# Deriving the Reissner-Nordström metric

• etotheipi
T_{ab}## tensor is a sum of three terms, the first of which is ##A^2##, the second of which is ##-A^2/4##, and the third of which is ##0##. So if I understand correctly, the first term is the energy density, the second term is the pressure in the direction of the radial coordinate, and the third term is the pressure in the direction of the angular coordinates?In summary, the conversation discusses the Maxwell equations and the Einstein equations in the context of tetrads and one forms. It is mentioned that the Maxwell equations can be written as ##\mathrm{d} \star F_{ab} = 0## and ##\mathrm
etotheipi
Homework Statement
This is following question 3), of chapter 6) of Wald.

Part a) is to argue that the general form of a Maxwell tensor in a static, spherically symmetric spacetime is ##F_{ab} = 2A(r) (e_0)_{[a} (e_1)_{b]} + 2B(r) (e_2)_{[a} (e_3)_{b]}##, then part b) is to show that in the case ##B(r) = 0##, the general solution has ##A(r) = -q/r^2##. And part c) asks to finally use the Einstein equation to derive the RN metric.
Relevant Equations
The general form of a static, spherically symmetric metric is$$\mathrm{d} s^2 = -f(r) \mathrm{d} t^2 + h(r) \mathrm{d} r^2 + r^2 \mathrm{d} \Omega^2$$and the following tetrad has already been defined:\begin{align*} (e_0)_a &= f^{1/2} (\mathrm{d} t)_a \\ (e_1)_a &= h^{1/2} (\mathrm{d} r)_a \\ (e_2)_a &= r(\mathrm{d} \theta)_a \\ (e_3)_a &= r\sin{\theta} (\mathrm{d} \phi)_a \\ \end{align*}[N.B. everything in this post will use abstract index notation]
I don't know how to do (a), so I decided to ignore it for now and just assume the result. Because ##j^a = 0## the Maxwell equations are ##\mathrm{d} \star F_{ab} = 0## and ##\mathrm{d} F_{ab} = 0##. For any two one forms, ##\frac{1}{2} \omega_a \wedge \eta_b = \omega_{[a} \eta_{b]}##, and so we may write ##F_{ab} = 2A(r) (e_0)_{[a} (e_1)_{b]} = A(r) (e_0)_{a} \wedge (e_1)_{b}##, so that the Maxwell equations are\begin{align*} \mathrm{d} \star F_{ab} = \mathrm{d} \left( A(r) \star ((e_0)_{a} \wedge (e_1)_{b}) \right) &= \mathrm{d} (A(r) (e_2)_a \wedge (e_3)_b) \\ &= \mathrm{d} ( A(r) \, r(\mathrm{d} \theta)_a \wedge r\sin{\theta} (\mathrm{d} \phi)_b ) \\ &= \mathrm{d} (r^2 A(r) \sin{\theta}) \wedge (\mathrm{d} \theta)_a \wedge (\mathrm{d} \phi)_b + 0 + 0 \end{align*}but since ##\mathrm{d} (r^2 A(r) \sin{\theta}) = \partial_r (r^2 A(r)) (\mathrm{d} r)_a \sin{\theta} + \partial_{\theta}(\sin{\theta}) (\mathrm{d} \theta)_a r^2 A(r)##, we just have$$\mathrm{d} \star F_{ab} = \partial_r (r^2 A(r)) (\mathrm{d} r)_a \wedge (\mathrm{d} \theta)_a \wedge (\mathrm{d} \phi)_a \equiv 0$$and that being identically zero implies ##\partial_r (r^2 A(r)) = 0 \implies r^2 A(r) := -q## is a constant.

For (c) we have the Einstein equations\begin{align*} 8 \pi T_{00} &= (rh^2)^{-1} h' + r^{-2}(1-h^{-1}) \\ 8 \pi T_{11} &= (rfh)^{-1} f' - r^{-2}(1-h^{-1}) \\ 8 \pi T_{22} &= \frac{1}{2} (fh)^{-1/2} \frac{d}{dr} [(fh)^{-1/2} f'] + \frac{1}{2} (rfh)^{-1} f' - \frac{1}{2}(rh^2) h'\end{align*}and we know that\begin{align*} T_{ab} &= \frac{1}{4\pi} \left\{ F_{ac} {F_b}^c - \frac{1}{4} g_{ab} F_{de} F^{de} \right\} \\ T_{ab} &= \frac{q^2}{\pi r^4} \left\{ (e_0)_{[a} (e_1)_{c]} (e_0)_{[b} (e_1)^{c]} + \frac{1}{4} g_{ab} (e_0)_{[d} (e_1)_{e]} (e_0)^{[d} (e_1)^{e]}\right\} \\ T_{ab} &= \frac{q^2 fh}{\pi r^4} \left\{ (\mathrm{d} t)_{[a} (\mathrm{d} r)_{c]} (\mathrm{d} t)_{[b} (\mathrm{d} r)^{c]} + \frac{1}{4} g_{ab} (\mathrm{d} t)_{[d} (\mathrm{d} r)_{e]} (\mathrm{d} t)^{[d} (\mathrm{d} r)^{e]}\right\} \\ \end{align*}How can you simplify that last expression? I'm not very comfortable yet with tetrads and my work is getting very messy

