Past exam of the kinetic theory course of the University of Oxford / 2023 / 1

Write $K_{ij}=K({\mathbf{x}}_i,{\mathbf{x}}_j)$ and $\mathbf{P}={\sum }_i{\mathbf{v}}_i$. The unit particle masses make this sum the total momentum. Exchange the two indices and use the symmetry $K_{ij}=K_{ji}$:Thus every interacting pair transfers equal and opposite momentum, even though the velocity alignment dissipates kinetic energy.

For the total kinetic energy $E_N=\frac{1}{2}{\sum }_i|{\mathbf{v}}_i{|}^2$, apply the same exchange of indices:The dissipated kinetic energy is that of relative motion. Since the total momentum is constant, $E_N-|\mathbf{P}{|}^2/(2N)=\frac{1}{2}{\sum }_i|{\mathbf{v}}_i-\mathbf{P}/N{|}^2$ also decreases. Equality holds when every pair with $K_{ij}>0$ has the same velocity; the hypotheses permit disconnected interacting groups, so they alone do not guarantee eventual global velocity alignment.

Let ${\mathbf{A}}_i={\sum }_{j\ne i}{K}_{ij}({\mathbf{v}}_j-{\mathbf{v}}_i)$. Conservation of probability in the full phase space gives the dissipative Liouville equationNormalize $\int \varrho {\prod }_i{d}^3{x}_i{d}^3{v}_i=1$. Decay at infinity makes all boundary fluxes vanish, so the equation preserves this normalization of a probability distribution.

Unlike the Hamiltonian Liouville equation, this transport has nonzero phase-space divergence. In three dimensions,The probability density increases along a contracting characteristic flow map; setting its material derivative to zero would lose conservation of probability. The $j=i$ force is identically zero, including its derivative, which is why the last sums exclude it.

Put $z_i=({\mathbf{x}}_i,{\mathbf{v}}_i)$ and $d{z}_i=d^3{x}_i{d}^3{v}_i$. Indistinguishability gives an exchangeable ensemble. Define unit-normalized marginal distributionsIntegrating the dissipative Liouville equation over $z_2,\dots ,z_N$ removes their boundary fluxes. The $N-1$ identical pair terms remain:This is the first equation of a BBGKY hierarchy, with no independence assumed yet.

The additional physical assumption is a mean-field closure for velocity alignment: neglect pair correlations, $p_2(z,z_{\star },t)\simeq p_1(z,t)p_1(z_{\star },t)$. Initial independence alone does not make this exact at later times, because the particles interact. In the usual number convention, $f=N{p}_1$ and the two-particle distribution function is $f_2=N(N-1)p_2$. Consequently the closed equation at finite $N$ isNeglecting the relative $1/N$ correction gives the intended mean-field alignment equation:Equivalently, the closure $f_2\simeq f{f}_{\star }$ incorporates this large-$N$ approximation. It cannot be an exact finite-$N$ identity with $\int f=N$, since $\int f_{2}d{z}_{\star }=(N-1)f$ whereas $\int f{f}_{\star }d{z}_{\star }=Nf$.

There is also a consistent finite-$N$ convention yielding the boxed equation directly under the independence closure: define $\tilde{f}=(N-1)p_1$ instead. Its total normalization of a probability distribution is $N-1$, rather than the physical particle number $N$. Thus the displayed source equation needs a stated normalization of a probability distribution convention, or the large-$N$ approximation, in addition to neglect of correlations. In the remaining parts we use the displayed mean-field alignment equation itself; its moment identities do not depend on the total normalization of a probability distribution.

Use $dz=d^{3}x{d}^{3}v$ and setFor unit particle mass, $\rho$ is both the number density and the mass density in the number convention of c. A fixed reference scale inside the logarithm changes $\mathcal{H}$ by a constant multiple of the conserved total mass, leaving its derivative unchanged.

Both prefactors in the source's part (d) need correction. Multiplication of the mean-field alignment equation by $|\mathbf{v}{|}^2/2$ and integration by parts givesExchange $z$ and $z_{\star }$, use the symmetry of $K$, and average the two expressions:The printed right-hand side omits $1/2$. This corrected velocity-alignment energy dissipation agrees with the pair-counting calculation in a.

For velocity-alignment entropy production, let $A(\mathbf{x},t)=\int K(\mathbf{x},{\mathbf{x}}_{\star })\rho ({\mathbf{x}}_{\star },t)d^3{x}_{\star }$. The velocity-space divergence is ${\nabla }_{\mathbf{v}}\cdot \mathbf{L}[f]=-3A$. Assuming $f>0$ and sufficient decay, integration by parts yieldsThe printed formula omits the factor $3$ from the three velocity components. In $d$ spatial dimensions, that factor would be $d$. Zeros of $f$ can be treated by a regularized entropy calculation when the required integrability holds.

A concrete smooth counterexample to both printed coefficients is $K\equiv 1$ and $f(z,0)=g(\mathbf{x})(2\pi {)}^{-3/2}{e}^{-|\mathbf{v}{|}^2/2}$, where $g$ is a positive normalized Gaussian distribution. The total mass is one, the total momentum is zero, and $\mathbf{L}[f]=-\mathbf{v}$ initially. The kinetic energy derivative is $-\int |\mathbf{v}{|}^{2}fdz=-3$, whereas the printed formula gives $-6$. The entropy functional derivative is $3$, whereas the printed formula gives $1$.

The signs describe cooling of relative velocities and contraction in phase space: $\mathcal{H}$ increases, so the fine-grained Gibbs entropy $-\mathcal{H}$ decreases. These dissipative particle dynamics are not an isolated Hamiltonian system; there is no contradiction with Liouville's theorem or the Boltzmann equation's usual entropy increase. The microscopic mechanism removing kinetic energy is outside this alignment model.

Define the local mean velocity and momentum flux byHere $\rho =\int f{d}^{3}v$ and $\mathbf{u}$ is defined where $\rho >0$. The pressure tensor separates random motion from the bulk velocity:Integrating the mean-field alignment equation over velocity space removes the force divergence and gives the continuity equationIts first velocity moment follows by multiplying by $v_i$ and using $\int v_i{\partial }_{v_j}(f{L}_j)d^{3}v=-\int f{L}_i{d}^{3}v$:Inserting the mean-field alignment equation's force and integrating each particle velocity separately givesThe right-hand side exchanges momentum between positions and aligns their bulk velocities. Its spatial integral vanishes by exchange symmetry, recovering total momentum conservation. No closure for the pressure tensor is needed for these exact velocity moment identities.