Quals grab bag: Jeans' Equations

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Last updated 2018-09-17 19:39:45 SGT

This is a collection of nice results I derived while studying for quals that I wanted to save somewhere.

“Jeans' Equation” is what (some) astronomers call taking the first moment of the velocity with respect to the collisionless Boltzmann equation, expressed in coordinates $(x^i, p_i)$ on the cotangent bundle as

${\partial f \over \partial t} + \dot{x}^i {\partial f \over \partial x^i} + {\partial \over \partial p_i}\left(F_i f\right)=0.$

Here, rather than being the component of a vector, $\dot{x}^i$ is understood to be ${\partial \mathcal{H} \over \partial p_i}$ , a function on the cotangent bundle.

Schematically, conserved currents in 4 dimensions have vanishing four-divergence. However, this is less easy to motivate without manifest relativistic covariance (as is the case with the CBE, which is defined without reference to Riemannian structure). In what follows we will work in flat coordinates and essentially pray that the results hold good when expressed in a covariant manner.¹ In principle it might be possible to construct a map to the tangent bundle that yields the desired result (it might be the usual Levi-Civita connection with respect to the Sasaki metric, but I am not sure about this).

Assuming such a map exists, we take advantage of this hand-waving to replace the final term with the equivalent expression on the tangent bundle in coordinates $(x^i, v^i)$ :

${\partial f \over \partial t} + {\partial \over \partial x^i} \left(v^i f\right) + {\partial \over \partial v^i}\left(F^i f\right)=0,$

pulling back $f$ as well. Note that now by construction $\partial_i v^j = 0$ , which was not necessarily the case on the cotangent bundle except for specific choices of Hamiltonian. We then proceed with the computations as follows:

Zeroth moment

Defining $\rho = \int f~\mathrm{d}^3v$ and correspondingly $\left<g(v)\right> = \int f g(v)~\mathrm{d}^3v$ , we observe that the velocity-volume integral of the last term has the structure of a volume integral of a 3-divergence. Under suitable regularity conditions (i.e. $f\to 0$ faster than any of the other multiplicands can grow), this vanishes, leaving us with

${\partial \rho \over \partial t} + {\partial \over \partial x^i}\left(\left<v^i\right> \rho \right) = 0.$

This is recognisably the classical fluid continuity equation. It is customary to define $j^i = \rho \left<v^i\right>$ . However, we will avoid doing so for reasons that we will soon explore.

First moment

Performing a similar integration after multiplying all terms against $v^j$ yields

${\partial \over \partial t}\left(\rho \left<v^i\right>\right) + {\partial \over \partial x^j}\left(\left<v^i v^j \right> \rho \right) = \rho\left< F^j {\partial \over \partial v^j} v^i\right>.$

Assuming that the force has no dependence on the velocity (as is the case for conservative forces), we obtain

$\rho F^j\left< {\partial \over \partial v^j} v^i\right> = \rho F^j\left< \delta^i_j\right> = \rho F^i,$

whereby we have

${\partial \over \partial t}\left(\rho \left<v^i\right>\right) + {\partial \over \partial x^i}\left(\left<v^i v^j \right> \rho \right) = \rho F^i.$

This is recognisably the classical fluid momentum equation. Again, it is customary to define a velocity dispersion tensor (in particular to massage it into a form where there is an overall factor of $\rho$ , which can then be cancelled out) but we will resist the temptation to do this.

It is also customary to construct a tensor $T^{ij} = \rho \left<v^i v^j\right>$ (manifestly symmetric), in terms of which the above can be written in 3 dimensions (+ time) as

${\partial \over \partial t}j^i + {\partial \over \partial x^j}T^{ij} = \rho F^i.$

Higher moments

Repeating the integration after multiplying by $\prod_n^N v^{i_n}$ , we obtain

${\partial \over \partial t}\left(\rho \left<\prod_n^N v^{i_n}\right>\right) + {\partial \over \partial x^j}\left(\rho\left<v^j \prod_n^N v^{i_n}\right> \right) = \rho F^j\left<{\partial \over \partial v^j} \prod_n^N v^{i_n}\right>.$

We expand the term in brackets on the RHS as

${\partial \over \partial v^j} \prod_n^N v^{i_n} = \sum_m^N\left(\delta_j^{i_m}\prod_{n\ne m}^N v^{i_n}\right),$

thereby yielding

${\partial \over \partial t}\left(\rho \left<\prod_n^N v^{i_n}\right>\right) + {\partial \over \partial x^j}\left(\rho\left<v^j \prod_n^N v^{i_n}\right> \right) = \rho \left<\sum_m^N\left(F^{i_m}\prod_{n\ne m}^N v^{i_n}\right)\right>.$

At first blush, this seems fairly unwieldy. However, this actually has a quite nice coordinate-free expression.

