2.5: Complications- Magnetic Fields, Internal Energy, and Rotation

The full power and utility of the virial theorem does not really become apparent until one realizes that we need not be particularly specific about the exact nature of the potential and kinetic energies that appear in the earlier derivations. Thus the presence of complicating forces can be included insofar as they are derivable from a potential. Similarly as long as the total kinetic energy can be expressed in terms of energies arising from macroscopic motions and internal thermal motions, it will be no trouble to express the virial theorem in terms of these more familiar parameters of the system. One may proceed in just this manner or return to the original Equations of motion for the system. We shall discuss both approaches.

In Chapter I, we derived the Euler-Lagrange Equations of hydrodynamic flow. These Equations of motion are completely general and are adequate to describe the effects of rotation and magnetic-fields if some care is taken with the coordinate frame and the ‘pressure tensor’ \(\mathfrak{P}\). With this in mind, we may rewrite Equation \ref{1.1.4}, noting that the left hand side is a total time derivative and that the pressure tensor can be explicitly split to include the presence of large scale electromagnetic fields. Thus

\[\frac{\partial \mathbf{u}}{\partial \mathrm{t}}=-\nabla \Phi-\frac{1}{\rho} \nabla \cdot \left(\mathfrak{P}_{\mathrm{g}}+\mathfrak{J}\right)-\frac{1}{\rho} \int \mathbf{S}(\mathbf{v}-\mathbf{u}) \mathrm{dv},\label{2.5.1}\]

where the tensor \(\mathfrak{P}_\mathrm{g}\) refers to the gas pressure alone and the tensor \(\mathfrak{J}\) represents the Maxwell stress tensor for the electromagnetic field which has components²⁷

\[\mathfrak{J}_{\mathrm{ij}}=\mathrm{E}_{\mathrm{i}} \mathrm{D}_{\mathrm{j}}-\frac{1}{2} \delta_{\mathrm{ij}} \sum_{\mathrm{k}} \mathrm{E}_{\mathrm{k}} \mathrm{D}_{\mathrm{k}}+\mathrm{H}_{\mathrm{i}} \mathrm{B}_{\mathrm{j}}-\frac{1}{2} \delta_{\mathrm{ij}} \sum_{\mathrm{k}} \mathrm{H}_{\mathrm{k}} \mathrm{B}_{\mathrm{k}},\label{2.5.2}\]

or in dyadic notation

\[\mathfrak{J}=\mathbf{E D}-\frac{1}{2} \mathbb{1}[\mathbf{E} \cdot \mathbf{D}]-\mathbf{H B}-\frac{1}{2} \mathbb{1}[\mathbf{H} \cdot \mathbf{B}].\label{2.5.3}\]

For almost all cases in astrophysics, it is appropriate to ignore electrostriction and magnetostriction effects which complicate the relationships between E and D and B and H. In the absence of these body forces, the divergence of Equation \ref{2.5.3} yields^2.4

\[\nabla \cdot \mathfrak{J}=\frac{1}{4 \pi}\left\{\mathbf{D} \rho_{\mathrm{e}}-\mathbf{D} \times(\nabla \times \mathbf{D})+\mathrm{c}^2[\mathbf{H} \times(\nabla \times \mathbf{H})]\right\},\label{2.5.4}\]

which is just the Lorentz force on the medium. It is useful when considering a configuration in uniform rotation to transform the problem into a co-rotating coordinate frame. This enables one to see the effect of macroscopic mass motion explicitly in the formalism and thus assess its interaction with other large scale properties of the system. In addition such systematic motion is represented by the stream velocity u in the collision term of the Boltzmann Equation making its meaning clearer. Therefore, just as we have separated the effects of the electromagnetic field from the pressure tensor, let us explicitly represent the effects of rotation.

In transforming the inertial coordinate frame to a non-inertial rotating frame attached to the system, we must allow for temporal changes in any vector seen in one frame but not the other. Goldstein²⁸ gives a particularly lucid account of how this is to be accomplished by use of the operator

\[\left(\frac{\mathrm{d}}{\mathrm{dt}}\right)_{\text {inertial }}=\left(\frac{\mathrm{d}}{\mathrm{dt}}\right)_{\text {non-inertial }}+\boldsymbol{w} \times,\label{2.5.5}\]

where \(\boldsymbol{w}\) is the angular velocity appropriate for the point function upon which the operator acts (i.e. the angular velocity of the rotating frame).

