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?

  • $\begingroup$ perhaps with this equation $\det \left( g\right) =e^{tr\left( \ln \left( g\right) \right) }$ tr is the trace $\endgroup$
    – Eli
    Jun 28, 2021 at 14:09
  • $\begingroup$ so $\partial _{\mu }\sqrt{-g}=\dfrac{1}{2}g^{\alpha \nu }\partial _{\mu }\left( g_{\alpha \nu}\right) \sqrt{-g}$ $\endgroup$
    – Eli
    Jun 28, 2021 at 14:17
  • $\begingroup$ I guess that the correct result is: $\partial_\mu \sqrt{-g} = -\frac{1}{2\sqrt{-g}} \partial_\mu g$ $\endgroup$
    – T. ssP
    Jun 28, 2021 at 14:26
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/101675/2451 $\endgroup$
    – Qmechanic
    Jun 28, 2021 at 15:29

1 Answer 1


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 $$\Box \phi = \nabla^{\mu}\nabla_{\mu}\phi = \partial^{\mu}\partial_{\mu}\phi - g^{\mu\nu}\Gamma^{\alpha}_{\mu\nu}\partial_{\alpha}\phi.$$

  • $\begingroup$ It should be as you say. But it is possible to demonstrate it using the expressions of the defined derivatives? $\endgroup$
    – T. ssP
    Jun 28, 2021 at 13:21
  • $\begingroup$ The last expression you wrote is wrong assuming $\phi$ is a scalar function on spacetime. $\endgroup$
    – Noone
    Jun 28, 2021 at 13:31
  • $\begingroup$ Okay, thank you very much. I have reformulate the question and fixed some errors! $\endgroup$
    – T. ssP
    Jun 28, 2021 at 13:46
  • $\begingroup$ The Box operator at the last equation is defined as the quantity with the derivatives in the first expression. $\endgroup$
    – Noone
    Jun 28, 2021 at 13:49
  • 1
    $\begingroup$ See physics.stackexchange.com/q/101675 $\endgroup$
    – Noone
    Jun 28, 2021 at 15:06

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