3.3: The Influence of Magnetic and Rotational Energy upon a Pulsating System

We shall now consider what the effect of introducing magnetic and rotational energies into a pulsating system will be upon the frequency of pulsation of that system. It is worth noting that solution of such a problem in terms of the force Equations would be difficult indeed as it would require detailed knowledge of the geometry of the magnetic field throughout the star. However, since our approach expresses the pulsation frequency in terms of volume integrals, only knowledge of the total magnetic and rotational energies will be required.

In order to simplify the mathematical development we shall make some of the assumptions which were made during previous sections. These assumptions are listed below:

1. A first order theory will be adequate. That is, all deviations from equilibrium shall be small.
2. Radiation pressure will be considered to be negligible (i.e., \(\Gamma_1=\gamma\)).
3. \(\gamma\) will be constant throughout the system.

We have already seen that it is possible to write Lagrange's identity so as to include the effects of rotational and magnetic energy. One of the points of that derivation that required some care was the inclusion of surface terms arising from the fact that stellar magnetic fields usually extend well beyond the normal surface of the star. However, for the moment let us neglect these terms since they usually will be small and as such will not affect the general character of the solution. Thus, as in Chapter II, we may write the scalar form of Lagrange's identity as follows:

\[\frac{1}{2} \frac{\mathrm{d}^2 \mathrm{I}}{\mathrm{dt}^2}=2 \mathrm{T}+\Omega+\mathscr{M},\label{3.3.1}\]

where T is the total kinetic energy including rotation and \(\mathscr{M}\) is the total magnetic energy. Now let us break up the total kinetic energy of the system into the sum of three energies \(\mathcal{J}_1\), \(\mathcal{J}_2\) and \(\mathcal{J}_3\) where \(\mathcal{J}_1\) is the kinetic energy of the pulsating system due to the pulsating motion, \(\mathcal{J}_2\) is the kinetic energy of the gas due to thermal energy, and \(\mathcal{J}_3\) is the kinetic energy of rotation. The contribution to the kinetic energy (of particle motion) due to an element of mass is

\[\mathrm{d} \mathcal{J}_2=\frac{3}{2} \mathrm{k} T \mathrm{dn}=\frac{3}{2} R T \mathrm{dm}=\frac{3}{2}\left(\mathrm{c}_{\mathrm{p}}-\mathrm{c}_{\mathrm{v}}\right) T \mathrm{dm},\label{3.3.2}\]

where \(T\) and \(R\) are the gas temperature and constant respectively. But the internal energy \(\mathrm{d}\mathcal{U}\) of the element of mass is

\[\mathrm{d} \mathcal{U}=\mathrm{c}_{\mathrm{v}} T \mathrm{dm},\label{3.3.3}\]

Combining Equation \ref{3.3.2} and Equation \ref{3.3.3}, with the definition of \(\gamma\), we have

\[\mathrm{d} \mathcal{J}_2=\frac{3}{2}(\gamma-1) \mathrm{d}\mathcal{U}.\label{3.3.4}\]

Integrating this over the entire system we obtain

\[\mathcal{2} \mathcal{J}_2=3(\gamma-1)\mathcal{U} .\label{3.3.5}\]

Thus, we may write the virial theorem for the system as

\[\frac{1}{2} \frac{\mathrm{d}^2 \mathrm{I}}{\mathrm{dt}^2}=2 \mathcal{J}_1+\mathcal{2} \mathcal{J}_3+3(\gamma-1) \mathcal{U}+\Omega+\mathscr{M}.\label{3.3.6}\]

The variational form becomes

\[\frac{1}{2} \frac{\mathrm{d}^2}{\mathrm{dt}^2}(\delta \mathrm{I})=2 \delta \mathcal{J}_3+3(\gamma-1) \delta \mathcal{U}+\delta \Omega+\delta \mathscr{M}.\label{3.3.7}\]

There is no term containing a variation in γ as it is assumed to be constant throughout the system. In section 2, it was shown that variation of the pulsational kinetic energy was of second order and could therefore be neglected. Thus

\[2 \delta \mathcal{J}_1=0.\label{3.3.8}\]

Since we have already computed the variation of the kinetic energy of the gas, we may most easily find the variation of the total internal energy (\(\delta \mathcal{U}\)) in terms of this quantity. As a result of the assumption of constant \(\gamma\), we may take the variation of Equation \ref{3.3.5}, and obtain

\[2 \delta \mathcal{J}_2=+3(\gamma-1) \delta \mathcal{U}.\label{3.3.9}\]