If we let

\[\mathbf{u}=\mathbf{w}+(\boldsymbol{w} \times \mathbf{r}),\label{2.5.6}\]

then the Equations of motion for uniform rotation (i.e., \(\boldsymbol{w}\) = constant) become

\[\frac{\mathrm{d} \mathbf{w}}{\mathrm{dt}}+2(\boldsymbol{w} \times \mathbf{w})+\boldsymbol{w} \times(\boldsymbol{w} \times \mathbf{r})=-\nabla \Phi-\frac{1}{\rho}[\nabla \cdot (\mathfrak{P}+\mathfrak{J})]-\frac{1}{\rho} \int_{\mathrm{v}} \mathbf{S}(\boldsymbol{w}) \mathrm{d} \mathbf{v}\label{2.5.7}\]

It can be shown that for uniform rotation the term^2.5.

\[\boldsymbol{w} \times(\boldsymbol{w} \times \mathbf{r})=\frac{1}{2} \nabla[(\boldsymbol{w} \times \mathbf{r}) \cdot (\boldsymbol{w} \times \mathbf{r})],\label{2.5.8}\]

so that it may be combined with the gravitational potential and \(\frac{1}{2}(\boldsymbol{w} \times \mathbf{r})^2\) may be considered a 'rotational potential'. The term \(2(\boldsymbol{w} \times \mathbf{w})\)

is known as the "coriolis force". Since both the pressure tensor and Maxwell tensor are normally "fixed" to the body, we should expect their formulation in the rotating frame to be simpler. It is difficult to proceed much further with the Equations of motion without making some simplifying assumptions. The most helpful and also reasonable of these is to assume that all collisions processes are isotropic in space. This has two results. Firstly, the collision term on the right hand side of the Equation of motion averages to zero when integrated over all velocity space. Secondly, the gas pressure tensor \(\mathfrak{P}_g\) becomes diagonal with all elements equal and can thus \(\nabla \cdot \mathfrak{P}_{\mathrm{g}}=\mathrm{P}\), where P is the scalar gas pressure. This effectively guarantees the existence of a scalar Equation of state which will be useful later when relating P to the internal energy. Since almost all astrophysical situations relate to plasmas, we may consider the configurations to have a very large conductivity and can therefore neglect the contribution to the Maxwell Field tensor of electric fields. Using these assumptions, and Equations \ref{2.5.4} and \ref{2.5.8}, the Equation of motion becomes

\[\frac{\mathrm{d} \mathbf{w}}{\mathrm{dt}}+2(\boldsymbol{w} \times \mathbf{w})=-\nabla\left[\Phi+(\boldsymbol{w} \times \mathbf{r})^2\right]-\nabla \mathbf{P} / \rho-\frac{1}{4 \pi \rho}[\mathbf{H} \times(\nabla \times \mathbf{H})].\label{2.5.9}\]

As we have done before we shall multiply the Equations of motion by ρr, and integrate over the volume and generate a general tensor expression for Lagrange's identity including rotation and magnetic fields.^2.6 Thus,

\[\frac{1}{2} \frac{\mathrm{d}^2}{\mathrm{dt}^2}(\mathfrak{I})+2 w \cdot \mathfrak{L}=2 \mathfrak{T}+\mathrm{n} \mathfrak{U}+w \cdot (\mathbb{1} w)-\omega^2 \mathbb{1}-\int_{\mathrm{v}} \mathbf{r} \nabla \mathrm{PdV}-\frac{1}{4 \pi} \int_{\mathrm{v}} \mathbf{r}[\mathbf{H} \times(\nabla \times \mathbf{H})] \mathrm{dV}.\label{2.5.10}\]

where \(\mathfrak{I}\), \(\mathfrak{T}\), and \(\mathfrak{U}\) are moment of inertia, kinetic and potential energy tensors respectively. \(\mathfrak{L}\) is an angular momentum-like tensor^2.6. In order to simplify the last two terms it is worth noting that they are derived from the divergence of two tensors. So, we may use the chain rule followed by the divergence theorem to simplify these terms. Thus