In particular, let us suppose that the equivalent coordinate-free expressions for the moments are valid tensorial quantities. That is to say,

$\begin{aligned}\mathbf{j} &= \rho \left<v^j\right>\partial_j,\\\mathbf{T}&=\rho\left<v^i v^j\right> \partial_i \otimes \partial_j,\end{aligned}$

and so on. By abuse of notation, we write this as moments of tensor products of the velocity as

$\begin{aligned}\mathbf{j} &= \rho \left<\mathbf{v}\right>,\\\mathbf{T}&=\rho\left<\mathbf{v} \otimes \mathbf{v}\right>,\end{aligned}$

and so on. Then we have, in coordinate-free quantities²,

$\begin{aligned}{\partial \rho \over \partial t} + \nabla \cdot \left(\rho\left<\mathbf{v}\right>\right) &= 0, \\ {\partial \over \partial t}\left(\rho\left<\mathbf{v}\right>\right) + \nabla \cdot \left(\rho\left<\mathbf{v} \otimes \mathbf{v}\right>\right) &= \rho \mathbf{F},\\{\partial \over \partial t}\left(\rho\left<\mathbf{v} \otimes \mathbf{v}\right>\right) + \nabla \cdot \left(\rho\left<\mathbf{v} \otimes \mathbf{v} \otimes \mathbf{v}\right>\right) &= \rho \left<\mathbf{F}\otimes\mathbf{v} + \mathbf{v}\otimes\mathbf{F}\right>,\end{aligned}$

and so on. Generalising in terms of the symmetric tensor product³, we obtain

$\boxed{{\partial \over \partial t}\left(\rho \left<\mathbf{v}^{\odot N}\right>\right) + \nabla \cdot \left(\rho \left<\mathbf{v}^{\odot (N+1)}\right>\right) = N \rho \left<\mathbf{v}^{\odot (N-1)}\right> \odot \mathbf{F}.}$

In the absence of forces, the RHS vanishes, and we are left with conservation laws. In particular, the $N+1$ ^th velocity moment (multiplied against $\rho$ ) is the tensorial current density in the conservation law for the $N$ ^th velocity moment considered as a charge density.

Finally, if we so wished, we could perform the trick with the dispersion tensor that is customarily done in astronomy. Explicitly, we eventually obtain

${\partial \over \partial t} \left<\mathbf{v}^{\odot N}\right> + \left<\mathbf{v}\right>\cdot\nabla\left<\mathbf{v}^{\odot N}\right> = N\left<\mathbf{v}^{\odot (N-1)}\right> \odot \mathbf{F} - {1 \over \rho}\mathrm{Tr}\left\{\nabla\left(\rho\left<\mathbf{v}^{\odot (N+1)}\right>-\rho\left<\mathbf{v}\right>\otimes\left<\mathbf{v}^{\odot N}\right>\right)\right\},$

with the trace taken with respect to the leftmost slot⁴. This has a meaningful interpretation only for $N=1$ (yielding the divergence of the momentum dispersion tensor on the RHS, which becomes the pressure gradient for an isotropic gas).

It is also important to note here that the momentum that appears in Liouville's equation is the canonical momentum $p_i = {\partial \mathcal{L} \over \partial \dot{x}^i}$ , rather than the usual covariant momentum $P_i=m g_{ij}\dot{x}^j$ , precluding things like velocity-dependent potentials of the form $A_i\dot{x}^i$ that we see in electromagnetism. The two are related by the relevant musical isomorphism only if the kinetic term in the Lagrangian is of the form $T={1 \over 2}g(\dot{\gamma}, \dot{\gamma})$ . In what follows I will also neglect this distinction. ↩
We abuse notation with the divergence. So far, this is OK because all the quantities involved are manifestly symmetric. ↩
Note that $\mathbf{v}^{\otimes N} = \mathbf{v}^{\odot N}$ . ↩
Now the final term in the trace is not symmetric (except for $N=1$ ), making it necessary to specify this. ↩