Now, if we further assume a periodic form for the pulsation and linearly increasing amplitude (ξ_o=const.), Equation \ref{3.2.15}, gives the following expression for the variation of the kinetic energy of the gas:

\[\begin{aligned}
2 \delta \mathcal{J}_2&=-3 \int_0^{\mathrm{M}} \frac{3 \mathrm{P}_0 \xi_0}{\rho_0}(\gamma-1) \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \mathrm{dm}\left(\mathrm{r}_0\right).\\[4pt]
\text{or}\quad 2 \delta \mathcal{J}_2&=-3(\gamma-1) \xi_0 \mathrm{e}^{\mathrm{i} \sigma t} \int_{\mathrm{v}} 3 \mathrm{P}_0 \mathrm{dV}.
\end{aligned}\label{3.3.10}\]

Combining Equation \ref{3.3.10} and Equation \ref{3.3.9}, we obtain

\[\mathrm{\delta \mathcal{U}=-\xi_0 e^{i \sigma t} \int_v 3 P_0 d V}.\label{3.3.11}\]

In order to express this variation in terms of the other energies present in the system, we shall assume that the system is in quasi-steady state. With this assumption, the relevant quantities are averaged over one pulsation period so that the \(<\mathrm{d}^2 \mathrm{I} / \mathrm{dt}^2>=0\). We shall assume that the remaining values are the equilibrium values of the configuration. The virial theorem as expressed in Equation \ref{3.3.6} then becomes

\[2 \mathcal{J}_1(0)+\mathcal{2}\mathcal{J}_2(0)+\mathcal{2} \mathcal{J}_3(0)+\Omega_0+\mathscr{M}=0 .\label{3.3.12}\]

Now, from the elementary kinetic theory of gases we have

\[\mathcal{J}_2=\frac{3}{2} \int_{\mathrm{v}} \mathrm{P}_0 \mathrm{dV}.\label{3.3.13}\]

Also, since the system is neither expanding or contracting,

\[\mathcal{J}_1(0)=<\mathcal{J}_1>=0.\label{3.3.14}\]

Making use of these two results and replacing the average values of the quantities in Equation \ref{3.3.12} with the equilibrium values we find

\[-3 \int_{\mathrm{v}} \mathrm{P}_0 \mathrm{dV}=2 \mathcal{J}_3(0)+\Omega_0+\mathscr{M}_0.\label{3.3.15}\]

Identifying the left side of Equation \ref{3.3.15} with the right side of Equation \ref{3.3.11}, we finally obtain

\[\delta \mathcal{U}=\xi_0 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}}\left(2 \mathcal{J}_3(0)+\Omega_0+\mathscr{M}_0\right).\label{3.3.16}\]

We have already determined expressions for the variations of the gravitational energy and moment of inertia in the previous section [Equation \ref{3.2.13}]. Under the assumption used above, of a constant perturbation \(\xi_o\), these Equations become

\[\left.\begin{array}{l}
\delta \mathrm{I}=2 \mathrm{e}^{\mathrm{i} \sigma t} \xi_0 \mathrm{I}_0 \\[4pt]
\delta \Omega=-\mathrm{e}^{\mathrm{i} \sigma t} \xi_0 \Omega_0
\end{array}\right\} .\label{3.3.17}\]

Thus, we need only obtain expressions for the variation of the magnetic and rotational energies in order to evaluate Equations Equation \ref{3.3.7}.

Consider first the rotational energy. Now an element of mass rotating about an axis with a velocity ω will possess an elemental angular momentum

\[\mathrm{d} \mathscr{L}=\omega\left(\mathrm{x}^2+\mathrm{y}^2\right) \mathrm{dm}(\mathrm{r})=\omega \mathrm{r}^2 \sin ^2 \theta \mathrm{dm}(\mathrm{r}).\label{3.3.18}\]

Here the x-y plane is the plane perpendicular to axis of rotation and θ is the polar angle measured from the axis of rotation. Such an elemental mass possessing such an angular momentum will have a rotational kinetic energy given by

\[\mathrm{d} \mathcal{J}_3=\frac{1}{2} \omega \mathrm{d} \mathscr{L}.\label{3.3.19}\]

Thus, the total rotational kinetic energy is just

\[\mathcal{J}_3=\frac{1}{2} \int_0^{\mathscr{L}} \omega \mathrm{d}\mathscr{L}.\label{3.3.20}\]

