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Maxwell's Equations in Curved Space-time

  1. Jun 26, 2013 #1
    Can one show that strict charge conservation ##\nabla_{a}J^{a} = 0## follows directly from ##\nabla_{a}F^{ab} = 4\pi J^{b}## alone?

    Also, how does ##d^{\star}F = 4\pi ^{\star}J## follow directly from that same equation where ##\star## is the Hodge dual operator?
     
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  3. Jun 26, 2013 #2

    WannabeNewton

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    Yes and it is extremely easy to show (almost trivial in fact). Starting with ##\nabla^{a}F_{ab} = 4\pi j_{b}##, we have ##\nabla^{b}\nabla^{a}F_{ab} = \nabla^{a}\nabla^{b}F_{ba} = -\nabla^{a}\nabla^{b}F_{ab}= 4\pi \nabla^{b}j_{b}## i.e. ##\nabla^{b}\nabla^{a}F_{ab} -\nabla^{a}\nabla^{b}F_{ab}= 8\pi \nabla^{b}j_{b}##. Now ##\nabla_{b}\nabla_{a}F^{ab} -\nabla_{a}\nabla_{b}F^{ab}= -R_{bae}{}{}^{a}F^{eb} - R_{bae}{}{}^{b}F^{ae} = -R_{be}F^{eb} + R_{ae}F^{ae} = 0## hence ##\nabla^{a}j_{a} = 0##.

    It is also very easy to show that ##d(^{\star }F) = 4\pi(^{\star }j)##. We have ##(^{\star}F)_{ab} = \frac{1}{2}\epsilon_{abcd}F^{cd}## so ##\epsilon^{abef}\nabla_{e}(^{\star}F)_{ab} = \frac{1}{2}\epsilon^{abef}\epsilon_{abcd}\nabla_{e}F^{cd} = -2\nabla_{e}F^{ef} = -8\pi j^{f}##. Hence ##\epsilon_{fjki}\epsilon^{feab}\nabla_{e}(^{\star}F)_{ab} = -6\nabla_{[j}(^{\star}F)_{ki]}= -8\pi\epsilon_{fjki} j^{f} = -8\pi(^{\star}j)_{jki}## therefore ##3\nabla_{[a}(^{\star}F)_{bc]} = d(^{\star}F)_{abc} = 4\pi(^{\star}j)_{abc}## i.e. ##d(^{\star}F) = 4\pi(^{\star}j)##.

    Note the implications of this. Because ##\nabla^{a}j_{a} = 0## in any space-time, we can apply Stokes' theorem to a space-time region ##\Omega \subseteq M## bounded by two space-like hypersurfaces ##\Sigma, \Sigma'## from a single foliation and find that ##\int _{\Omega}\nabla^{a}j_{a} = 0 = \int _{\Sigma}j_{a}n^{a} -\int _{\Sigma'}j_{a}n^{a}## i.e. the total charge ##Q = -\int _{\Sigma}j_{a}n^{a} ## is conserved (here ##n^{a}## is the outward unit normal field to the space-like foliation that ##\Sigma,\Sigma'## belong to; the negative sign is to compensate for the negative sign that comes out of the inner product in the integral).
     
  4. Jun 26, 2013 #3
    Thanks WannabeNewton! Your answer helps me a lot. I also have a related question. I read that a killing vector field always solves Maxwell's equations in curved (vacuum) space-time. Can you explain why this is the case?
     
    Last edited: Jun 26, 2013
  5. Jun 26, 2013 #4

    WannabeNewton

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    Well let's say we have a killing vector field ##\xi^{a}## and use it as a 4-potential so that ##F_{ab} = 2\nabla_{[a}\xi_{b]} = 2\nabla_{a}\xi_{b}##. Then ##\nabla^{a}F_{ab} =2\nabla^{a}\nabla_{a}\xi_{b}##.
    Since ##R_{abcd}\xi^{d} = \nabla_{a}\nabla_{b}\xi_{c} - \nabla_{b}\nabla_{a}\xi_{c} = -\nabla_{a}\nabla_{c}\xi_{b} + \nabla_{b}\nabla_{c}\xi_{a}\\ = R_{cabd}\xi^{d} - \nabla_{c}\nabla_{a}\xi_{b} + R_{bcad}\xi^{d} + \nabla_{c}\nabla_{b}\xi_{a} = 2\nabla_{c}\nabla_{b}\xi_{a} + R_{cabd}\xi^{d} + R_{bcad}\xi^{d}##
    we have ##2\nabla_{c}\nabla_{b}\xi_{a}=(R_{abcd} - R_{cabd} - R_{bcad})\xi^{d} ##. Using the first Bianchi identity, we find that ##R_{abcd} + R_{bcad} + R_{cabd} = 0\Rightarrow R_{abcd} =- R_{cabd} -R_{bcad} ## i.e. ##\nabla_{c}\nabla_{b}\xi_{a} = R_{abcd}\xi^{d}##.

    In particular, ##\nabla^{a}\nabla_{a}\xi_{b} = - R_{ab}{}{}^{a}{}{}_{d}\xi^{d} = -R_{bd}\xi^{d}##. If we are in vacuum space-time then ##R_{ab} = 0## so ##\nabla^{a}F_{ab} = \nabla^{a}\nabla_{a}\xi_{b} = 0##.

    Now ##\nabla_{[a}F_{bc]} = 2\nabla_{[a}\nabla_{b}\xi_{c]} = 2R_{[abc]d}\xi^{d} = 0## by virtue of the first Bianchi identity. Hence ##F_{ab} = 2\nabla_{a}\xi_{b}## solves Maxwell's equations in vacuum.

    Such solutions aren't always of physical interest but here is a paper that gives an example of a killing vector field solution to Maxwell's equations that is of physical interest: http://prd.aps.org/abstract/PRD/v10/i6/p1680_1
     
  6. Jul 2, 2013 #5
    Thanks. As always very helpful. You say in another thread (https://www.physicsforums.com/showpost.php?p=4359286&postcount=1)

    "I will spare you the details of the calculations involved in showing these two relations hold; you can, for now, take my word that I have indeed shown them to be true." Could you please show the details of that calculation? I am still new to these kind of tensor calculus manipulations and seeing more examples would be nice, especially another related to EM in curved space-time.
     
  7. Jul 2, 2013 #6

    WannabeNewton

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    Sure! Let's first show 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##, and that ##n_{[a}\nabla_{b}n_{c]} = 0## since the unit normal field is hypersurface orthogonal to the space-like foliation. 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##.
     
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