I What Is the Radius of Curvature of Space on Earth's Surface Due to Its Mass?

[Jaime Rudas]

"Matter tells spacetime how to curve, and curved spacetime tells matter how to move"

Taking the above into account, the space on the Earth's surface is slightly curved. What is the radius of curvature of space on the Earth's surface due to its mass?

 

[A.T.]

"Matter tells spacetime how to curve, and curved spacetime tells matter how to move"

Taking the above into account, the space on the Earth's surface is slightly curved. What is the radius of curvature of space on the Earth's surface due to its mass?

1) It's spacetime that is curved, not space.

2) Radius of curvature quantifies extrinsic curvature, but spacetime curvature is intrinsic.

3) Note that the spacetime curvature is related to tidal forces or gradient of the gravitational field strength, not to the field strength (gravitational acceleration) itself.

 

[Jaime Rudas]

1) It's spacetime that is curved, not space.

Yes, matter curves spacetime, but if spacetime is curved couldn't it imply that space also curves?

2) Radius of curvature quantifies extrinsic curvature, but spacetime curvature is intrinsic.

I understand that space can also have intrinsic curvature and, if this is homogeneous, its radius can be determined. Am I wrong?

 

[Nugatory]

I understand that space can also have intrinsic curvature and, if this is homogeneous, its radius can be determined. Am I wrong?

Not exactly, although the concept is of relatively little usefulness.
If we're working in the weak-field approximation (appropriate in the vicinity of the earth) we can consider the quantity $R=c^2/g$ to be the curvature of spacetime. This being an I-level thread, you might want to try deriving this formula for yourself: consider a ball thrown straight up into the air and falling back to earth on a geodesic trajectory, look for the radius of a circle that intersects the three events ball thrown, ball reaches maximum height, ball back on earth again.

 

[Jaime Rudas]

If we're working in the weak-field approximation (appropriate in the vicinity of the earth) we can consider the quantity $R=c^2/g$ to be the curvature of spacetime.

$R=c^2/g≈9.2\cdot 10^{15}$ m

Could it then be said that the radius of curvature of space on the surface of the Earth is approximately $9.2\cdot 10^{15}$ meters?

 

[Nugatory]

$R=c^2/g≈9.2\cdot 10^{15}$ m

Could it then be said that the radius of curvature of space on the surface of the Earth is approximately $9.2\cdot 10^{15}$ meters?

It's the curvature of spacetime, but yes, that's about the right order of magnitude. It's more or less one light-year.
No one can make someone else exercise, but that's never stopped anyone from handing out advice to do so: trying to derive that formula would be a good exercise that will tell more than any number of questions here.

 

[Jaime Rudas]

It's the curvature of spacetime, but yes, that's about the right order of magnitude. It's more or less one light-year.

Yes, I understand that it is the curvature of spacetime, but doesn't that imply that space is also curved?

No one can make someone else exercise, but that's never stopped anyone from handing out advice to do so: trying to derive that formula would be a good exercise that will tell more than any number of questions here.

Yes, I'm trying to derive the formula, but I'm afraid I don't have enough knowledge.

 

[Nugatory]

Yes, I understand that it is the curvature of spacetime, doesn't that imply that space is also curved?

Kinda sorta, yes. But spacetime can be split into time and space in multiple more or less arbitrary ways, leading to different values for the "curvature of space".

 

[A.T.]

Yes, matter curves spacetime, but if spacetime is curved couldn't it imply that space also curves?

The spatial curvature for a uniform mass density sphere, in the rest frame of the sphere, is described here:
https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric#Visualization

At the surface of the sphere the spatial scalar curvature goes from positive (sphere interior) to negative (sphere exterior), so it is zero.

its radius can be determined.

A radius of curvature describes the extrinsic curvature of 1-dimensional object (a curve). With higher dimensional objects you don't have a single extrinsic radius of curvature, but several of them. You can combine them, for example buy multiplying, and relate that to intrinsic curvature measures:

https://en.wikipedia.org/wiki/Gaussian_curvature

 

[Dale]

At the surface of the sphere the spatial scalar curvature goes from positive (sphere interior) to negative (sphere exterior), so it is zero.

Oh, is it smooth even with the discontinuity of the stress energy tensor?

Of course, this is assuming that you are defining space as orthogonal to the timelike Killing vector field.