We may use this expression to obtain the variation δT3. The first term on the right of Equation \ref{3.3.7} becomes

\[2 \delta \mathcal{J}_3=\delta \int_0^{\mathscr{L}} \omega \mathrm{d} \mathscr{L}=\int_0^{\mathscr{L}} \delta \omega \mathrm{d} \mathscr{L}+\int_0^{\mathscr{L}} \omega \mathrm{d}(\delta \mathscr{L}).\label{3.3.21}\]

However, the conservation of angular momentum requires that \(\mathscr{L}\) remain constant during the pulsation. Thus, the variation of \(\mathscr{L}\) is zero and the last integral on the right of Equation \ref{3.3.21} vanishes, so that

\[2 \delta \mathcal{J}_3=\int_0^{\mathscr{L}} \delta \omega \mathrm{d} \mathscr{L}+\int_0^{\mathscr{L}} \frac{\delta \omega}{\omega_0} \omega_0 \mathrm{d} \mathscr{L}.\label{3.3.22}\]

where ω_o is the rotational velocity of the equilibrium configuration.

Now, again making use of the conservation of angular momentum we see that

\[\mathrm{\omega r^2 \sin \theta=\text {const.}}\label{3.3.23}\]

Since we are only considering radial pulsation so that δθ is zero, the variation Equation \ref{3.3.23} yields

\[\mathrm{\delta \omega r^2 \sin \theta+2 \omega r \delta r \sin \theta=0}.\label{3.3.24}\]

This is equivalent to "conserving angular momentum in shells."

If we evaluate Equation \ref{3.3.24} at the equilibrium position, we obtain an expression of first order accuracy,

\[\frac{\delta \omega_0}{\omega_0}=\frac{2 \delta \mathrm{r}}{\mathrm{r}_0}=-2 \xi_0 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} .\label{3.3.25}\]

Substitution of this expression into Equation \ref{3.3.22} yields

\[2 \delta \mathcal{J}_3=-\int_0^{\mathscr{L}}\left(2 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \xi_0 \omega_0\right) \mathrm{d} \mathscr{L}_0.\label{3.3.26}\]

If, for simplicity, we further assume that the rotational velocity is a constant throughout the configuration. We obtain a very simple form of the variation of the rotational energy.

\[2 \delta \mathcal{J}_3=-\left(2 \mathrm{e}^{\mathrm{i} \sigma t} \xi_0 \omega_0\right) \mathscr{L}_0,\label{3.3.27}\]

where \(\mathscr{L}_o\) is the total angular momentum for the system. Thus, only the variation of the magnetic energy remains to be determined.

In order to determine the variation of the total magnetic energy it is necessary to establish a coordinate system appropriate to the geometry of the field and to the geometry of the configuration. Although the configuration is spherically symmetric, the geometry of the magnetic field present is not known. Thus, we shall consider the variations in Cartesian coordinates and later reduce our result to a form which is compatible with our previous results. Now the total magnetic energy of the configuration is defined (in c.g.s. units).

\[\mathscr{M}=\int_0^{\mathrm{M}} \frac{|\mathbf{H}|^2}{8 \pi \rho} \mathrm{dm}(\mathrm{r}).\label{3.3.28}\]

Thus, denoting the Cartesian coordinates as x₁, x₂, and x₃, the variational form of the magnetic energy is

\[\delta \mathscr{M}=\frac{1}{4 \pi} \int_0^{\mathrm{M}} \frac{\mathbf{H} \cdot \delta \mathbf{H}}{\rho} \mathrm{dm}(\mathrm{r})-\frac{1}{8 \pi} \int_0^{\mathrm{M}}|\mathbf{H}|^2 \frac{\delta \rho}{\rho^2} \mathrm{dm}(\mathrm{r}),\label{3.3.29}\]

or in Cartesian coordinates

\[\delta \mathscr{M}=\frac{1}{4 \pi} \iiint \sum_{\mathrm{i}} \mathrm{H}_{\mathrm{i}} \delta \mathrm{H}_{\mathrm{i}} \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3-\frac{1}{8 \pi} \iiint|\mathbf{H}|^2 \frac{\delta \rho}{\rho} \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3.\label{3.3.30}\]

Although we have already obtained an expression for δρ/ρ in section 3, due to the introduction of Cartesian coordinates it is convenient to express this variation in terms of the variation of the coordinates \(\eta_{\mathrm{i}}\) ^3.5, namely

\[\frac{\delta \rho}{\rho}=-\sum_{\mathrm{i}=1}^3 \frac{\partial \eta_{\mathrm{i}}}{\partial \mathrm{x}_{\mathrm{i}}},\label{3.3.31}\]

