<p>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 <em>surface</em> density as free parameters to achieve a fit to a few globular clusters (<a href="https://articles.adsabs.harvard.edu/pdf/1962AJ.....67..471K" rel="noreferrer">https://articles.adsabs.harvard.edu/pdf/1962AJ.....67..471K</a>). 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.</p>
<p>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 &amp; Tremaine (2008), which is the reference for the discussion below. B&amp;T also has the definition for <span class="math-container">$r_0$</span> that you call <span class="math-container">$r_c$</span> in your equation (1).</p>
<p>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 <span class="math-container">$\sigma^2$</span>:</p>
<p><span class="math-container">\begin{equation}
\rho(r) = \frac{\sigma^2}{4\pi G r^2}
\end{equation}</span></p>
<p>although this definition leads to a non-physical infinite density at the center of the cluster.</p>
<p>You can get around this by instead expressing the density and radius as dimensionless quantities which are divided by the (observed) central density <span class="math-container">$\rho_0$</span> and the core radius <span class="math-container">$r_0$</span> (I will use <span class="math-container">$r_0$</span> to match the B&amp;T notation rather than the <span class="math-container">$r_c$</span> 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.</p>
<p>If you define things this way, you get a Poisson Equation for <span class="math-container">$\rho(r)$</span> of the form</p>
<p><span class="math-container">\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}</span></p>
<p>This DE has to be solved numerically.</p>
<p>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 <span class="math-container">$r_0$</span> was first used, but I suspect that the factor of 9 was empirically added because it causes <span class="math-container">$\rho(r_0) \approx \rho_0/2$</span> to be roughly true.</p>
<p>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 &quot;natural&quot; definition of <span class="math-container">$r_0$</span>:</p>
<p><span class="math-container">\begin{equation}
r_0 = \sqrt{\frac{\sigma^2}{4\pi G \rho_0}}.
\end{equation}</span></p>
<p>However, <span class="math-container">$\rho(r_0) \approx \rho_0/2$</span> is no longer true for this &quot;natural&quot; definition of <span class="math-container">$r_0$</span>, which could make it less appealing to observers.</p>
<p>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!</p>
<p>As for your question about why <span class="math-container">$r_0\neq r_c$</span>, this is because the observational definition of <span class="math-container">$r_c$</span> is the half-light radius i.e. <span class="math-container">$\Sigma(r_c) = \Sigma(0)/2$</span>, whereas the definition for the truncated isothermal sphere scale radius <span class="math-container">$r_0$</span> appears to be arbitrary and coincidentally roughly corresponds to <span class="math-container">$\rho(r_0) \approx \rho(0)/2$</span>. These two definitions are not the same.</p>

CC BY-SA 4.0
2023-08-19T09:58:45.133Z
This is such a clear and detailed explanation that it must have taken some time, but thank you so so much!
user374355
CC BY-SA 4.0
2023-08-18T21:35:34.107Z
2023-08-18T21:39:06.493Z
2023-08-18T21:39:06.493Z
Tom Donlon
<p>I've been learning about the King's model for globular clusters and have been taught that it assumes a core radius:
<span class="math-container">$$r_c = \sqrt \frac{9\sigma^2}{4\pi \rho_0 G} \ (1)$$</span>
where <span class="math-container">$r_c$</span> is the core radius (the radius at which the density <span class="math-container">$\rho$</span> is half the central density value <span class="math-container">$\rho_0$</span>) and <span class="math-container">$\sigma$</span> is the velocity dispersion. But there are 2 things I'm confused about:</p>
<ol>
<li><p>How is this equation derived? When I attempt to derive it (given that <span class="math-container">$\rho(r_c) = \frac{\rho_0}{2}$</span>), I just keep getting this equation:
<span class="math-container">$$r_c = \sqrt \frac{6\sigma^2}{4\pi \rho_0 G}\ (2)$$</span></p>
</li>
<li><p>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 (<span class="math-container">$r_0$</span>) not <span class="math-container">$r_c$</span>, and that <span class="math-container">$r_0 \neq r_c$</span>. Is this true? And if so, what is the difference between <span class="math-container">$r_0$</span> and <span class="math-container">$r_c$</span> then?</p>
</li>
</ol>
<p>Thanks for any help anyone can give!</p>
<p>[EDIT]</p>
<p>References:</p>
<ol>
<li><p><a href="https://iopscience.iop.org/article/10.1088/0004-637X/778/1/57" rel="nofollow noreferrer">https://iopscience.iop.org/article/10.1088/0004-637X/778/1/57</a> , pg. 1 of PDF</p>
<p>This is the first article I came across that had the equation I
was looking for, but here the equation involves <span class="math-container">$r_0$</span> instead of
<span class="math-container">$r_c$</span>. It also says (on line 12 of the paragraph below Equations
1, 2 and 3) that <span class="math-container">$r_c$</span> is not equivalent to <span class="math-container">$r_0$</span>.</p>
</li>
<li><p><a href="https://people.ast.cam.ac.uk/%7Evasily/Lectures/SDSG/sdsg_7_clusters.pdf" rel="nofollow noreferrer">https://people.ast.cam.ac.uk/~vasily/Lectures/SDSG/sdsg_7_clusters.pdf</a> , pg. 40-41</p>
<p>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.</p>
</li>
</ol>

CC BY-SA 4.0
2023-08-18T11:40:33.013Z
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.
Peter Erwin
CC BY-SA 4.0
2023-08-18T12:29:25.610Z
@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.
user374355
CC BY-SA 4.0
2023-08-18T16:08:53.143Z
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  https://ui.adsabs.harvard.edu/abs/1986RvMP...58....1S/abstract.  Stop when you see $r_c \equiv 3\beta$
eshaya
<astrophysics><globular-clusters>
Core radius of King's model
user374355
# Core radius of King's model

(Asked by: username_0 on 2023-08-18)

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!

[EDIT]

References:

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](https://people.ast.cam.ac.uk/%7Evasily/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.

**Tags:** astrophysics, globular-clusters
> **Comments:**
> **username_1**: 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.  
> **username_0**: @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.  
> **username_2**: 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  https://ui.adsabs.harvard.edu/abs/1986RvMP...58....1S/abstract.  Stop when you see $r_c \equiv 3\beta$  

---

## 1 answers

### Answer by username_3 on 2023-08-18. Score: 5 | Accepted answer

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.

> **Comments:**
> **username_0**: This is such a clear and detailed explanation that it must have taken some time, but thank you so so much!  

---

