In the review Foundations of Black Hole Accretion Disk Theory, the authors defines an effective potential for Kerr geometry as (Chap. 2, eqn. 23) $$\mathcal{U}_{eff}=-\frac{1}{2}\ln\left|g^{tt}-2lg^{t\phi}+l^2g^{\phi\phi}\right|$$ where $l=\dfrac{\mathcal{L}}{\mathcal{E}}=-\dfrac{u_\phi}{u_t}$ is the specific angular momentum, $\mathcal{L}=p_\phi$ is the angular momentum and $\mathcal{E}=-p_t$ is the energy.

It is mentioned that this form of the potential is chosen because using the potential $\mathcal{U}_{eff}$, the rescaled energy $\mathcal{E}^*=\ln\mathcal{E}$ and $V=u^ru_r+u^\theta u_\theta<<u^\phi u_\phi$, slightly non-circular motion can be characterized by the the equation $$\frac{1}{2}V^2=\mathcal{E}^*-\mathcal{U}_{eff}$$

The form of this equation is indeed similar to that of the Newtonian equation. But there were nothing mentioned in the paper regarding the derivation of the effective potential. Also, I couldn't understand why they re-scaled the energy as $\mathcal{E}^*=\ln\mathcal{E}$.

My Questions:

  1. How to derive the effective potential $\mathcal{U}_{eff}$? Hints to the derivation would be sufficient.
  2. What is the logic behind the scaling of the conserved energy $\mathcal{E}^*=\ln\mathcal{E}$?

1 Answer 1


Take the four-velocity normalization $u^\mu u_\mu = -1$ and write it out as $$1 + u^r u_r + u^\vartheta u_\vartheta = -g^{tt}u_t^2 - 2 g^{t \varphi} u_t u_\varphi - g^{\varphi \varphi} u_\varphi^2 $$ Now rewrite this in terms of $\mathcal{E} = -u_t, \ell = -u_\varphi/u_t$, take a logarithm of the equation, and use the properties $\ln(xy) = \ln(x) + \ln(y)$ and $\ln(1 + x) = x + \mathcal{O}(x^2)$. The quantity $\mathcal{E}^*$ is used just because it is additive and is actually the specific Newtonian energy without the rest-mass term in the Newtonian limit.

I will leave it as an exercise to the dear reader that the minima of the potential $\mathcal{U}_{\rm eff}$ actually also correspond to circular orbits.

  • $\begingroup$ Thanks. I had derived the result. However when I used this effective potential to find the marginally bound and marginally stable orbits, I found that the marginally bound orbit matches exactly with the exact Kerr potential but the marginally stable orbit is slightly different. What might be the possible reason for this? $\endgroup$
    – Richard
    Commented Sep 23, 2019 at 11:40
  • $\begingroup$ They should be exactly the same, I believe the mistake could be in typing the analytical formula for the ISCO (it is a little bit involved) or some kind of numerical error. I had checked that the formula of Abramowicz & Fragile in the living review is correct. $\endgroup$
    – Void
    Commented Sep 23, 2019 at 12:00

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