\[\int_{\mathrm{v}} \mathbf{r} \nabla \cdot (\mathfrak{P}+\mathfrak{J}) \mathrm{dV}=\int_{\mathrm{v}} \nabla \cdot \left[\mathbf{r}(\mathfrak{P}+\mathfrak{J}) \mathrm{dV}-\int_{\mathrm{v}}(\nabla \mathbf{r}) \cdot (\mathfrak{P}+\mathfrak{J}) \mathrm{dV}\right..\label{2.5.11}\]

The divergence theorem guarantees that the first integral can be written as a surface integral and if the volume is taken to be large enough can always be made to be zero. However, since magnetic fields always extend beyond the surface of what one normally considers the surface of the configuration, we will keep these terms for the moment. Thus

\[\mathrm{\int_v \nabla \cdot \left[\mathbf{r}(\mathfrak{P}+\mathfrak{J}) d V=\int_s \mathbf{r}(\mathfrak{P}+\mathfrak{I}) \cdot d \mathbf{S} \equiv \mathfrak{S},\right.}\label{2.5.12}\]

or, in terms of the components, the surface terms can be written as

\[\mathrm{{\mathfrak{S}}_{i j}=\frac{1}{4 \pi} \int_s x_j\left(H_i \sum_k H_k d S_k-\frac{1}{8 \pi} \int_s x_j H^2 d S_i-\int_s P_0 x_j d S_i\right.}.\tag{2.5.12a}\label{2.5.12a}\]

Keeping in mind that \(\nabla \mathbf{r}=\mathbb{1}\) (i.e., a second rank tensor with components \(\delta_{\mathrm{ij}}\)), the second integral in Equation \ref{2.5.11} becomes

\[\mathrm{\int_v \mathbb{1} \cdot (\mathfrak{P}+\mathfrak{J}) d V=\int_v \mathbb{1}\left(P+H^2 / 8 \pi\right) d V+\frac{1}{4 \pi} \int_v(\mathbf{H H}) d V}.\label{2.5.13}\]

Defining

\[\mathfrak{M} \equiv \frac{1}{8 \pi} \int_{\mathrm{v}}(\mathbf{H H}) \mathrm{dV},\label{2.5.14}\]

we arrive at the final tensor form for Lagrange's Identity, including rotation and magnetic fields.

\[\mathrm{\begin{aligned}
& \quad \frac{1}{2} \frac{d^2 \mathfrak{I}}{d t^2}+2 w \cdot \mathfrak{L}=2[\mathfrak{T}-\mathfrak{M}]+n \mathfrak{U}+w \cdot (\mathbb{1} w)-\omega^2 \mathbb{1}+\mathbb{1}\left(\int_\mathrm{v}\left(P+H^2 / 8 \pi\right) d V\right)+\mathfrak{S}, \\[4pt]
& \text { where } \mathfrak{L}_{i j k}=\int_\mathrm{v} \rho\left(r_k r_i \varpi_j-\varpi_k r_i r_j\right) d V.
\end{aligned}}\label{2.5.15}\]

We have suffered through the tensor derivation in order to show the complete generality of this formalism. The tensor component Equations are essential for investigating non-radial oscillations and other such phenomena which cannot be represented by a simple scalar approach. However, it is easier to appreciate the physical significance of this approach by looking at the scalar counterpart of Equation \ref{2.5.15}. In section 1, we pointed out that the scalar form is derived by taking "inner" products of the position vector with the Equation of motion while the tensor virial theorem involves "outer" or tensor products. One may either repeat the derivation of Equation \ref{2.5.15} taking "inner" products or contract, the component form of Equation \ref{2.5.15} over indices \(i\) and \(j\).

