# Has the Klein-Gordon equation in curved spacetimes the same form as in flat ones? [duplicate]

The KG equation in curved geometries has the following form: $$\frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}~g^{\mu \nu}\partial_\nu\phi) + m^2\phi = 0,$$

where $$g$$ is the determinant of the metric tensor $$g^{\mu \nu}$$ and $$m$$ the mass of the field. Is it possible to reexpress the last equation as the usual KG equation in flat spacetimes defining a covariant derivative $$\nabla_\mu$$ in such a way that the equation reads:

$$(\square + m^2)\phi(x^\mu) = 0?$$

I don't arrive at the desired form, any help or hint?

• perhaps with this equation $\det \left( g\right) =e^{tr\left( \ln \left( g\right) \right) }$ tr is the trace
– Eli
Jun 28 at 14:09
• so $\partial _{\mu }\sqrt{-g}=\dfrac{1}{2}g^{\alpha \nu }\partial _{\mu }\left( g_{\alpha \nu}\right) \sqrt{-g}$
– Eli
Jun 28 at 14:17
• I guess that the correct result is: $\partial_\mu \sqrt{-g} = -\frac{1}{2\sqrt{-g}} \partial_\mu g$ Jun 28 at 14:26
• Possible duplicate: physics.stackexchange.com/q/101675/2451 Jun 28 at 15:29

A covariant derivative acting upon a scalar will reduce to a partial derivative. So you will have $$\nabla_{\mu} \phi = \partial_{\mu} \phi$$ The second covariant derivative will now act on a covector $$\partial_{\mu}\phi$$ so you will have $$\Box \phi = \nabla^{\mu}\nabla_{\mu}\phi = \partial^{\mu}\partial_{\mu}\phi - g^{\mu\nu}\Gamma^{\alpha}_{\mu\nu}\partial_{\alpha}\phi.$$
• The last expression you wrote is wrong assuming $\phi$ is a scalar function on spacetime. Jun 28 at 13:31