Theory of relativity/Reissner-Nordström

The Reissner-Nordström solution is the exact solution to General relativity/Einstein equations for the stress-energy tensor for the electric field from a point charge. It works as a charged nonrotating black hole solution.

$d{s}^2=(1-\frac{2GM}{r{c}^2}+\frac{k_{e}G{q}^2}{r^2{c}^4})dc{t}^2-\frac{d{r}^2}{(1-\frac{2GM}{r{c}^2}+\frac{k_{e}G{q}^2}{r^2{c}^4})}-r^{2}d{\theta }^2-r^2{\sin }^2\theta d{\varphi }^2$

Working out the solution

Prior to the discovery of the Reissner-Nordström solution Schwarzschild had discovered the noncharged nonrotating black hole solution, the Schwarzschild metric

$d{s}^2=(1-\frac{2GM}{r{c}^2})dc{t}^2-\frac{d{r}^2}{(1-\frac{2GM}{r{c}^2})}-r^{2}d{\theta }^2-r^2{\sin }^2\theta d{\varphi }^2$

which is an exact vacuum solution to Einstein's field equations. Given the Schwarzschild solution as known it is not as difficult as one might assume to arrive at the Reissner-Nordström solution. The Schwarzschild solution takes the equation form of

$d{s}^2=(1+f\left(r\right))dc{t}^2-\frac{d{r}^2}{(1+f\left(r\right))}-r^{2}d{\theta }^2-r^2{\sin }^2\theta d{\varphi }^2$

So it is only natural that one would start with that as a trial solution in looking for the exact point charge solution for the field equations. To do this one finds the Einstein tensor for that line element and then looks to see if there is an $f\left(r\right)$ that yields the right Einstein tensor corresponding to the stress-energy tensor of the electric field from a point charge. The Einstein tensor one arrives at for that line element is

$||G_{\nu }^{\mu }||=\left[\begin{array}{cccc}\frac{-(f+r\frac{df}{dr})}{r^2}& 0& 0& 0\\ 0& \frac{-(f+r\frac{df}{dr})}{r^2}& 0& 0\\ 0& 0& \frac{-(r^2\frac{d^{2}f}{d{r}^2}+2r\frac{df}{dr})}{2r^2}& 0\\ 0& 0& 0& \frac{-(r^2\frac{d^{2}f}{d{r}^2}+2r\frac{df}{dr})}{2r^2}\end{array}\right]$

The stress-energy tensor for a field mediated by massless particles such as the electric field from a point charge corresponds to a zero Ricci-scalar. This means tells us that $T_{\mu }^{\mu }=0$ and that therefor $G_{\mu }^{\mu }=0$. For this to be the case we must have

$\frac{-(f+r\frac{df}{dr})}{r^2}+\frac{-(r^2\frac{d^{2}f}{d{r}^2}+2r\frac{df}{dr})}{2r^2}=0$

which gives us the following second order differential equation to solve

$r^2\frac{d^{2}f}{d{r}^2}+4r\frac{df}{dr}+2f=0$

Inserting the trial solution $f=A{r}^n$ one finds that $n=-1,-2$ yielding

$f=\frac{A}{r}+\frac{B}{r^2}$

We may recognize from the Schwarzschild solution that $A=\frac{-2GM}{c^2}$, but to find $B$ we insert the solution for $f$ into the equation for the Einstein tensor and look for what corresponds to the exact solution from the electric field. We get

$||G_{\nu }^{\mu }||=\left[\begin{array}{cccc}\frac{B}{r^4}& 0& 0& 0\\ 0& \frac{B}{r^4}& 0& 0\\ 0& 0& -\frac{B}{r^4}& 0\\ 0& 0& 0& -\frac{B}{r^4}\end{array}\right]$

The stress-energy tensor for an electric field from a point charge is

$||T_{\nu }^{\mu }||=\left[\begin{array}{cccc}\frac{1}{2}{ϵ}_0{\left(k_e\frac{q^2}{r^2}\right)}^2& 0& 0& 0\\ 0& \frac{1}{2}{ϵ}_0{\left(k_e\frac{q^2}{r^2}\right)}^2& 0& 0\\ 0& 0& -\frac{1}{2}{ϵ}_0{\left(k_e\frac{q^2}{r^2}\right)}^2& 0\\ 0& 0& 0& -\frac{1}{2}{ϵ}_0{\left(k_e\frac{q^2}{r^2}\right)}^2\end{array}\right]$

