Lecture 11
Grad’s Equilibrium Analysis
Overview
This lecture is about equilibrium with a gyrotropic stress tensor, not about a
particular closure.
-
1.
- Starting from the static limit of the MHD momentum equation, we project force
balance parallel and perpendicular to \(\B \).
-
2.
- The parallel projection gives the mirror-force constraint; the perpendicular projection
shows how anisotropy modifies field-line tension.
-
3.
- In the isotropic limit one recovers \(p=p(\psi )\), which is the doorway to the Grad–Shafranov
equation derived in the next lecture; in the slender-tube limit one recovers paraxial
mirror balance.
We now stop evolving the full MHD system and instead ask what remains when the net force vanishes.
The central point is that static equilibrium requires only a stress tensor and a magnetic field; it does not
require the specific CGL evolution laws. This is why it is useful to separate the present lecture from the
earlier closure lecture. The equilibrium problem is broader: any gyrotropic distribution produces a tensor
of the form \[ \tens {P}=p_\perp \tens {I} + (p_\parallel -p_\perp )\vect {b}\vect {b}, \qquad \vect {b}\equiv \frac {\B }{B}, \] and the question is then how this tensor can balance magnetic pressure and magnetic
tension.
Historical Perspective
The isotropic equilibrium problem led to the great toroidal-equilibrium literature and,
ultimately, to the Grad–Shafranov equation. The anisotropic extension developed along
a somewhat different path, through guiding-center and mirror-machine physics. Grad
emphasized that weakly collisional plasmas should be formulated directly in terms of
a gyrotropic pressure tensor, while Newcomb clarified the connection between global
equilibrium and the paraxial mirror limit Grad [1967a,?], Newcomb [1981]. This is one
of the places where magnetic confinement, kinetic theory, and classical MHD visibly
overlap.
Caution
A conceptual boundary. This lecture treats static balance. It is not the place to do the
full mirror and firehose wave calculations. Those dynamical instability problems belong
later, when anisotropic MHD waves and the drift-kinetic appendix are in hand.
11.1 Static balance with a gyrotropic tensor
Start from the static limit of the momentum equation.
The momentum equation stated in Eq. (1.8) becomes, when \(\vect {u}=0\),
\[-\divergence \tens {P}+\J \times \B =0. \tag{11.1}\]
It is often cleaner to collect the magnetic force into the Maxwell stress tensor \[\tens {T}_B = -\frac {B^2}{2\muo }\tens {I} + \frac {\B \B }{\muo },\]
so that static equilibrium may be written as \[\boxed { \divergence \left (\tens {P}-\tens {T}_B\right )=0. } \tag{11.3}\]
This says that plasma stress and magnetic stress must balance exactly.
Write out the divergence of a gyrotropic tensor once, slowly.
Let \[ f \equiv p_\parallel -p_\perp , \qquad \tens {P}=p_\perp \tens {I}+f\,\vect {b}\vect {b}. \] Then
\[\begin {aligned} \divergence \tens {P} &= \grad p_\perp + \divergence \left (f\,\vect {b}\vect {b}\right ) \\ &= \grad p_\perp + (\vect {b}\cdot \grad f)\vect {b} + f\,\divergence \left (\vect {b}\vect {b}\right ). \end {aligned}\]
Now use \[\divergence \left (\vect {b}\vect {b}\right ) = (\vect {b}\cdot \grad )\vect {b} + \vect {b}\,\divergence \vect {b} \tag{11.5}\]
and define \[\vect {\kappa } \equiv (\vect {b}\cdot \grad )\vect {b}, \qquad \pp {}{\ell } \equiv \vect {b}\cdot \grad . \tag{11.6}\]
Because \(\divergence \B =0\), \[ 0=\divergence (B\vect {b})=\vect {b}\cdot \grad B + B\,\divergence \vect {b}, \] so \[\divergence \vect {b}=-\frac {1}{B}\pp {B}{\ell }. \tag{11.7}\]
Substituting Eqs. (11.5) and (11.7) gives \[\begin {aligned} \divergence \tens {P} &= \grad p_\perp + \vect {b}\pp {f}{\ell } + f\,\vect {\kappa } - \vect {b}\,\frac {f}{B}\pp {B}{\ell } \\ &= \grad _\perp p_\perp + (p_\parallel -p_\perp )\vect {\kappa } + \vect {b} \left [ \pp {p_\parallel }{\ell } - \frac {p_\parallel -p_\perp }{B}\pp {B}{\ell } \right ]. \end {aligned} \tag{11.8}\]
The last line uses \[ \pp {f}{\ell }=\pp {p_\parallel }{\ell }-\pp {p_\perp }{\ell }, \qquad \grad p_\perp = \grad _\perp p_\perp + \vect {b}\pp {p_\perp }{\ell }. \]
Kinetic examples.
Appendix C gives three explicit kinetic realizations of the functions \(p_\perp (\psi ,B)\) and \(p_\parallel (\psi ,B)\): a fixed-pitch sloshing-ion
distribution, a logarithmic mirror distribution generated by pitch-angle scattering, and the isotropic
Maxwellian limit. The appendix makes the magnetic-field dependence of these pressures completely
explicit and provides a useful bridge between the equilibrium formulas derived here and the underlying
kinetic physics.
Write the magnetic force in the same geometric language.