Before we can evaluate the expression for the variation of the magnetic energy, we must first determine the variation of the magnetic field δH_i. A rather lengthy argument^3.6 shows we can express this in terms of the coordinate variations, so

\[\delta \mathrm{H}_{\mathrm{i}}=\sum_{\mathrm{j}}\left(\mathrm{H}_{\mathrm{j}} \frac{\partial \eta_{\mathrm{i}}}{\partial \mathrm{x}_{\mathrm{j}}}-\mathrm{H}_{\mathrm{i}} \frac{\partial \eta_{\mathrm{j}}}{\partial \mathrm{x}_{\mathrm{i}}}\right).\label{3.3.32}\]

If we substitute Equation \ref{3.3.32} and Equation \ref{3.3.31} into Equation \ref{3.3.29}, we have

\[\delta \mathscr{M}= \frac{1}{4 \pi} \iiint \sum_{\mathrm{i}} \sum_{\mathrm{j}} \mathrm{H}_{\mathrm{i}} \mathrm{H}_{\mathrm{j}} \frac{\partial \eta_{\mathrm{i}}}{\partial \mathrm{x}_{\mathrm{j}}} \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3-\frac{1}{4 \pi} \iiint \sum_{\mathrm{i}} \sum_{\mathrm{j}} \mathrm{H}_{\mathrm{i}}^2 \frac{\partial \eta_{\mathrm{j}}}{\partial \mathrm{x}_{\mathrm{j}}} \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3 +\frac{1}{8 \pi} \iiint \mathrm{H}^2 \sum_{\mathrm{j}} \frac{\partial \eta_{\mathrm{j}}}{\partial \mathrm{x}_{\mathrm{j}}} \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3.\label{3.3.33}\]

The second and third terms combine to yield

\[\delta \mathscr{M}=\frac{1}{4 \pi} \iiint \sum_{\mathrm{i}} \sum_{\mathrm{j}} \mathrm{H}_{\mathrm{i}} \mathrm{H}_{\mathrm{j}} \frac{\partial \eta_{\mathrm{i}}}{\partial \mathrm{x}_{\mathrm{j}}} \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3+\frac{1}{8 \pi} \iiint \mathrm{H}^2 \sum_{\mathrm{j}} \frac{\partial \eta_{\mathrm{j}}}{\partial \mathrm{x}_{\mathrm{j}}} \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3.\label{3.3.34}\]

Now, if we assume that \(\eta_{\mathrm{i}}\) is only a function of x_i then the sum on j in the first term collapses and the remaining terms result in

\[\delta \mathscr{M}=\frac{1}{8 \pi} \iiint\left(\sum_{\mathrm{i}} 2 \mathrm{H}_{\mathrm{i}}^2 \frac{\partial \eta_{\mathrm{j}}}{\partial \mathrm{x}_{\mathrm{j}}}-|\mathbf{H}|^2 \frac{\partial \eta_{\mathrm{i}}}{\partial \mathrm{x}_{\mathrm{i}}}\right) \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3.\label{3.3.35}\]

At this point, it is appropriate to re-introduce the assumption concerning the nature of the variation \(\eta_\mathrm{i}\). It was earlier assumed that \(\xi_\mathrm{o}\) was constant. The equivalent assumption for the \(\eta_\mathrm{i}\)'s is that

\[\eta_{\mathrm{i}}=\text { const } \mathrm{x}_{\mathrm{i}}.\label{3.3.36}\]

Substitution of this explicit variation into Equation \ref{3.3.35} yields

\[\delta \mathscr{M}=-\frac{\text { const. }}{8 \pi} \iiint|\mathbf{H}|^2 \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3.\label{3.3.37}\]

Now, since we wish to consider the same type of pulsation from the \(\eta_\mathrm{i}\)'s, as we have assumed in the earlier section, we require that

\[\boldsymbol{\eta}=\delta \mathbf{r}.\label{3.3.38}\]

or

\[\frac{\boldsymbol{\eta}}{\mathrm{r}}=\boldsymbol{\xi}=\xi_0 \mathrm{e}^{\mathrm{i} \sigma t} \hat{\mathbf{r}}.\label{3.3.39}\]

Making use of our definition for \(\eta_\mathrm{i}\)'s, Equation \ref{3.3.36}, we have

\[\mathrm{\operatorname{const} .\left(\frac{x_i+x_j+x_k}{r}\right)=\xi=\xi_0 e^{i \sigma t}.}\label{3.3.40}\]

This relation can only be true if the pulsation in the three coordinates (\(\eta_\mathrm{i}\)) are in phase and of equal amplitude, and if

\[\text { const. }=\xi_0 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}}.\label{3.3.41}\]

