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Replace the nonlinear Boltzmann collision operator by relaxation to the local Maxwellian $\mathcal{M}[f]$ with the same mass density, mean velocity and random kinetic energy as $f$. Matching those velocity moments ensures vanishing collision contributions to mass, momentum and kinetic energy balance. The collision relaxation time $\tau$ is often treated as a constant, although microscopic collision rates generally depend on the state and particle velocity.

The timescale on which a kinetic relaxation model damps departures from a matching local Maxwellian. In the spatially homogeneous BGK model with constant $\tau$, conserved velocity moments fix $\mathcal{M}$, and $f(t)=\mathcal{M}+[f(0)-\mathcal{M}]e^{-t/\tau }$. The parameter is of the order of an effective mean time between collisions rather than an exact universal microscopic collision time.

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.

Let $\kappa =k{V}_e/{\omega }_{\mathrm{p}e}$ and $s=\omega /(k{V}_e)$. The long-wavelength electron branch has $|s|\gg 1$. The convergent power seriesturns the degenerate-plasma dielectric response intoSubstitute ${\omega }^2={\omega }_{\mathrm{p}e}^2+a{k}^2{V}_e^2+O(k^4{V}_e^4/{\omega }_{\mathrm{p}e}^2)$. The coefficient of $k^2$ is $a-3/5$, so the degenerate-plasma Langmuir dispersion isThe Langmuir wave therefore has a finite frequency at zero wavenumber, with the first dispersive correction set by the Fermi velocity. Indeed the occupied Fermi sphere has $⟨v_{\parallel }^2⟩=V_e^2/5$, so the coefficient $3⟨v_{\parallel }^2⟩=3V_e^2/5$ agrees with the high-phase velocity velocity moment expansion.

Screening-length convention. The source defines ${\lambda }_{\mathrm{D}e}^{\mathrm{exam}}=\sqrt{3}{V}_e/{\omega }_{\mathrm{p}e}$, so its long-wavelength condition is equivalent to $\kappa \ll 1$. We retain that definition when expressing answers in its symbols. The actual electron Thomas-Fermi screening length obtained from the static response is insteadFor a concrete comparison, at $k={\omega }_{\mathrm{p}e}/V_e$, the static dielectric function is $4$; using the printed length as the physical Thomas-Fermi screening length would give $1+1/3=4/3$. Calling the printed length the Debye length thus differs from the physical degenerate-electron screening convention by a factor of three. It does not affect the boxed dimensional dispersion relation.

Put $z_i=({\mathbf{x}}_i,{\mathbf{v}}_i)$ and $d{z}_i=d^3{x}_i{d}^3{v}_i$. Normalize the symmetric $N$-particle phase-space distribution function $D_N$ to one. Use factorial normalization for its one-particle marginal and two-particle distribution function:Thus $\int fdz=N$ and $\int f_2(z_1,z_2)d{z}_2=(N-1)f(z_1)$. For unit particle mass, $f$ also carries mass density. These conventions explain why no extra factor $N-1$ occurs in the result.

Separate the Hamiltonian as $H=H_1+H_{\mathrm{rest}}+H_{\mathrm{int}}$, whereand $H_{\mathrm{rest}}$ contains all terms involving only particles $2,\dots ,N$. The stated Poisson bracket gives Hamilton's equationsConservation of probability along the Hamiltonian flow gives the Hamiltonian Liouville equationAssume decay of the required velocity moments and either decay or periodic boundary conditions in position. Integrating over $z_2,\dots ,z_N$ eliminates the $H_{\mathrm{rest}}$ terms by integration by parts. The terms in $H_{\mathrm{int}}$ differentiating an integrated velocity also vanish. Indistinguishability makes the remaining $N-1$ terms equal, soRelabeling the two particles gives the first BBGKY hierarchy equation:This is exact for a differentiable pair potential; no independent-particle closure has been made.

Take the zeroth velocity moment of a. The right side is a velocity space divergence with vanishing boundary contribution, giving the continuity equationFor the first velocity moment, use integration by parts componentwise:Write $\mathbf{c}=\mathbf{v}-\mathbf{u}$; then $\int \mathbf{c}f{d}^{3}v=0$, and the momentum flux splits into bulk transport and the pressure tensor:Thereforewhere repeated Cartesian indices are summed and $n_2=\int f_2{d}^{3}v{d}^3{v}_{\star }$. The negative sign is the attractive or repulsive force $-\nabla \varphi$, rather than an additional sign convention for the pressure tensor.

Spatial homogeneity makes $\rho$ independent of position and makes the two-particle number density depend only on the separation. Isotropy removes its directional dependence. ConsequentlyHere $g_2$ is a radial distribution function, with the finite-$N$ normalization inherited from a; statistical independence is unnecessary. In particular, the translated pair in c has separation $\mathbf{R}$ for every $s$, so its $s$ integral is simply ${\rho }^2{g}_2(R,t)$.

Isotropy of the random velocity $\mathbf{c}=\mathbf{v}-\mathbf{u}$ makes its second velocity moment proportional to the identity. Taking its trace fixes the proportionality constant:Thus $\theta$ is the one-component random velocity variance; for unit mass in thermal equilibrium, $\theta =k_{B}T$. A Maxwell-Boltzmann velocity distribution is not required for the pressure tensor to be isotropic.

