# Intuition for the Ricci tensor of a Reissner-Nordström black hole

The Reissner-Nordström metric is given by $$g = -h(r)\,{\rm d}t^2 + h(r)^{-1}\,{\rm d}r^2 + r^2\,{\rm d}\Omega^2,$$where $${\rm d}\Omega^2$$ is the round metric on a unit sphere $$\Bbb S^2$$ and $$h(r) = 1-2mr^{-1}+qr^{-2}$$, where $$m\geq 0$$ is a mass and $$q\in \Bbb R$$ is an electric charge ($$q=0$$ gives a Schwarzschild black hole, $$q=m=0$$ gives empty Minkowski space). One can compute that $$R_{ij} = -qr^{-4}g_{ij}$$ for $$i,j\in \{t,r\}$$ and $$R_{ij} = qr^{-4}g_{ij}$$ for $$i,j\in \{\theta,\varphi\}$$.

Is there a physical interpretation for the fact that we get different signs for different sets of indices? Or in other words, why should we expect this to happen?

Actually, the Reissner-Nordström metric has $$h(r)=1-2m/r+q^2/r^2\tag1.$$ (The charge is squared.)

The Ricci tensor is diagonal with $$R_{tt}=-\frac{q^2}{r^4}g_{tt}\tag{2a},$$ $$R_{rr}=-\frac{q^2}{r^4}g_{rr}\tag{2b},$$ $$R_{\theta\theta}=\frac{q^2}{r^4}g_{\theta\theta}\tag{2c},$$ $$R_{\phi\phi}=\frac{q^2}{r^4}g_{\phi\phi}\tag{2d}.$$

This can be understood based on the form of the energy-momentum tensor $$T_{\mu\nu}$$. It is purely electromagnetic, with

$$4\pi T_{\mu\nu}=F_{\mu\alpha}F_\nu{}^\alpha-\frac14g_{\mu\nu}F^2\tag{3}$$

where $$F_{\mu\nu}$$ is the electromagnetic field tensor due to the hole’s charge $$q$$.

The electromagnetic field of a Reissner-Nordström hole consists of a radial electric field and no magnetic field. Its nonzero tensor components are only $$F_{tr}$$ and $$F_{rt}$$. In this case one finds that the nonzero components of (3) are

$$4\pi T_{tt}=\frac14g_{tt}F^2\tag{4a},$$ $$4\pi T_{rr}=\frac14g_{rr}F^2\tag{4b},$$ $$4\pi T_{\theta\theta}=-\frac14g_{\theta\theta}F^2\tag{4c},$$ $$4\pi T_{\phi\phi}=-\frac14g_{\phi\phi}F^2\tag{4d}$$

where

$$F^2=F_{\mu\nu}F^{\mu\nu}=-2\frac{q^2}{r^4}\tag{5}.$$

Finally, the Ricci scalar $$R$$ vanishes (which happens because the electromagnetic energy-momentum tensor is traceless), so the Einstein field equations simplify to

$$R_{\mu\nu}=8\pi T_{\mu\nu}\tag{6}.$$

Combining (6), (4), and (5) gives (2).

So the short answer is that the relationship you found is because the electric field is radial and the magnetic field is zero. Such an electromagnetic field has a mixed energy-momentum tensor of the form $$T^\mu{}_\nu\propto\text{diag}(-1,-1,1,1)$$ in spherical coordinates $$(t,r,\theta,\phi)$$, and so the mixed Ricci tensor has the same form.

• This answer is great. Not only confirmed the results I got, but the last paragraph hit the nail on the head. Thank you! – Ivo Terek Sep 6 at 19:22
• I had fun figuring it out. I had a hunch that $T_{\mu\nu}$ would explain it, but I wasn’t sure. – G. Smith Sep 6 at 19:24
• I thought the same thing... If we had the same signs throughout, then Schur's theorem would say that $q^2/r^4$ is constant, and this would force $q=0$. So I knew the explanation should come from physics and not mathematics -- $T_{\mu\nu}$ was the first thing to come to mind. The explanation made perfect sense to me, and I'm not a physicist :P – Ivo Terek Sep 6 at 19:27