Editing Stellar dynamics (section)

=== A worked example on moments of distribution functions in a uniform spherical cluster ===

We can find out various moments of the above distribution function, reformatted as with the help of three Heaviside functions, 
<math display="block"> f(|\mathbf{r}|,V_r,V_\theta,V_\varphi) = 
{C_0 \over V_0^3} \left.{\text{H}(1-x) 
\over \left(1-x^2\right)^{1 \over 2}}\right|_{x \equiv {|\mathbf{r}| \over r_0}} { \text{H}(V_\varphi) \text{H}(1-q) \over (1-q)^{1 \over 2} } , 
~~ q(\mathbf{r},\mathbf{V}) \equiv {V_r^2 \over V_e(|\mathbf{r}|)^2} + {V_\theta^2  \over V_0^2} +  {V_\varphi^2 \over V_0^2}, </math> 
once we input the expression for the earlier potential <math>\Phi(r) </math> inside <math> r \le r_0 </math>, or even better the speed to "escape from r to the edge" <math> r_0 </math> of a uniform sphere <math> V_e(r)=V_0 \sqrt{1-{r^2 \over r_0^2}}. </math>  
Clearly the factor <math> {V_e(|\mathbf{r}|) \over \sqrt{2Q} } = \sqrt{\max[0,{1 \over 1-q}]}</math> in the DF (distribution function) is well-defined only if <math>Q \ge 0 \rightarrow q\le 1  </math>, which implies a narrow range on radius <math> 0 \le |\mathbf{r}|<r_0 </math> and excludes high velocity particles, e.g., <math> V_t > V_0 > V_e(r) </math>, from the distribution function (DF, i.e., phase space density).

In fact, the positivity carves the (<math> V_\varphi \ge 0 </math>) left-half of an ellipsoid in the <math>[V_r, V_\theta, V_\varphi]</math> velocity space ("velocity ellipsoid"), 
<math display="block">  
q(\mathbf{r},\mathbf{V}) \equiv {V_r^2 \over V_0^2 (1-r^2/r_0^2)}  +  \left({V_\theta^2  \over V_0^2} + {V_\varphi^2 \over V_0^2} \right) \equiv u_r^2 + u_\theta^2 + u_\varphi^2 \le 1,  
</math>
where <math> (u_r,u_\theta,u_\varphi)</math> is <math> (V_r, V_\theta,V_\varphi)</math> rescaled by the function <math> V_e(r)=V_0 \sqrt{1-r^2/r_0^2} </math> or <math> V_0 </math> respectively.

The velocity ellipsoid (in this case) has rotational symmetry around the r axis or <math> V_r </math> axis.  It is more squashed (in this case) away from the radial direction, hence more tangentially anisotropic because everywhere <math> V_e(r) < V_0 </math>, except at the origin, where the ellipsoid looks isotropic. Now we compute the moments of the phase space.
 
E.g., the resulting density (moment) is
<math display="block">\begin{align}
 \rho(r,\theta,\varphi) & = 
\int_{-V_e(r)}^{V_e(r)} dV_r \int_{-V_0}^{V_0} dV_\theta \int_{0}^{V_0} dV_\varphi
{C_0 \over V_0^3} \left({2Q \over V_0^2}\right)^{-1/2} \\
& = \int_{-1}^{1} \int_{-1}^{1} \int_0^1 { (V_e du_r) (V_0 du_\theta) (V_0 du_\varphi) C_0 
\over V_0^3 (1- r^2/r_0^2)^{1/2} (1 - q)^{1/2} }\left.\right|_{q=u_r^2 + u_\theta^2 + u_\varphi^2}\\
& = C_0  { \int_0^1 (1-u^2)^{-1/2} (2\pi u^2 du)} = \rho_0
\end{align}</math>
is indeed a spherical (angle-independent) and uniform (radius-independent) density inside the edge, where the normalisation constant <math> C_0 =2 \pi^{-2} \rho_0 </math>.

The streaming velocity is computed as the weighted mean of the velocity vector
<math display="block">\begin{align}
 \langle\mathbf{V}\rangle (\mathbf{x}) & \equiv 
{\int f d\mathbf{V}^3 \mathbf{V} \over \int f d\mathbf{V}^3 } \\
& = {1 \over \rho} \int f d\mathbf{V}^3  [V_r, V_\theta, V_\varphi] {C_0 V_0^2 (2Q)^{-1/2}} \\
& = \left[{ 
  \int_{-1}^{1} \!\!u_r...du_r, 
~~\int_{-1}^{1} \!\!u_\theta...du_\theta , 
~~\int_0^1 (2du_r) \int_{0}^{\sqrt{1-u_r^2}} \!\!(2du_\theta) \int_{0}^{\sqrt{1-u_r^2-u_\theta^2}} \!\!\!\!\!\!\!\!\!\!{du_\varphi  u_\varphi V_0 \over (1 - u_r^2 - u_\theta^2 - u_\varphi^2)^{1/2} } 
\over \int_{0}^1 (2\pi U dU) \int_{0}^{\sqrt{1-U^2}}  du_\varphi (1 - U^2 - u_\varphi^2)^{-1/2}  }\right] \\
& = \left[0,0,{4 V_0 \over 3\pi}\right] = \overline{\mathbf{V}(\mathbf{x})}, 
\end{align}</math>
where the global average (indicated by the overline bar) of flow implies uniform pattern of flat azimuthal rotation, but zero net streaming everywhere in the meridional <math> (r,\theta)</math> plane.

