In Zee's book, Einstein Gravity in a Nutshell, p. 6 + p. 77, he says that

\begin{equation} R = \text{lim}_{\text{radius} \rightarrow 0} \frac{6}{(\text{radius})^2} \left(1 - \frac{\text{circumference}}{2\pi \text{ radius}} \right) \end{equation}

Then proceeds to use that on two 2d metrics in appendix 1 and that all makes sense. My question is how does this generalize to 3d an beyond? For 3d is it a

\begin{equation} R = \text{lim}_{\text{radius} \rightarrow 0} \frac{6}{(\text{radius})^2} \left(1 - \frac{\text{Surface Area}}{2\pi \text{ radius}} \right) \end{equation}

like object? Then generalized,

\begin{equation} R = \text{lim}_{\text{radius} \rightarrow 0} \frac{6}{(\text{radius})^2} \left(1 - \frac{\text{Spatial Measure in N-1 D}}{2\pi \text{ radius}} \right) \end{equation}

I would love if someone had a better word than "Spatial Measure in N-1 D" for the analogue of Circumference to 2D, Surface area to 3D and on.


1 Answer 1


On p. 6 + p. 77 Ref. 1 is apparently talking about the Gaussian curvature in $d=2$, which is half the scalar curvature in $d=2$. Later on p. 345 + p. 350 Ref. 1 is talking about the scalar curvature $S=g_{ij}R^{ij}$, so let's do the same here.

The Wikipedia page lists that in $d$ dimensions, the scalar curvature is $$ S~=~ \text{lim}_{r\to 0} \frac{6d}{r^2} \left(1 - \frac{{\rm Vol}(\partial B(0,r)\subset M)}{{\rm Vol}(\partial B(0,r)\subset\mathbb{R}^d)}\right), $$ with the volume $${\rm Vol}(\partial B(0,r)\subset\mathbb{R}^d)~=~2\frac{\pi^{\frac{d}{2}}r^{d-1}}{\Gamma(\frac{d}{2})}$$ of a $(d\!-\!1)$-sphere $\partial B(0,r)$ [which is the boundary of a $d$-ball $B(0,r)$].


  1. A. Zee, Einstein Gravity in a Nutshell, 2013; p. 6 + 77 + 345 + 350.

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