# Klein-Gordon equation in curved space time

The Klein-Gordon equation in curved spacetime has the following form:

$$\left (\square+m^2 \right)\Phi=\left[\frac{1}{\sqrt{-g}} \partial_{\mu}\left(\sqrt{-g}g^{\mu\nu} \partial_{\nu} \right)+m^2\right]\Phi=0$$

In the case of the Schwarzschild Metric, $g_{00}$ and $g_{11}$ are dimensionless, while $g_{22}$ and $g_{33}$ are not. However, in the equation we need every term to have the same dimensions. What I have missed?

• There is more than one way to think about units in general relativity. For a detailed discussion, see section 5.11 of my GR book, lightandmatter.com/genrel . in the equation we need every term to have the same dimensions It's not clear to me why you think this is in danger of being violated. Note that the derivatives also have units. – Ben Crowell Feb 23 '18 at 4:44
• This form of the Klein-Gordon equation seems ugly to me. I could be wrong, but it seems to me that if you just take the flat-spacetime version and replace partial derivatives with covariant derivatives, you ought to get something much prettier. – Ben Crowell Feb 23 '18 at 4:46

Every component of the metric is dimensionless, if you use rectilinear coordinates. $g_{22}$ and $g_{33}$ only have dimensions if you are using curvilinear coordinates (probably spherical, in this case). In that case, the $\partial_2$ and $\partial_3$ also have correspondingly different dimensions than $\partial_0$ and $\partial_1$.
• @BekaModrekiladze The difference in units between the derivatives and metric components do indeed cancel in spherical coordinates. Remember that the raised indices on the metric tensor indicate that the inverse matrix to the metric is being used, so the units of $g_{22}$ and $g_{33}$ (which are length, in spherical coordinates) should be inverted when plugging $g^{22}$ and $g^{33}$ into the equation. – jawheele Mar 28 '19 at 1:10