Is space curved at the center of gravity?

[King Solomon]

TL;DR Summary

At the core of the earth, or sun, there is no net force of gravity...so

At the core of the earth, or sun, there is no net force of gravity...every direction is up...does this mean that space is not curved, or more generally, becomes less curved from the surface of an object towards its center?

 

[Ibix]

It isn't space that's curved, but spacetime. Yes, it is still curved at the center of a spherically symmetric mass.

Curvature is related to the second derivative of the metric - you seem to be attempting to visualise the first derivatives, which can be arranged to be zero by an appropriate choice of coordinates.

 

[Dale]

At the core of the earth, or sun, there is no net force of gravity...every direction is up...does this mean that space is not curved,

Curvature is related to tidal effects rather than gravitational acceleration. So in the center of a solid spherical mass you do still have curvature.

However, inside a hollow spherical shell spacetime is flat (there are no tidal effects).

more generally, becomes less curved from the surface of an object towards its center?

Curvature is a tensor, so “less curved” doesn’t really make sense. I guess you could ask if it has a lower Ricci scalar. I don’t know the answer to that.

 

[pervect]

Summary:: At the core of the earth, or sun, there is no net force of gravity...so

At the core of the earth, or sun, there is no net force of gravity...every direction is up...does this mean that space is not curved, or more generally, becomes less curved from the surface of an object towards its center?

If you managed to make a hollow at the center of an idealized, homogeneous, non-rotating sphere, the space-time in the hollow would not be curved, because the Riemann tensor would be zero.

I'd have to do some calculations for a rotating sphere (or a non-spherical object with an equatorial bulge). I suspect there might be a non-zero Riemann tensor, due to frame dragging effects, but I'm not really sure. It wouldn't be a very big effect, though.

Since the Earth rotates, I suppose it would be important to know to answer your question fully.

I'm not sure of the effects of inhogoneities or mass concentrations. I rather suspect that they could cause the Riemann tensor to be non-zero, but I haven't done any calculations.

"The Riemann tensor" is rather abstract and may not be familiar, though it is what is what people familiar with the theory would assume you meant when you said "curved space-time". For the most part, it can be thought of as tidal forces. This is technically only part of the Riemann tensor, it doesn't address the frame-dragging "magnetic" part of the tensor, an issue I alluded to.

 

[mitochan]

Einstein's gravity equation says where momentum-energy is, Einstein tensor is. If I take your "curved" is non zero Einstein tensor, the center of the Earth or the sun is "curved" if no void of matter and energy exist in core.

The component of metric $g_{00}$ plays the role of gravity potential and at the center $\nabla g_{00}=0$ as you expect.

 

[Ibix]

If I take your "curved" is non zero Einstein tensor,

Note that this is a restrictive definition of curvature - Ricci curvature. It is indeed non-zero inside a source of gravity and zero outside. That's sufficient to answer the question posed here, but we should note that Ricci curvature is only part of the story of curvature, and there are other components of the full story that are (or can be) non-zero in vacuum. This is how gravity is able to work in vacuum.

OP - as others have noted, a perfectly spherical hollow perfectly concentric with a perfectly spherical Earth would give zero curvature at the centre. With no hollow (the situation I imagined in my response) there would be curvature. With a less symmetric Earth or hollow, there may be curvature - it would depend on the mass distribution.

 

[mitochan]

we should note that Ricci curvature is only part of the story of curvature, and there are other components of the full story that are (or can be) non-zero in vacuum.

Do you mean cosmological term ? If not I am interested in mathematics I do not know yet.

 

[Ibix]

Do you mean cosmological term ?

No - that also produces Ricci curvature. The point is that the Riemann tensor completely describes curvature, and the Ricci tensor only includes part of the information in the Riemann tensor. You are correct that Ricci curvature is non-zero only where the stress-energy tensor is non-zero. However, this can't be the full story or there'd be no gravity in space.

The bit that gets left out when extracting the Ricci tensor from the Riemann is called Weyl curvature, and can be non-zero in vacuum. For example, a Schwarzschild black hole is vacuum everywhere, but there's lots of curvature.

 

[King Solomon]

Note that this is a restrictive definition of curvature - Ricci curvature. It is indeed non-zero inside a source of gravity and zero outside. That's sufficient to answer the question posed here, but we should note that Ricci curvature is only part of the story of curvature, and there are other components of the full story that are (or can be) non-zero in vacuum. This is how gravity is able to work in vacuum.

