= Virial theorem
{title2=$2\langle K\rangle+\langle U\rangle=0$}
{wiki}
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 $2\langle K\rangle+\langle U\rangle=0$: 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 $G_{\!N}$:
$$
G(t)=\sum_i\mathbf p_i(t)\cdot\mathbf r_i(t),\qquad
\langle A\rangle_T=\frac1T\int_0^T A(t)\,dt,\qquad
2\langle K\rangle_T+\left\langle\sum_i\mathbf r_i\cdot\mathbf F_i\right\rangle_T
=\frac{G(T)-G(0)}{T}.
$$
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 $(G(T)-G(0))/T\to0$. 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>.<Cosmology by David Tong/1 The Expanding Universe>[Cosmology by David Tong, §1.4.3, “The Virial Theorem and Galaxy Clusters”: bounded positions and velocities, existence of averages, equation (1.74) and endpoint limit]{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> gives
$$
U(\lambda\mathbf r_1,\ldots,\lambda\mathbf r_N)=\lambda^qU(\mathbf r_1,\ldots,\mathbf r_N),\qquad
\sum_i\mathbf r_i\cdot\mathbf F_i=-qU,\qquad
2\langle K\rangle_T-q\langle U\rangle_T=\frac{G(T)-G(0)}T.
$$
Thus a <harmonic oscillator> has $q=2$ and $\langle K\rangle=\langle U\rangle$. An <isolated system> with <Newtonian gravitational potential energy> has
$$
U=-G_{\!N}\sum_{i<j}\frac{m_im_j}{|\mathbf r_i-\mathbf r_j|},\qquad q=-1,\qquad
2\langle K\rangle+\langle U\rangle=0,\qquad
E=\langle K\rangle+\langle U\rangle=-\langle K\rangle.
$$
Each 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.
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