I've been learning about the King's model for globular clusters and have been taught that it assumes a core radius: $$r_c = \sqrt \frac{9\sigma^2}{4\pi \rho_0 G} \ (1)$$ where $r_c$ is the core radius (the radius at which the density $\rho$ is half the central density value $\rho_0$) and $\sigma$ is the velocity dispersion. But there are 2 things I'm confused about:

  1. How is this equation derived? When I attempt to derive it (given that $\rho(r_c) = \frac{\rho_0}{2}$), I just keep getting this equation: $$r_c = \sqrt \frac{6\sigma^2}{4\pi \rho_0 G}\ (2)$$

  2. When I was searching about the core radius in my attempt to derive the equation, I found that Equation 1 is actually the equation for the King radius ($r_0$) not $r_c$, and that $r_0 \neq r_c$. Is this true? And if so, what is the difference between $r_0$ and $r_c$ then?

Thanks for any help anyone can give!



  1. https://iopscience.iop.org/article/10.1088/0004-637X/778/1/57 , pg. 1 of PDF

    This is the first article I came across that had the equation I was looking for, but here the equation involves $r_0$ instead of $r_c$. It also says (on line 12 of the paragraph below Equations 1, 2 and 3) that $r_c$ is not equivalent to $r_0$.

  2. https://people.ast.cam.ac.uk/~vasily/Lectures/SDSG/sdsg_7_clusters.pdf , pg. 40-41

    This is the 2nd and only other reference to this equation that I have found so far, which also has the same above information as the 1st reference.

  • $\begingroup$ Could you provide some references to where this is coming from? In particular, the original King (1962) profile for GCs was a purely photometric description, where the core radius was a free parameter and the velocity dispersion was not involved. $\endgroup$ Aug 18, 2023 at 11:40
  • $\begingroup$ @PeterErwin Of course! I got Equation 1 from one of my university lectures which I can’t link, but I’ve added the other references I used to the post. $\endgroup$
    – user374355
    Aug 18, 2023 at 12:29
  • $\begingroup$ Note $\sigma$ should be $\sigma_r$, the radial component of the velocity dispersion and $r_c$ is radius where surface density is half of central surface density ($\sigma_0$), not density ($\rho_0$). See section on Models for galaxy distrtibutions P 12 in X-ray emission from clusters of galaxies by Sarazin, Craig L. at ui.adsabs.harvard.edu/abs/1986RvMP...58....1S/abstract. Stop when you see $r_c \equiv 3\beta$ $\endgroup$
    – eshaya
    Aug 18, 2023 at 16:08

1 Answer 1


The first thing you should know (as Peter Erwin and eshaya have implied) is that the King Model is notorious for being a mishmash of observational and theoretical definitions. From what I understand, the original King Model was defined as an empirical tool that used the core radius, tidal radius, and central surface density as free parameters to achieve a fit to a few globular clusters (https://articles.adsabs.harvard.edu/pdf/1962AJ.....67..471K). I don't believe that this definition of a King Model is easily or cleanly extendable to the 3D density of the cluster or the potential of the cluster.

I think where the confusion appears is that there is a related but distinct concept, which is also called a King Model, but is better described as a truncated isothermal sphere model. This model is the one present in Vasily Belokurov's course notes that you shared above, and his treatment of it is nearly identical to the discussion of King Models in Binney & Tremaine (2008), which is the reference for the discussion below. B&T also has the definition for $r_0$ that you call $r_c$ in your equation (1).

The isothermal sphere version of the King Model has a simple analytical form for the density of the singular isothermal sphere given the velocity dispersion of the cluster $\sigma^2$:

\begin{equation} \rho(r) = \frac{\sigma^2}{4\pi G r^2} \end{equation}

although this definition leads to a non-physical infinite density at the center of the cluster.

You can get around this by instead expressing the density and radius as dimensionless quantities which are divided by the (observed) central density $\rho_0$ and the core radius $r_0$ (I will use $r_0$ to match the B&T notation rather than the $r_c$ you use above). The choice of core radius is completely arbitrary as long as it has units of length. It seems like the common definition is the one you share in your equation (1) above.

If you define things this way, you get a Poisson Equation for $\rho(r)$ of the form

\begin{equation} \frac{\textrm{d}}{\textrm{d}r}\left( r^2 \frac{\textrm{d} \ln \rho}{\textrm{d}r} \right) = -9 r^2 \rho. \end{equation}

This DE has to be solved numerically.

Why was 9 chosen for this definition? Your guess as good as mine. I don't think this is a quantity that has been derived from anything in particular from first principals. I couldn't find the original place that this definition for $r_0$ was first used, but I suspect that the factor of 9 was empirically added because it causes $\rho(r_0) \approx \rho_0/2$ to be roughly true.

To illustrate this, you could easily replace the 9 in the above equation with 1 (or any other number, it doesn't change the physics at all) with a more "natural" definition of $r_0$:

\begin{equation} r_0 = \sqrt{\frac{\sigma^2}{4\pi G \rho_0}}. \end{equation}

However, $\rho(r_0) \approx \rho_0/2$ is no longer true for this "natural" definition of $r_0$, which could make it less appealing to observers.

If anyone knows more about this, or if there actually is a clean way to derive this extra factor of 9, I would appreciate seeing it!

As for your question about why $r_0\neq r_c$, this is because the observational definition of $r_c$ is the half-light radius i.e. $\Sigma(r_c) = \Sigma(0)/2$, whereas the definition for the truncated isothermal sphere scale radius $r_0$ appears to be arbitrary and coincidentally roughly corresponds to $\rho(r_0) \approx \rho(0)/2$. These two definitions are not the same.

  • $\begingroup$ This is such a clear and detailed explanation that it must have taken some time, but thank you so so much! $\endgroup$
    – user374355
    Aug 19, 2023 at 9:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .