Deriving the Reissner-Nordström metric

[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

 

[JD_PM]

I did not check your computation. However, section 31.1 here might help

 

[etotheipi]

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

 

[etotheipi]

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...

 

[TSny]

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).

 

[etotheipi]

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

 

[TSny]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

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.)

 

[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!