# Effective Potential in General Relativity

I would like to clarify a concept about the Effective Potential in General Relativity when the kinetic energy term is not unitary.

Suppose (in spherical coordinates) one has a generic line element of the form

$$ds^{2}=-e^{\nu(r)}dt^{2}+e^{-\nu(r)}f(r)dr^{2}+r^{2}d\Omega^{2}$$

being $d\Omega^{2}=d\theta^{2}+\sin^{2}\theta\;d\phi^{2}$ the usual solid angle element, and the functions $\nu(r)$ and $f(r)$ are continuous functions depending only on the radial coordinate $r$, such that if $r\gg r_{0}$, being $r_{0}$ some specific length scale: $f(r\gg r_{0})=1$.

For a massive particle, freely moving in such space-time, the energy conservation equation is written as

$$\dfrac{1}{2}E^{2}=\dfrac{1}{2}f(r)\;\dot{r}^{2}+\dfrac{1}{2}e^{\nu(r)}\left(1+\dfrac{L^{2}}{r^{2}}\right).$$

How to read from here the effective potential $V_{\text{eff}}(r)$ given that the kinetic term is not unitary due to the presence of the factor $f(r)$ ?

• Comment to the question (v1): One may show that Einstein's equations in vacuum imply that the function $f(r)$ is a constant independent of $r$, thereby resolving OP's question within vacuum sectors of spacetime. – Qmechanic Jul 13 '16 at 10:01
• @Qmechanic thanks for the comments, but I'm asking in general. Suppose that one has to describe the state of motion of a generic massive particle freely moving in that space-time, where the metric is measured with the given line element. This is a typical geometric problem which doesn't involve any theory of gravity, at least a priori. – user115376 Jul 13 '16 at 10:13

The answer is simply that not every space-time has a corresponding effective potential in the sense that we have a coordinate $x$ such that $\dot{x}=\sqrt{2(E-V_{eff})}$.
But this is true even in Newtonian mechanics, consider a problem with a Lagrangian $$L = \frac{m}{2}(\dot{r}^2 + r^2 \dot{\varphi}^2) - V(\varphi)$$ Obviously, $p_r\equiv m \dot{r}$ is an integral of motion, and the resulting motion is effectively one-dimensional, but we will be unable to express $\dot{\varphi}$ as $\sim \sqrt{E - V_{eff}}$, we will rather obtain $$\dot{\varphi}= \frac{\sqrt{2(E-V_{eff})}}{r}$$ where $V_{eff}=p_r^2/(2m) + V(\varphi)$, and $E$ is of course the conserved Hamiltonian.
Here, the difference $E-V_{eff}$ can be used to investigate allowed regions of motion because $r$ is always positive.
The same is true in the example you mention, at least if the function $f(r)$ is always positive. You can define an effective potential as $\mathcal{V}_{eff} = e^\nu(1 + L^2/r^2)$, and your radial velocity will then be $$\dot{r} = \sqrt{\frac{E^2 - \mathcal{V}_{eff}}{f(r)}}$$ I.e., by plotting $E^2 - \mathcal{V}_{eff}$ and finding where it is above zero, you will be able to tell the allowed regions of motion for the particle. (The extrema of $\mathcal{V}_{eff}$ will also give you circular orbits etc.)