For the interaction contribution, the angular integral is $\int d\Omega {\hat{R}}_i{\hat{R}}_j=(4\pi /3){\delta }_{ij}$, by isotropy and its trace. With $d^{3}R=R^{2}dRd\Omega$, c becomesCombining the two pressure tensors gives the virial pressure from pair correlations:For a repulsive pair potential, the outward radial force is $-{\varphi }^{\prime }(R)\ge 0$. Since $g_2\ge 0$, the second term in $p$ is nonnegative, and positive when occupied separations feel nonzero repulsion. It expresses additional momentum transfer through interactions. The displayed integral assumes a differentiable pair potential and convergence; a hard core requires a contact formula or a limiting argument.

To obtain the Boltzmann equation, take a dilute gas with a short-range pair potential. Its interaction range is small compared with the mean free path and the spatial scale on which the phase-space distribution function changes. The binary collision approximation treats encounters as isolated elastic collisions; simultaneous encounters involving three or more particles can be neglected. The encounter duration is short compared with the collision relaxation time and macroscopic evolution time, and the external force changes negligibly during an encounter. This permits local, effectively instantaneous scattering at one position $\mathbf{x}$.

The essential statistical closure is molecular chaos for the Boltzmann equation: the distribution of incoming pairs factorizes as $f{f}_{\star }$. Outgoing pairs need not be independent. Follow each incoming pair through its two-body scattering, and use conservation of momentum and kinetic energy, together with time reversal and the invariant scattering measure, to express both gain and loss in the same variables. This givesHere $d\Omega =\sin \theta d\theta d\phi$; when the measure is written as $d\theta d\phi$, its Jacobian is included in the Boltzmann scattering kernel. In a scattering-angle convention, that kernel contains relative speed times differential scattering cross-section. The hemisphere prevents double counting equivalent outcomes for indistinguishable particles; the precise factor is included in $B$. The primed velocities are connected to the unprimed ones by the reversible elastic collision map. This closure is appropriate for a dilute short-range gas, rather than a generic long-range interacting system with an important collective mean-field theory contribution.

The further replacement by the BGK model is a phenomenological relaxation closure. Replace the full nonlinear Boltzmann collision operator by relaxation towards a local Maxwellian, with a single, velocity-independent collision relaxation time $\tau >0$:It models relaxation towards a local Maxwellian without resolving the detailed scattering. Treating $\tau$ as constant is an additional simplification; it is of the order of a mean time between effective collisions. Near a local Maxwellian, departures in the nonconserved velocity moments decay on this timescale.

Choose $f^{(0)}$ to match the collision invariants $1$, $\mathbf{v}$ and $|\mathbf{v}{|}^2$ of $f$. For unit particle mass, this meansHere $\rho$ is the mass density, $\mathbf{u}$ is the mean velocity, and $\Theta$ is the temperature in squared-velocity units, $\Theta =k_{B}T/m$ before setting $m=1$. In particular,Thus relaxation preserves local mass, momentum and kinetic energy rather than relaxing them to externally prescribed values.

Use the BGK model from b. Multiply it by $v_i$ and integrate over velocity space. Matching the collision invariants makes the collision contribution vanish. For the external force term, integration by parts givesDefine the momentum flux and force density byThe resulting momentum conservation equation isThe pressure tensor is the second central velocity moment,The zeroth velocity moment also supplies the continuity equation ${\partial }_t\rho +{\partial }_j(\rho u_j)=0$, which will be needed in d.

Write $c_i=v_i-u_i$ and introduce the raw third velocity moment $M_{ijk}$ and the third central velocity moment $Q_{ijk}$:The expansion of their product givesThe heat flux is $q_i=Q_{ijj}/2$ for unit mass.

Multiply the BGK model by $v_i{v}_j$. The external force term isTherefore the raw second velocity moment satisfiesThe matching local Maxwellian has ${\Pi }_{ij}^{(0)}=\rho u_i{u}_j+\rho \Theta {\delta }_{ij}$, soTo remove the bulk motion, let $D_t={\partial }_t+u_k{\partial }_k$ be the material derivative. The continuity equation and the momentum conservation equation implyand henceSubtract this identity from the raw second velocity moment equation and insert the expansion of $M_{ijk}$. The terms $u_i{\partial }_k{P}_{jk}$ and $u_j{\partial }_k{P}_{ik}$ cancel by the product rule. The external force terms also cancel, since a common acceleration changes the local mean velocity without directly changing the relative velocities. What remains isThe trace of the relaxation term is zero because $P_{ii}=3\rho \Theta$. Thus the BGK model relaxes anisotropy of the pressure tensor while preserving the local random kinetic energy. The evolution is still a velocity-moment hierarchy, since $Q_{ijk}$ requires further information about $f$.

For unit particle mass, this symmetric third-order tensor transports the pressure tensor in a kinetic fluid description. Its contraction gives the heat flux $q_i=Q_{ijj}/2$. It vanishes for a local Maxwellian, whose odd central velocity moments vanish, but it need not vanish for a general anisotropic or skewed phase-space distribution function.

Taking velocity moments of a kinetic transport equation couples the equation for a moment of order $r$ to a spatial derivative of a moment of order $r+1$. A finite fluid description therefore requires closure. For example, the pressure tensor equation involves the third central velocity moment.