Lecture 15 The Energy Principle

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Lecture 15
The Energy Principle

Overview

The energy principle is the organizing theorem of ideal-MHD stability.
1.
Linearize the ideal equations about a static equilibrium satisfying the momentum balance in Eq. (1.8).
2.
Use the frozen-in constraint, Eq. (4.8), to express the perturbed magnetic field entirely in terms of a fluid displacement \(\vect {\xi }\).
3.
The equilibrium is linearly stable in ideal MHD if the quadratic functional \(\delta W[\vect {\xi }]\) is positive for every admissible displacement.

The importance of this lecture is hard to overstate. The earlier equilibrium lectures asked whether magnetic forces, pressure gradients, and geometry can balance. The energy principle asks the next question: if they do balance, does the plasma return when nudged, or does it find a lower-energy distortion and run away? The remarkable answer is that, for static ideal-MHD equilibria, the entire linear stability problem can be recast as a variational problem.

Historical Perspective

Few ideas shaped fusion stability theory more strongly than the 1958 papers of Bernstein, Frieman, Kruskal, and Kulsrud, together with Kruskal and Oberman, which showed that ideal-MHD stability can be decided from the sign of a quadratic energy functional Bernstein et al. (1958); Kruskal and Oberman (1958). Newcomb soon turned this into a field-line-based framework for diffuse pinches and interchange physics Newcomb (1960, 1961), and Frieman–Rotenberg later generalized the formalism to equilibria with flow Frieman and Rotenberg (1960). It is one of the classic moments in plasma theory: a messy set of coupled PDEs became a geometrically transparent statement about whether a displacement raises or lowers the total potential energy.

This lecture stays with static equilibria. That is the setting in which the force operator is self-adjoint and the ideal-MHD stability question becomes especially clean. The price of that cleanliness is that dissipation, kinetic resonances, and equilibrium flow are all set aside for the moment.

15.1 Displacement Formulation of Linearized Ideal MHD

We linearize about a time-independent equilibrium with background fields \(\rho (\vect {x})\), \(p(\vect {x})\), \(\B (\vect {x})\), and \(\J (\vect {x})\) satisfying Eq. (1.8). Introduce a Lagrangian displacement

\vect {\xi }(\vect {x},t) = \xi _\parallel \,\vect {b} + \vect {\xi }_\perp , \qquad \vect {b} \equiv \frac {\B }{B}. \tag{15.1}

Only the perpendicular part bends field lines. The parallel part matters for compressibility and for the bookkeeping of admissible perturbations, but it does not directly perturb the magnetic topology at first order.

Perturbed velocity. The Eulerian velocity perturbation is simply

\uvec _1 = \pp {\vect {\xi }}{t}. \tag{15.2}

Perturbed magnetic field. Starting from the ideal Ohm law, Eq. (4.8), and Faraday’s law, Eq. (1.10), we have \[ \pp {\B _1}{t} = \curl (\uvec _1\times \B ). \] Using (15.2) and integrating once in time gives

\vect {Q} \equiv \B _1 = \curl (\vect {\xi }\times \B ) = \curl (\vect {\xi }_\perp \times \B ), \tag{15.3}

because \(\xi _\parallel \vect {b}\times \B =0\).

Now expand the curl using \[ \curl (\vect {A}\times \vect {C}) = (\vect {C}\cdot \grad )\vect {A} - (\vect {A}\cdot \grad )\vect {C} + \vect {A}\,\divergence \vect {C} - \vect {C}\,\divergence \vect {A}, \] with \(\vect {A}=\vect {\xi }_\perp \) and \(\vect {C}=\B \). Since \(\divergence \B =0\),

\boxed { \vect {Q} = (\B \cdot \grad )\vect {\xi }_\perp - (\vect {\xi }_\perp \cdot \grad )\B - \B \,\divergence \vect {\xi }_\perp . } \tag{15.4}

This is the basic magnetic perturbation formula used throughout ideal-MHD stability theory.

