For an isolated final stellar population with isotropic one-dimensional velocity dispersion $\sigma$, kinetic energy is $K=3M{\sigma }^2/2$. The virial theorem gives $W=-2K$ and total energy $E=-K$. Collapse from a diffuse nearly unbound gas reservoir therefore requires removal of approximately $K$ of energy. The radiated fraction can differ if there are outflows, appreciable initial binding energy, or boundary stresses.

An isolated Newtonian system in virial equilibrium has total energy $E=K+W=-K$. Along an equilibrium sequence at fixed particle number, define the kinetic temperature by $K=3N{k}_B\Theta /2$. Its heat capacity is therefore negative, as displayed: losing total energy increases the kinetic temperature. A collisionless stellar system need not have a thermodynamic temperature; this is a kinetic-temperature statement unless thermal equilibrium is additionally specified. Boundary pressure and external confinement can change the energy relation.

For a differentiable total pair potential energy homogeneous of degree $q$ in all particle positions, the Euler theorem for homogeneous functions gives ${\sum }_i{\mathbf{r}}_i\cdot {\nabla }_{i}W=qW$. Since ${\mathbf{F}}_i=-{\nabla }_{i}W$, the force virial is $-qW$. The bounded-motion time-averaged virial theorem consequently gives the displayed equation. For $q\ne 0$, total energy is $E=(q+2)⟨K⟩/q$. A positive kinetic-temperature coefficient makes its heat capacity negative exactly when $-2<q<0$. For a pair potential proportional to $r^{p+1}$ this means $-3<p<-1$.

For an isolated equilibrium of equal-mass points, the virial theorem gives $⟨v^2⟩=-W/M$. A particle's instantaneous escape velocity satisfies $v_{\mathrm{esc}}^2=-2\Phi$ with zero potential at infinity. Since $W=\frac{1}{2}{\sum }_{i}m{\Phi }_i$, averaging gives $⟨v_{\mathrm{esc}}^2⟩=-4W/M$. These are three-dimensional mean-square speeds, not one-component velocity dispersions.

The proof is a single time derivative. Apply the product rule to each dot product and use Newton's second law:Integrating from $0$ to $T$ gives the endpoint identity; the stated sublinear condition removes the endpoint term only in the infinite-time limit. Equivalently, for the scalar inertia $I=\frac{1}{2}{\sum }_i{m}_i|{\mathbf{r}}_i{|}^2$ used in the pairwise moment-of-inertia identity, $\dot{I}=G$ and $\ddot{I}=2K+{\sum }_i{\mathbf{r}}_i\cdot {\mathbf{F}}_i$. Specifying this factor of one half avoids confusion with another common inertia convention.

In On a Mechanical Theorem applicable to Heat by R. Clausius (1870, pp. 123-124), the author defines stationary motion through bounded positions and velocities, includes planetary orbits and elastic vibrations, and names the average negative half of the force-position sum the “virial”. That historical convention differs from the auxiliary $G(t)$ defined above. His proof on pp. 126-127 keeps an endpoint term, then explains why it vanishes over a complete period or becomes small in a long bounded irregular motion. The same paper separates internal and external forces and includes uniform boundary pressure on pp. 125-126: confinement was already part of the formulation, rather than a reason to omit external stresses.^[ref]

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]

Motion can weigh a system whose mass is otherwise inaccessible. For a self-gravitating isolated system in virial equilibrium, let $K$ and $U$ denote equilibrium averages and use its gravitational radius $R_g=-G_N{M}^2/U$ and a mass-weighted mean squared speed relative to its centre of mass. Then $K=M⟨v^2⟩/2$ gives $M=R_g⟨v^2⟩/G_N$. If random velocities are isotropic and mean streaming has been removed, $⟨v^2⟩=3{\sigma }_{\mathrm{los}}^2$, where ${\sigma }_{\mathrm{los}}$ is the one-dimensional line-of-sight velocity dispersion. Observations sample many objects at one epoch; replacing a time average by that sample requires a representative, approximately stationary population, beyond the trajectory identity itself.^[ref]