Now, using the definition for the mass in a given volume in Cartesian coordinates and the value for the constant in Equation \ref{3.3.37}, we may re-write the variation of the magnetic energy as follows:

\[\delta \mathscr{M}=-\xi_0 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}}\left(\frac{1}{8 \pi} \int_0^{\mathrm{M}} \frac{\mathrm{H}^2}{\rho} \mathrm{dm}(\mathrm{r})\right).\label{3.3.42}\]

Making use of Equation \ref{3.3.28} we may rewrite the variation in terms of the total magnetic energy of the equilibrium configuration

\[\delta \mathscr{M}=-\xi_0 \mathrm{e}^{\mathrm{i} \sigma t} \mathscr{M}_0.\label{3.3.}43\]

Thus, we have obtained an expression for the last variation required to evaluate the variational form of the virial theorem (Equations 3.3.7). We may, therefore, substitute Equations Equation \ref{3.3.16}, Equation \ref{3.3.27}, and Equation \ref{3.3.43}, into Equations Equation \ref{3.3.7}, and obtain

\[\frac{1}{2} \frac{\mathrm{d}^2}{\mathrm{dt}^2}\left(2 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \xi_0 \mathrm{I}_0\right)=-2 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \xi_0 \omega_0 \mathscr{L}_0+3(\gamma-1)\left(\xi_0 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}}\right)\left(\omega_0 \mathscr{L}_0+\Omega_0+\mathscr{M}_0\right)-\xi_0 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \Omega_0-\xi_0 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \mathscr{M}_0.\label{3.3.44}\]

Simplifying Equation \ref{3.3.44}, we find that

\[\sigma^2=\frac{-(3 \gamma-4)\left(\Omega_0+\mathscr{M}_0\right)+(5-3 \gamma) \omega_0 \mathscr{L}_0}{\mathrm{I}_0} .\label{3.3.45}\]

Although we have made some strict assumptions in deriving Equation \ref{3.3.45}, one should not feel that they are all of paramount importance. The assumption of constancy of \(\xi_\mathrm{o}\) and \(\gamma\) were only made so that the resultant integrals could be integrated in terms of the original parameters. If necessary, these assumptions may be omitted and an integral expression similar to Equation \ref{3.2.18} may be derived. However, the work required to obtain this expression is non-trivial and in order for it to be useful one must have a detailed model in mind. One must also know the detailed geometry of the magnetic fields and of the star in order to evaluate the integrals that result. Also, to study the behavior of \(\sigma^2\), a great deal of numerical work will be required. F\)(\r purposes of studying the effects of changes in \(\gamma\), \(\Omega_\mathrm{o}\), \(\mathrm{M}_\mathrm{o}\), \(\omega_0\) and \(\mathscr{L}_0\), Equation \ref{3.3.43} will be quite adequate and is much easier to handle. Equation \ref{3.3.45} contains many aspects which one may check for 'reasonableness'. If we let \(\eta_\mathrm{o}\) and be zero, then Equation \ref{3.3.45} becomes identical to the previously derived Equation \ref{3.2.19}. Letting only \(\omega_0\) be zero we obtain an expression identical to one arrived at by Chandrasekhar and Limber⁵ (1954). If \(\omega_0\) is non-zero, while \(\mathscr{M}_\mathrm{o}\) is zero, the expression is that of Ledouxl (1945). It should not be surprising to find the magnetic energy entering in an additive manner to the gravitational energy. Both are potential energies and since the basic Equations are scalar in nature, we should expect the final result to merely 'modify' the gravitational energy. However, the rotational energy is kinetic in nature and hence would not enter into the final result in the same manner as the magnetic energy. Let us briefly investigate the effect upon \(\sigma^2\), and hence on the pulsational period of the presence of magnetic and rotational energy. Since,

\[T=2 \pi / \sigma,\label{3.3.46}\]

an increase in σ indicates a decrease in the period and vice-versa. Now, since \(\gamma>4/3\) and the gravitational potential energy is defined as being negative, the first term in the numerator of Equation \ref{3.3.45} will be positive only if

\[||\Omega_0|>\mathscr{M}_0.\label{3.3.47}\]

Thus, the introduction of magnetic fields only serves to reduce \(\sigma^2\) and thereby lengthen the period of pulsation. However, the addition of rotational energy \((\omega_0\mathscr{L}_0)\) will tend to increase \(\sigma^2\) as long as \(\gamma>5/3\). When \(\gamma=5/3\), the introduction of rotation has no effect on the period of pulsation. If \(\gamma>5/3\), the influence of rotation is similar to that of magnetic energy.