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Charge conservation in curved spacetime

  1. Jan 9, 2009 #1
    I'm reading some texts on general relativity and I am wondering how one can mathematically proof that the covariant derivative (wrt mu) of the four-vector j^mu equals zero.
    I know that the covarient derivative (wrt nu) of F^mu^nu equals the four-current times some costant and that you should use this to obtain the final result, but I can't find in any text how you should do this exactly.

    Anyone who knows the proof?
     
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  3. Jan 9, 2009 #2

    CompuChip

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    The easiest way is using the vector potential.
    [tex]\partial_\mu j^\mu \propto \partial_\mu (\partial_\nu F^{\mu\nu}) = \partial_\mu \partial_\nu( \partial^\mu A^\nu - \partial^\nu A^\mu) [/tex]

    then rename some dummy indices and use equality of mixed derivatives ([itex]\partial_\mu \partial_\nu = \partial_\nu \partial_\mu[/itex]).

    [edit]Now that I come to think of it, you don't even need that. You can just use that F is anti-symmetric:
    [tex]\partial_\mu \partial_\nu F^{\mu\nu} = \frac12\left( \partial_\mu \partial_\nu F^{\mu\nu} + \partial_\mu \partial_\nu F^{\mu\nu} \right) = \frac12(\partial_\mu \partial_\nu F^{\mu\nu} - \partial_\mu \partial_\nu F^{\nu\mu} \right)[/tex];
    then swap the dummy labels mu, nu in the second term and use equality of mixed partial derivatives.
     
  4. Jan 9, 2009 #3
    For the case of ordinary erivatives that is true, because these derivatives commute. But I am looking for the proof in de case of covariant derivatives, and these don't necessarily commute...:(
     
  5. Jan 9, 2009 #4

    Stingray

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    It just requires a couple more steps:

    [tex]
    4 \pi \nabla_a j^a = \nabla_{[a} \nabla_{b]} F^{ab} = - \frac{1}{2} ( R_{abc}{}^{a} F^{cb} + R_{abc}{}^{b} F^{ac} ) = \frac{1}{2} (R_{bc} F^{cb} - R_{ac} F^{ac} ) = 0.
    [/tex]
     
  6. Jan 9, 2009 #5
    Thanks! I've spent almost the whole day on it, because I didn't know which way to go...

    Are you allowed to take only the asymmetric part of the covariant derivatives

    [tex]
    \nabla_{[a} \nabla_{b]} F^{ab}
    [/tex]

    because the symmetric part gives zero in combination with the anti symmetric F tensor?
     
  7. Jan 9, 2009 #6
    I think there is still a mistake, because

    [tex]
    4 \pi \nabla_a j^a = \nabla_{[a} \nabla_{b]} F^{ab} = - \frac{1}{2} ( R_{abc}{}^{a} F^{cb} + R_{abc}{}^{b} F^{ac} ) = \frac{1}{2} (R_{bc} F^{cb} - R_{ac} F^{ac} ) = \frac{1}{2} (R_{bc} F^{cb} + R_{ac} F^{ca} )[/tex] so they don't cancel. Is there a minussign wrong?
     
  8. Jan 9, 2009 #7

    Stingray

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    That's right.

    The last terms I wrote down are individually zero. That follows from noticing that the Ricci tensor is symmetric while the field tensor is antisymmetric.
     
  9. Jan 9, 2009 #8

    robphy

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    Interesting... it appears no metric is needed.
    The next generalization is presumably the case with torsion [so Ricci is not symmetric].
     
  10. Jan 10, 2009 #9
    :blushing:
    You're right, I didn't think of that!
     
  11. Jan 10, 2009 #10

    CompuChip

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    Also note that if you swap the indices on the final F, then you get
    [tex]\frac12( R_{bc} F^{cb} - R_{ac} F^{ac} )[/tex]
    which is precisely the expression posted by Stingray, so the both of you did derive the same result.
     
  12. Jan 10, 2009 #11

    CompuChip

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    That's a nice reasoning, but you can also just do the math by swapping (s) and renaming (r) indices:
    [tex]\nabla_a \nabla_b F^{ab} = \frac12\left( \nabla_a \nabla_b F^{ab} + \nabla_a \nabla_b F^{ab} \right) \stackrel{\mathrm s}{=} \frac12\left( \nabla_a \nabla_b F^{ab} - \nabla_a \nabla_b F^{ba} \right) \stackrel{\mathrm r}{=} \frac12\left( \nabla_a \nabla_b F^{ab} - \nabla_b \nabla_a F^{ab} \right) = \left( \frac12 (\nabla_a \nabla_b - \nabla_b \nabla_a ) \right) F^{ab}[/tex]
    where the bracketed part in the last identity is the definition of [itex]\nabla_{[a} \nabla_{b]}[/itex].
     