A practical measurement route uses an optical telescope to locate members and a spectrograph to measure Doppler shifts of identified spectral lines. Their shifts relative to the system's mean give radial velocities; angular separations plus a distance estimate constrain its size. Subtracting instrumental broadening and measurement variance, testing membership, and looking for coherent rotation or subgroups help control the inference. The projected luminous size need not equal $R_g$, and luminous tracers need not have the same distribution as all gravitating mass. Unknown velocity-anisotropy parameters, interlopers, mergers, external tidal forces and appreciable boundary stresses can bias a naive estimate. A large velocity dispersion alone does not establish virial equilibrium.

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.

The velocity gradient tensor is $A_{ij}={\partial }_j{u}_i$. Its symmetric part is the rate-of-strain tensor $S$ and its antisymmetric part describes local rotation. For incompressible flow, $\mathrm{tr}A=\mathrm{tr}S=0$. The sign convention for a cubic invariant must be specified explicitly.

A secondary flow is a usually smaller velocity component superposed on a primary flow selected by symmetry or a simplified force balance. Examples include meridional circulation around a rotating sphere, or transverse circulation in a curved channel. Its direction depends on the forcing mechanism: elastic normal stresses and inertial effects can produce opposite circulations.

Use the quantum harmonic oscillator in units with reduced Planck constant $\hslash =1$, mass $m=1$, and angular frequency $\omega =1$. Its Hamiltonian operator and trial Gaussian wave packet areThis is the same Gaussian wave packet used for the Heisenberg uncertainty relation: $|{\psi }_{\sigma }{|}^2$ is a normal distribution with mean zero and variance ${\sigma }^2$. The Gaussian integral gives wavefunction normalization and $⟨x^2⟩={\sigma }^2$. These smooth, rapidly decaying trials lie in the one-dimensional harmonic oscillator form domain and even in the operator domain.

Since ${\psi }_{\sigma }^{\prime }=-x{\psi }_{\sigma }/(2{\sigma }^2)$, integration by parts with vanishing boundary terms gives the kinetic energy and potential energy:Narrow trials cost kinetic energy, while broad trials cost potential energy. The exact differenceshows that the unique optimal width is ${\sigma }^2=1/2$, with $E=1/2$. At $\sigma =1/2$ and $\sigma =1$, the trial energies are both $5/8=0.625$; the kinetic energy and potential energy contributions exchange places.

Minimizing within this trial family alone establishes $E_0\le 1/2$. To establish exactness, use the Heisenberg uncertainty relation to obtain $E_{\psi }\ge 1/2$ for every admissible quantum state, as in its existing oscillator example. Together the bounds give $E_0=1/2$. Direct substitution also yields $H{\psi }_{1/\sqrt{2}}={\psi }_{1/\sqrt{2}}/2$, agreeing with the Gaussian eigenstate for a quadratic potential. This is a useful distinction: the variational method supplies the upper bound, while the Heisenberg uncertainty relation supplies the complementary lower bound.

For constant long-wave speed and $S_0\propto e^{\beta x}$, a standing free-surface amplitude obeys ${\eta }^{\prime \prime }+\beta {\eta }^{\prime }+k^2\eta =0$. The transformation $\eta =e^{-\beta x/2}Y$ gives $Y^{\prime \prime }+(k^2-{\beta }^2/4)Y=0$. Impermeable ends at $0,L$ impose ${\eta }^{\prime }=0$, yielding the displayed nonzero-frequency eigenvalues and ${\eta }_n\propto e^{-\beta x/2}[\cos (n\pi x/L)+\beta L\sin (n\pi x/L)/(2n\pi )]$. A separate zero-frequency mode changes the uniform water level.