OP - as others have noted, a perfectly spherical hollow perfectly concentric with a perfectly spherical Earth would give zero curvature at the centre. With no hollow (the situation I imagined in my response) there would be curvature. With a less symmetric Earth or hollow, there may be curvature - it would depend on the mass distribution.

The question concerns the center of gravity, not the center of an object.

The center of gravity of a tennis racket is not the geometric center of the tennis racket, hence the intermediate axis theorem for such objects.

Likewise if someone glued together a hemisphere of lead and hemisphere of styrofoam, the combined sphere wouldn't have zero gravity at i's geometric center, but rather somewhere within the lead hemisphere.

 

[Ibix]

The question concerns the center of gravity, not the center of an object.

You specifically asked about the Sun or Earth, for which this is a distinction without a difference.

If the center of gravity is inside the object, curvature is non-zero because Ricci curvature is trivially non-zero anywhere where there is matter. In other cases, it's certainly possible to have zero curvature, as in the case of a perfectly spherical hollow in the exact center of a spherical body. In less symmetric circumstances, though, I think curvature will be non-zero.

 

[King Solomon]

There is also the second question: Assuming the the curvature of space is continuous function, if there are any regions within the interior of an object that have zero curvature, do these occur abruptly, or does space become less curved (begins to flatten) towards these regions with no curvature inside of an an object?

Which of the above two pictures best represent gravity for a mostly homogeneous object.

 

[Dale]

Did you even read my post 3?

 

[Ibix]

Curvature can be discontinuous in GR. The metric needs to be continuous but its derivatives needn't be. For example, a solid sphere of constant density with radius $r_g$ and Schwarzschild radius $r_S$ has Kretschmann scalar ($K=R^{\mu\nu\rho\sigma}R_{\mu\nu\rho\sigma}$, a one-number summary of the curvature tensor) $K=15r_S^2/r_g^6$ at the surface using the interior metric and $K=12r_S^2/r_g^6$ using the exterior metric.

 

[King Solomon]

Did you even read my post 3?

The moment you introduce tides my brain explodes. The answer Ibix gave is something I can relate to though, where space is continuous, while having abrupt changes.

Overall, I just wanted to know if the 2D profile of a gravity well of common celestial object (a massive homogeneous spherical/ellipsoidal object) is shaped like a U or W.

Red Line = U; Blue Line = W

 

[Dale]

The moment you introduce tides my brain explodes.

Curvature is tidal effects. You cannot discuss one without the other. They are just different names for the same thing. If you think that your brain is fine with curvature but explodes with tides then you are making a mistake with one of those.

 

[Nugatory]

Overall, I just wanted to know if the 2D profile of a gravity well of common celestial object (a massive homogeneous spherical/ellipsoidal object) is shaped like a U or W.

”Gravity well” is a property of the potential, so can be described without involving curvature and tides.

You can calculate the potential using just classical gravity ignoring relativistic effects; for bodies like the sun and the Earth that’s enough to determine whether it’s a U or a W.

 

[A.T.]

.does this mean that space is not curved,

Overall, I just wanted to know if the 2D profile of a gravity well

Gravity well and space geometry are different things. You seem to confuse the two.

 

[Ibix]

The moment you introduce tides my brain explodes. The answer Ibix gave is something I can relate to though, where space is continuous, while having abrupt changes.

Unfortunately you were asking about curvature, which is the origin of tidal gravity. As I noted in post #2, you did not seem to be thinking about tidal gravity, but rather gravitational force. However you asked about curvature so I answered on that basis.

The metric, which I mentioned in passing in my last post, is not a gravity well. Gravity well is not a term that works well in general relativity in general, although it works in the particular cases you asked about. As @Nugatory says, it's essentially the potential, which is related to the metric but is not the same thing as the metric. In particular, note that it is not the same as the "rubber sheet" model, which is an embedding of a spacelike slice of the spacetime into Euclidean space (edit: as @A.T. noted while I was writing this). That has a lot to do with the difference between relativistic and Newtonian gravity, but does not explain gravity particularly well (despite a lot of misleading popsci).

So I do not know if I've answered your question. Certainly both the potential and the embedding of the spacelike slice are both more like a U than a W. Tidal effects are more like a W. Certainly the metric is continuous, as is its first derivative (which more or less corresponds to what Newtonian physics would call gravitational force), but its second derivative (curvature, tidal gravity) need not be.