Perturbed pressure. For adiabatic perturbations, the conservative pressure law derived earlier in Eq. (3.11) gives

p_1 = -\vect {\xi }\cdot \grad p -\gamma p\,\divergence \vect {\xi }. \tag{15.5}

The first term is advection of the equilibrium pressure profile. The second is ordinary adiabatic compression.

Perturbed density. Starting from continuity, Eq. (1.7), \[ \pp {\rho _1}{t} + \divergence (\rho \uvec _1)=0, \] substitute \(\uvec _1=\pp {\vect {\xi }}{t}\) and integrate once in time:

\rho _1 = -\divergence (\rho \vect {\xi }). \tag{15.6}

Expanding the divergence gives

\begin{aligned}\rho _1 &= -\vect {\xi }\cdot \grad \rho -\rho \,\divergence \vect {\xi }.\end{aligned} \tag{15.7}

Again the structure is simple: advection of the equilibrium density plus compressional change.

15.2 The Linearized Force Operator

Start from the momentum equation in Eq. (1.8), written here with an optional gravitational potential \(\Phi _g\),

\rho \pp {\uvec }{t} = \J \times \B - \grad p - \rho \grad \Phi _g. \tag{15.8}

Linearization about the static equilibrium gives

\rho \pp {^2\vect {\xi }}{t^2} = \underbrace {\J _1\times \B }_{(1)} + \underbrace {\J \times \B _1}_{(2)} - \underbrace {\grad p_1}_{(3)} - \underbrace {\rho _1\grad \Phi _g}_{(4)}. \tag{15.9}

The perturbed current is

\J _1 = \frac {1}{\muo }\curl \B _1 = \frac {1}{\muo }\curl \vect {Q}. \tag{15.10}

Substituting (15.3), (15.5), and (15.7) into (15.9) yields

\rho \pp {^2\vect {\xi }}{t^2} = \vect {F}(\vect {\xi }), \tag{15.11}

with

\boxed { \vect {F}(\vect {\xi }) = \frac {1}{\muo }(\curl \vect {Q})\times \B + \J \times \vect {Q} + \grad \bigl (\vect {\xi }\cdot \grad p + \gamma p\,\divergence \vect {\xi }\bigr ) + \divergence (\rho \vect {\xi })\,\grad \Phi _g. } \tag{15.12}

This operator is linear in \(\vect {\xi }\). For static ideal-MHD equilibria it is also self-adjoint under the usual ideal boundary conditions, and that is the key structural fact behind the entire lecture.

Caution

What is and is not being assumed. The clean variational structure in this lecture belongs to static ideal MHD. Resistive MHD, kinetic theory, or equilibria with flow require additional terms and, in general, destroy the simple Hermitian structure used below. The Frieman–Rotenberg generalization is the right extension for stationary equilibria with flow Frieman and Rotenberg (1960).

15.3 From the Force Operator to a Potential-Energy Functional

Assume a normal-mode time dependence \[ \vect {\xi }(\vect {x},t)=\vect {\xi }(\vect {x})e^{-i\omega t}. \] Then Eq. (15.11) becomes an eigenvalue equation of the form

-\omega ^2\rho \vect {\xi } = \vect {F}(\vect {\xi }), \tag{15.13}

which can be solved in its own right. This suggests defining the bilinear "energy" form

\delta W(\vect {\eta },\vect {\xi }) \equiv -\frac 12\int _P \vect {\eta }^{*}\cdot \vect {F}(\vect {\xi })\,dV, \tag{15.14}

where \(P\) denotes the plasma region. The actual potential energy is the quadratic form

\delta W[\vect {\xi }]\equiv \delta W(\vect {\xi },\vect {\xi }). \tag{15.15}

Using (15.13),

\delta W[\vect {\xi }] = \frac 12\omega ^2\int _P \rho |\vect {\xi }|^2\,dV. \tag{15.16}

The kinetic energy of the mode is therefore

K = \frac 12\omega ^2\int _P \rho |\vect {\xi }|^2\,dV = \delta W[\vect {\xi }]. \tag{15.17}

Equivalently,

\boxed { \omega ^2 = \frac {2\,\delta W[\vect {\xi }]}{\displaystyle \int _P \rho |\vect {\xi }|^2\,dV}. } \tag{15.18}

This is the Rayleigh quotient for ideal-MHD stability.

