1.4: Continuum-Field Representation of the Virial Theorem

Although nearly all derivations of the virial theorem consider collections of mass-points acting under forces derivable from a potential, it is useful to look at this formalism as it applies to a continuum density field of matter. This is particularly appropriate when one considers applications to stellar structure where a continuum representation of the material is always used.

In the interests of preserving some rigor let us pass from Equation \ref{1.2.1} to its analogous representation in the continuum. Let the mass m_i be obtained by multiplying the density \(ρ(r)\) by an infinitesimal volume \(∆V\) so that 1.2.1 becomes

\[\mathbf{f}_{\mathrm{i}}=\frac{\mathrm{d}}{\mathrm{dt}}(\rho \mathbf{v} \Delta \mathrm{V})=\mathbf{v} \Delta \mathrm{V} \frac{\mathrm{d} \rho}{\mathrm{dt}}+\rho \Delta \mathrm{V} \frac{\mathrm{d} \mathbf{v}}{\mathrm{dt}}+\rho \mathbf{v} \frac{\mathrm{d}(\Delta \mathrm{V})}{\mathrm{dt}}.\label{1.4.1}\]

Conservation of mass requires that

\[\frac{\mathrm{dm}_{\mathrm{i}}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\rho \Delta \mathrm{V})=\Delta \mathrm{V} \frac{\mathrm{d} \rho}{\mathrm{dt}}+\rho \frac{\mathrm{d}(\Delta \mathrm{V})}{\mathrm{dt}}=0 .\label{1.4.2}\]

Multiplying this expression by v we see that the first and last terms on the right hand side of Equation \ref{1.4.1} are of equal magnitude and opposite sign. Thus, if we define a "force density", \(f\), so that \(f \Delta V=\mathbf{f}_\mathrm{i}\), we can pass to this continuum representation of Equation \ref{1.2.1}:

\[f(\mathbf{r})=\rho(\mathbf{r}) \frac{\mathrm{d}}{\mathrm{dt}}[\mathbf{v}(\mathbf{r})]=\dot{\boldsymbol{p}}(\mathbf{r}),\label{1.4.3}\]

where p(r) by analogy to 1.2.1 is just the local momentum density.

We can now define G in terms of the continuum variables so that

\[\mathrm{G}=\int_{\mathrm{V}} \boldsymbol{p} \cdot \mathbf{r}\mathrm{dV}=\int_{\mathrm{V}} \rho \frac{\mathrm{d} \mathbf{r}}{\mathrm{dt}} \cdot \mathbf{r} \mathrm{dV}=\frac{1}{2} \int_{\mathrm{V}} \rho \frac{\mathrm{d}}{\mathrm{dt}}(\mathbf{r} \cdot \mathbf{r}) \mathrm{dV}=\frac{1}{2} \int_{\mathrm{V}} \rho \frac{\mathrm{dr}^2}{\mathrm{dt}} \mathrm{dV},\label{1.4.4}\]

so that

\[\mathrm{G}=\frac{1}{2} \int_{\mathrm{V}} \frac{\mathrm{d}}{\mathrm{dt}}\left(\rho \mathrm{r}^2\right) \mathrm{dV}-\frac{1}{2} \int_{\mathrm{V}} \mathrm{r}^2 \frac{\mathrm{d} \rho}{\mathrm{dt}} \mathrm{dV}.\label{1.4.5}\]

Once again, one uses conservation of mass requiring that the mass within any sub volume V' is constant with time so that \(\mathrm{dm}\left(\mathrm{V}^{\prime}\right) / \mathrm{dt}=0\) with that sub-volume V' defined such that

\[\frac{\mathrm{d}}{\mathrm{dt}}\left(\int_{\mathrm{V}^{\prime}} \rho \mathrm{dV}\right)=\int_{\mathrm{V}^{\prime}} \frac{\mathrm{d} \rho}{\mathrm{dt}} \mathrm{dV}=0.\label{1.4.6}\]

Thus, the second integral in Equation \ref{1.4.5} after integration by parts is zero. If we take the original volume V to be large enough so as to always include all the mass of the object, we may write Equation \ref{1.4.5} as

\[\mathrm{G}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dt}} \int_{\mathrm{V}}\left(\rho \mathrm{r}^2\right) \mathrm{dV}=\frac{1}{2} \frac{\mathrm{dI}}{\mathrm{dt}}.\label{1.4.7}\]

