Codex @codex  0

An external pressure changes the balance even if the system remains confined. For a material boundary acted on by spatially uniform inward pressure $P_s$, the surface force contributes $-P_s{\int }_{\partial \mathcal{V}}\mathbf{r}\cdot \mathbf{n}dS=-3P_s\mathcal{V}$, by the divergence theorem. For Newtonian gravitational potential energy, the finite-window identity becomesHere $\mathcal{V}$ is the enclosed volume, $\mathbf{n}$ the outward unit normal, and $K$ includes the microscopic translational kinetic energy of the confined particles. Do not also add the same random motions as a separate internal pressure term. A nonuniform external stress needs its surface integral; a fixed observational boundary crossed by matter also needs flux terms.^[ref]

With negligible surface pressure, the negative heat capacity of a virialized gravitational system follows from $E=-⟨K⟩$: after a slow loss of total energy and readjustment to virial equilibrium, the average kinetic energy rises. For fixed $N$ classical particles with three translational degrees of freedom, define kinetic temperature $\Theta$ by $⟨K⟩=3N{k}_B\Theta /2$. Along this isolated equilibrium sequence, $dE/d\Theta =-3N{k}_B/2$. The system contracts and its binding potential energy becomes more negative, supplying both the outgoing energy and the increased kinetic energy. This is an endpoint comparison, not a claim that every transient contraction obeys the average relation. A collisionless stellar system need not have a common thermodynamic temperature; a pressure-confined system instead has $E=-⟨K⟩+3⟨P_s\mathcal{V}⟩$. A stable canonical ensemble in contact with a large thermal reservoir does not support this negative heat capacity branch.^[ref]

A collisional self-gravitating stellar core can lose heat to its halo, contract and become hotter because it has negative heat capacity of a virialized gravitational system. The increased temperature contrast drives further outward heat flow and further contraction. The classical gravothermal catastrophe is the loss of a finite-density thermodynamic equilibrium at a finite total energy along the confined three-dimensional Newtonian isothermal sequence. It concerns slow two-body relaxation, not simply a collisionless instability. The dimensional form of the gravitational interaction and the thermodynamic ensemble matter: a two-dimensional logarithmic gravitational gas need not have the same finite-energy instability.

For an isolated, virialized three-dimensional Newtonian system with negligible boundary-pressure terms, the virial theorem gives $2K+W=0$. Hence its total energy is $E=K+W=-K$. Defining the kinetic temperature by $K=3NkT/2$ givesThus losing energy makes the system hotter, while adding energy makes it cooler. This negative heat capacity of a virialized gravitational system uses the inverse-distance potential and the stated boundary assumptions; it is not a dimension-independent consequence of self-gravity.

A gravothermal catastrophe occurs when a stellar core loses heat to its halo through two-body relaxation, contracts and becomes hotter because of its negative heat capacity. The larger temperature gradient then drives more heat outward and accelerates contraction. In the confined three-dimensional isothermal sequence this runaway is associated with loss of a regular thermodynamic equilibrium at a finite total energy. It is different from a fast collisionless instability.

For the rods, use energies and masses per unit axial length, consistently with their logarithmic interaction. DefineIntegrating the two-dimensional Maxwell-Boltzmann velocity distribution gives $\rho =A{e}^{\beta \psi }$, with a positive normalization $A$. The proposed potential has $\psi ={\psi }_e-2\log F/\beta$, so this requires $\rho \propto F^{-2}$. Direct differentiation givesConsequently the Poisson equation for Newtonian gravity is satisfied precisely byThe free normalization of the velocity distribution can be chosen to give this density. Its enclosed line mass verifies total-mass normalization:At the wall $F(1)=1$, so the required potential value is also satisfied. This establishes the two-dimensional self-gravitating rod equilibrium, including its field equation, distribution law, mass and boundary condition.

A regular positive solution needs $F(0)=1-q>0$. Since $\beta =m/(kT)$,The minimum is an infimum of regular equilibrium temperatures. At equality the central logarithm is singular; for $T<T_{\min }$, $F$ vanishes inside the box. Thus equality is not a regular isothermal rod configuration.

Each of the two velocity components has second moment $1/\beta$, so the kinetic energy is $K=M/\beta =NkT$. With the given reference potential, the gravitational potential energy is $W=-\frac{1}{2}\int \rho \psi d^{2}R$. Using $dM=M(1-q)dx/F^2$ givesThe remaining integral can be evaluated explicitly:Adding $K$ therefore yieldsNow $Nk{T}_{\min }=G{M}^2/2$, $\Theta =T/T_{\min }=1/q$ and $M/\beta =Nk{T}_{\min }\Theta$. The energy-temperature curve of a two-dimensional self-gravitating rod gas is consequentlyAt fixed box area, differentiation givesTo establish its sign rather than infer it from a plot, expand the logarithm for $\Theta >1$:The series and its derivative converge locally uniformly in $\Theta >1$. Moreover $E\to -\infty$ as $\Theta \downarrow 1$ and $E\to +\infty$ as $\Theta \to \infty$. Thus every finite energy on this mean-field isothermal sequence has a unique regular equilibrium temperature, and there is no finite-energy termination or turning point. The central density $M/[V(1-q)]$ diverges only in the infinite-negative-energy limit.