The logic is now transparent. If \(\delta W[\vect {\xi }]>0\) for every admissible displacement, then every eigenvalue has \(\omega ^2>0\) and the equilibrium is linearly stable. If one can find even a single admissible trial function for which \(\delta W<0\), then the Rayleigh quotient is negative and the equilibrium is unstable.

15.4 General Expression for \(\delta W\)

To derive the explicit form, it is convenient to suppress complex conjugation temporarily and work with a real displacement. The Hermitian form used for Fourier amplitudes is recovered at the end by replacing one factor by its complex conjugate.

From (15.14),

2\delta W = -\int _P \vect {\xi }\cdot \vect {F}(\vect {\xi })\,dV. \tag{15.19}

Substituting (15.12) gives

\begin{aligned}2\delta W &= -\int _P dV\,\vect {\xi }\cdot \left [ \frac {1}{\muo }(\curl \vect {Q})\times \B + \J \times \vect {Q} +\grad \bigl (\vect {\xi }\cdot \grad p + \gamma p\,\divergence \vect {\xi }\bigr ) +\divergence (\rho \vect {\xi })\,\grad \Phi _g \right ].\end{aligned} \tag{15.20}

We now reduce this term by term.

Magnetic term. Use the scalar triple-product identity:

\begin{aligned}-\vect {\xi }\cdot \bigl [(\curl \vect {Q})\times \B \bigr ] &= (\curl \vect {Q})\cdot (\vect {\xi }\times \B ).\end{aligned}

Now apply \[ \divergence \bigl [\vect {Q}\times (\vect {\xi }\times \B )\bigr ] = (\vect {\xi }\times \B )\cdot \curl \vect {Q} - \vect {Q}\cdot \curl (\vect {\xi }\times \B ), \] and recall from (15.3) that \(\vect {Q}=\curl (\vect {\xi }\times \B )\). Therefore

\begin{aligned}-\vect {\xi }\cdot \bigl [(\curl \vect {Q})\times \B \bigr ] &= |\vect {Q}|^2 + \divergence \bigl [\vect {Q}\times (\vect {\xi }\times \B )\bigr ].\end{aligned} \tag{15.22}

So the magnetic piece contributes a positive-definite volume term plus a surface term.

Pressure term. Let \[ S\equiv \vect {\xi }\cdot \grad p + \gamma p\,\divergence \vect {\xi }, \qquad \text {so that} \qquad p_1=-S. \] Then

\begin{aligned}-\vect {\xi }\cdot \grad S &= -\divergence (S\vect {\xi }) + S\,\divergence \vect {\xi } \nonumber \\ &= -\divergence (S\vect {\xi }) + (\vect {\xi }\cdot \grad p)\,\divergence \vect {\xi } + \gamma p\,(\divergence \vect {\xi })^2.\end{aligned} \tag{15.23}

Again the bulk part separates into a mixed pressure-gradient term and a positive compression term.

Gravity term. Using (15.7),

\begin{aligned}-\vect {\xi }\cdot \Bigl [\divergence (\rho \vect {\xi })\,\grad \Phi _g\Bigr ] &= -(\vect {\xi }\cdot \grad \Phi _g)\,\divergence (\rho \vect {\xi }).\end{aligned} \tag{15.24}

This is the term that later produces the gravitational interchange and magnetic-buoyancy contributions. For most of the lecture we will set \(\Phi _g=0\) to recover the standard ideal-MHD form.

