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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 $f(r)$ ?

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  • $\begingroup$ 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. $\endgroup$ – Qmechanic Jul 13 '16 at 10:01
  • $\begingroup$ @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. $\endgroup$ – user115376 Jul 13 '16 at 10:13
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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.)

However, the moral is that in relativity (or for geodesics on Lorentzian manifolds, if you wish) the concept of the effective potential becomes increasingly frail and conventional. To see how the concept of the effective potential can be introduced in the more complicated case of the Kerr space-time, I recommend the respective chapters in Misner, Thorne and Wheeler.

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