3N: Notes to Chapter 3

3.1

Remember the defining expression for the gravitational potential energy is

\[\Omega=-\mathrm{G} \int_0^{\mathrm{M}} \frac{\mathrm{m}(\mathrm{r}) \mathrm{dm}(\mathrm{r})}{\mathrm{r}}\label{N3.1.1}\]

We may again use the fact that the variation of \(m(r)\) and \(dm(r)\) are both zero, to obtain

\[\delta \Omega=-\mathrm{G} \int_0^{\mathrm{M}} \frac{\delta \mathrm{r}}{\mathrm{r}^2} \mathrm{m}(\mathrm{r}) \mathrm{dm}(\mathrm{r}),\label{N3.1.2}\]

which can be written as an energy integral by noting from Equation \ref{N3.1.1} that

\[\frac{\mathrm{d} \Omega}{\mathrm{dm}(\mathrm{r})}=-\frac{\mathrm{Gm}(\mathrm{r})}{\mathrm{r}}.\label{N3.1.3}\]

Evaluating the above expression at \(\mathrm{r=r_o}\) and using the result in Equation \ref{3.2.11} to first order accuracy we get

\[\delta \Omega=-\int_0^{\Omega_0} \frac{\delta \mathrm{r}}{\mathrm{r}_0} \mathrm{d} \Omega.\label{N3.1.4}\]

3.2

We may write the total energy as the sum of two energies \(\mathcal{J_1}\) and \(\mathcal{J_2}\), where \(\mathcal{J_1}\) is the kinetic energy due to the mass motion of the gas arising from the pulsations themselves. Now the total kinetic energy of mass motion is given by

\[\mathcal{J}_1=\frac{1}{2} \int_0^{\mathrm{R}} 4 \pi \mathrm{r}^2 \rho\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)^2 \mathrm{dr}=\frac{1}{2} \int_0^{\mathrm{M}}\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)^2 \mathrm{dm}(\mathrm{r}).\label{N3.2.1}\]

Thus, the variation of \(\mathcal{J_1}\)

\[\delta \mathcal{J}_1=\frac{1}{2} \int_0^{\mathrm{M}}\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)\left(\frac{\mathrm{d} \delta \mathrm{r}}{\mathrm{dt}}\right) \mathrm{dm}(\mathrm{r}).\label{N3.2.2}\]

However, from the definition of δr we see that

\[\mathrm{r=r_o+\delta r}.\label{N3.2.3}\]

Since the equilibrium point r₀ cannot vary with time by definition, Equation \ref{N3.2.2} becomes

\[\delta \mathcal{J}_1=\frac{1}{2} \int_0^{\mathrm{M}}\left(\frac{\mathrm{d}(\delta \mathrm{r})}{\mathrm{dt}}\right)^2 \mathrm{dm}(\mathrm{r}).\label{N3.2.4}\]

The largest term in the integral of Equation \ref{N3.2.4} is second order in \(δr\) and may be neglected with respect to the first order terms of Equation \ref{3.2.13}, and Equation \ref{3.2.6}. Thus, to first order we have

\[\delta \mathrm{T}=\delta \mathcal{J}_1+\delta \mathcal{J}_2 \cong \delta \mathcal{J}_2.\label{N3.2.5}\]

Now consider the kinetic energy of a small volume of an ideal gas.

\[\mathrm{d} \mathcal{J}_2=\frac{3}{2} \mathrm{Nk} T \mathrm{dV}.\label{N3.2.6}\]

However, the gas pressure is given by \(\mathrm{P}_{\mathrm{g}}=\mathrm{Nk} T\) and \(\mathrm{dm(r)=\rho dV}\). Therefore,

\[\mathrm{d} \mathcal{J}_2=\frac{3}{2} \frac{\mathrm{P}_{\mathrm{g}}}{\rho} \mathrm{dm}(\mathrm{r}).\label{N3.2.7}\]

Thus, twice the total kinetic energy of the gas sphere arising from thermal sources is

\[2 \mathcal{J}_2=3 \int_0^{\mathrm{M}} \frac{\mathrm{P}_{\mathrm{g}}}{\rho} \mathrm{dm}(\mathrm{r}).\label{N3.2.8}\]

