# Does the Reissner-Nordstrom Metric Necessarily Represent a Charged Black-hole?

The Reissner-Nordstrom metric

$$ds^2 = - \bigg(1-\dfrac{2M}{r}+\dfrac{4\pi Q^2}{r^2}\bigg)dt^2 + \dfrac{1}{\bigg(1-\dfrac{2M}{r}+\dfrac{4\pi Q^2}{r^2}\bigg)} dr^2 + r^2 d\Omega_2^2$$

is a solution of Einstein equations if the non-trivial components of the stress-energy tensor are the following:

$$T_{tt} = \dfrac{Q^2\bigg(1-\dfrac{2M}{r}+\dfrac{4\pi Q^2}{r^2}\bigg)}{r^4}$$

$$T_{rr} = \dfrac{-Q^2}{r^4\bigg(1-\dfrac{2M}{r}+\dfrac{4\pi Q^2}{r^2}\bigg)}$$

$$T_{\theta\theta} = \dfrac{Q^2}{2 r^2}.$$

Here, the origin of these energy densities can be easily explained if we identify the parameter $$Q$$ as an electric charge. The energy densities will then be just the energy densities of the electric field of such an electric charge as obtained by solving general relativistic Maxwell's equations. And thus, the Reissner-Nordstrom metric is generally regarded as a solution of coupled Einstein-Maxwell equations and is also regarded to represent a charged black hole with charge $$Q$$ and mass $$M$$.

Now, clearly, the metric of a charged black hole will be a Reissner-Nordstrom metric, but my doubt is that can this metric also represent something else? I mean can the energy densities to which the metric is a solution of Einstein equations have a different origin? I cannot think of a reason why the same gravitational situation should imply the same electromagnetic situation. Is there any argument which dictates that these energy densities can only arise out of a radially falling static electric field?

• Nice question ! Jun 12, 2017 at 12:27

## 3 Answers

It is the only possible electrovac (stress energy only due to electromagnetic field, otherwise vacuum) solution (i.e., an external solution) that is static, as the generalized Birkhoff's theorem has it. It is stated in wiki at https://en.m.wikipedia.org/wiki/Birkhoff%27s_theorem_(relativity), also in standard textbooks like Wald.

There's a dual solution that represents a magnetic monopole, but that gives a different stress energy (and Einstein) tensor also.

The Harvard paper referenced below from the 1960's depicts an interior solution of a charged gravitating body in equilibrium, with the outside metric the Reissner Nordstrom metric. Thus, it's not only black holes, but any static charged body. http://adsabs.harvard.edu/full/1965MNRAS.129..443B

It is true that Q is normally interpreted as charge. As you go to infinity you get that M is the total mass, so M doesn't have other interpretations.

The question of whether you can get the same stress energy tensor with something other than electromagnetism on the right side is not obvious. But once you get the metric clearly Q is something other than mass, and it is clearly some scalar constant. The equations of motion of a charged particle with mass will exactly predict what you could measure. At infinity you'd measure that total charge inside a spherical volume is indeed Q.

I suppose this is not proof. I suspect one should be able to prove that it could only arise from a Maxwell source, but I am not certain, and could be wrong. Possibly someone else can provide a definitive answer.

• "There's a dual solution that represents a magnetic monopole, but that gives a different stress energy (and Einstein) tensor also." - I'm pretty sure the metric (and stress-energy etc.) is identical if you use the same units for electric and magnetic charge. Jul 26 at 19:07

There's no fundamental reason within GR itself that it would stem from an electromagnetic field. Any matter field that would have the same stress energy tensor would also give us that metric.

The trick is of course that I don't think that any non-EM field could satisfy it, except for another vector field with the same dynamic. I think you can obtain the solution from magnetic monopoles but I think it might be slightly different.

The Reissner-Nordstrom type of metrics given by

$$\begin{eqnarray} && ds^2 = -f(r)dt^2 + \frac{dr^2}{f(r)} + r^2 d\Omega^2~, \\ && f(r) = 1-\frac{2M}{r} + \frac{Q^2}{r^2}~, \end{eqnarray}$$ arise as a solution of the field theory $$$$\mathcal{S} = \int d^4x \sqrt{-g} \left(\frac{R}{2} - \frac{1}{4}F_{\mu\nu}F^{\mu\nu}\right)~.$$$$ This particular spacetime element however could be a solution of a $$f(R)$$ gravity theory with constant curvature see equation 2.10 in f(R) Black holes.

It is also possible that this solution might be a solution of the full Horndeski theory for some particular potentials (i don't know if this is true though).

To give another example, the BBMB black hole (equation 12 in) Exact Black Hole Solutions with a Conformally Coupled Scalar Field and Dynamic Ricci Curvature in f(R) Gravity Theories is given by the same spacetime form but with

$$$$f(r) = \left(1-\frac{m}{r}\right)^2$$$$

and it gravitationally undistinguisable from the case of a Reissner-Nordstrom black hole with a charge equal to the mass (extremal black hole). However, the BBMB metric arises from the action

$$$$\mathcal{S} = \int d^4x \sqrt{-g} \left(\frac{R}{2} - \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi - \frac{1}{12} R\phi^2\right)~.$$$$

Besides the fact that the BBMB black hole is unphysical for many reasons (see the second cited article), if someone would use the metric function $$f(r) = \left(1-m/r\right)^2$$ to probe an astrophysical scenario, none could reassure that this spacetime metric comes from an extremal Reissner-Nordstrom black hole and not from a BBMB black hole.

In conclusion, at the level of a spacetime element one cannot make concrete arguments about the matter or theory that could source a particular geometry, only speculations.