  13. Jan 10, 2009 #12
    You don't need covariant derivatives. Way messy. Maxwell's equations are canonically covariant on the pseudo Riemann manifold of General Relativity.

    I have a proof of the charge continuity equation here somewhere, but I can't just paste it, because it's not in real-valued fields like it should be--I could lose a negative sign somewhere...

    With that disclaminer, (it washes out 'cause +0=-0) in differential forms, Maxwell's equations are:

    d*F = -*J , and dF=0.

    J is the dual of the 4-current, and F is the Maxwell Tensor dual. Charge continuity immediately follows.
    Taking the exterior derivative of the first equation,

    -d*J = dd*F = 0, the charge continuity equation. (dd(anything)=0. All exact forms are closed.)

    d*J = 0
     
    Last edited: Jan 10, 2009
  14. Jan 10, 2009 #13

    Vanadium 50

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    I realize that people here are looking for a more mathematical way to do this, but can't one invoke symmetry arguments? If charge isn't conserved, in what was is it not conserved? How does one pick between becoming more positive and becoming more negative?
     
  15. Jan 10, 2009 #14
    What do you mean, Vanadium?
     
  16. Jan 10, 2009 #15

    Vanadium 50

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    Essentially, we're trying to prove [itex]\partial_\mu j^\mu = 0[/itex], right? If it's not zero, it has to be something else. (Obviously!) The idea would be to look at what the properties are of this something else. For example, that something else has to be a pure number. We know it can't be a non-zero constant, because we know the statement is true in flat space.

    So it must be some combination of F's, R's and possibly g's such the indices contract to give a pure number. My question/suggestion is wouldn't it be easier to show that there is no combination that does this - probably because of the symmetry properties of F?
     
  17. Jan 10, 2009 #16
    Edit: total rewrite after running off half cocked.

    G=*F
    dG = -*J
    *J is an exact form.
    ddG=0=-d*J
    (*J is closed)
    d*J=0.

    Are you're hinting at something else?

    * is the Hodge duality operator. G is the Hodge dual of F.
    [tex]G_{\mu\nu} = (1/2)\epsilon_{\mu\nu}\\ ^{\sigma\rho}F_{\sigma\rho}}[/tex]
    *J is the Hodge dual of J
    [tex](*J)_{\mu\nu\sigma} = (1/6)\epsilon_{\mu\nu\sigma}\\ ^{\rho}J_{\rho}}[/tex]
     
    Last edited: Jan 10, 2009
  18. Jan 10, 2009 #17
    The only requirement on F to obtain charge continuity is that it be antisymmetric in it's indices.
    [tex]F_{\mu \nu} = -F_{\nu \mu} = F_{[\mu \nu]} [/tex]​
    Then, automatically, the connection terms cancel, for any connection. So what is true in Minkowski space is true in a space with arbitrary connections. (It need not be metric compatible nor torsion free such as required in general relativity.)

    The simplest example of this is the exterior derivative of a vector field.

    [tex] V_{\mu} = g_{\mu\nu}V^{\nu}[/tex]

    [tex]dV = (dV)_{\mu\nu} \equiv \partial_{[\mu}V_{\nu]} \equiv 2(\partial_{\mu}V_{\nu} - \partial_{\nu}V_{\mu}) = 2(\nabla_{\mu}V_{\nu} - \nabla_{\nu}V_{\mu})[/tex]

    Where F is defined as the derivative of the 4-vector potential, F is manifestly antisymmetric. Essencially, charge is a mathematical property of the vector potential, and likewise, charge continuity is a mathematical property of the vector potential--a direct mathematical consequence of imposing a vector field on a smooth manifold.

    In fact it's possible to eliminate every equal sign out of Maxwell's formulation, and simply associate physically measurable parameters with elements of the vector potential and it's derivatives.
     
    Last edited: Jan 11, 2009
  19. Jan 11, 2009 #18
    No comments, Vanadium?

    I guess differential forms are not as commonly studied as required to understand all this rhetoric I've put out. I didn't do a lot of translating beteen forms and tensors because it's very tedious, and frankly, I've gotten very rusty at it.
     
    Last edited: Jan 11, 2009
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