Wave equation in curved spacetime

1. Apr 7, 2008

Pacopag

Does anyone know how to derive the wave equation in curved spacetime?
$$(-g)^{-1\over 2}\partial_\mu((-g)^{1\over 2}g^{\mu \nu}\partial_\nu \phi) = 0$$

A reference, or an outline of the derivation would be very helpful. Thanks.

Last edited: Apr 7, 2008
2. Apr 7, 2008

Pacopag

It seems that just writing the d'Alembertian in covariant form
$$\Delta \phi = g^{\mu \nu}\phi_{;\mu \nu}=0$$ 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.

3. Apr 7, 2008

shoehorn

Given a vector $V^\mu$, can you think of any cute expressions for $\nabla_\mu V^\mu$? (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.)

4. Apr 7, 2008

Antenna Guy

Just a guess, but : $\nabla_\mu V^\mu=V^\mu$?

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

Regards,

Bill

Last edited: Apr 7, 2008
5. Apr 7, 2008

shoehorn

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 $(M,g)$, 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 $M$. The relationship between the covariant divergence of a vector and the partial derivative of the vector is given by

$$\nabla_\mu V^\mu = \frac{1}{\sqrt{\textrm{det}g_{\rho\sigma}}}\partial_\mu(\sqrt{\textrm{det}g_{\rho\sigma}}V^\mu)$$

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