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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.

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.

The long-wavelength limit in d is resonant heating. Most electrons screen the slowly varying spatial field. The noise nevertheless excites collective Langmuir waves, whose sharp temporal spectrum lies near $\omega =\pm {\omega }_{pe}$. Electrons with $v\simeq \omega /k$ remain in phase with a wave and exchange energy with it. A stable tail absorbs net wave energy through Landau damping. The heating is initially deposited in narrow tail regions of velocity space, even though the volume-integrated rate equals that in c.

The short-wavelength limit in c is stochastic heating in the terminology intended here. With $ϵ\simeq 1$, the imposed electric field is essentially unscreened. Its temporally uncorrelated kicks cause a random walk of electron velocity, so $⟨(\Delta v{)}^2⟩=2D_0\Delta t$. Electrons throughout the thermal population contribute to the resulting kinetic energy increase; no narrow collective wave peak selects a small tail population.

Both mechanisms are driven by white noise, and the formal condition $\omega =kv$ enters the stationary quasilinear diffusion calculation in both cases. The physical distinction is the sharp collective resonance in d versus the broad direct random forcing in c, rather than the mere presence of randomness. In collisionless theory this heating is an increase of the averaged kinetic energy; subsequent collisions can establish thermal equilibrium.

Normalize the symmetric $N$-particle phase-space distribution function $D_N$ to one and write $z_i=({\mathbf{x}}_i,{\mathbf{v}}_i)$, $d{z}_i=d^3{x}_i{d}^3{v}_i$. The number-normalized one- and two-particle distribution functions areThus $\int fdz=N$ and $\int f_2(z,z_{\star })d{z}_{\star }=(N-1)f(z)$. These conventions absorb the otherwise explicit factor $N-1$ in the first BBGKY hierarchy equation.

Separate the Hamiltonian into kinetic energy, external potential energy, and pair potential terms. Their Hamilton's equations giveThe sign follows directly from the stated Poisson bracket: ${\dot{v}}_{i\alpha }=\{v_{i\alpha },H\}=-{\partial }_{x_{i\alpha }}H$. The Hamiltonian Liouville equation is consequentlyIntegrate over $z_2,\dots ,z_N$ and multiply by $N$. Assume sufficient decay in velocity space and either decay or periodic boundary conditions in position, so that integration by parts removes all transport terms with $i\ge 2$. The surviving interaction term iswhere indistinguishability makes all $N-1$ terms equal. Relabeling $(z_1,z_2)$ as $(z,z_{\star })$ yields the exact resultNo assumption of molecular chaos for the Boltzmann equation has entered: the dependence on the two-particle distribution function is still exact.

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.

Use a mass-normalized galactic distribution function $F$, so that $\rho =\int F{d}^{3}v$ and $\rho ⟨v_i{v}_j⟩=\int v_i{v}_{j}F{d}^{3}v$. Multiplying the stationary Collisionless Boltzmann equation by $v_j$ and integrating over velocity space givesAssume the galactic distribution function decays sufficiently rapidly for the velocity boundary term to vanish. Integration by parts then gives $\int v_j{\partial }_{v_i}F{d}^{3}v=-{\delta }_{ij}\rho$, and therefore the Jeans equations areThe second moment includes both ordered motion and velocity dispersion; no assumption of zero mean velocity was needed.

To obtain the tensor virial theorem, multiply the $j$th Jeans equation by $x_i$ and integrate over position. For an isolated, finite system with a vanishing spatial surface term,Since the left side is $-2K_{ij}$, $2K_{ij}+W_{ij}=0$. This convention puts all stellar second moments into $K_{ij}$; it does not separate ordered and random kinetic energies.

For a self-gravitating system, the trace $W$ is the Newtonian gravitational potential energy. Indeed, symmetrizing the pair integral givesThus the trace of the tensor virial theorem is $2K+W=0$, and the total energy obeysThese identities require self-gravity without an additional external potential or an omitted confining boundary pressure.

Let $⟨v_I^2⟩$ now denote a mass-weighted average over the entire initial system. Then $K_I=M_I⟨v_I^2⟩/2$. The gravitational radius is defined by $R_I=-G{M}_I^2/W_I$, so the virial theorem immediately yieldsThe gravitational radius measures total binding energy, rather than a particular geometric edge or half-mass radius.

For the accreted systems define the mass-weighted internal mean-square speed by $M_A⟨v_A^2⟩={\sum }_s{M}_s⟨v_s^2⟩$. If each satellite initially satisfies the virial theorem, its internal energy contributes to $E_A=-M_A⟨v_A^2⟩/2$. In parabolic dry-merger energy accounting, the orbital energy at large separation is zero. Assume a dry galaxy merger, no loss of mass or energy through escaping stars, no external work, and a final relaxed system satisfying the virial theorem. Then conservation of energy gives $E_F=E_I+E_A$ and $M_F=M_I+M_A$. Energy transferred by Chandrasekhar dynamical friction remains part of the total energy under these assumptions. Consequently,Applying the final virial theorem and dividing by the initial relation givesFor a mass doubling, $\eta =1$. A merger of two identical equilibrated galaxies has $ϵ=1$, hence $R_F/R_I=2$. For accretion through many minor galaxy mergers whose satellites have much smaller internal mean-square speeds, $ϵ\ll 1$, hence $R_F/R_I\simeq 4$, tending to $4$ in the cold-satellite limit. Equivalently, nearly fixed total energy makes the gravitational radius scale as $M^2$ during cold minor-merger size growth. Small satellite mass alone does not logically imply $ϵ\ll 1$: the factor four also requires this weak-binding assumption. Likewise, equal masses require comparable internal binding to give the factor two. These are the physical limits behind the two stated merger comparisons.

Random changes of particle velocity described by a diffusion equation in velocity space. For constant $D$, $⟨(\Delta v{)}^2⟩=2D\Delta t$. With zero boundary flux it conserves number. For a number-normalized one-dimensional distribution, its kinetic energy production is $d\mathcal{K}/dt=-m\int vD(v)f^{\prime }(v)dv$, which reduces to $mnD$ for constant $D$. Variable $D$ must remain inside the derivative.