Moments of the Boltzmann equation

Although the collisionless Boltzmann equation (CBE for short) provides us with a complete description of the stellar dynamics problem, it is rather unwieldy for practical applications. The Jeans equations, which give a clearer physical picture of what is going on, are moments of the Boltzmann equation. Here is how to obtain them.

Suppose we integrate Eq. (8.19) over all velocities. The first term will become

$$\int {\partial f(\hbox{$\bf r$},\hbox{$\bf v$},t)\over \partial t... ... v$},t)\,d\hbox{$\bf v$}\equiv {\partial n(\hbox{$\bf r$},t)\over\partial t}\,.$$ (8.21)

The integral of the distribution function over velocities, $n\equiv \int\,f\,d\hbox{$\bf v$}$ is just the density of stars. And we're allowed to interchange the integral over velocities, with the derivative wrt time, since the range of velocities over which we integrate does not depend on $t$.

Integrating the second term, gives

$$\int v_i{\partial f\over\partial x_i}\,d\hbox{$\bf v$}= {\partial... ..., \int v_i\,f\,d\hbox{$\bf v$}\equiv {\partial( n\bar v_i)\over\partial x_i}\,,$$ (8.22)

where I've introduced the mean velocity $\bar v_i$ as

$$n\,\bar v_i=\int v_i\,f\,d\hbox{$\bf v$}\,.$$ (8.23)

Now the last term is zero, since

$$\int_{-\infty}^\infty {\partial f\over\partial v_i} dv_i=f\vert _{v_i=-\infty}^{v_i=+\infty}\,,$$ (8.24)

since there are no stars with infinite velocity.

And so the first moment of the CBE is a continuity equation for the density,

$${\partial n\over\partial t}+{\partial ( n\bar v_i)\over\partial x_i} = 0\,.$$ (8.25)

Now the second moment is what we're after. The trick is to first multiply the CBE with $v_j$, and then integrate over all velocities. The calculation proceeds exactly as for the density, and the result is after a bit of algebra:

$$n\,{\partial\bar v_j\over\partial t} + n\,\bar v_i\,{\partial \ba... ...artial\Phi\over\partial x_j} - {\partial n\,\sigma^2_{ij}\over \partial x_i}\,.$$ (8.26)

where

$$\sigma^2_{ij}\equiv \overline{v_i v_j}-\bar v_i\bar v_j\,.$$ (8.27)

This was a lot of mathematics to come to a simple equation: suppose the system is in a steady state, $\partial/\partial t$=0, and there are no streaming motions, so that $\bar v_i=0$. Then the previous equation simplifies to

$${\partial\Phi\over\partial x_j} = -{1\over n}{\partial n\sigma^2_{ij}\over \partial x_i}\,.$$ (8.28)

This is a more general version for the equation of hydrostatic equilibrium. To see this, note that the matrix $\sigma_{ij}$ is symmetric: $\sigma_{ij}=\sigma_{ji}$. For such symmetric tensors, we can always rotate the coordinate system such that the tensor becomes diagonal, i.e. $\sigma^2_{ij}=0$ for $i\ne j$. Now, suppose that the velocities are isotropic, so that $\sigma_{xx}=\sigma_{yy}=\sigma_{zz}\equiv\sigma^2$. Then we find that

$${\partial\Phi\over\partial x_j} = -{1\over n}{\partial n\sigma^2\over \partial x_j}\,.$$ (8.29)

which is exactly the same as the equation for hydrostatic equilibrium, if $n\sigma^2$ is identified with the pressure $p$. Because of this analogy, the tensor $n\,\sigma^2_{ij}$ is called the pressure tensor.

In conclusion: the behaviour of collisionless stellar systems is similar to that of self-gravitating gas spheres. The role of temperature is taken over by that of the stellar velocity dispersion. It is the high velocity dispersion of the stars in an elliptical (and also in the bulge of spirals), which balances the gravitational pull. Since the required velocities are so high, such systems are called `hot stellar systems'. Recall that in disk galaxies, it is the ordered motion of the stars - i.e. the rotation of the disk - which supports the system against gravity. Such systems are dynamically cold, i.e. the stellar velocity dispersion is small compared to the streaming motions. For disks, the `velocity dispersion' refers to the small velocities that stars have with respect to their `local standard of rest' (10s of km s$^{-1}$), versus the rotational velocity of the disk, 200km s$^{-1}$.

In a gas the pressure is isotropic and hence stars are spherical. In contrast in stellar systems, the pressure tensor $n\,\sigma^2_{ij}$ is in general not isotropic, and so the pressure gradient can be larger in one direction ($x$, say) than in another direction ($y$). So even if the potential is spherical, the stellar distribution may be more extended in $x$ than in $y$, and the system will be elliptical in shape.