I'm trying to run some calculations on Kerr-Newman black holes, but I'm having two major difficulties. First, most equations I've been able to find are only for Kerr black holes. Second, essentially all equations are in one of several incompatible varieties of "natural units" (this is a big problem for me, since my background is not in Physics, and so I don't magically know when to multiply things by $c$, $G$, $k_B$, $\hslash$, $M$, etc.).

In approximately increasing order of my confusion:

  • "$J$": I assume this is angular momentum, with the usual $\text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}$ units?

  • "$J_{max}(M)$": Presumably the maximum allowable value of $J$ for a given mass $M$ (in $\text{kg}$). I found an SI formula for a Kerr black hole. Would it be different for a Kerr-Newman black hole?

  • "$\Omega$": Angular velocity (from an observer at infinity, with units $\text{rad}\cdot\text{s}^{-1}$, I assume). I found this formula for Kerr, not Kerr-Newman, and I couldn't get the units to make sense.

  • "$M$": I assume this is mass in $\text{kg}$. But, I've also seen it rescaled to $\text{m}$, somehow.

  • "$a$": Some kind of spin parameter? Unclear how exactly this relates to $J$. AFAICT, there are multiple definitions, with dimensionless units, meters (e.g. as $J/(M c)$ from Abramowicz and Fragile 2013 $\S 3$), and possibly other units. Is the dimensionless version supposed to be $a/M$, with $M$ rescaled to length (by $GM/c^2$) or something? I also found a dimensionless version here as a ratio of $J$ to $J_{max}(M)$. How does that tie in?

  • "$a/M$": Specific angular momentum? Seen in e.g. Bardeen et al. 1972 (fig 1.). Also seems to be called $\chi$, "dimensionless spin". Does this have anything to do with spin?

For each of these quantities (and anything else relevant), I'm looking for (1) their scientific names (so I can research them), (2) their SI definitions (so I can do dimensional analysis on them), and (3) their SI equations for Kerr-Newman black holes (so I can calculate them). At least a self-consistent explanation of $a$, $a/M$ and how they relate to $J$, as treated in the literature, so that I'm not completely lost when reading it anymore.

  • $\begingroup$ May I ask what you are doing this for? If you can't convert from natural units, understanding GR, which is much more subtle, is going to be a giant bear. $\endgroup$ Aug 4, 2017 at 21:15
  • $\begingroup$ @JerrySchirmer My problem is that there are different varieties of natural units, and when coupled with the fact that sources disagree about the equations (due to errors), I can't use dimensional analysis to make an educated guess to corroborate them. Required for this project. $\endgroup$
    – geometrian
    Aug 7, 2017 at 2:39
  • $\begingroup$ In the GR community, the converstion is always taking meters to be the base unit, and using factors of $G$ and $c$ (and $k$, if there are temperatures) along with dimensional analysis. $\endgroup$ Aug 7, 2017 at 14:35

2 Answers 2


On the wikipedia page https://en.wikipedia.org/wiki/Kerr%E2%80%93Newman_metric the Kerr-Newman (KN) metric is given in SI units and all metric parameters are defined. $J$ is the black holes (BH) angular momentum, $Q$ its charge and $M$ its mass.

$a=J/(M c)$ is called "Kerr parameter" or angular momentum per unit mass it has the dimension of length. One can get this dimensionless by e.g. $\chi=a/(M G)c^2$. I am not aware of an interpretation of $\chi$ as a quantum mechanical spin but I know that the a KN BHhas a Dirac gyromagnetic ratio of $\gamma_{KN}=Q a c$. For more information on that topic I would recommend the references in section "Interpretive issues" of http://www.scholarpedia.org/article/Kerr-Newman_metric .

The a KN BH has two horizons $r_\pm$: $$r_\pm=\frac{M G}{c^2}\pm\sqrt{\left(\frac{M G}{c^2}\right)^2-a^2-r_Q^2},\tag{1}$$ with $r_Q\equiv \frac{Q^2 G}{4\pi\epsilon_0 c^4}$ as the length scale corresponding to the black holes charge Q. A Kerr-Newman black hole is extremal for $r_+=r_-$ which means for $a_{ext}$ $$a_{ext}=\pm\sqrt{\left(\frac{M G}{c^2}\right)^2-r_Q^2}.$$

The angular velocities on the horizons of the Kerr metric are given by


so using eq. (1) would lead to a charge contribution. Which makes sense, since the KN metric has electromagnetically induced frame-dragging but I am not certain if (2) is correct for a KN BH.

Maybe after that one short note on geometrical units in General relativity $GR$: they usually use only $G=c=1$ and use that to convert all dimensions to powers of length: a second can be converted to a meter by multiplying with $c$ and a kilogram can be converted to meter by multiplying it with $G/c^2$. For currents or charges there are a few conventions out there. One example would be $\mu_0/(4\pi)=c=G=1$ which makes currents dimension less. One Ampere can be converted multiplying it with $\sqrt{G \mu_0/(4\pi)}/c^2$. If temperatures are involved $k_B=1$ can be used to convert temperatures into energies. Energies can be converted to length by multiplying with $G/c^4$. $\hbar$ is not common in GR since $c=G=\hbar=k_B=\mu_0=1$ would result in completely dimensionless Planck units. But in principle $\hbar$ can be used instead of $G$ to convert masses or with $c=G=k_B=\mu_0=1$ to eliminate the last dimension.

Geometrized units are not hard to understand and extremely useful in theoretical physics, since they make equations less cluttered and numerical computations easier.

  • $\begingroup$ I am still properly processing this answer. But for now, shouldn't it be $(M G / c^2)^2$ in the radicals? $\endgroup$
    – geometrian
    Jun 15, 2017 at 5:49
  • $\begingroup$ Yes. I was to focused about adding the right constants and forgot those squares. I have corrected it in the answers. $\endgroup$
    – N0va
    Jun 16, 2017 at 8:34
  • $\begingroup$ Thought so. Anyway, this answer is really great; thanks! Re: $\Omega_{\pm}$, the formula for the surface area of the outer horizon, $A_H=4\pi(r_+^2+a^2)$, is similar in form, and is correct for Kerr-Newman. $\endgroup$
    – geometrian
    Jun 19, 2017 at 8:02
  • $\begingroup$ I think it should be "$r_Q^2 \equiv$", also. $\endgroup$
    – geometrian
    Jun 19, 2017 at 11:09
  • $\begingroup$ . . . and $r_\pm^2$ in the calculation of $\Omega_\pm$. $\endgroup$
    – geometrian
    Jun 22, 2017 at 0:45

To get the dimensionless spin parameter use

$$\rm \bar a=\frac{J c}{G M^2}$$

where J is the angular momentum of the black hole (the angular momentum J can be expressed in terms of GM²/c, so divide by that to get rid of the dimensions). This always gives a number between 0 and 1 for black holes (regular objects also can have spin parameters larger than 1, but those have to eject some of their angular momentum before collapsing to a black hole). If for some reasons you want to keep the M in the equations instead of setting G=M=c=K=1, just divide by M instead of M².


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.