- #1

valesdn

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## \Box \phi = \frac{1}{\sqrt{g}} \frac{ \partial}{\partial x^{\mu}} \sqrt(g)g^{(\mu)(\nu)} \frac{\partial}{\partial x^{\nu}} \phi=0##

I used the lagrangian for a scalar field: ## L = \nabla_{\mu}\phi \nabla_{\nu} \phi g^(\mu \nu) ##

The density lagrangian is ## L= \sqrt(-g) R ##, where R is the curvature scalar.

The variation of the Einstein-Hilbert action can be write as follows:

## \delta S= \delta \int (\sqrt (-g) R d^4 x= \int d^4 x \delta (\sqrt(-g)g^{ab}R_{ab}= \int d^4 x \sqrt(-g)g^{ab} \delta R_{ab} + \int d^4 x \sqrt(-g) R_{ab} \delta g^{ab} + \int d^4 x R \delta (\sqrt(-g))##

I've calculated the single terms, respectively.

I found:

- for the first term ## \int d^4 x \sqrt(-g)g^{ab} \delta R_{ab}##:

## \int d^4 x \sqrt(-g)g^{ab} \delta R_(ab)=## ##\int d^4 x \sqrt(-g) \nabla_c## ##[g^{ab}## ##\delta (\Gamma^c)_{ab} - g^{ac} \delta ## ##(\Gamma ^{b})####_{ab}##]

I can't calculate the second term ##\int d^4 x \sqrt(-g) R_{ab} \delta g^{ab}##.

The third term ##\int d^4 x R \delta (\sqrt(-g))## is equal to ##- \frac{1}{2} \int d^4 x R \sqrt(-g) g_{ab} \delta (g^{ab})##.

So, I have ## \delta S =## ##\int d^4 x \sqrt(-g) \nabla_c## ##[g^{ab}## ##\delta (\Gamma^c)_{ab} - g^{ac} \delta ## ##(\Gamma ^{b})####_{ab}##] + ##\int d^4 x \sqrt(-g) R_{ab} \delta g^{ab}## + ##- \frac{1}{2} \int d^4 x R \sqrt(-g) g_{ab} \delta (g^{ab})##

Is it correct this calculations to find the equation for a scalar field? Is there an other way to find it? How can I demonstrate this equation?

Thank you in advance for your help.