<p>Using unit analysis, I've found that a natural unit of angular momentum should be
\begin{equation}\tag{1}
L_0 = \frac{G M_{\odot}}{c} \approx 8.816 \times 10^{41} \, \text{kg} \, \text{m}^2/\text{s}.
\end{equation}</p>

<p>In the case of a <em>bounded</em> system of two equal masses, circularly moving around their center of mass, the total angular momentum (relative to the center of mass) is
\begin{equation}\tag{2}
L_{\text{bound}} = 2 M \, v \, r = \frac{2M \, v^2 \, r}{v}.
\end{equation}
Now, using Newton's equation and the gravitationnal force, we have
\begin{equation}
M \, v^2 = \frac{G M^2}{4 \, r}.
\end{equation}
Thus, equation (2) becomes :
\begin{equation}\tag{3}
L_{\text{bound}} = \frac{G M^2}{2 \, v}.
\end{equation}
The classical theory cannot be trusted when $v \sim c$, and since $v &lt; c$ (forgetting the 2 factor) :
\begin{equation}\tag{4}
L_{\text{bound}} \approx \frac{G M^2}{v} &gt; \frac{G M^2}{c} \equiv L_0.
\end{equation}
So $L_0$ appears to be a good unit.  When $L \sim L_0$, we then need relativity.  We're in a classical regime when $L \gg L_0$.</p>

<p>The system isn't bounded when $L \gg L_{\text{bound}}$, which implies $L \ggg L_0$.</p>

<p>I don't know if this is a good way to introduce the $L_0$ unit defined by (1).  Any opinion/suggestions/comments on this ?</p>

<p>Do astronomers use $G M^2 / c$ as natural unit of angular momentum, when they study star clusters and globulars ?</p>

CC BY-SA 3.0
2017-06-03T12:33:22.907Z
2017-06-03T12:47:38.460Z
2017-06-03T12:47:38.460Z
Cham
<p>I'm calculating the total angular momentum of a bunch of theoretical stars of a few solar masses, separated by a few light years, moving at velocities of about a few dozens of <strong>km / sec</strong>.</p>

<p>Mechanical angular momentum relative to an arbitrary origin (the cluster's center of mass is a good choice) is this :
\begin{equation}
\vec{L} = \sum_i \vec{r}_i \times m_i \, \vec{v}_i,
\end{equation}
so the <em>SI</em> units are $\mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{sec}$.  But in real astronomy, what should be the most simple/natural units for star clusters ?  $M_{\odot} \cdot (\mathrm{ly})^2 / \mathrm{year} \equiv M_{\odot} \, c^2 \cdot \mathrm{year}$ ?</p>

<p>What are the angular momentum units used by astronomers ?</p>

<p>And what about energy units for a cluster of stars ?  I'm interested in kinematical energy and gravitationnal potential energy of the whole cluster.  Certainly not <strong>joules</strong> !?  And $M_{\odot} \, c^2$ would certainly be too much, I suppose !</p>

CC BY-SA 3.0
2017-06-03T12:08:12.957Z
I think I've found a simple answer.  I'll add an answer to my own query.
Cham
<star><newtonian-gravity><units><star-cluster><globular-clusters>
Natural units of star clusters angular momentum and energy
Cham
# Natural units of star clusters angular momentum and energy

(Asked by: username_0 on 2017-06-02)

I'm calculating the total angular momentum of a bunch of theoretical stars of a few solar masses, separated by a few light years, moving at velocities of about a few dozens of **km / sec**.

Mechanical angular momentum relative to an arbitrary origin (the cluster's center of mass is a good choice) is this :
\begin{equation}
\vec{L} = \sum\_i \vec{r}\_i \times m\_i \, \vec{v}\_i,
\end{equation}
so the *SI* units are $\mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{sec}$. But in real astronomy, what should be the most simple/natural units for star clusters ? $M\_{\odot} \cdot (\mathrm{ly})^2 / \mathrm{year} \equiv M\_{\odot} \, c^2 \cdot \mathrm{year}$ ?

What are the angular momentum units used by astronomers ?

And what about energy units for a cluster of stars ? I'm interested in kinematical energy and gravitationnal potential energy of the whole cluster. Certainly not **joules** !? And $M\_{\odot} \, c^2$ would certainly be too much, I suppose !

**Tags:** star, newtonian-gravity, units, star-cluster, globular-clusters

> **Comments:**
> **username_0**: I think I've found a simple answer.  I'll add an answer to my own query.  

---

## 1 answers

### Answer by username_0 on 2017-06-03. Score: 2

Using unit analysis, I've found that a natural unit of angular momentum should be
\begin{equation}\tag{1}
L\_0 = \frac{G M\_{\odot}}{c} \approx 8.816 \times 10^{41} \, \text{kg} \, \text{m}^2/\text{s}.
\end{equation}

In the case of a *bounded* system of two equal masses, circularly moving around their center of mass, the total angular momentum (relative to the center of mass) is
\begin{equation}\tag{2}
L\_{\text{bound}} = 2 M \, v \, r = \frac{2M \, v^2 \, r}{v}.
\end{equation}
Now, using Newton's equation and the gravitationnal force, we have
\begin{equation}
M \, v^2 = \frac{G M^2}{4 \, r}.
\end{equation}
Thus, equation (2) becomes :
\begin{equation}\tag{3}
L\_{\text{bound}} = \frac{G M^2}{2 \, v}.
\end{equation}
The classical theory cannot be trusted when $v \sim c$, and since $v < c$ (forgetting the 2 factor) :
\begin{equation}\tag{4}
L\_{\text{bound}} \approx \frac{G M^2}{v} > \frac{G M^2}{c} \equiv L\_0.
\end{equation}
So $L\_0$ appears to be a good unit. When $L \sim L\_0$, we then need relativity. We're in a classical regime when $L \gg L\_0$.

The system isn't bounded when $L \gg L\_{\text{bound}}$, which implies $L \ggg L\_0$.

I don't know if this is a good way to introduce the $L\_0$ unit defined by (1). Any opinion/suggestions/comments on this ?

Do astronomers use $G M^2 / c$ as natural unit of angular momentum, when they study star clusters and globulars ?


---

