Editing Stellar dynamics (section)

=== A worked example of Jeans equation in a thick disk ===

Consider again the thick disk potential in the above example. 
If the density is that of a gas fluid, then the pressure would be zero at the boundary <math> z=\pm z_0 </math>.  To find the peak of the pressure, we note that 
<math display="block"> P(R,z) = \int^{z_0}_z \partial_z\Phi \rho(R) dz = \rho(R) [\Phi(R,z_0) - \Phi(R,z)].</math>

So the fluid temperature per unit mass, i.e., the 1-dimensional velocity dispersion squared would be 
<math display="block"> \sigma^2(R,z) = {P(R,z) \over \rho(R)},  ~~ |z| \le z_0  </math>

<math display="block"> \sigma^2= {G M_0 \over 2z_0} \log{Q(z) Q(-z) \over Q (z_0)Q(-z_0) },  ~~ Q(z) \equiv R_0 + z_0 + z +\sqrt{R^2+(R_0+z_0+z)^2}. </math>

Along the rotational z-axis, 
<math display="block"> \sigma^2(0,z) = {G M_0 \over 2z_0} \log {4(R_0+z_0+z) (R_0+z_0-z) \over 4R_0 (R_0+2z_0) } </math>
<math display="block"> \sigma(0,z) = \sqrt{G M_0 \over 2z_0} \sqrt{\log { (R_0+z_0)^2-z^2 \over (R_0+z_0)^2-z_0^2 } } , </math>
which is clearly the highest at the centre and zero at the boundaries <math> z=\pm z_0 </math>.  Both the pressure and the dispersion peak at the midplane <math> z=0 </math>.  In fact the hottest and densest point is the centre, where <math> P(0,0) = { M_0 \over 4 \pi R_0^2 z_0 } {-G M_0 \log [1- (1+R_0/z_0)^{-2}] \over 2 z_0} . </math>