Now neglecting radiation pressure, so that the total pressure is equal to the gas pressure, and remembering that the variation of dm(r) is zero, we have

\[2 \delta \mathrm{T}=2 \delta \mathcal{J}_2=3 \int_0^{\mathrm{M}} \delta\left(\frac{\mathrm{P}_{\mathrm{g}}}{\rho}\right) \mathrm{dm}(\mathrm{r}).\label{N3.2.9}\]

We shall now assume that the pulsations are adiabatic so that

\[\frac{\delta \mathrm{P}}{\mathrm{P}}=\gamma \frac{\delta \rho}{\rho},\label{N3.2.10}\]

where \(\gamma\) is the ratio of specific heats \(\mathrm{c_p/c_v}\). Now

\[\delta\left(\frac{\mathrm{P}}{\rho}\right)=\frac{\rho \delta \mathrm{P}-\mathrm{P} \delta \rho}{\rho^2}=\frac{(\delta \mathrm{P} / \mathrm{P}) \rho-\delta \rho}{\left(\rho^2 / \mathrm{P}\right)}=\frac{\mathrm{P}}{\rho^2\left[(\gamma-1) \delta \rho_0\right]}.\label{N3.2.11}\]

Therefore, again evaluating at the equilibrium position, substituting into Equation \ref{N3.2.9}, And keeping only terms up to first order, we have

\[2 \delta \mathrm{T} \cong 3 \int_0^{\mathrm{M}} \frac{\mathrm{P}_0}{\rho_0}(\gamma-1) \frac{\delta \rho}{\rho_0} \mathrm{dm}(\mathrm{r}).\label{N3.2.12}\]

3.3

In out attempt to obtain an expression for \(\delta \rho / \rho\) and thereby determining the variation energy, we shall invoke the following argument. From the conservation of mass, we have

\[\delta \mathrm{m}\left(\mathrm{r}^{\prime}\right)=0=\int_0^{\mathrm{r}^{\prime}} 4 \pi \mathrm{r}^2 \rho \mathrm{dr}=\int_0^{\mathrm{r}^{\prime}} 4 \pi(2 \mathrm{r} \delta \mathrm{r}) \rho \mathrm{dr}+\int_0^{\mathrm{r}^{\prime}} 4 \pi \mathrm{r}^2 \delta \rho \mathrm{dr}+\int_0^{\mathrm{r}^{\prime}} 4 \pi \mathrm{r}^2 \rho \mathrm{d}(\delta \mathrm{r}).\label{N3.3.1}\]

Rewriting Equation \ref{3.2.28} and evaluating at \(\mathrm{r=r_0}\), we have

\[\mathrm{\int_0^{r^{\prime}} r_0^2 \delta \rho d r_0=-\int_0^{r^{\prime}} \rho_0 r_0^2\left[\frac{2 \delta r d r_0}{r_0}+d(\delta r)\right]}.\label{N3.3.2}\]

From the definition of \(\xi\) (Equation 3.2.12), we have

\[\frac{\mathrm{d} \xi}{\mathrm{dr}_0}=\frac{\mathrm{r}_0 \mathrm{d}(\delta \mathrm{r}) / \mathrm{dr}-\mathrm{dr}}{\mathrm{r}_0^2} \text { or, } \mathrm{d} \xi=\frac{\mathrm{d}(\delta \mathrm{r})}{\mathrm{r}_0}-\frac{\delta \mathrm{rdr}_0}{\mathrm{r}_0^2}.\label{N3.3.3}\]

Eliminating \(\mathrm{d(\delta r)}\) from Equation \ref{N3.3.2) with the aid of Equation \ref{N3.3.3}, we have again to the first

\[\int_0^{\mathrm{r}^{\prime}} \mathrm{r}_0^2 \delta \rho \mathrm{dr}_0=-\int_0^{\mathrm{r}^{\prime}} \rho_0 \mathrm{r}_0^2\left[3+\mathrm{r}_0 \frac{\mathrm{d} \xi}{\mathrm{dr}_0}\right] \mathrm{dr}_0.\label{N3.3.4}\]

Equation \ref{N3.3.4} must hold for all values of \(r'\). This can only be true if the integrands equal.