For a spherical drop of constant mass density $\rho$, radius $a>0$ and downward velocity $v>0$, suppose its volume grows at rate $c\pi a^{2}v$ by collecting stationary mist and its upward quadratic drag is $k\rho \pi a^2{v}^2$, with $c,k>0$. The drop mass is $4\pi \rho a^3/3$. Its incoming mass accretion rate and the variable-mass system momentum law give the displayed equations. The accreted matter must be accelerated from rest, producing the additional $c$ term alongside drag.

For a small bounded mass perturbation $m(v)=M_0+\mu (v)$ that becomes constant in the future, write the event horizon radius as $r_H(v)=2M_0+\rho (v)$. The outgoing radial null geodesics of the Vaidya metric give ${\rho }^{\prime }-\rho /(4M_0)=-\mu /(2M_0)$ to first order. Future stationarity selects the displayed solution and removes the exponentially growing homogeneous solution. If ${\mu }^{\prime }\ge 0$, integration by parts gives $\rho -2\mu =2{\int }_v^{\infty }{e}^{-(s-v)/(4M_0)}{\mu }^{\prime }(s)ds\ge 0$. Thus the event horizon can grow before local matter arrives, lying outside the instantaneous spherical apparent horizon $r=2m(v)$. This is a consequence of the event horizon's global definition, not propagation of a local signal into the past.

The difference of two classical Poisson equation solutions with the same Dirichlet boundary conditions is a harmonic function vanishing on the boundary. Under sufficient boundary smoothness, Green's first identity gives ${\int }_V|\nabla ({\varphi }_1-{\varphi }_2){|}^2=0$. More generally the maximum principle for harmonic functions proves the same uniqueness for functions continuous up to the boundary and twice continuously differentiable in the bounded interior.

This integral domain embeds in $\mathbb{C}$ but has the inequivalent factorizations $6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$. The multiplicative norm $N(a+b\sqrt{-5})=a^2+5b^2$ takes neither the value two nor three, proving that the four displayed factors are irreducible elements. Its only units are $\pm 1$, so those two factorizations cannot be related by units and reordering.

A spline is a piecewise polynomial function whose polynomial pieces obey specified smoothness conditions at their joins, called spline knots. For example, a cubic spline is cubic on each interval and commonly has two continuous derivatives across interior knots. Spline approximation and spline interpolation use these finite-dimensional function spaces to represent data or approximate functions.

Let $f$ be a real continuous function on a nondegenerate compact interval $[a,b]$, and let $p$ be a real polynomial of degree at most $m$. Then $p$ is a best uniform approximation from the polynomials of degree at most $m$ if and only if there are $m+2$ ordered pointswith $\sigma \in \{-1,1\}$. The best polynomial is unique. The real scalar field and ordered interval are part of this alternation statement.

For sufficiency with nonzero error, a strictly better polynomial $q$ would make $q-p$ have alternating strict signs at these points, forcing at least $m+1$ distinct zeros by the intermediate value theorem. A nonzero degree-at-most-$m$ polynomial cannot have so many zeros. Necessity follows from the Kolmogorov criterion for uniform approximation and the zero-counting property of the polynomial Chebyshev system. If the error is zero, optimality is immediate.

The uniform approximation closure of the integer-coefficient polynomials on $[0,1]$ consists precisely of the continuous functions taking integer values at both endpoints. Necessity follows because evaluations at zero and one are integers and convergent integer sequences are eventually constant. For sufficiency, the rounded Bernstein polynomial has integer monomial coefficients and differs uniformly by a vanishing amount from the Bernstein polynomial, which converges uniformly to the target.

For a compact convex set $K\subset \mathbb{C}$ and $\zeta \notin K$, strict separation gives a unit vector $u$ with $\mathrm{Re}(\bar{u}\zeta )>{\max }_K\mathrm{Re}(\bar{u}z)$. Taking $c=-Ru$ for large $R$ makes $q={\max }_K|z-c|/|\zeta -c|<1$. The polynomialhas uniform error at most $q^{N+1}/(|\zeta -c|(1-q))$. This elementary geometric series argument proves exterior-pole approximation without a general complex approximation theorem.