 

[Nugatory]

Yes, I'm trying to derive the formula, but I'm afraid I don't have enough knowledge.

You have three points (in spacetime not space!). For any three points there will be only one circle that passes through all three. What is the radius of that circle?

 

[Jaime Rudas]

Kinda sorta, yes. But spacetime can be split into time and space in multiple more or less arbitrary ways, leading to different values for the "curvature of space".

I see, but for an observer at rest on the surface of the Earth, what would be the curvature of space?

 

[A.T.]

Oh, is it smooth even with the discontinuity of the stress energy tensor?

Good point. Not sure about that.

 

[PeterDonis]

If we're working in the weak-field approximation (appropriate in the vicinity of the earth) we can consider the quantity $R=c^2/g$ to be the curvature of spacetime.

Kinda sorta. One can't exactly say this kind of hand-waving argument is wrong, since even Misner, Thorne, & Wheeler give it in an early chapter of their classic GR textbook, but it's still a hand-waving argument that doesn't really get where it needs to go. It is best viewed as a vague heuristic that can be used to motivate a more detailed investigation.

As was already said, spacetime curvature is tidal gravity. The tidal gravity near the surface of the Earth is approximately $M / R^3$ in geometric units, or $GM / c^2 R^3$ in conventional units. We can of course use $g = GM / R^2$ to rewrite this as $g / c^2 R$, and then we can wave our hands and say that $c^2 / g$ is sort of like a radius of curvature, but there's still that other factor of $R$ there which is the radius of the Earth. (And there's also a factor of $2$ that appears if you're talking about radial tidal gravity as opposed to tangential.)

 

[PeterDonis]

is it smooth even with the discontinuity of the stress energy tensor?

Yes. A proper model will make use of the appropriate junction conditions at the boundary that enforce the smoothness.

 

[A.T.]

I see, but for an observer at rest on the surface of the Earth, what would be the curvature of space?

https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric#Visualization

 

[PeterDonis]

if spacetime is curved couldn't it imply that space also curves?

The key point is that "space curvature" doesn't predict anything. Spacetime curvature is what makes physical predictions.

 

[Ted Jacobson]

A geometry of two dimensions, like the surface of a sphere or an egg, has a single number that characterizes it's intrinsic curvature at each point. The "radius of curvature" is defined as the inverse of the square root of the curvature. In three or higher dimensions, to characterize the curvature in different directions you need more numbers.

One intuitive way to visualize this is in terms of what are called "sectional curvatures". To define a sectional curvature at a point, you choose a 2d tangent plane localized at that point, and shoot out geodesics in all directions emanating from that point and tangent to that plane. This generates a 2d surface in the higher dimensional space or spacetime.

That 2d surface has an intrinsic metric, inherited (i.e. induced) from the metric of the ambient space, and the curvature of that surface at the origin point is the sectional curvature associated with that plane. The collection of sectional curvatures associated with all the 2d planes at a point determines all the information contained in the Riemann curvature tensor at that point.

Now, we can interpret your question as asking what are the sectional curvatures at a point on the surface of the Earth. It we choose, as you suggest, a frame at rest with respect to the surface of the Earth, we could consider for example the sectional curvature associated with a plane tangent to the surface of the Earth, or a vertical plane.

The latter is -2 times the former, and the former is GM/r^3c^2 = g/(c^2r), where r is the radius of the Earth, so the vertical-horizontal sectional curvature is positive, and its curvature radius is√c^2r/g ~ 2 x 10^11 meters, while the horizontal-horizontal sectional curvature is negative, and its curvature radius is 4 x 10^11 meters. [If you add two orthogonal vertical-horizonal curvatures to the horizontal-horizontal one you get zero. This means that the curvature scalar of the spatial geometry in the rest frame of the Earth is zero at the surface, which is correct since the stress tensor vanishes outside the matter of the Earth.]

 

[A.T.]

The key point is that "space curvature" doesn't predict anything.

I would not be that harsh to spatial curvature. Hypothetically, if somebody wanted to build some giant structure around and/or inside a massive object, he would have to factor the non-Euclidean geometry of space into the design.

 

[PeterDonis]

Hypothetically, if somebody wanted to build some giant structure around and/or inside a massive object, he would have to factor the non-Euclidean geometry of space into the design.

