tommyj said:
Sorry to get your hopes up but I made mistakes in both parts! How did you do part a) may i ask? I can't get terms to dissappear.
Let me start by showing that Gauss's law for electricity holds on the space-like Cauchy surface. Keep in mind that ##n^{a}n_{a} = -1##, which implies that ##n^{a}\nabla_{b}n_{a} = 0##; also keep in mind that ##n_{[a}\nabla_{b}n_{c]} = 0## since the unit normal field is hypersurface orthogonal to the space-like foliation ##\Sigma_t## (c.f. Theorem 8.3.14).
I will denote the derivative operator associated with the spatial metric ##h_{ab}## by ##\tilde{\nabla}_{a}##.
We have ##\tilde{\nabla}_{a}E^{a} = h_{a}{}{}^{b}h^{a}{}{}_{c}\nabla_{b}(F^{c}{}{}_{d}n^{d})\\ = (\delta^{bc} + n^{b}n^{c})(n^{d}\nabla_{b}F_{cd} + F_{cd}\nabla_{b}n^{d})\\ = n^{d}\nabla^{c}F_{cd} + F_{cd}\nabla^{c}n^{d} + n^{b}n^{c}n^{d}\nabla_{b}F_{cd} + n^{b}n^{c}F_{cd}\nabla_{b}n^{d}##.
Now ##n^{c}n^{d}\nabla_{b}F_{cd} = n^{d}n^{c}\nabla_{b}F_{dc} = -n^{c}n^{d}\nabla_{b}F_{cd}\Rightarrow n^{c}n^{d}\nabla_{b}F_{cd} = 0##
and ##n_{[a}\nabla_{b}n_{c]} = 0\Rightarrow n^{b}n^{c}F_{cd}\nabla_{b}n^{d} - n^{b}n^{d}F_{cd}\nabla_{b}n^{c}= 2n^{b}n^{c}F_{cd}\nabla_{b}n^{d}\\ = F_{cd}\nabla^{d}n^{c} - F_{cd}\nabla^{c}n^{d} = - 2F_{cd}\nabla^{c}n^{d}##
thus ##\tilde{\nabla}_{a}E^{a}= n^{d}\nabla^{c}F_{cd} = -4\pi j_{d}n^{d} = 4\pi\rho## by virtue of the inhomogeneous Maxwell equations.
Showing Gauss's law for magnetism holds on the spacelike Cauchy surface is very similar.
We have ##\tilde{\nabla}_{a}B^{a} =-\frac{1}{2}\epsilon^{cdef} h_{a}{}{}^{b}h^{a}{}{}_{c}\nabla_{b}(F_{de}n_{f})\\ = -\frac{1}{2}\epsilon^{cdef} (n_{f}\nabla_{c}F_{de} + F_{de}\nabla_{c}n_{f} + n^{b}n_{c}n_{f}\nabla_{b}F_{de} + n^{b}n_{c}F_{de}\nabla_{b}n_{f})##.
Now ##\epsilon^{cdef}n_{c}n_{f} = 0## because the volume form is totally antisymmetric and just as before we have ##n_{[a}\nabla_{b}n_{c]} = 0 \Rightarrow \epsilon^{cdef} n^{b}n_{c}F_{de}\nabla_{b}n_{f} - \epsilon^{cdef} n^{b}n_{f}F_{de}\nabla_{b}n_{c}= 2\epsilon^{cdef} n^{b}n_{c}F_{de}\nabla_{b}n_{f}\\ = \epsilon^{cdef} F_{de}\nabla_{f}n_{c} - \epsilon^{cdef}F_{de}\nabla_{c}n_{f} = -2 \epsilon^{cdef}F_{de}\nabla_{c}n_{f}##
so we are left with ##\tilde{\nabla}_{a}B^{a} = -\frac{1}{2}\epsilon^{cdef} n_{f}\nabla_{c}F_{de}##. But ##\epsilon^{cdef}\nabla_{c}F_{de} = -\epsilon^{cdef}\nabla_{d}F_{ce} = \epsilon^{cdef}\nabla_{e}F_{cd}## hence ##3\epsilon^{cdef}\nabla_{c}F_{de} = 3\epsilon^{cdef}\nabla_{[c}F_{de]} = 0 ## by virtue of the homogeneous Maxwell equations thus we have the desired result ##\tilde{\nabla}_{a}B^{a} = 0##.
tommyj said:
Also, for part b) both constraints only seem to imply that F^{ab}\nabla _an_b=0 which don't really seem to help to solve for the time derivative of the initial conditions. I've spent ages looking and can't find anything on it!
I'm stuck on that calculation as well. I have to finish my particle physics HW but after that I'll take another jab at it.