- #1

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

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

\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

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