Thus,

\[\frac{\delta \rho}{\rho_0}=-\left(3 \xi+\mathrm{r}_0 \frac{\mathrm{d} \xi}{\mathrm{dr}_0}\right).\label{N3.3.5}\]

3.4

In an attempt to simplify Equation \ref{3.2.15}, let us consider the second integral.

\[3 \int_0^{\mathrm{M}} \frac{\mathrm{P}_0}{\rho_0}(\gamma-1) \mathrm{r}_0 \frac{\mathrm{d} \xi_0}{\mathrm{dr}_0} \mathrm{e}^{\mathrm{i} \sigma t} \mathrm{dm}(\mathrm{r})=3 \int_0^{\mathrm{R}} \frac{\mathrm{P}_0}{\rho_0}(\gamma-1) \mathrm{r}_0 \frac{\mathrm{d} \xi_0}{\mathrm{dr}_0} \mathrm{e}^{\mathrm{i} \sigma t} 4 \pi \mathrm{r}_0^2 \mathrm{dr}_0.\label{N3.4.1}\]

Integrating the right hand side by parts we obtain

\[12 \pi \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \int_0^{\mathrm{R}_0}\left(\frac{\mathrm{d} \xi_0}{\mathrm{dr_0}}\right) \frac{\mathrm{P}_0}{\rho_0}(\gamma-1) \mathrm{r}_0^3 \mathrm{dr} _0=\left.12 \pi \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \mathrm{P}_0(\gamma-1) \mathrm{r}_0^3 \xi_0\right|_0 ^{\mathrm{R}}-12 \pi \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \int_0^{\mathrm{R}} \xi_0 \frac{\mathrm{d}}{\mathrm{dr}}\left[\mathrm{P}_0(\gamma-1) \mathrm{r}_0^3\right] \mathrm{dr}_0.\label{N3.4.2}\]

At this point we shall impose the boundary conditions that \(\mathrm{P}_{\mathrm{o}} \rightarrow 0\) at \(\mathrm{r_o=R}\) and \(\mathrm{\xi_o}\rightarrow0\) at \(\mathrm{r_o=0}\). The first of these conditions is the familiar condition of stellar structure which essentially defines the surface of the gas sphere. The second condition is required by assuming that the radial pulsation is a continuous function. With these conditions, the integrated part of Equation \ref{N3.4.2) vanishes and the remaining integral becomes:

\[\int_0^{\mathrm{R}} \xi_0 \frac{\mathrm{d}}{\mathrm{dr}}\left[\mathrm{P}_0(\gamma-1) \mathrm{r}_0^3\right] \mathrm{dr}_0=\int_0^{\mathrm{R}} \mathrm{P}_0 \xi_0 \mathrm{r}_0^3+\int_0^{\mathrm{R}}(\gamma-1) \xi_0 \frac{\mathrm{dP}_0}{\mathrm{dr}_0} \mathrm{r}_0^3 \mathrm{dr}_0+3 \int_0^{\mathrm{R}} \mathrm{P}_0(\gamma-1) \xi_0 \mathrm{r}_0^2 \mathrm{dr}_0.\label{N3.4.3}\]

Conservation of momentum (i.e., hydrostatic equilibrium) requires that

\[\frac{\mathrm{dP}_0}{\mathrm{dr}_0}=-\frac{\mathrm{Gm}\left(\mathrm{r}_0\right) \rho}{\mathrm{r}_0^2}.\label{N3.4.4}\]

Therefore,

\[4 \pi \mathrm{r}_0^3 \frac{\mathrm{dP}_0}{\mathrm{dr}_0}=-\frac{4 \pi \mathrm{Gr}_0^2 \mathrm{m}\left(\mathrm{r}_0\right) \rho}{\mathrm{r}_0}=-\frac{\mathrm{Gm}\left(\mathrm{r}_0\right)}{\mathrm{r}_0} \frac{\mathrm{dm}\left(\mathrm{r}_0\right)}{\mathrm{dr}_0}=\frac{\mathrm{d} \Omega_0}{\mathrm{dr}_0}.\label{N3.4.5}\]