Collecting (15.22)–(15.24) gives

\begin{aligned}2\delta W &= \int _P dV\Bigg [ \frac {|\vect {Q}|^2}{\muo } - \vect {\xi }\cdot (\J \times \vect {Q}) + \gamma p\,(\divergence \vect {\xi })^2 + (\vect {\xi }\cdot \grad p)\,\divergence \vect {\xi } - (\vect {\xi }\cdot \grad \Phi _g)\,\divergence (\rho \vect {\xi }) \Bigg ] \nonumber \\ &\qquad + \int _S dS\,\hat {\vect {n}}\cdot \left [ \frac {1}{\muo }\vect {Q}\times (\vect {\xi }\times \B ) - \bigl (\vect {\xi }\cdot \grad p + \gamma p\,\divergence \vect {\xi }\bigr )\vect {\xi } \right ].\end{aligned} \tag{15.25}

Here \(S\) is the plasma boundary and \(\hat {\vect {n}}\) is the outward normal.

On the interface, \[ \hat {\vect {n}}\cdot \bigl [\vect {Q}\times (\vect {\xi }\times \B )\bigr ] = (\vect {Q}\cdot \B )\,\xi _n - (\vect {Q}\cdot \hat {\vect {n}})(\vect {\xi }\cdot \B ), \qquad \xi _n\equiv \vect {\xi }\cdot \hat {\vect {n}}. \] For an ideal plasma–vacuum interface, the perturbed normal magnetic field vanishes on the moving boundary, so \(\vect {Q}\cdot \hat {\vect {n}}=0\). The surface term therefore reduces to

2\delta W = \int _P dV\, (\cdots ) + \int _S dS\,\xi _n \left [ \frac {\vect {Q}\cdot \B }{\muo } - \vect {\xi }\cdot \grad p - \gamma p\,\divergence \vect {\xi } \right ]. \tag{15.26}

This is the general quadratic form before the usual geometric rearrangement.

15.5 The Intuitive Form of the Plasma, Surface, and Vacuum Terms

The expression above is correct but not yet very transparent. The standard “intuitive” form isolates the pieces associated with field-line bending, magnetic compression, plasma compression, current drive, and curvature. From here on we set \(\Phi _g=0\) and return to that physics later in the interchange lecture.

Write

\delta W = \delta W_P + \delta W_S + \delta W_V, \tag{15.27}

where \(\delta W_P\) is the plasma contribution, \(\delta W_S\) the moving-boundary term, and \(\delta W_V\) the vacuum magnetic energy.

Useful identities. Introduce the field-line curvature

\vect {\kappa } \equiv \vect {b}\cdot \grad \vect {b} = -\vect {b}\times \curl \vect {b}. \tag{15.28}

For a static isotropic equilibrium, the perpendicular force balance implies

\grad p = -\grad _\perp \!\left (\frac {B^2}{2\muo }\right ) + \frac {B^2}{\muo }\vect {\kappa }. \tag{15.29}

Decompose the perturbed field as

\vect {Q}=\vect {Q}_\perp + Q_\parallel \vect {b}, \tag{15.30}

with

\vect {Q}_\perp = (\B \cdot \grad )\vect {\xi }_\perp . \tag{15.31}

We now derive the standard formula for \(Q_\parallel \).