This implicitly defines "space" the way @Dale did in post #10--but that definition is not frame-dependent because the timelike Killing vector field, in spacetimes that have one, is an invariant geometric property of the spacetime. So the "space" defined this way does actually have a frame-independent geometric meaning. "Space" in general in GR does not.

 

[A.T.]

So the "space" defined this way does actually have a frame-independent geometric meaning.

I objected to the statement that spatial curvature doesn't predict anything. Now we are already talking about frame-dependent vs. frame-independent predictions.

Also, I would say that definitions which imply a reference frame, are actually frame-independent.

Like in: "velocity" is frame-dependent, but "velocity relative to ..." is frame-independent.

In this case, the amount of building material needed to build a structure near/inside a given massive object is a frame-independent prediction.

 

[PeterDonis]

I objected to the statement that spatial curvature doesn't predict anything.

Yes, to be more precise I should have added a qualifier that the "space" whose curvature is being evaluated is not picked out by any invariant geometric property of the spacetime.

Now we are already talking about frame-dependent vs. frame-independent predictions.

To be more precise, we are talking about predictions that can be grounded in invariant geometric properties of the spacetime, vs. predictions that can't.

In this case, the amount of building material needed to build a structure near/inside a given massive object is a frame-independent prediction.

Yes, because the structure itself, as well as the massive object, are static--which means their worldlines are orbits of a timelike Killing vector field that is hypersurface orthogonal. That is the invariant geometric property of the spacetime that makes the prediction invariant.

 

[Demystifier]

Yes, I understand that it is the curvature of spacetime, but doesn't that imply that space is also curved?

It doesn't. A counterexample is Robertson-Walker metric with $k=0$.

 

[martinbn]

Just to add one more thing. There is no such thing as "the space". Given a spacetime there are many ways to pick a space and a time (excluding those where you cannot at all). So a question about the curvature of the space is meaningless without additional information. And one more, there are different notions of curvature and in general it is not just a number, so any question about "the curvature" is also meaningless without more information.

 

[PeroK]

"Space, the coordinate-dependent frontier..."

 

[Jaime Rudas]

It doesn't. A counterexample is Robertson-Walker metric with $k=0$.

Yes, but I am referring to the particular case of space on the surface of the Earth.

 

[martinbn]

Yes, but I am referring to the particular case of space on the surface of the Earth.

And what do you mean by that?

 

[cianfa72]

This implicitly defines "space" the way @Dale did in post #10--but that definition is not frame-dependent because the timelike Killing vector field, in spacetimes that have one, is an invariant geometric property of the spacetime. So the "space" defined this way does actually have a frame-independent geometric meaning. "Space" in general in GR does not.

As @Dale said, the "space" we are talking about consists of "pieces of local spacelike planes" orthogonal at each point to the timelike Killing vector field (assuming it exists for the given spacetime). However such "local spacelike planes" in general will not combine/match up giving rise to spacelike hypersurfaces.

 

[PeterDonis]

I am referring to the particular case of space on the surface of the Earth.

It still depends on how you pick "space". In Painleve coordinates on Schwarzschild spacetime, for example, "space" is flat (Euclidean 3-space) even though spacetime is curved. In other words, just saying "space on the surface of the Earth" still does not pick out a unique "space".

 

[JimWhoKnew]

[If you add two orthogonal vertical-horizonal curvatures to the horizontal-horizontal one you get zero. This means that the curvature scalar of the spatial geometry in the rest frame of the Earth is zero at the surface, which is correct since the stress tensor vanishes outside the matter of the Earth.]

If we take $x$ , $y$ , as the horizontal directions, and $z$ as the vertical one, then the horizontal-horizontal sectional curvature should be $R_{xyxy}$ (up to notation convention; I take it from do-Carmo's book on Riemannian geometry, English translation). The vertical-horizontal sectional curvatures are $R_{xzxz}$ and $R_{yzyz}$ . How does the vanishing¹ of the stress tensor infer the vanishing of $R_{xyxy} +R_{xzxz}+R_{yzyz}$ ?

¹ The Earth has an atmosphere, but who cares?

 

[PeterDonis]

How does the vanishing¹ of the stress tensor infer the vanishing of $R_{xyxy} +R_{xzxz}+R_{yzyz}$ ?