Last edited by a moderator:
JD_PM
I did not check your computation. However, section 31.1 here might help

etotheipi
JD_PM said:
I did not check your computation. However, section 31.1 here might help

Thanks! At first glance it looks like he took a bit of a different approach to how I was trying to do it

JD_PM
I think I've made some progress. I realized I was not helping myself by converting from the tetrad basis back into the non-normalised coordinate basis, and actually doing that caused me to make a few errors. Let's start from\begin{align*}
T_{ab} = \frac{1}{4\pi} \left\{ F_{ac} {F_{b}}^c - \frac{1}{4} g_{ab} F_{de} F^{de} \right\}
\end{align*}and consider the first term ##F_{ac} {F_{b}}^c ##, and use that ##(e_0)^a = -(e_0)_a## and ##(e_1)^a = (e_1)_a## as well as the orthonormality of the tetrad, ##(e_{\mu})^a##, i.e. ##(e_{\mu})_a (e_{\nu})^a = \eta_{\mu \nu}##\begin{align*}

F_{ac} {F_{b}}^c &= 4A^2(e_1)_{[a} (e_1)_{c]} (e_0)_{[b} (e_1)^{c]} \\

&= A^2 [(e_0)_a (e_1)_c - (e_0)_c (e_1)_a][(e_0)_b (e_1)^c - (e_0)^c (e_1)_b] \\ \\

&= A^2 [(e_0)_a (e_1)_c (e_0)_b (e_1)^c + 0 + 0 + (e_0)_c (e_1)_a (e_0)^c (e_1)_b]\\ \\

&= A^2 \left[ (e_0)_a (e_0)_b - (e_1)_a (e_1)_b \right]

\end{align*}Since ##(e_0)^0 = -(e_0)_0 = 1## and ##(e_1)^0 = (e_1)^0 = 1##, it follows that ##F_{0 \nu}{F_{0}}^{\nu} = A^2##, then ##F_{1 \nu}{F_{1}}^{\nu} = -A^2## and then ##F_{2 \nu}{F_{2}}^{\nu} = 0##. Now let's consider ##F_{de} F^{de}##, and again use the orthonormality,
\begin{align*}

F_{de}F^{de} &= 4A^2 (e_0)_{[d} (e_1)_{e]} (e_0)^{[d} (e_1)^{e]} \\

&= A^2 [(e_0)_d (e_1)_e - (e_0)_e (e_1)_d][(e_0)^d (e_1)^e - (e_0)^e (e_1)^d] \\

&= A^2 [ -1 + 0 + 0 - 1 ] = -2A^2

\end{align*}The ##T_{\mu \nu}## are then\begin{align*}

T_{00} = \frac{1}{4\pi} \left\{ A^2 +\frac{1}{4}(-2A^2) \right\} = \frac{A^2}{8\pi} = \frac{q^2}{8 \pi r^4} \\

T_{11} = \frac{1}{4\pi} \left\{ -A^2 -\frac{1}{4}(-2A^2) \right\} = -\frac{A^2}{8\pi} = \frac{-q^2}{8 \pi r^4} \\

T_{22} = \frac{1}{4\pi} \left\{ 0 -\frac{1}{4}(-2A^2) \right\} = \frac{A^2}{8\pi} = \frac{q^2}{8 \pi r^4}
\end{align*}thus we need to solve the equations\begin{align*}

\frac{q^2}{r^4} &= (rh^2)^{-1} h' + r^{-2}(1-h^{-1}) \\

\frac{-q^2}{r^4} &= (rfh)^{-1} f' - r^{-2}(1-h^{-1}) \\

\frac{q^2}{r^4} &= \frac{1}{2} (fh)^{-1/2} \frac{d}{dr} [(fh)^{-1/2} f'] + \frac{1}{2} (rfh)^{-1} f' - \frac{1}{2}(rh^2)^{-1} h'
\end{align*}Do these look right? I'm not sure how to go about solving that...

