Undergrad Show Maxwell's Eqns. on a Cauchy Surface (Wald Ch. 10 Pr.2)

[ergospherical]

This problem is Wald Ch. 10 Pr. 2.; it asks us to show that $D_a E^a = 4\pi \rho$ and $D_a B^a = 0$ on a spacelike Cauchy surface $\Sigma$ (with normal vector $n^a$) of a globally hyperbolic spacetime $(M, g_{ab})$. Using the expression $E_a = F_{ab} n^b$ for the electric field gives $D_a E^a = {h^a}_b {h_a}^c \nabla_c E^b$. I replace ${h^a}_b {h_a}^c = (\delta^a_b + n^a n_b)(\delta_a^c + n_a n^c) = \delta_b^c+ n_b n^c = {h_b}^c$, thereby obtaining\begin{align*}
D_a E^a &= {h_b}^c \nabla_c (F^{bd} n_d) = \nabla_b (F^{bd} n_d) + n_b n^c \nabla_c (F^{bd} n_d) \\
&= n_d \nabla_b F^{bd}+ F^{bd} \nabla_b n_d + n^c n_b n_d \nabla_c F^{bd} + n_b n^c F^{bd} \nabla_c n_d
\end{align*}The third term vanishes because $n_b n_d \nabla_c F^{bd} = n_{(b} n_{d)} \nabla_c F^{[bd]} = 0$. Also, since $n^a$ is orthogonal to $\Sigma$, the condition $n_{[b} \nabla_c n_{d]}$ holds i.e. $n_b \nabla_c n_d$ is totally antisymmetric, and the sum of the second and fourth terms is $F^{bd} \nabla_b n_d + n_b n^c F^{bd} \nabla_c n_d = F^{bd} \nabla_b n_d - n_c n^c F^{bd} \nabla_b n_d = 2F^{bd} \nabla_b n_d$. Using Maxwell's equation $\nabla^a F_{ab} = -4\pi j_b$ on the first term gives\begin{align*}
D_a E^a &= -4\pi n_d j^d + 2F^{bd} \nabla_b n_d \\
&= 4\pi \rho + 2F^{bd} \nabla_b n_d
\end{align*}Does $2F^{bd} \nabla_b n_d = 0$, i.e. is $\nabla_b n_d$ symmetric? I can't see why this should be so. (I thought about $0 = \nabla_b(-1) = \nabla_b(n_a n^a) = n^a \nabla_b n_a$, but this doesn't seem to help).

 

[martinbn]

Does $2F^{bd} \nabla_b n_d = 0$, i.e. is $\nabla_b n_d$ symmetric? I can't see why this should be so. (I thought about $0 = \nabla_b(-1) = \nabla_b(n_a n^a) = n^a \nabla_b n_a$, but this doesn't seem to help).

This is up to a sign the second fundamental form, which is symmetric. With mathematical notations you can do the following. Let $n$ be the normal, and $X,Y$ tangent to the surface. Then differentiating $g(n,Y)=0$ you get $g(\nabla _X n,Y)+g(n,\nabla _XY)=0$. So your expression is the negative of $g(n,\nabla _XY)$, which is symmetric because $g(n,\nabla _XY)=g(n,\nabla _YX)+g(n,[X,Y])$. The last term is zero because the comutator is also tangent to the surface.