The lesson here is to write a couple of sentences upfront on why you are asking. You may get a whole bunch of clarifying questions, but the chances of getting misleading answers because you aren't asking the question you think you are asking are much lower.

And finally, do tidal forces really make your head explode? Your feet are closer to the center of the Earth than your head, so the force on them is different. That means you are being very slightly pulled apart - that's tidal force.

 

[A.T.]

In particular, note that it is not the same as the "rubber sheet" model, which is an embedding of a spacelike slice of the spacetime into Euclidean space (edit: as @A.T. noted while I was writing this).

The confusion between the spacelike slice of the spacetime and the potential well is one of several reasons, why the rubber sheet analogy for GR is so bad:

- Rolling balls on a rubber sheet can be used as a qualitative analogy for the gravity well (gravitational potential). That's why it gives the correct qualitative result. But that has nothing to do with explaining General Relativity and curved space-time, because it applies equally to Newtonian Gravity.

- The indented rubber sheet can be used as a qualitative visualization of the space (not space-time) distortion in General Relativity (Flamm's paraboloid). But that has has nothing to do with explaining how masses attract each other in General Relativity, which requires including the time dimension (space-time distortion). Flamm's paraboloid represents a distortion of spatial distances between coordinates, and could just as well be shown with the funnel upwards, so the rolling balls would give a wrong result. Therefore rolling some balls on the curved surface representing Flamm's paraboloid makes no sense.

- The local intrinsic curvature of space-time you are asking about is primarily related to tidal-effects, or gravity gradient.

 

[Vanadium 50]

Gravity well and space geometry are different things.

And space geometry and spacetime geometry are also different things.

 

[King Solomon]

Don't think I'm ignoring the above responses, I just haven't had time to let them sink in.

But I've understood the rubber sheet model to be an orthogonal projection of 4D spacetime into a 3D subspace, and thus a true representation of reality.

 

[PeterDonis]

I've understood the rubber sheet model to be an orthogonal projection of 4D spacetime into a 3D subspace

No, it's not a projection. It's just picking out one particular 3D subspace and showing the geometry of that subspace by itself.

and thus a true representation of reality

No, because, first, which 3D subspace you pick is coordinate-dependent; there is no invariant notion of "space" in GR. And second, representing just a 3D subspace leaves out the all-important element of time.

 

[Ibix]

But I've understood the rubber sheet model to be an orthogonal projection of 4D spacetime into a 3D subspace, and thus a true representation of reality.

No, it isn't a projection. It's actually a pretty abstract representation when you get into the nuts and bolts - but it happens to yield a diagram which is incredibly easy to interpret intuitively. Unfortunately, that intuitive interpretation is pretty much wrong.

If you build a ring around a planet and measure its circumference, $C$, and then drill a hole through the planet so you can measure the ring's radius, $r$, you will find that $C\neq 2\pi r$ because of the non-Euclidean nature of spacetime. Imagine building a whole lot of rings of different radii lying in the same plane around the planet. And to get from one ring to the next you build stepladders.

Now let's try to build a scale model of that on a table. We know the circumferences of each ring, so we build scale models of the rings and lay them on the table. Then we build scale models of the ladders - but they are too long to fit between the rings. The table is Euclidean, so the distance between rings of diameter $C_1$ and $C_2$ is $C_2/2\pi-C_1/2\pi$. But the length of the ladder is modeled on the non-Eucludian reality, and the real distance between the rings is not $C_2/2\pi-C_1/2\pi$ - it's slightly greater. To make the slightly-too-long ladder actually fit, I have to lift the larger ring slightly off the table so that the ladder can be on a diagonal. If I had four ladders at 90⁰ intervals round the ring I could actually prop the outer ring up with them. Then I could get a third ring and prop it up a bit higher on the ladders connecting it to the second ring, and so on. That would build up a bowl structure if I kept on adding rings.

That bowl is what the embedding diagram, the "rubber sheet", shows. It's representing something real enough - the distance along the rings and ladders is to scale with what it would be in the real thing. But we've pushed a planar structure in reality into a bowl in the model so the junctions are all distorted. And does it really "explain gravity" in any way?

The reality is that this embedding has very little to do with gravity. It's not completely unconnected, but we actually dropped the most relevant aspects of curvature along the way. But the resulting structure is very similar to a model of potential which you can roll balls along so that they follow conic section trajectories, which is a different kind of abstraction, and people get misled. It's important to understand that all such models are abstractions, useful models that illustrate some aspect of reality, and not take them too literally.

