homework and exercises - Derivation of Klein-Gordon equation in General Relativity

I am trying to derive the Klein-Gordon equation for the case of GR using the action:

$$S\left[ {\varphi ,{g_{\mu \nu }}} \right] = \int {\sqrt g {d^4}x\left( { - {1 \over 2}{g^{\mu \nu }}{\nabla _\mu }\varphi {\nabla _\nu }\varphi - {1 \over 2}{m^2}{\varphi ^2}} \right)} \tag{1}$$

So in the Euler - Lagrange equation for the SR case which is:

$${\partial _\mu }\left( {{{\partial L} \over {\partial \left( {{\partial _\mu }\varphi } \right)}}} \right) = {{\partial L} \over {\partial \varphi }}\tag{2}$$ I use the correspondence $${\partial \over {\partial \left( {{\partial _\mu }\varphi } \right)}} \to {\nabla _{{\nabla _\mu }\varphi }},{\partial \over {{\partial _\mu }\varphi }} \to {\nabla _\mu } \tag{3}$$ and so, $${\nabla _{{\nabla _\mu }\varphi }}L = - {1 \over 2}\sqrt g {g^{\mu \nu }}{\nabla _\nu }\varphi - {1 \over 2}\sqrt g {g^{\mu \nu }}{\nabla _\mu }\varphi {\nabla _{{\nabla _\mu }\varphi }}\left( {{\nabla _\nu }\varphi } \right) \tag{4}$$ and using this dubious relation that I am unable to prove (extending the SR case of partial derivatives),

$${\nabla _{{\nabla _\mu }\varphi }}\left( {{\nabla _\nu }\varphi } \right) = \delta _\nu ^\mu \tag{5}$$ I finally get the desired KG equation, $${\nabla _\mu }\left( { - \sqrt g {g^{\mu \nu }}{\nabla _\nu }\varphi } \right) = - \sqrt g {m^2}\varphi$$ However, I am very uncomfortable with my assumption 5 that I am unable to prove. Is my analysis correct? I am a 60y old doing this as retirement fun so please don't cut me down too brutally :-)

Answer

Note: technically, everywhere we've been using $\sqrt{g}$ it should be $\sqrt{|g|}$, but I'll let the former denote the latter in a slight abuse of notation like the rest of this page.

gj255 has already provided the textbook approach, in which we keep to partial derivatives. You can actually do it a different way; the same from-first-principles argument that derives Eq. (2) can be modified to obtain something that manifestly works with covariant derivatives. Write the Lagrangian density as $L=\sqrt{g}L_0$ with $L_0=\frac{1}{2}\nabla_\mu\varphi\nabla^\mu\varphi-\frac{1}{2}\varphi^2$, so

$$0=\delta S=\int d^4x \sqrt{g}\left(\delta\varphi \frac{\partial L_0}{\partial\varphi}+\nabla_\mu\delta\varphi \frac{\partial L_0}{\partial\nabla_\mu\varphi}\right)$$ ($\nabla_\mu\varphi$ is actually just $\partial_\mu\varphi$, because $\varphi$ is a scalar field), where we have written $\delta\nabla_\mu\varphi=\nabla_\mu\delta\varphi$. But $$\int d^4x\sqrt{g}\nabla_\mu\left(\delta\varphi \frac{\partial L_0}{\partial\nabla_\mu\varphi}\right)=\int d^4x\partial_\mu\left(\sqrt{g}\delta\varphi \frac{\partial L_0}{\partial\nabla_\mu\varphi}\right)$$ is a boundary term, so we can rewrite $\delta S=0$ as $$0=\int d^4 x\sqrt{g}\delta\varphi \left(\frac{\partial L_0}{\partial\varphi}-\nabla_\mu\frac{\partial L_0}{\partial\nabla_\mu\varphi}\right),$$ giving us an Euler-Lagrange function more in the spirit of what you wanted, viz. $$0=\frac{\partial L_0}{\partial\varphi}-\nabla_\mu\frac{\partial L_0}{\partial\nabla_\mu\varphi}=-m^2\varphi-\nabla_\mu\nabla^\mu\varphi.$$ Note: I don't know what the standard symbol is for $L/\sqrt{g}$, but I do know it's sometimes called the scalar Lagrangian density.