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Im having some confusion regarding the gravitational field equations we can get from a certain action. If Im given an action and they ask me to obtain the gravitational field equations from it I would only vary w.r.t. the metric. If they tell me also that the theory is coupled with, say, two scalar fields $\phi$ and $\psi$ I would have to vary the action w.r.t. those two as well? Im having trouble in identifying which fields do I have to vary w.r.t. in order to obtain the gravitational field equations for that certain action.

Edit:

Einstein´s theory coupled with $\phi$ and $\psi$:

$$S=\int d^4x\sqrt{-g}\left({}{\kappa^2 (R-2 \Lambda)}+\frac{1}{2}(\partial \phi)^2 + \frac{1}{2}(\partial \psi)^2-V(\phi,\psi)\right).$$

And we want the gravitational field equations. I already did the variation w.r.t. the metric but I dont know if I have to vary w.r.t. to both fields or just one or none. $k^2$ and $\Lambda$ are constants.

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2 Answers 2

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The gravitational field equations are the equations one obtains varying with respect to the metric field. These equations determine the form of spacetime. Now, if your theory is coupled to some other field(s) say a scalar, variation with respect to the other field(s) yields the equation of motion for the other field(s). These equations (gravitational and other field(s) equations) should be solved together in order to obtain the form of spacetime and a consistent form for the other field(s). Yes you have to vary with respect to the metric tensor in order to obtain the gravitational (Einstein) field equation.

Edit 1:

Assuming that $\phi$, $\psi$ are matter fields then you only need to vary with respect to the metric.

Edit 2:

Lets say we have Einstein's theory of gravity and a scalar field as a matter field:

$$ S = \int d^4x \sqrt{-g} \Big(R/2 - \cfrac{1}{2}g^{μν}\partial_{μ}\phi\partial_{ν}\phi\Big) $$

By variation with respect to the metric field we obtain:

$$G_{\mu\nu} = \partial_{μ}\phi\partial_{ν}\phi - \cfrac{1}{2}g_{μν}g^{αβ}\partial_{α}\phi\partial_{β}\phi $$

where the right hand side is the energy momentum tensor. (You can see for yourself that $(\partial \phi)^2 = g^{μν}\partial_{μ}\phi\partial_{ν}$) To obtain this equation (gravitational equation) you vary with respect to the metric everywhere the metric appears!

There are theories where a scalar field is a "gravitational" scalar field (a part of the theory of gravitation) (Scalar Tensor Gravity and Brans Dicke for example). There, to obtain the gravitrational field equations you vary with respect to the metric tensort and the scalar field. In theories where we consider Einstein gravity and a scalar field as a matter field you vary only with respect to the metric.

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  • $\begingroup$ Can one vary with respect to the connection instead of the metric? $\endgroup$ Mar 15, 2021 at 13:16
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remeber that the action for the fields also depends on the metric. You have hidden that dependence in the way you write your action. Writing the dependence explicitly gives $$ S_{\rm fields}= \int d^dx \sqrt{-g}L=\int d^dx \sqrt{-g}\left( \frac 12 g_{\mu\nu} \partial^\mu \phi \partial^\nu \phi -V(\phi)\right) $$ so $$ \frac{\delta S_{\rm fields}}{\delta g_{\mu \nu}}= \partial^\mu \phi \partial^\nu \phi- g^{\mu\nu} L $$

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  • $\begingroup$ Yes I did that for both fields, considering they are both scalar fields. My confusion is do we get the gravitational field equations just by varying the metric everywhere it appears or do we need to know the equations of motion for $\phi$ and $\psi$ ? $\endgroup$ Nov 2, 2020 at 15:06
  • $\begingroup$ The derivative wrt the metric of the field action is just the energy-momentum tensor that is the source of the gravitational field. Of course if you want to solve the resulting equations you need to also simultansously solve the matter field equations -- as is done in the Resiesser-Nordstrom charged black hole for example. $\endgroup$
    – mike stone
    Nov 2, 2020 at 15:10

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