Last edited by a moderator:
vanhees71
etotheipi said:
The ##T_{\mu \nu}## are then\begin{align*}

T_{00} = \frac{1}{4\pi} \left\{ A^2 +\frac{1}{4}(-2A^2) \right\} = \frac{A^2}{8\pi} = \frac{q^2}{8 \pi r^4} \\

T_{11} = \frac{1}{4\pi} \left\{ -A^2 -\frac{1}{4}(-2A^2) \right\} = -\frac{A^2}{8\pi} = \frac{-q^2}{8 \pi r^4} \\

T_{22} = \frac{1}{4\pi} \left\{ 0 -\frac{1}{4}(-2A^2) \right\} = \frac{A^2}{8\pi} = \frac{q^2}{8 \pi r^4}
\end{align*}thus we need to solve the equations\begin{align*}

\frac{q^2}{r^4} &= (rh^2)^{-1} h' + r^{-2}(1-h^{-1}) \\

\frac{-q^2}{r^4} &= (rfh)^{-1} f' - r^{-2}(1-h^{-1}) \\

\frac{q^2}{r^4} &= \frac{1}{2} (fh)^{-1/2} \frac{d}{dr} [(fh)^{-1/2} f'] + \frac{1}{2} (rfh)^{-1} f' - \frac{1}{2}(rh^2) h'
\end{align*}Do these look right?
These look good to me.

I'm not sure how to go about solving that...
Look at how Wald solves equations (6.135), (6.1.36), and (6.137).

vanhees71 and etotheipi
TSny said:
Look at how Wald solves equations (6.135), (6.1.36), and (6.137).
Ah, thanks! The equations to solve, numbered:
\begin{align*}
\frac{q^2}{r^4} &= (rh^2)^{-1} h' + r^{-2}(1-h^{-1}) \quad(1) \\
\frac{-q^2}{r^4} &= (rfh)^{-1} f' - r^{-2}(1-h^{-1}) \quad (2) \\
\frac{q^2}{r^4} &= \frac{1}{2} (fh)^{-1/2} \frac{d}{dr} [(fh)^{-1/2} f'] + \frac{1}{2} (rfh)^{-1} f' - \frac{1}{2}(rh^2)^{-1} h' \quad (3)
\end{align*}
Adding ##(1)## and ##(2)##,$$0 = \frac{h'}{rh^2} + \frac{f'}{rfh} \implies \frac{h'}{h} + \frac{f'}{f} = 0 \implies fh' + hf' = (fh)' = 0$$so that ##fh = A##, and using Wald's trick of re-scaling ##t## such as to make ##A=1## we then have ##h=1/f##. Making that substitution in ##(3)## yields$$\frac{q^2}{r^4} = \frac{1}{2} f'' + \frac{1}{2r}f' + \frac{1}{2r}f' = \frac{1}{2} f'' + \frac{1}{r} f'$$which can be re-written as$$\frac{q^2}{r^2} = \frac{1}{2}r^2 f'' + r f' = \frac{d}{dr} \left( \frac{1}{2}r^2f' \right) \, \implies \, c-\frac{q^2}{r} = \frac{1}{2}r^2 f'$$so we get$$f' = \frac{c'}{r^2} - \frac{2q^2}{r^3} \implies f = - \frac{c'}{r} + \frac{q^2}{r^2} + c''$$and then, I guess we're supposed to set ##c' = 2M## and ##c'' = 1## [I can't totally see why we're allowed to do that, it doesn't seem to preserve generality...]? The ##g_{rr}## term is just ##h = 1/f##.

Thanks again

vanhees71 and TSny
etotheipi said:
I guess we're supposed to set ##c' = 2M## and ##c'' = 1## [I can't totally see why we're allowed to do that, it doesn't seem to preserve generality...]?

Maybe one can argue that if ##q## is small enough, or ##r## large enough, such that the ##q^2/r^2## term in ##f## is negligible, then ##f## should agree with the Schwarzschild solution.

vanhees71 and etotheipi
etotheipi said:
I don't know how to do (a)

You might consider the following:

First, that the Killing vector field ##\partial_t## is orthogonal to the ##\theta##, ##\phi## subspace (i.e., the 2-spheres--this is a general result for stationary, spherically symmetric spacetimes, and is easily shown from the metric you give).