First, you are looking at the wrong components. The correct components to look at are $R_{r t r t}$, $R_{\theta t \theta t}$, and $R_{\phi t \phi t}$. Those are the components that describe tidal gravity. (Note that I have written them in spherical coordinates--the radial component is twice the tangential components with the sign flipped.)

As for why the sum of all three vanishes if the stress-energy tensor vanishes, this is because of the vanishing of the Ricci tensor; the sum of the three components given above (strictly speaking, you have to raise the first index before summing) is the Ricci tensor component $R_{tt}$. This discussion by John Baez and Emory F. Bunn might help:

https://arxiv.org/abs/gr-qc/0103044

 

[PeterDonis]

If you add two orthogonal vertical-horizonal curvatures to the horizontal-horizontal one you get zero. This means that the curvature scalar of the spatial geometry in the rest frame of the Earth is zero at the surface

As I said in post #31 just now, these are the wrong components. The relevant curvature components that get linked to the stress-energy tensor are the ones I gave in that post. Note that those are not pure "space-space" components, they are "time-space" components. Your statement about adding the three components to get zero applies to those components, since the radial one is twice the two tangential ones with the sign flipped.

All the above is about the Riemann tensor of spacetime, which is not the same thing as the Riemann tensor of "the spatial geometry in the rest frame of the Earth". Nor can you just lop off the components involving $t$ from the spacetime Riemann tensor. There is no Einstein Field Equation that relates "the spatial geometry in the rest frame of the Earth" to the stress-energy tensor. There is only an Einstein Field Equation that relates the spacetime geometry (more particularly its Einstein tensor) to the stress-energy tensor.

 

[Jaime Rudas]

And what do you mean by that?

I mean, more or less, the same thing that Feynman describes in his Lectures on Physics, in particular, what he describes in 42-3

 

[JimWhoKnew]

First, you are looking at the wrong components.

Thanks. I've read B&B's paper before, and by your advice, took a fresh peep at it just now.

The relation between the stress tensor and the Ricci tensor is of the very basics of GR. No reason to post a question here. @Ted Jacobson was talking about sectional curvatures, so I wonder whether he has a new insight (i.e. new to me). In this context, I think I referred correctly to the components that match his description.

 

[PeterDonis]

I mean, more or less, the same thing that Feynman describes in his Lectures on Physics, in particular, what he describes in 42-3

In that section he is not giving a full description of GR. You can tell that because he says: "Einstein said that space is curved and that matter is the source of the curvature." No, actually Einstein did not say that. He said that spacetime is curved and that stress-energy (a more accurate term than "matter" although the difference is negligible for this discussion) is the source of the curvature. In the case of the Earth, if we use the Earth's rest frame, the spatial part of the spacetime curvature (the "radius excess" that Feynman talks about in that section) is actually a very small effect; all of the familiar phenomena we associate with gravity in the Earth's vicinity involve the components of spacetime curvature that include time.

 

[Jaime Rudas]

In the case of the Earth, if we use the Earth's rest frame, the spatial part of the spacetime curvature (the "radius excess" that Feynman talks about in that section) is actually a very small effect

Yes, that was the answer I was waiting for. That the spatial curvature on the Earth's surface is that corresponding to the Earth having a radius excess of 1.5 millimeters.

 

[A.T.]

Yes, that was the answer I was waiting for. That the spatial curvature on the Earth's surface is that corresponding to the Earth having a radius excess of 1.5 millimeters.

The radius excess doesn't quantify the local spatial curvature on the Earth's surface, but rather the combined effect of spatial curvature throughout the entire Earth's interior. And radius excess is not a "radius of curvature" that you have been asking about.

 

[JimWhoKnew]

@Ted Jacobson was talking about sectional curvatures, so I wonder whether he has a new insight (i.e. new to me). In this context, I think I referred correctly to the components that match his description.

It seems that @Ted Jacobson was right. I was reminded by section 11.2 in Feynman's "Lectures on Gravitation" that $$G^0_{~~0}=R^{12}_{~~~~ 12}+R^{13}_{~~~~13}+R^{23}_{~~~~23}\quad.$$Up to a possible sign, this is also equal to the curvature scalar $R^{(3)}$ in three spatial dimensions, and therefore can be used in Eq. 11.2.5 to derive the relation $$r_{excess}=\frac{m_{inside}}3\quad.$$For some reason, the excess term in Eq. 11.2.5 differs by a factor of 2 from the second equation here.