitational force this is impossible to disprove experimentally. But any reasonable dependence on the force instead of the potential is completely inconsistent with measurements done in places with a non-zero force.

 

[DaveC426913]

As there is no point with zero gravitational force this is impossible to disprove experimentally.

I think 'approaching zero' would surely suffice

But any reasonable dependence on the force instead of the potential is completely inconsistent with measurements done in places with a non-zero force.

I think that's part of what's got him going: the curve of gravitational force and the curve of time dilation look similar in all the places we've measured so far. He probably figures why would it stop being similar at the Earth's surface?

 

[A.T.]

He probably figures why would it stop being similar at the Earth's surface?

Because naive reasoning based on "looking similar" doesn't always work. Because a function and it's derivatives can "look similar" in some ranges, but not in others.

 

[stevendaryl]

I think that's part of what's got him going: the curve of gravitational force and the curve of time dilation look similar in all the places we've measured so far. He probably figures why would it stop being similar at the Earth's surface?

I'm not sure what you mean by "similar", but the formula for proper time in terms of coordinate time involves the gravitational potential, rather than the gravitational force. The relation works whether you are above or below the Earth's surface.

The approximate relationship, good when gravitational fields are weak and velocities are low compared to the speed of light is (using Schwarzschild coordinates):

##\delta \tau = (1 + \frac{m \Phi_g - \frac{1}{2} m v^2}{mc^2}) \delta t##

where ##\Phi_g## is the gravitational potential, and ##v## is the velocity of the clock measuring ##\tau##. This approximate formula works above and below the surface.

Note: the potential ##\Phi_g## should be chosen so that ##\Phi_g \rightarrow 0## as ##r \rightarrow \infty##. For a different choice, the above formula needs to be adjusted slightly.

 

[stevendaryl]

The approximate relationship, good when gravitational fields are weak and velocities are low compared to the speed of light is (using Schwarzschild coordinates):

##\delta \tau = (1 + \frac{m \Phi_g - \frac{1}{2} m v^2}{mc^2}) \delta t##

The approximate time-dilation factor, ##K \equiv 1 + \frac{m \Phi_g - \frac{1}{2} m v^2}{mc^2}##, depends on the gravitational potential, not the gravitational force. Since the force is the negative derivative of the potential, you have: ##\frac{\partial K}{\partial x} = \frac{1}{c^2} \frac{\partial \Phi_g}{\partial x} = - \frac{g}{c^2}##, where ##g## is the local acceleration due to gravity (in Schwarzschild coordinates---coordinate acceleration is a coordinate-dependent quantity). So ##g## being zero as it is at the center of the Earth doesn't mean that the time dilation is zero, it just means that the time dilation is approximately constant. So near the center of the Earth, there would not be a measurable location-dependence to time dilation. In contrast, on the surface of the Earth, where ##g## is nonzero, clocks that are higher run faster (as measured in Schwarzschild coordinates) tha clocks that are lower.

 

[DaveC426913]

Since he has a naive understanding of GR in the first place, he's not going to follow the logic of equations.

But he has produced some diagrams, at my behest. Here is one of a hollow sphere, which has some unexplainable cusps and nigh-infinite slopes:

I have started showing him the logical implications of his idea that lead to paradoxes. (Such as a watch on one wrist running at a dramatically different rate than the watch on the other wrist).

He has, predictably, begun to lose interest and wander away, mumbling about 'paradoxes' in Einsteinian models.

If anyone is interested in the dialogue, it is here:
http://www.sciforums.com/threads/where-is-most-gravity-inside-or-out.155498/page-8#post-3512088
poster's name is nebel.

 

[A.T.]

Since he has a naive understanding of GR in the first place...

He should start with Newtonian gravity, and the difference between gravitational force and potential, as shown here:

http://teacher.pas.rochester.edu/PHY235/LectureNotes/Chapter05/Chapter05.htm

Gravitational time dilation in GR is related to the potential (upper plot, note the inverted y-axis), and is maximal at the center where gravity in Newtonian sense is zero (lower plot).

 

[Pencilvester]

What about hovering at the midpoint between two identical, non-rotating black holes? My intuition at least tells me (if the black holes are close enough, even momentarily) that there will be significant time dilation, even though the net “gravitational force” will be 0 (let’s also say we’re relatively small enough that the tidal forces are negligible, just to make things simpler). But then again, my intuition also tells me time dilation will be greater at the center of the Earth than at the surface, so who knows, this might be useless.

 

[Jonathan Scott]

By the shell theorem, the Newtonian potential outside a spherical body is the same as if the entire mass was concentrated at the central point, and the potential inside a spherical shell is the same everywhere. So as you go deeper, the contribution towards the potential of any shell you pass remains constant once you have passed it, and the potential due to the spherical shells that you haven't yet passed is steadily getting deeper due to the fact that you are getting closer to the middle.
[Edited to correct typo]

 

[A.T.]

What about hovering at the midpoint between two identical, non-rotating black holes? My intuition at least tells me (if the black holes are close enough, even momentarily) that there will be significant time dilation, even though the net “gravitational force” will be 0 (let’s also say we’re relatively small enough that the tidal forces are negligible, just to make things simpler). But then again, my intuition also tells me time dilation will be greater at the center of the Earth than at the surface, so who knows, this might be useless.

Both is correct.