Second, that the vector field ##\partial_r## in spacelike hypersurfaces of constant ##t## is also orthogonal to the 2-spheres, and therefore the spacetime as a whole can be split into two orthogonal pieces, the ##t##, ##r## piece and the ##\theta##, ##\phi## piece.

Third, that the geometric properties of the EM field must match the geometric properties of the spacetime, so the EM field, geometrically, should be capable of being split into two pieces in the same way as the spacetime geometry.

vanhees71 and etotheipi
etotheipi said:
I guess we're supposed to set ##c' = 2M## and ##c'' = 1##

You shouldn't have to. For any static, spherically symmetric spacetime, the ##00## component of the EFE, which is your Equation (1), is exactly solvable with the function

$$h(r) = \frac{1}{1 - 2m(r) / r}$$

with

$$m(r) = \int_0^r 4 \pi R^2 \rho(R) dR + M$$

where ##M## is an undetermined constant, and ##\rho## is just ##T_{00}##. In the vacuum case, ##\rho = 0## and the integral vanishes, so we have ##m(r) = M## (the Schwarzschild solution). But solving the integral is simple enough for the electrovacuum case you are considering here. The key point is that the specific form of ##h(r)## with the ##m(r)## you get from the integral can be shown to solve your Equation (1) by just plugging it in, and the only freedom of choice remaining is the choice of the constant ##M##; the ##1## and the ##2## in ##1 - 2m(r) / r## are not free to vary.

(Note, btw, that all this works because Equation (1) is an equation in ##h## only--##f## does not appear. So it can be solved as an ODE in ##h##. Your analysis did not make use of that fact.)

vanhees71, TSny and etotheipi
Thanks guys, I see why we can fix the constants of integration now! From post #8 it seems more intuitive also that, given that the spacetime can be split into two orthogonal pieces, so must the Maxwell tensor. And further, the static requirement implies the coefficients cannot depend on time, and the spherically symmetric requirement implies the coefficients cannot depend on ##\theta## and ##\phi##.

Another way might be to look at symmetry transformations of ##F_{ab}##, explained in appendix C. Let ##\zeta: \mathcal{M} \longrightarrow \mathcal{M}## be a diffeomorphism corresponding to an [active] rotation in the ##\theta##-##\phi## plane by angle ##\varphi## about any chosen point ##p \in \mathcal{M}##, then\begin{align*}
(\zeta_{*} \bar{e}_{2})_a \big{|}_p &= (e_2)_a \big{|}_p \cos{\varphi} + (e_3)_a \big{|}_p \sin{\varphi} \\
(\zeta_{*} \bar{e}_{3})_a \big{|}_p &= - (e_2)_a \big{|}_p \sin{\varphi} + (e_3)_a \big{|}_p \cos{\varphi}
\end{align*}whilst ##(\zeta_{*} \bar{e}_{0})_a \big{|}_p = (e_0)_a \big{|}_p## and ##(\zeta_{*} \bar{e}_{1})_a \big{|}_p = (e_1)_a \big{|}_p## are unchanged. The most general possible form of the Maxwell tensor, dropping the ##|_p## notation for brevity, is \begin{align*}F_{ab} = &2A(r) (e_0)_{[a} (e_1)_{b]} + 2B(r) (e_2)_{[a} (e_3)_{b]} + 2C(r) (e_0)_{[a} (e_2)_{b]} + 2D(r) (e_0)_{[a} (e_3)_{b]} \\

&+ 2E(r) (e_1)_{[a} (e_2)_{b]} + 2F(r) (e_1)_{[a} (e_3)_{b]}
\end{align*}Now let's consider the transformed Maxwell tensor after this symmetry transformation. In particular, let's first consider the following part of it:\begin{align*}
\bar{Q}_{ab} := 2C(\bar{r}) (\zeta_{*} \bar{e}_{0})_{[a} (\zeta_{*} \bar{e}_{2})_{b]} + 2D(\bar{r}) (\zeta_{*} \bar{e}_{0})_{[a} (\zeta_{*} \bar{e}_{3})_{b]}
\end{align*}The first term is just\begin{align*}
2C(\bar{r}) (\zeta_{*} \bar{e}_{0})_{[a} (\zeta_{*} \bar{e}_{2})_{b]} &= C(\bar{r}) \left\{ ( \zeta_{*} \bar{e}_{0})_{a} (\zeta_{*} \bar{e}_{2})_{b} - (\zeta_{*} \bar{e}_{0})_{b} (\zeta_{*} \bar{e}_{2})_{a} \right\} \\