And Einstein's field equations relate the two tensors by

$G_{\nu }^{\mu }=\frac{8\pi G}{c^4}{T}_{\nu }^{\mu }$

where G is the gravitational constant of the universe. From this we get

$\frac{B}{r^4}=\frac{8\pi G}{c^4}\frac{1}{2}{ϵ}_0{\left(k_e\frac{q^2}{r^2}\right)}^2$

Recall that $k_e=\frac{1}{4\pi {ϵ}_0}$ to arrive at

$B=\frac{k_{e}G{q}^2}{c^4}$

So we find that the exact solution to Einstein's field equations for the stress-energy tensor from the electric field of a spinless point charge is

$d{s}^2=(1-\frac{2GM}{r{c}^2}+\frac{k_{e}G{q}^2}{r^2{c}^4})dc{t}^2-\frac{d{r}^2}{(1-\frac{2GM}{r{c}^2}+\frac{k_{e}G{q}^2}{r^2{c}^4})}-r^{2}d{\theta }^2-r^2{\sin }^2\theta d{\varphi }^2$

It is common for authors to either use geometric-plank units where $k_e=G=1$ or to convert the charge to units of distance by $e=\frac{\sqrt{k_{e}G}}{c^2}q$ and the mass to distance units by $m=\frac{GM}{c^2}$ in which case you will more often find it written something like

$d{s}^2=(1-\frac{2m}{r}+\frac{e^2}{r^2})d{t}^2-\frac{d{r}^2}{(1-\frac{2m}{r}+\frac{e^2}{r^2})}-r^{2}d{\theta }^2-r^2{\sin }^2\theta d{\varphi }^2$

Geodesic Motion

The equations of geodesic motion for a neutral test mass with respect to proper time are:

$\frac{dt}{d\tau }=\frac{\gamma }{(1-\frac{2GM}{r{c}^2}+\frac{k_{e}G{q}^2}{r^2{c}^4})}$

where $\gamma$ is a constant of the motion called the energy parameter, the energy per $m{c}^2$ for the test mass.

$\frac{d\varphi }{d\tau }=\frac{l_z}{r^2{\sin }^2\theta }$

where $l_z$ is the conserved angular momentum per mass $m$ for the test mass.

$\frac{d}{d\tau }\left(r^2\frac{d\theta }{d\tau }\right)=r^2\sin \theta \cos \theta {\left(\frac{d\varphi }{d\tau }\right)}^2$

and finally

$\frac{{\gamma }^2-1}{2}{c}^2=\frac{1}{2}{\left(\frac{dr}{d\tau }\right)}^2-\frac{GM}{r}+\frac{k_{e}G{q}^2}{2r^2{c}^2}+\frac{1}{2}(\frac{l_z^2}{r^2{\sin }^2\theta }+r^2{\left(\frac{d\theta }{d\tau }\right)}^2)(1-\frac{2GM}{r{c}^2}+\frac{k_{e}G{q}^2}{r^2{c}^4})$

For radial motion this reduces to

$\frac{dt}{d\tau }=\frac{\gamma }{(1-\frac{2GM}{r{c}^2}+\frac{k_{e}G{q}^2}{r^2{c}^4})}$

$\frac{{\gamma }^2-1}{2}{c}^2=\frac{1}{2}{\left(\frac{dr}{d\tau }\right)}^2-\frac{GM}{r}+\frac{k_{e}G{q}^2}{2r^2{c}^4}$

Wormhole Structure

Penrose diagram for coordinate extension of a charged or rotating black hole

Penrose diagram for coordinate extension of a charged or rotating black hole

Above we see a Penrose diagram representing a coordinate extension (1) for a charged or rotating black hole. The known charged rotating black hole solution is the Kerr–Newman metric. The same way as mapping Schwarzschild coordinates onto Kruskal-Szekeres coordinate reveals two separate external regions for the Schwarzschild black hole, such a mapping done for a charged or rotating hole reveals an even more multiply connected region for charged and rotating black holes. Lets say region I represents our external region outside a charged black hole. In the same way that the other external region is inaccessible as the wormhole connection is not transversible, external region II is also not accessible from region I. The difference is that there are other external regions VII and VIII which are indeed accessible from region I by transversible paths at least one way. One should expect this as the radial movement case of geodesic motion for a neutral test particle written above leads back out of the hole without intersecting the physical singularity at r=0.

References

(1)Black Holes-Part 4 pp 26-35

• Relativity

See also

• Schwarzschild metric
• Kerr–Newman metric