 

[A.T.]

But I've understood the rubber sheet model to be an orthogonal projection of 4D spacetime into a 3D subspace, and thus a true representation of reality.

The rubber sheet is a surface, so it just a 2D subspace of 4D space-time. It is embedded in an Euclidean 3D "illustration space" to visualize its non-Euclidean geometry.

The rubber sheet has only 2 spatial dimensions, but for gravity the time dimension is key. As a wrote earlier:

- The indented rubber sheet can be used as a qualitative visualization of the space (not space-time) distortion in General Relativity (Flamm's paraboloid). But that has has nothing to do with explaining how masses attract each other in General Relativity, which requires including the time dimension (space-time distortion). Flamm's paraboloid represents a distortion of spatial distances between coordinates, and could just as well be shown with the funnel upwards, so the rolling balls would give a wrong result. Therefore rolling some balls on the curved surface representing Flamm's paraboloid makes no sense.

Here is what it looks like with time included locally.

For a more global picture follow the links in the video description on youtube.

 

[King Solomon]

No, it isn't a projection. It's actually a pretty abstract representation when you get into the nuts and bolts - but it happens to yield a diagram which is incredibly easy to interpret intuitively. Unfortunately, that intuitive interpretation is pretty much wrong.

If you build a ring around a planet and measure its circumference, $C$, and then drill a hole through the planet so you can measure the ring's radius, $r$, you will find that $C\neq 2\pi r$ because of the non-Euclidean nature of spacetime. Imagine building a whole lot of rings of different radii lying in the same plane around the planet. And to get from one ring to the next you build stepladders.

Now let's try to build a scale model of that on a table. We know the circumferences of each ring, so we build scale models of the rings and lay them on the table. Then we build scale models of the ladders - but they are too long to fit between the rings. The table is Euclidean, so the distance between rings of diameter $C_1$ and $C_2$ is $C_2/2\pi-C_1/2\pi$. But the length of the ladder is modeled on the non-Eucludian reality, and the real distance between the rings is not $C_2/2\pi-C_1/2\pi$ - it's slightly greater. To make the slightly-too-long ladder actually fit, I have to lift the larger ring slightly off the table so that the ladder can be on a diagonal. If I had four ladders at 90⁰ intervals round the ring I could actually prop the outer ring up with them. Then I could get a third ring and prop it up a bit higher on the ladders connecting it to the second ring, and so on. That would build up a bowl structure if I kept on adding rings.

That bowl is what the embedding diagram, the "rubber sheet", shows. It's representing something real enough - the distance along the rings and ladders is to scale with what it would be in the real thing. But we've pushed a planar structure in reality into a bowl in the model so the junctions are all distorted. And does it really "explain gravity" in any way?

The reality is that this embedding has very little to do with gravity. It's not completely unconnected, but we actually dropped the most relevant aspects of curvature along the way. But the resulting structure is very similar to a model of potential which you can roll balls along so that they follow conic section trajectories, which is a different kind of abstraction, and people get misled. It's important to understand that all such models are abstractions, useful models that illustrate some aspect of reality, and not take them too literally.

I found this to be the most useful answer to the question.I do want to partition the response though.

Suppose we had five rings around a star. Let the radius of the star (center to surface) = A

The first ring occurs r = 2A, the second ring occurs are r = A, the third ring occurs at r = 1/2 A and the final ring occurs at r = 1/100A, and fifth ring occurs at 100A.

Which of these rings has it true radius most distorted (proportionally).

I would would seems rings 4 and 5, at r= 1/100 A and r=100A would have their radii least affected (assuming the W model), and ring 3 at r = A would suffer the most distortion.

But under the U model, ring 5 at r = 100A would have the least distortion and ring 4 at r = 1/100A would suffer the most distortion of its true radius.

Whatever the answer is to this question solves my wonders about whether the U or the W is correct interpolation of the measures of curvature about a homogeneous ellipsoid.

 

[Ibix]

You are asking how the ratio of $C/2\pi$ to the measured-with-a-ruler radius of your rings varies with the size. That's exactly the kind of question you can answer with the embedding diagram. $C/2\pi$ is the straight line diameter of a ring, while the measured distance is the distance following the surface.

