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

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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).
 
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ergospherical said:
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.
 
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