Jeans's theorem

Consider the collisionless Boltzmann equation for the distribution function ${\displaystyle f(\mathbf {x} ,\mathbf {v} ,t)}$

${\displaystyle {\frac {\partial f}{\partial t}}+\mathbf {v} \cdot \nabla f+{\frac {1}{m}}\mathbf {F} \cdot \nabla _{v}f=0.}$

Consider the Lagrangian approach to the particle's motion in which case, the required equations are

${\displaystyle {\frac {d\mathbf {x} }{dt}}=\mathbf {v} }$

${\displaystyle {\frac {d\mathbf {v} }{dt}}={\frac {\mathbf {F} }{m}}.}$

Let the solutions of these equations be

${\displaystyle \mathbf {x} =\mathbf {x} (\alpha _{1},\dots ,\alpha _{6},t)}$

${\displaystyle \mathbf {v} =\mathbf {v} (\alpha _{1},\dots ,\alpha _{6},t)}$

where ${\displaystyle \alpha _{i}}$s are the integration constants. Let us assume that from the above set, we are able to solve ${\displaystyle \alpha _{i}}$, that is to say, we are able to find

${\displaystyle \alpha _{i}=\alpha _{i}(\mathbf {x} ,\mathbf {v} ,t).}$

Now consider an arbitrary function of ${\displaystyle \alpha _{i}}$'s,

${\displaystyle f=f(\alpha _{1},\dots ,\alpha _{6}).}$

Then this function is the solution of the collisionless Boltzmann equation, as can be verified by substituting this function into the collisionless Boltzmann equation to find[2][3]

${\displaystyle \sum _{i=1}^{6}{\frac {\partial f}{\partial \alpha _{i}}}\left[{\frac {\partial \alpha _{i}}{\partial t}}+\mathbf {v} \cdot \nabla \alpha _{i}+{\frac {1}{m}}\mathbf {F} \cdot \nabla _{v}\alpha _{i}\right]=\sum _{i=1}^{6}{\frac {\partial f}{\partial \alpha _{i}}}{\frac {d\alpha _{i}}{dt}}=0.}$

This proves the theorem.

A trivial set of integration constants are the initial location ${\displaystyle \mathbf {x} _{0}}$ and the initial velocities ${\displaystyle \mathbf {v} _{0}}$ of the particle. In this case, any function

${\displaystyle f=f(\mathbf {x} _{0}(\mathbf {x} ,\mathbf {v} ,t),\mathbf {v} _{0}(\mathbf {x} ,\mathbf {v} ,t))}$

is a solution of the collisionless Boltzmann equation.

• Jeans equations