The contraction of the tensors \(\mathfrak{I}\), \(\mathfrak{T}\), \(\mathfrak{M}\), \(\mathfrak{U}\), and \(\mathbb{1}\) yield the moment of inertia about the origin of the coordinate system I, the kinetic energy due to internal motion T, the total magnetic energy \(\mathscr{M}\), the total potential energy Φ respectively. Some care must be taken in contracting the tensors \(\mathfrak{L}\), and \(\mathbb{1}\boldsymbol{w}\). From the definition of \(\mathfrak{L}\), in Equation (N2.6.3), it is clear that the contracted form of that expression can be written as

\[2 \int_{\mathrm{v}} \rho \mathbf{r} \cdot (\boldsymbol{w} \times \mathbf{w}) \mathrm{dV}=-2 \boldsymbol{w} \cdot \int_{\mathrm{v}} \rho(\mathbf{r} \times \mathbf{w}) \mathrm{dV}=2 \boldsymbol{w} \cdot \int_{\mathrm{v}} l \mathrm{dV},\label{2.5.16}\]

where \(\boldsymbol{l}\) is the net volume angular momentum density on the material due to coriolis forces. We can again choose our rotating frame so that \(\int \boldsymbol{l} \mathrm{dV}\) is zero and this term must vanish from the contracted Equation. The simplest method for deriving the value of the contracted form of \(w \cdot (\mathbb{1} w-w \mathbb{1})\) in Equation \ref{2.5.15} is again to examine the contracted form of the term giving rise to it. Since \(\mathbf{w}=w \mathbf{x} \mathbf{r}\), we can expand the left hand side of Equation \ref{2.5.16} by means of identities relating to the vector triple product [see note 2.6, specifically Equation (N2.6.4)], and obtain

\[\int_{\mathrm{v}} \rho \mathbf{r} \cdot [\boldsymbol{w} \times(\boldsymbol{w} \times \mathbf{r})] \mathrm{dV}=\int_{\mathrm{v}} \rho \mathbf{r} \cdot [\boldsymbol{w} \times \mathbf{v}] \mathrm{dV}=\int_{\mathrm{v}} \rho[\mathbf{r} \times \boldsymbol{w}] \cdot \mathbf{v d V}=\int_{\mathrm{v}} \rho \mathrm{v}^2 \mathrm{dV}=2 \Re,\label{2.5.17}\]

where \(\mathfrak{R}\) is just the total energy due to rotation. Substitution of the contraction of these tensors into Equation \ref{2.5.15} yields a much simpler result

\[\frac{1}{2} \frac{\mathrm{d}^2 \mathrm{I}}{\mathrm{dt}^2}=2(\mathrm{T}+\Re)+\mathscr{M}-\Omega+\mathrm{n} \mathcal{U}+3 \int_{\mathrm{v}} \mathrm{PdV}+\sum_{\mathrm{i}} \mathfrak{S}_{\mathrm{ii}}\label{2.5.18}\]

The contraction of the surface terms yield the scalar integrals

\[\sum_{\mathrm{i}} \mathfrak{S}_{\mathrm{ii}}=\frac{1}{4 \pi} \int_{\mathrm{s}}\left(\mathbf{r}_0 \cdot \mathbf{H}_0\right)\left(\mathbf{H}_0 \cdot \mathrm{d} \mathbf{S}\right)-\int_{\mathrm{s}}\left(\mathrm{P}_0+\mathrm{H}_0^2 / 8 \pi\right) \mathbf{r} \cdot \mathrm{d} \mathbf{S}\label{2.5.19}\]

Here, the subscript "0" indicates the value of the variables on the surface. In practice these integrals are usually small compared with the magnitude of the volume integrals found in the remainder of the expression. So far, we have said little about the contribution of the volume integral of the pressure. Since even in the tensor representation this term appears as a scalar, there was no loss of generality in deferring the evaluation of the integral until now. Dimensional analysis will lead one to the result that the pressure integral is an 'energy-like' integral. From thermodynamic consideration, we can write the internal "heat energy" \(\mathcal{U}\) as

\[\mathcal{U}=\int_{\mathrm{v}} \rho \mathrm{c}_{\mathrm{v}} \mathcal{J} \mathrm{dV},\label{2.5.20}\]

where \(\mathrm{c_v}\) is the specific heat of constant volume and is the temperature. We also know that the kinetic energy associated with the material is

\[\mathscr{E}=\frac{3}{2} \mathrm{Nk} \mathcal{J}=\frac{3}{2} \rho\left(\mathrm{c}_{\mathrm{p}}-\mathrm{c}_{\mathrm{v}}\right) \mathcal{J}-\frac{3}{2} \mathrm{P}.\label{2.5.21}\]

