# Ricci scalar in Scalar Field in Curved Space-time

I was recently looking at a Lagrangian of a scalar field in curved space-time at http://www.unc.edu/~mgood/research/Carroll_QFT_CS.pdf on page 8. I am not a physicist, and I am currently studying introductory physics in high school, but from my knowledge a Lagrangian is basically the kinetic energy minus the potential energy. In this Lagrangian, I understand the kinetic part, but I do not understand the potential part. Is it $-1/2m^2\phi^2 -(1/6)R\phi^2$ or just $-1/2m^2\phi^2$? Can a Ricci scalar be simply coupled to a scalar field like that?

Also, since the Ricci scalar is equal to the Ricci curvature tensor multiplied by the inverse of the metric tensor, and using some identities, the Einstein Field Equation can be rearranged into $R(u,v)=k(T(u,v)-(1/2)Tg(u,v))$- how is the Ricci scalar exactly related to the energy momentum tensor, in other words, can it be expressed in terms of the energy momentum tensor, say the diagonal of the energy momentum tensor (multiplying the energy density, T00, by the momentum terms T11, T22, T33? Please be aware that I am not an expert in the field, but would like to know more about the field.

Yes, you can regard the expression

$$-\frac12m^2\phi^2-\frac16R\phi^2$$

as potential energy. Compare it to the harmonic oscillator: its potential energy is given by a term quadratic in position. The Ricci scalar can then be interpreted as simply contributing to the square of the mass of the scalar.

To answer your question whether this is possible or not: when constructing a field theory, one always tries to write down a Lagrangian consisting of terms that respect symmetries one demands from the theory. The Ricci scalar is coordinate-invariant and simple, and as such a good choice for a term in the Lagrangian.

Regarding your last question, consider the Einstein equations:

$$R_{\mu\nu}-\frac12Rg_{\mu\nu}=8\pi T_{\mu\nu},$$

which relates the stress-energy tensor component-wise to the Ricci tensor $R_{\mu\nu}$, the metric $g_{\mu\nu}$ and the Ricci scalar. You can now take the trace on both sides, in $d$ dimensions given by

$$\frac{2-d}{2}R=8\pi T_{\mu\nu}g^{\mu\nu}.$$

In 4 dimensions, this reduces to

$$-R=8\pi(T_{00}g^{00}+T_{11}g^{11}+T_{22}g^{22}+T_{33}g^{33}),$$

which might be the relation you are looking for.