Making use of the second form of Equation \ref{N3.4.5} to simplify the second integral in Equation \ref{N3.4.3}, and the definition of dm(r) to simplify the other two integrals in Equation \ref{N3.4.3}, we obtain the following expression for the variation of the kinetic energy [i.e. Equation \ref{3.2.15}].

\[2 \delta \mathrm{T} \cong-9 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \int_0^{\mathrm{M}} \frac{\mathrm{P}_0 \xi_0}{\rho_0}(\gamma-1) \mathrm{dm}\left(\mathrm{r}_0\right)+3 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \int_0^{\mathrm{M}} \frac{\mathrm{P}_0 \xi_0 \mathrm{r}_0}{\rho_0} \frac{\mathrm{d} \gamma}{\mathrm{dr_0}} \mathrm{dm}\left(\mathrm{r}_0\right)+3 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \int_0^{\Omega_0} \xi_0(\gamma-1) \mathrm{d} \Omega_0
+9 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \int_0^{\mathrm{M}} \frac{\mathrm{P}_0 \xi_0}{\rho_0}(\gamma-1) \mathrm{dm}\left(\mathrm{r}_0\right).\label{N3.4.6}\]

The above expression is obtained by making use of the previously mentioned definition and substituting it into Equation \ref{N3.4.3}, then into Equation \ref{N3.4.2}, and finally into Equation \ref{3.2.15}. Since the first and last integrals are identical except for the difference in sign, they vanish from the expression and

\[2 \delta \mathrm{T} \cong 3 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \int_0^{\mathrm{M}} \frac{\mathrm{P}_0 \xi_0 \mathrm{r}_0}{\rho_0} \frac{\mathrm{d} \gamma}{\mathrm{dr}_0} \mathrm{dm}\left(\mathrm{r}_0\right)+3 \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} \int_0^{\Omega_0} \xi_0(\gamma-1) \mathrm{d} \Omega_0.\label{N3.4.7}\]

3.5

Let us define

\[\eta_{\mathrm{i}} \equiv \delta \mathrm{x}_{\mathrm{i}}.\label{N3.5.1}\]

Now the definition of the mass within a given volume in Cartesian coordinates becomes

\[\mathrm{m}(\mathrm{V})=\iiint\rho \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3.\label{N3.5.2}\]

The conservation of mass requires, as it did in section 2, that the variation of the mass is zero. Thus taking the variations of Equation \ref{N3.5.2}, we obtain

\[\delta \mathrm{m}(\mathrm{V})=\iiint \delta \rho \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3+\iiint \rho \mathrm{d}\left(\delta \mathrm{x}_1\right) \mathrm{dx}_2 \mathrm{dx}_3+\iiint \rho \mathrm{dx}_1 \mathrm{d}\left(\delta \mathrm{x}_2\right) \mathrm{dx}_3 \iiint \rho \mathrm{d} \mathrm{x}_1 \mathrm{dx}_2 \mathrm{d}\left(\delta \mathrm{x}_3\right).\label{N3.5.3}\]

Rewriting the last three integrals, we have

\[\iiint \delta \rho \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3=-\iiint \rho \sum_{\mathrm{i}=1}^3 \frac{\mathrm{d}\left(\delta \mathrm{x}_{\mathrm{i}}\right)}{\mathrm{dx}_{\mathrm{i}}} \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3.\label{N3.5.4}\]

Since \(\mathrm{\delta x_i=\eta_i}\), we have by the chain rule

\[\frac{\mathrm{d}\left(\delta \mathrm{x}_{\mathrm{i}}\right)}{\mathrm{dx}_{\mathrm{i}}}=\frac{\mathrm{d} \eta_{\mathrm{i}}}{\mathrm{dx}_{\mathrm{i}}}=\sum_{\mathrm{j}=1}^3 \frac{\mathrm{dx}_{\mathrm{i}}}{\mathrm{dx}_{\mathrm{j}}} \frac{\partial \eta_{\mathrm{i}}}{\partial \mathrm{x}_{\mathrm{j}}},\label{N3.5.5}\]

However, since the x_i's are linearly independent and the η_i's are just the variation of these coordinates, not only is the second term in the product [under the summation sign of Equation \ref{N3.5.5)] zero if i= j but, so is the first term.