Starting from (15.4), take the component along \(\vect {b}\):

\begin{aligned}Q_\parallel &= \vect {b}\cdot \vect {Q} \nonumber \\ &= \vect {b}\cdot \curl (\vect {\xi }_\perp \times \B ) \nonumber \\ &= -B\,\divergence \vect {\xi }_\perp -\vect {\xi }_\perp \cdot \grad B -B\,\vect {\xi }_\perp \cdot \vect {\kappa }.\end{aligned} \tag{15.32}

Now use (15.29). Since \[ \vect {\xi }_\perp \cdot \grad p = -\vect {\xi }_\perp \cdot \grad \!\left (\frac {B^2}{2\muo }\right ) + \frac {B^2}{\muo }\vect {\xi }_\perp \cdot \vect {\kappa } = -\frac {B}{\muo }\vect {\xi }_\perp \cdot \grad B + \frac {B^2}{\muo }\vect {\xi }_\perp \cdot \vect {\kappa }, \] we obtain

\vect {\xi }_\perp \cdot \grad B = -\muo \,\frac {\vect {\xi }_\perp \cdot \grad p}{B} + B\,\vect {\xi }_\perp \cdot \vect {\kappa }. \tag{15.33}

Substituting (15.33) into (15.32) gives

\boxed { Q_\parallel = -B\left (\divergence \vect {\xi }_\perp + 2\,\vect {\kappa }\cdot \vect {\xi }_\perp \right ) + \muo \,\frac {\vect {\xi }_\perp \cdot \grad p}{B}. } \tag{15.34}

Current term. Decompose the equilibrium current into parallel and perpendicular pieces,

\J = J_\parallel \vect {b} + \J _\perp , \qquad \J _\perp \times \B = \grad p. \tag{15.35}

Thus

\J _\perp = \frac {\B \times \grad p}{B^2}. \tag{15.36}

Now expand

\begin{aligned}-\vect {\xi }_\perp \cdot (\J \times \vect {Q}) &= -\vect {\xi }_\perp \cdot \bigl [(J_\parallel \vect {b})\times \vect {Q}\bigr ] - \vect {\xi }_\perp \cdot (\J _\perp \times \vect {Q}) \nonumber \\ &= -\frac {J_\parallel }{B}(\vect {\xi }_\perp \times \B )\cdot \vect {Q}_\perp - \frac {Q_\parallel }{B}(\vect {\xi }_\perp \cdot \grad p).\end{aligned} \tag{15.37}

The first piece is the current-driven contribution that later becomes the kink term. The second will combine with the \(Q_\parallel ^2\) and pressure-gradient pieces.

Algebraic reduction of the compressional terms. Define

D\equiv \divergence \vect {\xi }_\perp + 2\,\vect {\kappa }\cdot \vect {\xi }_\perp , \qquad A\equiv \vect {\xi }_\perp \cdot \grad p. \tag{15.38}

Then (15.34) becomes

Q_\parallel = -BD + \muo \frac {A}{B}. \tag{15.39}

Now collect the pieces of (15.25) containing \(Q_\parallel \):

\begin{aligned}\frac {Q_\parallel ^2}{\muo } - \frac {Q_\parallel }{B}A + A\,\divergence \vect {\xi }_\perp &= \frac {1}{\muo }\left (-BD+\muo \frac {A}{B}\right )^2 - \left (-D+\muo \frac {A}{B^2}\right )A + A\,\divergence \vect {\xi }_\perp \nonumber \\ &= \frac {B^2}{\muo }D^2 -DA + A\,\divergence \vect {\xi }_\perp \nonumber \\ &= \frac {B^2}{\muo }\left (\divergence \vect {\xi }_\perp + 2\,\vect {\kappa }\cdot \vect {\xi }_\perp \right )^2 - 2(\vect {\xi }_\perp \cdot \grad p)(\vect {\xi }_\perp \cdot \vect {\kappa }).\end{aligned} \tag{15.40}

This is the key algebraic step. The mixed terms rearrange themselves into a positive-definite magnetic-compression piece plus the pressure–curvature coupling.