&= C(\bar{r}) \left\{ (e_0)_a [(e_2)_b \cos{\varphi} + (e_3)_b \sin{\varphi}] - (e_0)_b [(e_2)_a \cos{\varphi} + (e_3)_a \sin{\varphi}] \right\} \\
&= 2C(\bar{r}) (e_0)_{[a} (e_2)_{b]} \cos{\varphi} + 2C(\bar{r})(e_0)_{[a} (e_3)_{b]} \sin{\varphi}
\end{align*}whilst the second term is\begin{align*}
2D(\bar{r}) (\zeta_{*} \bar{e}_{0})_{[a} (\zeta_{*} \bar{e}_{3})_{b]} &= D(\bar{r}) \left\{ ( \zeta_{*} \bar{e}_{0})_{a} (\zeta_{*} \bar{e}_{3})_{b} - (\zeta_{*} \bar{e}_{0})_{b} (\zeta_{*} \bar{e}_{3})_{a} \right\} \\

&= D(\bar{r}) \left\{ (e_0)_a [-(e_2)_b \sin{\varphi} + (e_3)_b \cos{\varphi}] - (e_0)_b [-(e_2)_a \sin{\varphi} + (e_3)_a \cos{\varphi}] \right\} \\

&= -2D(\bar{r}) (e_0)_{[a} (e_2)_{b]} \sin{\varphi} + 2D(\bar{r})(e_0)_{[a} (e_3)_{b]} \cos{\varphi}
\end{align*}so we obtain\begin{align*}

\bar{Q}_{ab} := (e_0)_{[a} (e_2)_{b]} \left\{ 2C(\bar{r}) \cos{\varphi} - 2D(\bar{r}) \sin{\varphi} \right\} + (e_0)_{[a} (e_3)_{b]} \left\{

2D(\bar{r})\cos{\varphi} + 2C(\bar{r})\sin{\varphi}

\right\}
\end{align*}But we can compare this to\begin{align*}

Q_{ab} := 2C(r)(e_0)_{[a} (e_2)_{b]} + 2D(r)(e_0)_{[a} (e_3)_{b]}
\end{align*}and in order for the symmetry transformation to leave ##F_{ab}## unchanged for any ##\varphi##, we must have ##C(r) = 0## and ##D(r) = 0##. The same argument applies to the similar looking terms involving ##E## and ##F##, and hence we deduce that the four final terms in the Maxwell tensor vanish and we're left with what Wald wrote down!

Last edited by a moderator:
vanhees71, PeterDonis and TSny

## 1. What is the Reissner-Nordström metric?

The Reissner-Nordström metric is a solution to Einstein's field equations in general relativity that describes the spacetime curvature around a charged, non-rotating black hole. It was first derived by Hans Reissner and Gunnar Nordström in 1916.

## 2. How is the Reissner-Nordström metric derived?

The Reissner-Nordström metric is derived by solving the Einstein field equations for a spherically symmetric, static, and electrically charged mass distribution. This involves using the Gauss's law for gravity and the Maxwell's equations for electromagnetism.

## 3. What are the assumptions made in deriving the Reissner-Nordström metric?

The Reissner-Nordström metric is derived under the assumptions of spherical symmetry, staticity, and electrostatics. This means that the spacetime around the black hole is symmetric in all directions, does not change with time, and has no magnetic fields.

## 4. What are the physical implications of the Reissner-Nordström metric?

The Reissner-Nordström metric predicts that the spacetime around a charged black hole is curved, and that the curvature increases with the charge of the black hole. It also predicts the existence of an event horizon, beyond which nothing, including light, can escape the black hole's gravitational pull.

## 5. How does the Reissner-Nordström metric relate to other black hole solutions?

The Reissner-Nordström metric is a special case of the Kerr-Newman metric, which describes the spacetime around a rotating, charged black hole. It is also related to the Schwarzschild metric, which describes the spacetime around a non-charged, non-rotating black hole. The Reissner-Nordström metric is important in understanding the properties and behavior of charged black holes.

Replies
1
Views
978
Replies
1
Views
718
Replies
2
Views
994
Replies
10
Views
2K
Replies
1
Views
1K
Replies
9
Views
1K
Replies
3
Views
1K
Replies
6
Views
2K