Quals grab bag: Jeans' Equations
CommentsLast updated 20180917 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 fourdivergence. 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 LeviCivita connection with respect to the Sasaki metric, but I am not sure about this).
Assuming such a map exists, we take advantage of this handwaving 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
This is recognisably the classical fluid continuity equation. It is customary to define [j^i = \rho \left
First moment
Performing a similar integration after multiplying all terms against [v^j] yields
[{\partial \over \partial t}\left(\rho \leftAssuming 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 \leftThis 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
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\leftWe 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\leftAt first blush, this seems fairly unwieldy. However, this actually has a quite nice coordinatefree expression.
In particular, let us suppose that the equivalent coordinatefree expressions for the moments are valid tensorial quantities. That is to say,
[\begin{aligned}\mathbf{j} &= \rho \leftand 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 coordinatefree quantities^{2},
[\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 product^{3}, 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 (N1)}\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 (N1)}\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 slot^{4}. 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 velocitydependent 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. ↩