For very small rings near the center of the bowl the distances are similar, for the same reason you regard your floor as flat even though it's on the curved surface of the Earth. For very large rings the distances are similar because the bowl flattens out at large radii and you get a large (nearly) flat surface with a relatively small dent in the middle. So the ratio of the distances tends to 1 for small and large rings, but will be different at in between sizes. Exact numbers, like what size of ring has the maximum ratio, would depend on the internal structure of the star. Even if I were to assume a structure that would give me a differential equation that may or may not be soluble without numerical integration.

The absolute difference between the two radii will increase with ring size, of course.

 

[King Solomon]

Ok, so given everything that I know, and that light travels in a straight line geodesic, then the path of light from the exterior object towards its center has the following parametric equation in 4D space.

x = a1(t) + 0
y = a2(t) + 0
z = a3(t) + 0 (without gravity we'd stop here)
w = a4(t) + 0

Where the constant a4 is determined by some function of the mass of the object, f(M).

Also please be kind. I'm nearly 80 years old and I wish to have some understanding of how this world works before I die.

 

[Ibix]

I have no idea what you are trying to express here. The geodesic equations for the Schwarzschild metric reduce to equations 7.43, 7.44, 7.47 and 7.48 in chapter 7 of Carroll's GR notes. Set $\epsilon=0$ for light. You will need a numerical integrator for anything other than pure radial or circular motion.

 

[A.T.]

Suppose we had five rings around a star.

Are all outside of the star? Then you can use the external Schwarzshild metric:

From: https://www.researchgate.net/figure/Cone-like-embedding-diagram-visualizing-the-equatorial-plane-curvature-of-the_fig2_335210694

The distance ds on the upper diagram is the real physical radial distance (proper distance). The distance dr on the lower is calculated from the circumference or surface area using Euclidean formulas (assuming flat space).

But if you go inside it looks like this the round tip at the bottom here:

From: https://en.wikipedia.org/wiki/Interior_Schwarzschild_metricSo the spatial "distortion" expressed as the ratio ds/dt is greatest at the surface, and goes towards zero at the center and towards infinity. But note that:

- This spatial "distortion" is not related to gravitational pull itself. In fact it doesn't affect objects initially at rest at all.

- The spatial curvature is not zero at the center (its a spherical geometry), and neither is space-time curvature (not shown here).

 

[King Solomon]

Are all outside of the star? Then you can use the external Schwarzshild metric:

View attachment 268220
From: https://www.researchgate.net/figure/Cone-like-embedding-diagram-visualizing-the-equatorial-plane-curvature-of-the_fig2_335210694

The distance ds on the upper diagram is the real physical radial distance (proper distance). The distance dr on the lower is calculated from the circumference or surface area using Euclidean formulas (assuming flat space).

But if you go inside it looks like this the round tip at the bottom here:

View attachment 268221
From: https://en.wikipedia.org/wiki/Interior_Schwarzschild_metricSo the spatial "distortion" expressed as the ratio ds/dt is greatest at the surface, and goes towards zero at the center and towards infinity. But note that:

- This spatial "distortion" is not related to gravitational pull itself. In fact it doesn't affect objects initially at rest at all.

- The spatial curvature is not zero at the center (its a spherical geometry), and neither is space-time curvature (not shown here).

Ok, so the first derivative is zero at the center of the object, which makes sense. So the effect on the curvature from the surface of an object to its center decreases, as noted by inflection point located at the boundary between the grey and black shade ( I assume this boundary represents the actual surface of the object); however, the effects due to gravity remain at the center.

This is making sense of the original responses I got from this topic.

I assume that the evaluation of this function (Ein Field Equ) goes to infinity with a finite distance from the center of the object given sufficient density of the generating object (Schwarzschild Radius)?

Also I'm assuming a non-rotating object (so I don't have a stroke).
--------------------------

For the record rings 1 and 5 were outside the star; ring 3 was the surface, rings 3 and 4 were inside.

 

[A.T.]

So the effect on the curvature from the surface of an object to its center decreases, as noted by inflection point located at the boundary between the grey and black shade ( I assume this boundary represents the actual surface of the object);

Curvature is related the 2nd derivatives. It is positive inside and negative outside (for a uniform density sphere).

however, the effects due to gravity remain at the center.

Gravity has nothing to do with these diagrams. You need to include the time dimension for that. See my earlier posts, and chapter 2 of this thesis:
http://www.relativitet.se/Webtheses/tes.pdf