Plasma contribution. Restoring the Hermitian form for complex Fourier amplitudes, we obtain

\begin{aligned}\delta W_P = \frac 12\int _P dV\Bigg [ & \underbrace {\frac {|\vect {Q}_\perp |^2}{\muo }}_{\text {field-line bending}} + \underbrace {\frac {B^2}{\muo } \left |\divergence \vect {\xi }_\perp + 2\,\vect {\xi }_\perp \cdot \vect {\kappa }\right |^2}_{\text {magnetic compression}} + \underbrace {\gamma p\,|\divergence \vect {\xi }|^2}_{\text {plasma compression}} \nonumber \\[0.5em] & - \underbrace {2\,(\vect {\xi }_\perp \cdot \grad p) (\vect {\xi }_\perp ^{*}\cdot \vect {\kappa })}_{\text {pressure--curvature coupling}} - \underbrace {\frac {J_\parallel }{B} (\vect {\xi }_\perp ^{*}\times \B )\cdot \vect {Q}_\perp }_{\text {current-driven term}} \Bigg ].\end{aligned} \tag{15.41}

This is the form one usually stares at when trying to guess what kind of instability is possible.

Surface contribution. Now let the plasma region \(P\) be surrounded by a vacuum region \(V_{\rm vac}\), with interface \(S\) and outward normal \(\hat {\vect {n}}\) pointing from plasma into vacuum. Define \(\xi _n\equiv \vect {\xi }_\perp \cdot \hat {\vect {n}}\). The plasma surface term from (15.26) is

2\delta W_{S,{\rm pl}} = \int _S dS\,\xi _n \left ( p_{1,{\rm pl}} + \frac {\B \cdot \vect {Q}_{\rm pl}}{\muo } \right ). \tag{15.42}

In the vacuum region there is no pressure or current, so the corresponding magnetic contribution is

2\delta W_{\rm vac} = \int _{V_{\rm vac}} dV\,\frac {|\B _1|^2}{\muo } - \int _S dS\,\xi _n\,\frac {\B \cdot \vect {Q}_{\rm vac}}{\muo }. \tag{15.43}

The minus sign appears because the outward normal of the vacuum region is \(-\hat {\vect {n}}\) on the plasma boundary. The key boundary condition is continuity of the Lagrangian total pressure across the moving interface:

\left [\left [ \Delta \!\left (p+\frac {B^2}{2\muo }\right ) \right ]\right ]_{\rm plasma}^{\rm vacuum} = 0, \qquad \Delta f = f_1 + \xi _n\,\hat {\vect {n}}\cdot \grad f. \tag{15.44}

Therefore,

p_{1,{\rm pl}} + \frac {\B \cdot \vect {Q}_{\rm pl}}{\muo } - \frac {\B \cdot \vect {Q}_{\rm vac}}{\muo } = \xi _n\,\hat {\vect {n}}\cdot \left [\left [ \grad \!\left (p+\frac {B^2}{2\muo }\right ) \right ]\right ]_{\rm plasma}^{\rm vacuum}. \tag{15.45}

Substituting this into the combined plasma-plus-vacuum surface energy gives

\boxed { \delta W_S = \frac 12\int _S dS\,\xi _n^2\, \hat {\vect {n}}\cdot \left [\left [ \grad \!\left (p+\frac {B^2}{2\muo }\right ) \right ]\right ]_{\rm plasma}^{\rm vacuum}. } \tag{15.46}

Equivalently, since \(d\vect {S}=\hat {\vect {n}}\,dS\) and \(\vect {\xi }_\perp =\xi _n\hat {\vect {n}}\) on the interface,

\delta W_S = \frac 12\int _S (d\vect {S}\cdot \vect {\xi }_\perp )\; \vect {\xi }_\perp \cdot \left [\left [ \grad \!\left (p+\frac {B^2}{2\muo }\right ) \right ]\right ]_{\rm plasma}^{\rm vacuum}. \tag{15.47}

Vacuum contribution. Finally,

\boxed { \delta W_V = \frac 12\int _{V_{\rm vac}} dV\;\frac {|\B _1|^2}{\muo }, \qquad \curl \B _1=0 \quad \text {in vacuum}. } \tag{15.48}

This term is always positive. Vacuum magnetic energy cannot by itself drive an ideal instability; it can only add stabilizing cost or alter the balance when combined with the surface term.

