# Effective Potential of Einstein Cluster

I was reading this paper (PDF), "On Einstein Clusters", and in equation (34) they write the effective potential of a particle moving in this system as $$V_{eff}(r)=e^{\nu/2}\sqrt{1+\frac{\tilde{L}^2}{r^2}}.\tag{34}$$ I'm trying to understand where they get this result from, especially since it seems the rest of their stability analysis is based off this effective potential.

In this other paper, they clarify that this effective potential is derived from the normalized 4-velocity equation, which makes sense since that's how the effective potential is arrived at for the Schwarzschild metric. However, starting with the metric, $$ds^2=-e^{\nu}dt^2+e^{\lambda}dr^2+r^2d\Omega.$$ I ultimately get the equation, $$\tilde{E}^2=e^{\nu+\lambda}(\frac{dr}{d\tau})^2+e^{\nu}(1+\frac{\tilde{L}^2}{r^2}),$$ where $$\tilde{E}=e^{\nu}\dot{t}$$ and $$\tilde{L}=r^2\dot{\phi}$$. Unlike in the Schwarzschild metric, the $$e^\lambda$$ and $$e^\nu$$ don't just cancel out, as shown in eq(25) of their paper. Also, they seem to take the square of the last term for a reason I don't understand.

I find that the constants of motion are given by: $$\frac{dt}{d\tau} = e^{-\nu} \bar{E}\ ,$$ $$\frac{d\phi}{d\tau} = \frac{\bar{L}}{r^2}\ .$$

Substituting these definitions into the metric interval $$d\tau^2 = e^{\nu} dt^2 - e^{\lambda} dr^2 - r^2 d\phi^2\ ,$$ I find $$\frac{dr^2}{d\tau^2} = e^{-\lambda} \left[ e^{-\nu} \bar{E}^2 - \left(1 + \frac{\bar{L}^2}{r^2}\right) \right]\ .$$ $$\frac{dr^2}{d\tau^2} = e^{-\lambda-\nu} \left[\bar{E}^2 - e^{\nu}\left(1 + \frac{\bar{L}^2}{r^2}\right) \right]\$$ and therefore $$V_{\rm eff} = e^{\nu/2}\left( 1 + \frac{\bar{L}^2}{r^2}\right)^{1/2}\ .$$

Alternatively, you could just call the effective potential $$V_{\rm eff} = e^{\nu}\left( 1 + \frac{\bar{L}^2}{r^2}\right)$$ and you still have that $$dr/d\tau$$ depends on the difference between an "energy term" and an effective potential. It just depends whether you want an equation of the form $$\frac{dr^2}{dt^2} = \epsilon^2 - V_{\rm eff}^2\ ,$$ where $$\epsilon$$ is the "energy term", or $$\frac{dr^2}{dt^2} = \epsilon - V_{\rm eff}\ ,$$ which is more in line with a "Newtonian" definition.

I don't think there is a right or wrong way and both approaches can be adopted in the Schwarzschild metric too.

• Writing my answer I missed apparently your correct (1) answer. However, it looks like nobody is interested in that topic.
– JanG
Commented Mar 7 at 15:56

The definition of the effective potential in GR is a matter of convention. Apparently, the author uses geodesics equation in the form $$\tilde{E}^2=e^{\nu+\lambda}(\frac{dr}{d\tau})^2+V^{2}_{eff} \tag{1}$$ trying to treat the classical energy conservation formula relativistically. However, there is no physical meaning behind it.

Some time ago I have posted similar question: Different definitions for effective potential in static spherically symmetric spacetimes – which is right? but get no answer to it. Thus, I would like to quote again professor Tiberiu Harko who kindly answered my question in private communication:

The potential in static general relativity is rather arbitrary. As opposed to general relativity, the potential is an effective quantity, which does not have a direct physical meaning, like in Newtonian gravity. If you can write the equation of motion for $$r$$ in the form $$\frac{\it{1}}{\it{2}}~\dot{r}^2+something=constant$$, you can define something as an effective potential, by analogy with classical mechanics, $$V\equiv something$$. This would be a kind of "standard" definition. But, if for mathematical or other reason, it is more convenient to define the effective quantity in another way, I don't think any problem in this. The effective potential in general relativity is mostly a mathematical tool. However, the definition may be important, because some physical quantities, like the marginally stable circular orbits, are obtained from it. But even in this case, once you impose the condition $$\dot{r}=0$$, you can handle the problem in various ways.

I think that answers the posted question.