# Wald Chapter 10 Problem 2

1. Feb 2, 2014

### tommyj

EDIT dw i figured it out, not sure how to remove it though!

Last edited: Feb 2, 2014
2. Feb 2, 2014

### WannabeNewton

Could you show me how you did part (b) of problem 10.2?

Some while back I did problem 10.2 and I thought I did part (b) of it right (https://www.physicsforums.com/showthread.php?t=687641#post4359286) but a few weeks later I found a mistake in my solution. Since then I forgot about the problem but you bringing it up has reminded me I have yet to still fix the mistake in my solution to part (b)!

3. Feb 3, 2014

### tommyj

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.

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!

4. Feb 3, 2014

### WannabeNewton

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 $\tilde{\nabla}_{a}B^{a} = 0$.

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.

5. Feb 4, 2014

### tommyj

ah man I even noted the hypersurface orthogonal relation as I knew it would be useful but I forgot about it. You live and learn as they say. thanks alot, I know how much effort it is to write tensor equations on here so I really appreciate it!

I cannot see how to do the second part. Its completely different from the example in the book, the Ricci Tensor term just messes everything up for the second constraint part