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Wave equation in curved spacetime

  1. Apr 7, 2008 #1
    Does anyone know how to derive the wave equation in curved spacetime?
    [tex](-g)^{-1\over 2}\partial_\mu((-g)^{1\over 2}g^{\mu \nu}\partial_\nu \phi) = 0 [/tex]

    A reference, or an outline of the derivation would be very helpful. Thanks.
     
    Last edited: Apr 7, 2008
  2. jcsd
  3. Apr 7, 2008 #2
    It seems that just writing the d'Alembertian in covariant form
    [tex]\Delta \phi = g^{\mu \nu}\phi_{;\mu \nu}=0[/tex] does the trick.
    This form is giving me the results I want, but I still don't know how to put it in the form written in my original post.
     
  4. Apr 7, 2008 #3
    Given a vector [itex]V^\mu[/itex], can you think of any cute expressions for [itex]\nabla_\mu V^\mu[/itex]? (Hint: you can write the covariant four-divergence of a vector in terms of the coordinate four-divergence much like the expression in the first post.)
     
  5. Apr 7, 2008 #4
    Just a guess, but : [itex]\nabla_\mu V^\mu=V^\mu[/itex]?

    [addendeum: or is it [itex]\nabla_\mu V^\mu=\frac{\delta V^\mu}{\delta t}[/itex]?]

    Regards,

    Bill
     
    Last edited: Apr 7, 2008
  6. Apr 7, 2008 #5
    Nope. That can't work because the left-hand side is a scalar quantity, while the right-hand side is a vector.

    Not this either. Perhaps it will be simpler if I just state the result and leave the proof as an exercise. Given an orientable Riemannian manifold [itex](M,g)[/itex], one has a preferred idea of a connection in the form of the Levi-Civita connection. This allows one to define, for example, a straightforward notion of covariant differentiation on tensor fields over [itex]M[/itex]. The relationship between the covariant divergence of a vector and the partial derivative of the vector is given by

    [tex]\nabla_\mu V^\mu = \frac{1}{\sqrt{\textrm{det}g_{\rho\sigma}}}\partial_\mu(\sqrt{\textrm{det}g_{\rho\sigma}}V^\mu)[/tex]
     
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