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Wald Chapter 10 Problem 2

  1. Feb 2, 2014 #1
    EDIT dw i figured it out, not sure how to remove it though!
     
    Last edited: Feb 2, 2014
  2. jcsd
  3. Feb 2, 2014 #2

    WannabeNewton

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    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)!
     
  4. Feb 3, 2014 #3
    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 [itex]F^{ab}\nabla _an_b=0[/itex] 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!
     
  5. Feb 3, 2014 #4

    WannabeNewton

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    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.
     
  6. Feb 4, 2014 #5
    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
     
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