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Curved-space Maxwell equations by differential forms?

  1. Feb 5, 2015 #1
    The flat-space source-free Maxwell equations can be written in terms of differential forms as
    $$d F = 0; \ \ d \star F = 0.$$
    And in the theory of gauge fields, one can introduce a connection one-from A from which one can formulate general Maxwell equations (for Yang-Mills fields) by
    $$ dF + A \wedge F = 0; \ \ d \star F + A \wedge \star F = 0,$$
    or if one introduces the covariant exterior derivative, just
    $$D F = 0; \ \ D\star F = 0.$$

    Is it also possible to introduce a curved-space covariant exterior derivative and express the curved-space Maxwell equations in terms of this? If so I would very much appreciate some links to literature where I could read about it.
     
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  3. Feb 5, 2015 #2

    Ben Niehoff

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    You've already written down Maxwell's equations (and the Yang-Mills equations) in curved space. That is the power of forms.
     
  4. Feb 6, 2015 #3
    How come? Does the anti-symmetry kill off the connection coefficients? (Do you have reference that deal with Maxwell equations in curved spacetime in terms of forms)?

    Also, one expects the Riemann tensor to appear in second derivatives. Does it also do that in the form language?
     
  5. Feb 6, 2015 #4

    rubi

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    ##A## and ##F## are Lie algebra-valued forms. In the case of electrodynamics, they take values in the Lie algebra of ##U(1)##, which is abelian and therefore the Lie brackets in the definition of ##A\wedge F## (which is to be understood as the wedge product of Lie algebra valued forms) vanish, so the exterior covariant derivative reduces to the standard exterior derivative.
     
  6. Feb 6, 2015 #5

    Ben Niehoff

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    This is really the wrong way to think about this, and I think books like MTW have done people a disservice here. The exterior derivative is already covariant under coordinate transformations by construction. You don't even need to define a connection (or a metric, for that matter).

    If you do stick in connection coefficients, then yes, they vanish due to antisymmetry of the forms. At least, they do in the absence of torsion. In the presence of torsion, I seem to remember something might go wrong. But in any case, the ##d## operator is the correct way to do electromagnetism.

    Nakahara.

    No, it doesn't.

    In a non-Abelian gauge theory, the field strength ##F## appears in second derivatives, because essentially ##F \equiv D^2##. But the Riemann tensor never appears, essentially because ##d^2 = 0##.

    Note: There is a sense in which you can talk about forms valued in the tangent bundle. Then the (Levi-Civita) connection can be treated like a gauge connection. Then you can use the language of gauge connections on vector bundles to do Riemannian geometry; if your vector potential is (essentially) the Christoffel symbols, then you can get Riemann tensors to appear (in the guise of matrix-valued 2-forms).
     
  7. Feb 6, 2015 #6
    If you do not have to define a connection or metric, how do the Maxwell equations for curved spacetime, as expressed in terms of forms, include information about the spacetime geometry?

    Furthermore, how would you derive the curved-space equations
    $$\nabla_\mu F^{\mu \nu} = j^\mu; \ \ \epsilon^{\alpha \beta \gamma} \nabla_{\alpha} F_{ \beta \gamma} = 0$$
    from the differential form equations.

    At what page of Nakahara? I seem to remember that the book only include Minkowski-spacetime electromagnetism.
     
  8. Feb 6, 2015 #7

    rubi

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    You seem to confuse the gauge connection with the Levi-Civita connection. The covariant exterior derivative contains the information about the gauge connection. The information about the spacetime geometry isn't contained in the (covariant) exterior derivative, but in the Hodge dual operator. The Christoffel symbols will pop up automatically if you apply the (covariant) exterior derivative to the dual of the field stenght tensor.
     
  9. Feb 7, 2015 #8

    Ben Niehoff

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    The Maxwell equations involve the Hodge dual, which contains information about the metric. The Hodge dual does not contain ALL the information about the metric, however; but the Maxwell field doesn't care about ALL the metric information. For example, in 4 dimensions, Maxwell's equations are conformally invariant (essentially, because the Hodge dual acting on 2-forms is insensitive to an overall conformal factor in the metric).

    In the first equation, the curvature information comes from the Hodge dual. Notice that you're only taking a covariant divergence, so the equation is not sensitive to every component of ##\Gamma^\mu_{\nu\rho}##. In the second equation, you can just as easily replace ##\nabla_\mu## by ##\partial_\mu##.

    Read the sections on differential forms, fiber bundles, and connections on principal bundles. A Yang-Mills potential is just a gauge connection on a principal bundle; the field strength is just the curvature of the connection.
     
  10. Feb 8, 2015 #9
    Well I certainly missed something while learning about forms. Thank you!

    P.S: would you happen to know how to derive the wave-equation in terms of the differential forms language? (How would we express the equivalent of the d'Lambertian?)
     
  11. Feb 8, 2015 #10

    Ben Niehoff

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    If ##\varphi## is a function (i.e. 0-form), then

    $$d \star d \varphi = 0$$
    is the Laplace equation on a Riemannian manifold; it is the wave equation on a pseudo-Riemannian manifold.

    More generally, you should read about "harmonic forms" and the Hodge decomposition. A harmonic form ##\omega## satisfies

    $$\star d \star d \omega + d \star d \star \omega = 0,$$
    which reduces to the above when ##\omega## is a 0-form.
     
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