Thus as might be expected from the orthogonality of the x_i's we have

\[\frac{\mathrm{d} \eta_{\mathrm{i}}}{\mathrm{dx}_{\mathrm{i}}}=\frac{\partial \eta_{\mathrm{i}}}{\partial \mathrm{x}_{\mathrm{i}}}.\label{N3.5.6}\]

Substitution of this into Equation \ref{N3.5.4), yields

\[\iiint \delta \rho \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3=-\iiint \rho \sum_{\mathrm{i}=1}^3 \frac{\partial \eta_{\mathrm{i}}}{\partial \mathrm{x}_{\mathrm{i}}} \mathrm{dx}_1 \mathrm{dx}_2 \mathrm{dx}_3.\label{N3.5.7}\]

Equation \ref{N3.5.7} must hold for any volume of integration. This can only be true if the integrands themselves are equal. Thus, we finally obtain

\[\frac{\delta \rho}{\rho}=-\sum_{\mathrm{i}=1}^3 \frac{\partial \eta_{\mathrm{i}}}{\partial \mathrm{x}_{\mathrm{i}}}.\label{N3.5.8}\]

3.6

Suppose a displacement n takes place with the slow continuous movement so that

\[\boldsymbol{u}=\frac{\mathrm{d} \boldsymbol{n}}{\mathrm{dt}}.\label{N3.6.1}\]

Then, if the electrical conductivity of the medium is infinite, the time-variation of the electric field is

\[\Delta \mathbf{E}=-\mathbf{u} \times \mathbf{H}.\label{N3.6.2}\]

However, Maxwell's Equations for an infinitely conducting medium require that

\[\nabla \times(\Delta \mathbf{E})=-\frac{\partial}{\partial \mathrm{t}} \Delta \mathbf{H}.\label{N3.6.3}\]

Combining Equation \ref{N3.6.2} and Equation \ref{N3.6.3}, we have

\[-\frac{\partial(\Delta \mathbf{H})}{\partial \mathrm{t}}=\nabla \times(-\boldsymbol{u} \times \mathbf{H}).\label{N3.6.4}\]

The integral form of Equation \ref{N3.6.4} is just

\[\Delta \mathbf{H}=\nabla \times(\boldsymbol{n} \times \mathbf{H}).\label{N3.6.5}\]

But, from the definition of the time and space variations and the Equation relating to the total time derivatives, we know that

\[\Delta \mathbf{H}=\frac{\partial \mathbf{H}}{\partial \mathrm{t}} \mathrm{dt},\label{N3.6.6}\]

and

\[\frac{\partial}{\partial \mathrm{t}}=\frac{\mathrm{d}}{\mathrm{dt}}-\mathbf{v} \cdot \nabla,\label{N3.6.7}\]

which results in

\[\Delta \mathbf{H}=\left(\frac{\mathrm{d} \mathbf{H}}{\mathrm{dt}}-\sum_{\mathrm{j}} \frac{\mathrm{d} \mathbf{x}_{\mathrm{j}}}{\mathrm{dt}} \cdot \nabla \mathbf{H}\right) \mathrm{dt}.\label{N3.6.8}\]

However,

\[\frac{\mathrm{d}}{\mathrm{dt}}=\sum_{\mathrm{i}} \frac{\partial}{\partial \mathrm{x}_{\mathrm{i}}} \frac{\mathrm{dx}_{\mathrm{i}}}{\mathrm{dt}},\label{N3.6.9}\]

while definition of the space variation \(\delta\mathbf{H}\) is

\[\delta \mathbf{H}=\sum_{\mathrm{i}} \frac{\partial \mathbf{H}}{\partial \mathrm{x}} \mathrm{dt} \cdot \nabla \mathbf{H}.\label{N3.6.10}\]