With these same constraints on V we may differentiate Equation \ref{1.4.4} with respect to time and obtain

\[\frac{\mathrm{dG}}{\mathrm{dt}}=\int_{\mathrm{V}}\left[\boldsymbol{p} \cdot \frac{\mathrm{d} \mathbf{r}}{\mathrm{dt}}+\mathbf{r} \cdot \frac{\mathrm{d} \boldsymbol{p}}{\mathrm{dt}}\right] \mathrm{dV}=\int_{\mathrm{V}}\left(\rho \mathrm{v}^2+\mathbf{r} \cdot \boldsymbol{f}\right) \mathrm{dV}.\label{1.4.8}\]

The first term under the integral is just kinetic energy density and hence its volume integral is .just the total kinetic energy of the configuration and

\[\frac{1}{2} \frac{\mathrm{d}^2 \mathrm{I}}{\mathrm{dt}^2}=2 \mathrm{T}+\int_{\mathrm{V}} \boldsymbol{f} \cdot \mathbf{r}\mathrm{dV}.\label{1.4.9}\]

Considerable care must be taken in evaluating the second term in Equation \ref{1.4.9} which is basically the virial of Claussius. In the previous derivation we went to some length [i.e., Equation \ref{1.2.10}] to avoid "double counting" the forces by noting that the force between any two particles A and B can be viewed as a force at A due to B, or a force at B resulting from A. The contributions to the virial, however, are not equal as they involve a 'dot' product with the position vector. Thus, we explicitly paired the forces and arranged the sum so pairs of particles were only counted once. Similar problems confront us within continuum derivation. Thus, each force at a field point \(\boldsymbol{f}(\mathbf{r})\) will have an equal and opposite counterpart at the source points \(\mathbf{r}’\).

After some algebra, direct substitution of the potential gradient into the definition of the Virial of Claussius yields ^1.2

\[\begin{aligned}
\int_{\mathrm{V}} \boldsymbol{f} \cdot \mathbf{r d V} & =-\frac{\mathrm{n}}{2} \int_{\mathrm{v}}\int_{\mathrm{v’}} \rho(\mathbf{r}) \rho\left(\mathbf{r}^{\prime}\right)\left(\mathbf{r}-\mathbf{r}^{\prime}\right) \cdot \left(\mathbf{r}-\mathbf{r}^{\prime}\right)\left(\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)^{\mathrm{n}-2} \mathrm{dV}^{\prime} \mathrm{dV} \\[4pt]
& =\mathrm{n}\left[\frac{1}{2} \int_{\mathrm{v}}\int_{\mathrm{v’}} \rho(\mathbf{r}) \rho\left(\mathbf{r}^{\prime}\right)\left(\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)^{\mathrm{n}} \mathrm{dV}^{\prime} \mathrm{dV}\right]
\end{aligned}\label{1.4.10}\]

Since V = V', the integrals are fully symmetric with respect to interchanging primed with non-primed variables. In addition the double integral represents the potential energy of ρ(r) with respect to ρ(r '), and ρ(r ') with respect to ρ(r); it is just twice the total potential energy. Thus, we find that the virial has the same form as Equation \ref{1.2.12}, namely,

\[\int_{\mathrm{V}} \boldsymbol{f} \cdot \mathbf{r}\mathrm{dV}=-\mathrm{n} \mathcal{U}.\label{1.4.11}\]

Substitution of this form into Equation \ref{1.4.9} and taking n = -1 yields the same expression for Lagrange's identity as was obtained in Equation \ref{1.2.13}, specifically,

\[\frac{1}{2} \frac{\mathrm{d}^2 \mathrm{I}}{\mathrm{dt}^2}=2 \mathrm{T}+\Omega\label{1.4.12}\]

Thus Lagrange's identity, the virial theorem and indeed the remainder of the earlier arguments, are valid for the continuum density distributions as we might have guessed.

Throughout this discussion it was tacitly assumed that the forces involved represented "gravitational" forces insofar as the force was -ρ∇Φ. Clearly, if the force depended on some other property of the matter (e.g., the charge density, ε(r) the evaluation of \(\int_{\mathrm{v}} \boldsymbol{f} \cdot \mathbf{r}\mathrm{dV}\) would go as before with the result that the virial would again be \(-\mathrm{n}\mathcal{U}\) where \(\mathcal{U}\) is the total potential energy of the configuration.