Caution

How to read the intuitive form. The positive-definite terms are field-line bending, magnetic compression, plasma compression, and vacuum magnetic energy. Instability must therefore come from one of the indefinite pieces: pressure–curvature coupling, parallel-current drive, or an unfavorable surface contribution. That bookkeeping is the real practical value of the energy principle.

15.6 Stability Criterion and the Variational Viewpoint

The formal criterion is now immediate:

\boxed { \begin {array}{ll} \delta W[\vect {\xi }]>0 & \text {for all admissible } \vect {\xi } \quad \Longrightarrow \quad \text {stable},\\[0.4em] \delta W[\vect {\xi }]=0 & \text {for some admissible } \vect {\xi } \quad \Longrightarrow \quad \text {marginal},\\[0.4em] \delta W[\vect {\xi }]<0 & \text {for some admissible } \vect {\xi } \quad \Longrightarrow \quad \text {unstable}. \end {array} } \tag{15.49}

The power of Eq. (15.18) is that one does not need the exact eigenfunction to prove instability. Any admissible trial displacement with \(\delta W<0\) is enough. Conversely, proving \(\delta W>0\) for all admissible displacements is often difficult, but when it can be done it gives a global ideal-MHD stability theorem.

This is why the energy principle became the natural language of so many “classic problems” that follow in these notes. Kruskal–Shafranov, interchange, ballooning, and external-kink arguments are all, at bottom, carefully chosen ways of estimating the sign of \(\delta W\).

15.7 Consequences of Self-Adjointness

Define the weighted inner product

\langle \vect {\eta },\vect {\xi }\rangle _\rho \equiv \int _P \rho \,\vect {\eta }^{*}\cdot \vect {\xi }\,dV. \tag{15.50}

For static ideal-MHD equilibria, the force operator is self-adjoint under the usual ideal boundary conditions:

\boxed { \int _P \vect {\eta }^{*}\cdot \vect {F}(\vect {\xi })\,dV = \int _P \vect {F}(\vect {\eta })^{*}\cdot \vect {\xi }\,dV. } \tag{15.51}

This is the structural reason why the stability problem reduces to a real quadratic form.

Reality of \(\omega ^2\). Take the inner product of (15.13) with \(\vect {\xi }\):

\int _P \vect {\xi }^{*}\cdot \vect {F}(\vect {\xi })\,dV = -\omega ^2\int _P \rho |\vect {\xi }|^2\,dV. \tag{15.52}

Complex conjugation gives

\int _P \vect {F}(\vect {\xi })^{*}\cdot \vect {\xi }\,dV = -\omega ^{*2}\int _P \rho |\vect {\xi }|^2\,dV. \tag{15.53}

By self-adjointness, the left-hand sides of (15.52) and (15.53) are equal. Therefore

\omega ^{*2}=\omega ^2, \tag{15.54}

so \(\omega ^2\) is real.

This does not mean that \(\omega \) itself must be real. Rather,

if \(\omega ^2>0\), the motion is oscillatory;
if \(\omega ^2=0\), the mode is marginal;
if \(\omega ^2<0\), then \(\omega =\pm i|\omega |\) and one obtains exponentially growing and decaying solutions.

In ideal MHD there is no damping in this static, self-adjoint problem.