Incidentally, the angular momentum global average of this flat-rotation sphere is 
<math display="block"> \overline{\mathbf{r} \times \langle\mathbf{V}\rangle} = \int_0^{r_0}  {(\rho 4\pi r^2 dr) \over M_0} [0,0,r \langle V_\varphi\rangle] = [0, 0, {3 r_0\over 4} \overline{V_\varphi}]. </math>
Note global average of centre of mass does not change, so <math> \overline{\mathbf{V}_i(\mathbf{x})} =0 </math> due to global momentum conservation in each rectangular direction <math> i=x,y,z</math>, and this does not contradict the global non-zero rotation.

Likewise thanks to the symmetry of <math> f(r,\theta,\varphi,V_r,V_\theta,V_\varphi) = f(r,\theta,\pm \varphi, \pm V_r, \pm V_\theta,V_\varphi) </math>, we have
<math>  \langle\mathbf{(\pm V_r) V_\varphi}\rangle  =0 </math>, <math>~ \langle \mathbf{(\pm V_\theta) V_\varphi}\rangle  =0 </math>, <math>~ 
\langle\mathbf{(\pm V_r) V_\theta}\rangle  =0 </math> everywhere}.

Likewise the rms velocity in the rotation direction is computed by a weighted mean as follows, E.g.,
<math display="block">\begin{align}
\langle\mathbf{V}_\varphi^2\rangle(|\mathbf{x}|)
&\equiv {\int f d\mathbf{V}^3 V_\varphi^2 \over \rho(|\mathbf{r}|)} \\
& = {\int_0^1 (2du_r) \int_{0}^{\sqrt{1-u_r^2}} (2du_\theta) \int_{0}^{\sqrt{1-u_r^2-u_\theta^2}} du_\varphi { (u_\varphi V_0)^2 \over (1 - q)^{1/2} } \over \int_0^1 { (2\pi u^2 du) (1 - u^2 )^{-1/2} } } \\
& = 0.25V_0^2 = 0.5 \langle V_t^2 \rangle \\
& = {\!\!\int_0^1 (2du_r)\!\! \int_{0}^{\sqrt{1-u_r^2}} (2du_\varphi) \!\!\int_{0}^{\sqrt{1-u_r^2-u_\varphi^2}} du_\theta { (u_\theta V_0)^2 \over (1 - q)^{1/2} }
\over \int_0^1 { (2\pi u^2 du) (1 - u^2 )^{-1/2} } } \\
& =\langle\mathbf{V}_\theta^2\rangle(|\mathbf{x}|),  \\
\end{align}</math>

Here
<math> \langle V_t^2 \rangle = \langle V_\theta^2 +  V_\varphi^2\rangle =0.5V_0^2. </math>

Likewise
<math display="block"> \langle\mathbf{V}_r^2\rangle(\mathbf{x}) = {\!\!\int_0^1 (du_\varphi) \int_{0}^{\sqrt{1-u_\varphi^2}} \!\!(2du_\theta) \!\!\int_{0}^{\sqrt{1-u_\varphi^2-u_\theta^2}} \!\!\!{ (2du_r)(u_r V_e(r))^2 \over (1 - q)^{1/2} } \over \int_0^1 { (2\pi u^2 du) (1 - u^2 )^{-1/2} } } = \left({V_0 \over 2} \sqrt{1-{r^2 \over r_0^2}} \right)^2. </math>

So the pressure tensor or dispersion tensor is
<math display="block"> \begin{align}
\sigma^2_{ij}(\mathbf{r})= &  {P_{ij}(\mathbf{r}) \over \rho(\mathbf{r})} \\
  =&\langle\mathbf{V}_i\mathbf{V}_j\rangle- \langle\mathbf{V}_i\rangle\langle\mathbf{V}_j\rangle \\
=  & \begin{bmatrix}
    \left[1-({r \over r_0})^2\right]\left({V_0\over 2}\right)^2 & 0 & 0 \\
    0 & \left({V_0\over 2}\right)^2 & 0 \\
    0 & 0 & \left[1- ({8 \over 3 \pi})^2\right]\left({V_0\over 2}\right)^2 
    \end{bmatrix} 
\end{align}
</math>
with zero off-diagonal terms because of the symmetric velocity distribution.  
Note while there is no Dark Matter in producing the previous flat rotation curve, 
the price is shown by the reduction factor <math> {8 \over 3 \pi} = 0.8488</math> in the random velocity spread in the azimuthal direction. Among the diagonal dispersion tensor moments, <math> \sigma_\theta \equiv \sqrt{\sigma^2_{\theta\theta}} = 0.5V_0</math> is the biggest among the three at all radii, and <math> \sigma_\varphi \equiv \sqrt{\sigma^2_{\varphi\varphi}} \ge \sigma_r \equiv \sqrt{\sigma^2_{rr}} </math> only near the edge between <math> 0.8488 r_0 \le r \le r_0 </math>.

The larger tangential kinetic energy than that of radial motion seen in the diagonal dispersions is often phrased by an anisotropy parameter
<math display="block"> \beta(r) \equiv 
1 -  { 0.5\langle {\mathbf V_t}^2(|\mathbf{r}|) \rangle \over \langle{\mathbf V_r}^2\rangle(|\mathbf{r}|) } = 1 -  {\langle {\mathbf V_\theta}^2(|\mathbf{r}|) \rangle \over \langle{\mathbf V_r}^2\rangle(|\mathbf{r}|) } = - {r^2 \over r_0^2 - r^2} \le 0; </math>
a positive anisotropy would have meant that radial motion dominated, and a negative anisotropy means that tangential motion dominates (as in this uniform sphere).