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

[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.