Combining Equations \ref{N3.6.10}, \ref{N3.6.9}, and Equation \ref{N3.6.8}, we have

\[\Delta \mathbf{H}=\delta \mathbf{H}-\sum_{\mathrm{i}} \frac{\mathrm{d} \mathbf{x}_{\mathrm{i}}}{\mathrm{dt}} \mathrm{dt} \cdot \nabla \mathbf{H}.\label{N3.6.11}\]

Noting that the variation of a linearly independent quantity may be interpreted as the total differential of the quantity

\[\mathrm{dx}_{\mathrm{i}}=\sum_{\mathrm{i}} \frac{\partial \mathrm{x}_{\mathrm{i}}}{\partial \mathrm{x}_{\mathrm{j}}} \mathrm{dx}_{\mathrm{j}}=\delta \mathrm{x}_{\mathrm{i}},\label{N3.6.12}\]

we obtain

\[\Delta \mathbf{H}=\delta \mathbf{H}-\boldsymbol{n} \cdot \nabla \mathbf{H}.\label{N3.6.13}\]

Combining Equation \ref{N3.6.13} and Equation \ref{N3.6.S}, we finally obtain an Equation for the variation of the magnetic field \((\delta\mathbf{H})\).

\[\delta \mathbf{H}=\nabla \times(\boldsymbol{n} \times \mathbf{H})-(\boldsymbol{n} \cdot \nabla) \mathbf{H}.\label{N3.6.14}\]

Now the curl of a cross-product may be written as

\[\nabla \times(\boldsymbol{n} \times \mathbf{H})=(\mathbf{H} \cdot \nabla) \boldsymbol{n}-\mathbf{H}(\nabla \cdot \boldsymbol{n})-(\boldsymbol{n} \cdot \nabla) \mathbf{H}+\boldsymbol{n}(\nabla \cdot \mathbf{H}).\label{N3.6.15}\]

Since the divergence of H is always zero, the last term vanishes. Combining this expression for the curl of a cross-product with Equation \ref{N3.6.14} yields

\[\delta \mathbf{H}=(\mathbf{H} \cdot \nabla) \boldsymbol{n}-\mathbf{H}(\nabla \cdot \boldsymbol{n})-(\boldsymbol{n} \cdot \nabla) \mathbf{H}+(\boldsymbol{n} \cdot \nabla) \mathbf{H}.\label{N3.6.16}\]

The last two terms are identical except for opposite signs and thus cancel out. The remaining expression may be written in component form as follows:

\[\mathrm{\delta H_i=\sum_j\left(H_j \frac{\partial \eta_i}{\partial x_j}-H_i \frac{\partial \eta_j}{\partial x_i}\right)}.\label{N3.6.17}\]

3.7

In the section 3.2 we went through an extensive argument (see Note 3.6, Equation \ref{N3.6.16}, to show that

\[\delta \mathbf{H}=(\mathbf{H} \cdot \nabla) \delta \mathbf{r}-\mathbf{H}(\nabla \cdot \delta \mathbf{r}),\label{N3.7.1}\]

where \(\delta \mathbf{r}\) plays the role of \(\eta\) in that discussion. The easiest way to evaluate the first term is in Cartesian coordinates, remembering that ξ₀ is constant. Then,

\[(\mathbf{H} \cdot \nabla) \delta \mathbf{r}=\sum_{\mathrm{i}} \mathrm{H}_{\mathrm{i}} \frac{\partial\left(\xi \mathrm{x}_{\mathrm{i}}\right)}{\partial \mathrm{x}_{\mathrm{i}}}=\xi \mathbf{H}.\label{N3.7.2}\]

The second term can be evaluated the same way, so that

\[\mathbf{H}(\nabla \cdot \delta \mathbf{r})=\mathrm{H}_{\mathrm{j}} \sum_{\mathrm{i}} \frac{\partial\left(\xi \mathrm{x}_{\mathrm{i}}\right)}{\partial \mathrm{x}_{\mathrm{i}}}=3 \xi \mathbf{H}.\label{N3.7.3}\]

So, the variation of the magnetic field has taken the particularly simple form

\[\delta \mathbf{H}=-2 \xi \mathbf{H}_0.\label{N3.7.4}\]