A Mathematical Exploration of the Virial Theorem

The virial theorem relates average kinetic energy to the forces acting on a system. Its essential qualification is an endpoint term: an average time derivative does not vanish merely because we call it an average.

The finite-time identity

Consider $N$ particles with constant masses $m_i$ , positions $\mathbf q_i(t)$ , and velocities $\mathbf v_i(t)$ in an inertial frame. Assume Newton's equations $m_i\dot{\mathbf v}_i=\mathbf F_i$ hold on the interval under consideration. Define

G(t)=\sum_{i=1}^{N}m_i\mathbf q_i\cdot\mathbf v_i, \qquad K(t)=\frac12\sum_{i=1}^{N}m_i\|\mathbf v_i\|^2.

Differentiating gives

\dot G=2K+\sum_{i=1}^{N}\mathbf q_i\cdot\mathbf F_i.

For $T>0$ , let $\langle f\rangle_T=T^{-1}\int_0^T f(t)\,dt$ . Integration yields the exact identity

2\langle K\rangle_T+ \left\langle\sum_{i=1}^{N}\mathbf q_i\cdot\mathbf F_i\right\rangle_T =\frac{G(T)-G(0)}{T}.

The right side is zero over a period for which $G(T)=G(0)$ . It tends to zero in a long-time limit if $G(T)/T\to0$ ; bounded positions and velocities are one sufficient condition. If the relevant averages also converge, the usual virial theorem follows:

2\langle K\rangle+ \left\langle\sum_{i=1}^{N}\mathbf q_i\cdot\mathbf F_i\right\rangle=0.

An escaping particle need not satisfy these conditions. For free motion with nonzero velocity, $G$ grows linearly, so dropping the endpoint term would incorrectly imply zero kinetic energy.

A homogeneous potential

Suppose the forces derive from a differentiable total potential $U(\mathbf q_1,\ldots,\mathbf q_N)$ , so $\mathbf F_i=-\nabla_iU$ . This formulation includes interactions between particles; we need not pretend each particle has an independent potential.

Assume simultaneous scaling gives

U(\lambda\mathbf q_1,\ldots,\lambda\mathbf q_N) =\lambda^\alpha U(\mathbf q_1,\ldots,\mathbf q_N) \quad(\lambda>0)

where the potential is defined. Differentiate with respect to $\lambda$ at $1$ :

\sum_{i=1}^{N}\mathbf q_i\cdot\nabla_iU=\alpha U.

Under the averaging conditions above, substitution gives

\boxed{2\langle K\rangle=\alpha\langle U\rangle.}

The potential's additive constant is fixed by the homogeneity condition. Adding an arbitrary constant usually destroys that condition, even though it leaves the forces unchanged.

Two examples

For a bound, collision-free Kepler orbit, $U=-k/r$ is homogeneous of degree $-1$ . Thus $2\langle K\rangle=-\langle U\rangle$ . If $E=K+U$ is constant, then $\langle K\rangle=-E$ and $\langle U\rangle=2E$ .

For a harmonic oscillator, $U=kx^2/2$ has degree $2$ , so $\langle K\rangle=\langle U\rangle$ over a period. Instantaneous kinetic and potential energy generally differ; the equality concerns their averages.