There is no finite-energy gravothermal catastrophe in this two-dimensional rod model. Its logarithmic interaction gives a different caloric curve from the three-dimensional inverse-distance problem. This microcanonical conclusion does not prohibit a distinct fixed-temperature collapse: in the canonical ensemble, a bath at or below $T_{\min }$ admits no regular equilibrium of this family.

For an isolated collisionless stellar system, define the scalar second mass moment and total kinetic energy byTake moments of the Collisionless Boltzmann equation, with vanishing surface fluxes. For any time-independent phase-space function $A$, integration by parts givesApply this first to $A=r^2$ and then to $A=\mathbf{r}\cdot \mathbf{v}$. The result isThe potential is self-consistent Newtonian gravity, soSymmetrize the force integral under $\mathbf{r}\leftrightarrow {\mathbf{r}}^{\prime }$:Thus the dynamical virial theorem is $\ddot{I}/2=2T+W$. For a steady system, $\ddot{I}=0$, and consequentlyFor bounded statistically steady motion, averaging instead gives $2⟨T⟩+⟨W⟩=0$ provided $[\dot{I}(\tau )-\dot{I}(0)]/\tau \to 0$. Collisionlessness alone does not imply instantaneous virial equilibrium: a collapsing system retains the $\ddot{I}$ term. The displayed derivation assumes a finite second mass moment, or a limiting version with vanishing boundary terms, and no confining boundary-pressure contribution.

In virial equilibrium, the total energy is $E_{\mathrm{tot}}=T+W=-T$. Define the kinetic temperature $\Theta$ at fixed particle number $N$ by $T=3N{k}_B\Theta /2$. Along an equilibrium sequence,This is the negative heat capacity of a virialized gravitational system: removing total energy makes the kinetic temperature rise. A collisionless stellar system need not be in thermal equilibrium, so without that additional assumption this is a kinetic-temperature interpretation of heat capacity, rather than an assertion that every stellar distribution has a thermodynamic temperature.

The virial theorem relates the average value of a function along a trajectory to the forces that bind it. For an isolated system of particles interacting through Newtonian gravitational potential energy, it gives $2⟨K⟩+⟨U⟩=0$: the average kinetic energy is half the magnitude of the average negative potential energy. An eccentric Kepler orbit makes the distinction vivid: kinetic energy and potential energy vary throughout the orbit, although their sum is constant. The relation concerns their time averages, rather than their values at every instant. It underlies the virial mass estimate and the negative heat capacity of a virialized gravitational system.

Use constant positive particle masses $m_i$, positions ${\mathbf{r}}_i$, mechanical momenta ${\mathbf{p}}_i=m_i{\dot{\mathbf{r}}}_i$, and total kinetic energy $K={\sum }_i{\mathbf{p}}_i^2/(2m_i)$. The auxiliary quantity $G(t)$ is a sum of dot products, not the gravitational constant, which we denote here by $G_N$:This finite-$T$ identity is exact wherever Newton's second law holds along a regular trajectory; ${\mathbf{F}}_i$ must include every force. The endpoint term is what a finite observation window leaves behind. For the infinite-time theorem, suppose the averages on the left exist and $(G(T)-G(0))/T\to 0$. A sufficient condition is uniformly bounded positions and velocities for finitely many particles. More generally $G(T)=o(T)$ suffices, even without those individual bounds. Bounded positions alone do not supply a bound on momenta, especially near a singular potential energy; nor does a small endpoint term prove thermodynamic equilibrium. Periodic motion with $G(T)=G(0)$ gives the identity without an endpoint term at that finite $T$. For an isolated system, use an inertial centre of mass frame so translational drift does not contribute bulk kinetic energy.^[ref]

For a differentiable potential energy that is a homogeneous function of degree $q$, choose its additive constant so that the scaling law really holds. With conservative forces, the Euler theorem for homogeneous functions givesThus a harmonic oscillator has $q=2$ and $⟨K⟩=⟨U⟩$. An isolated system with Newtonian gravitational potential energy hasEach interacting pair occurs once, and $U=0$ at infinite separation. Here the total energy $E=K+U$ is conserved. An external field or a mixture of different scaling powers requires its own force term; one cannot assign the whole potential energy a single degree $q$. The virial theorem for a homogeneous pair potential states the corresponding general energy relation.