# Effective potential in a time-dependent spacetime

My question is regarding an arbitrary time-dependent spherically symmetric spacetime with line-element, in co-moving coordinates, to be

$$ds^2 = -f(R) dt^2 + a(t)\bigg\lbrace\frac{dR^2}{f(R)} +R^2d\Omega^2 \bigg\rbrace.$$

Obviously, I'm working with something "Schwarzschild-like" and the $$f(R)$$ and $$a(t)$$ can be any user-defined function. I need help verifying if the procedure I do below is correct.

First, the geodesic equation for the $$R$$-coordinate can be written as

$$\frac{d^2R}{d\lambda^2}+\Gamma^{R}_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}=0$$

where $$\lambda$$ is an affine parameter. If I multiply both sides of the equation by $$\frac{dR}{d\lambda}$$ I then get

$$\frac{d^2R}{d\lambda^2}\frac{dR}{d\lambda}+\frac{dR}{d\lambda}\Gamma^{R}_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}=0$$

which I could then take to mean

$$\frac{1}{2}\left(\frac{dR}{d\lambda}\right)^2 + \displaystyle \int{d\lambda\frac{dR}{d\lambda}\Gamma^{R}_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}} = \textrm{constant}$$.

From this, can I read off the term in the integral sign to be the effective potential? Furthermore, since the terms inside the integral have no explicit dependence on the affine parameter, can I further reduce it to

$$U_{\textrm{effective}} =\displaystyle \int {\Gamma^{R}_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}dR }$$

$$\int {\Gamma^{R}_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}dR }$$