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

[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.