Combining these two Equations we get the total internal "heat energy" as

\[\mathcal{U}=\frac{2}{3} \int_\mathrm{v} \frac{\mathscr{E} \mathrm{dV}}{\left(\mathrm{c_p / c_v-1}\right)}=\int_\mathrm{v}\frac{\rho \mathrm{d V}}{\left(\mathrm{c_p / c_v-1}\right)}.\label{2.5.22}\]

It is traditional to let \(\gamma=\mathrm{c_p / c_v}\) and if we let this be constant throughout the volume and neglect the surface term in Equation \ref{2.5.18}, then we can write the scalar form of Lagrange's identity as

\[\frac{1}{2} \frac{\mathrm{d}^2 \mathrm{I}}{\mathrm{dt}^2}=2(\mathrm{T}+\Re)+\mathscr{M}-\Omega+3(\gamma-1) \mathcal{U}.\label{2.5.23}\]

At the beginning of this discussion, I said that one could either derive Lagrange's identity from the Equations of motion or from careful consideration of meanings of potential and kinetic energy. For example, the first and last terms on the right hand side of Equation \ref{2.5.23} are just the contribution to the total kinetic energy of the system from macroscopic motions, rotation and thermal motion respectively. The remaining terms are just a specification of the nature of the total potential energy. Thus, by realizing that \(3(\gamma-1)\mathcal{U}\) is just twice the kinetic energy due to thermal motion we could have written Equation \ref{2.5.23} down immediately. However, it is unlikely that this could have been done for Equation \ref{2.5.15}. Since the variational form of this Equation will be useful in the next chapter, our efforts have not been wasted.

Before leaving this section on complicating phenomena, there is one last aspect to be investigated. In Chapter I, section 3, we noted that the inclusion of velocity dependent forces such as frictional force do not alter the results of the virial theorem as they can be averaged to zero given sufficient time. This result was apparently first noted by E. H. Milne²⁹ in 1925. Although no modification is made to the virial theorem some effects can be seen in Lagrange's identity and so, let us take a moment to recapitulate these arguments. If we have a volume force which can be derived from a "friction" tensor, then one would add terms to the Equations of motion of the form

\[\boldsymbol{f}_{\mathrm{f}}=\rho \mathbf{w} \cdot \mathfrak{f},\label{2.5.24}\]

which make the following contribution to the tensor form of Lagrange's identity:

\[\int_\mathrm{v} \mathbf{r} \boldsymbol{f}_{\mathrm{f}} \mathrm{dV}=\int_\mathrm{v} \rho \mathbf{r w} \cdot \mathfrak{f}\mathrm{dV}=\frac{1}{2} \int_\mathrm{v} \rho\left[\frac{\mathrm{d}}{\mathrm{dr}}(\mathbf{r r})\right] \cdot \mathfrak{f}\mathrm{dV}.\label{2.5.25}\]

If \(\mathfrak{f}\) were indeed constant throughout the configuration, the right-hand side of Equation \ref{2.5.25} would just become

\[\frac{1}{2}\left(\frac{\mathrm{d} \mathfrak{I}}{\mathrm{dt}}\right): \mathfrak{f},\label{2.5.26}\]

and Lagrange's identity in its full generality would be

\[\frac{1}{2} \frac{\mathrm{d}^2 \mathfrak{I}}{\mathrm{dt}^2}+\frac{1}{2}\left(\frac{\mathrm{d} \mathfrak{I}}{\mathrm{dt}}\right): \mathfrak{f}+2 \boldsymbol{w} \cdot \mathfrak{L}=2(\mathfrak{T}-\mathfrak{M})+\mathrm{n} \mathfrak{U}+\boldsymbol{w} \cdot [\boldsymbol{l} \boldsymbol{w}-\boldsymbol{w} \boldsymbol{l}]+\boldsymbol{l}[(1-\gamma) \mathcal{U}+\mathscr{M}]+\mathfrak{S}.\label{2.5.27}\]