Orthogonality of distinct eigenmodes. Let \(\vect {\xi }_m\) and \(\vect {\xi }_n\) satisfy

\vect {F}(\vect {\xi }_n) = -\omega _n^2\rho \vect {\xi }_n, \qquad \vect {F}(\vect {\xi }_m) = -\omega _m^2\rho \vect {\xi }_m. \tag{15.55}

Take the inner product of the first equation with \(\vect {\xi }_m\) and of the second with \(\vect {\xi }_n\), then subtract:

\begin{aligned}-\omega _n^2\int _P \rho \,\vect {\xi }_m^{*}\cdot \vect {\xi }_n\,dV + \omega _m^2\int _P \rho \,\vect {\xi }_m^{*}\cdot \vect {\xi }_n\,dV &= \int _P \vect {\xi }_m^{*}\cdot \vect {F}(\vect {\xi }_n)\,dV - \int _P \vect {F}(\vect {\xi }_m)^{*}\cdot \vect {\xi }_n\,dV \nonumber \\ &=0,\end{aligned}

where the last equality follows from (15.51). Hence,

(\omega _m^2-\omega _n^2) \int _P \rho \,\vect {\xi }_m^{*}\cdot \vect {\xi }_n\,dV = 0. \tag{15.57}

If \(\omega _m^2\neq \omega _n^2\), then

\boxed { \int _P \rho \,\vect {\xi }_m^{*}\cdot \vect {\xi }_n\,dV = 0. } \tag{15.58}

Distinct ideal-MHD eigenmodes are therefore orthogonal under the density weight.

Takeaways

The energy principle replaces the linear stability PDEs by a quadratic functional \(\delta W[\vect {\xi }]\).

The sign of \(\delta W\) decides ideal-MHD stability for static equilibria.

Field-line bending, compression, and vacuum energy are stabilizing; pressure–curvature, parallel current, and unfavorable surface terms are the dangerous pieces.

The method is variational: a clever trial displacement can prove instability without solving the full eigenvalue problem.

Bibliography

I. B. Bernstein, E. A. Frieman, M. D. Kruskal, and R. M. Kulsrud. An energy principle for hydromagnetic stability problems. Proceedings of the Royal Society of London. Series A. Mathematical and Physical Sciences, 244(1236):17–40, 1958. doi:10.1098/rspa.1958.0023.

M. D. Kruskal and C. R. Oberman. On the stability of plasma in static equilibrium. Physics of Fluids, 1(4):275–280, 1958. doi:10.1063/1.1705885.

William A Newcomb. Hydromagnetic stability of a diffuse linear pinch. Annals of Physics, 10 (2):232–267, 1960. doi:10.1016/0003-4916(60)90023-3.

W. A. Newcomb. Convective instability induced by gravity in a plasma with a frozen-in magnetic field. Physics of Fluids, 4:391–396, 1961. doi:10.1063/1.1706342.

E. Frieman and Manuel Rotenberg. On hydromagnetic stability of stationary equilibria. Reviews of Modern Physics, 32(4):898–902, 1960. doi:10.1103/RevModPhys.32.898.

Problems

Problem 15.1. Recovering the shear-Alfvén wave from the energy principle

Consider a uniform plasma with constant \(\rho \), constant \(p\), and equilibrium field \( \B =B_0\ez \) in an unbounded domain. Let the displacement be incompressible and transverse, \( \divergence \vect {\xi }=0 \) and \( \vect {\xi }=\vect {\xi }_\perp (z)e^{-i\omega t} \).

(a): Starting from Eq. (15.3), show that \[ \vect {Q}=B_0\dd {\vect {\xi }_\perp }{z}. \]
(b): Evaluate \(\delta W\) explicitly and show that only the field-line-bending term survives.
(c): Using Eq. (15.18), recover the shear-Alfvén dispersion relation \[ \omega ^2 = k_\parallel ^2 \frac {B_0^2}{\muo \rho }. \]

Problem 15.2. Identifying the dangerous terms

Take the intuitive form in Eq. (15.41) and discuss the sign of each contribution for the following limits.

(a): A straight magnetic field with \(\vect {\kappa }=0\) and \(J_\parallel =0\).
(b): A curved equilibrium with finite pressure gradient but \(J_\parallel =0\).
(c): A force-free equilibrium with \(\grad p=0\) but finite \(J_\parallel \).

For each case, identify which instability class the energy principle suggests should be absent or possible.