Using Eq. (1.11) together with Eq. (1.12), the Lorentz force may be written as
\[\begin {aligned} \J \times \B &= \frac {1}{\muo }(\curl \B )\times \B \\ &= -\grad \left (\frac {B^2}{2\muo }\right ) + \frac {1}{\muo }(\B \cdot \grad )\B \\ &= -\grad _\perp \left (\frac {B^2}{2\muo }\right ) + \frac {B^2}{\muo }\vect {\kappa }. \end {aligned} \tag{11.9}\]
The first term is magnetic-pressure gradient; the second is field-line tension.
Now project the force balance.
Substitute Eqs. (11.8) and (11.9) into Eq. (11.1):
\[\begin {aligned} 0 &= -\grad _\perp \left (p_\perp + \frac {B^2}{2\muo }\right ) \\ &\quad + \left (\frac {B^2}{\muo }+p_\perp -p_\parallel \right )\vect {\kappa } \\ &\quad - \vect {b} \left [ \pp {p_\parallel }{\ell } + \frac {p_\perp -p_\parallel }{B}\pp {B}{\ell } \right ]. \end {aligned} \tag{11.10}\]
Since \(\vect {b}\cdot \vect {\kappa }=0\) and \(\vect {b}\cdot \grad _\perp =0\), the parallel and perpendicular projections follow immediately.
Parallel balance.
Taking \(\vect {b}\cdot \)Eq. (11.10) yields
\[\boxed { \pp {p_\parallel }{\ell } + \frac {p_\perp -p_\parallel }{B}\pp {B}{\ell } =0. } \tag{11.11}\]
This is the static mirror-force constraint.
Perpendicular balance.
Projecting Eq. (11.10) perpendicular to \(\B \) gives
\[Q \equiv \frac {B^2}{\muo }+p_\perp -p_\parallel , \tag{11.12}\]
so that \[\boxed { \grad _\perp \left (p_\perp +\frac {B^2}{2\muo }\right ) = Q\,\vect {\kappa }. } \tag{11.13}\]
The coefficient \(Q\) is the effective field-line tension. Pressure anisotropy changes the restoring force associated
with curvature, which is why it later reappears in anisotropic wave theory.
Caution
Why this is broader than CGL. Equations (11.11) and (11.13) follow from static
force balance plus gyrotropy alone. They do not require the CGL evolution laws. CGL
is one possible time-dependent closure that is compatible with these static relations, but
the equilibrium relations themselves are more general.
11.2 Flux surfaces, sign conditions, and the isotropic bridge
Assume flux surfaces exist.
Suppose there is a flux function \(\psi \) such that
\[\B \cdot \grad \psi =0.\]
If \(p_\parallel \) and \(p_\perp \) depend on position through \((\psi ,B)\), then along a field line \[ \pp {p_\parallel }{\ell } = \left .\pp {p_\parallel }{B}\right |_{\psi }\pp {B}{\ell }, \] because \(\vect {b}\cdot \grad \psi =0\). Insert this into Eq. (11.11):
\[\left [ \left .\pp {p_\parallel }{B}\right |_{\psi } + \frac {p_\perp -p_\parallel }{B} \right ] \pp {B}{\ell }=0.\]
For a generic mirror field with \(\pp {B}{\ell }\neq 0\), the bracket must vanish. Therefore \[\boxed { p_\perp = p_\parallel - B\left .\pp {p_\parallel }{B}\right |_{\psi }. } \tag{11.16}\]
An equivalent form, often convenient in mirror geometry, is \[\boxed { p_\perp = -B^2\left .\pp {}{B}\left (\frac {p_\parallel }{B}\right )\right |_{\psi }. } \tag{11.17}\]
The old label eq:cgl_constraint is preserved for continuity in these notes, but the relation is really a
gyrotropic equilibrium constraint, not a specifically CGL one.
The isotropic limit is the bridge to Grad–Shafranov theory.
If \(p_\perp =p_\parallel =p\), then Eq. (11.11) reduces to
\[\pp {p}{\ell }=0.\]
Thus the scalar pressure is constant along field lines and, in axisymmetry, becomes a flux function:
\[p=p(\psi ). \tag{11.19}\]
This is the key simplification that makes the isotropic axisymmetric equilibrium problem collapse to a
single elliptic equation for \(\psi \). The next lecture is devoted to that reduction.
Positive effective tension.
Equation (11.13) makes it clear that a bent field line only provides a restoring force when the coefficient \(Q\)
is positive. Thus one requires
\[\boxed { \frac {B^2}{\muo }+p_\perp -p_\parallel >0. } \tag{11.20}\]
This is the static sign condition behind firehose behavior.
Include steady parallel flow if needed.
A steady field-aligned flow adds an inertial curvature term \(-\rho u_\parallel ^2\vect {\kappa }\), so the effective coefficient becomes
\[Q_{\rm flow} = \frac {B^2}{\muo }+p_\perp -p_\parallel -\rho u_\parallel ^2.\]
The corresponding condition is \[\boxed { \frac {B^2}{\muo }+p_\perp -p_\parallel -\rho u_\parallel ^2>0. } \tag{11.22}\]
This is the form that later becomes relevant in the solar-wind context when the flow approaches the Alfvén
speed.
Positive compressive response.
A second useful sign condition comes from asking how the total perpendicular pressure responds to a local
increase in \(B\). Starting from Eq. (11.16),
\[Q = \frac {B^2}{\muo }+p_\perp -p_\parallel = \frac {B^2}{\muo }-B\left .\pp {p_\parallel }{B}\right |_{\psi },\]
so that \[\frac {Q}{B} = \frac {B}{\muo }-\left .\pp {p_\parallel }{B}\right |_{\psi }. \tag{11.24}\]
Differentiate Eq. (11.24) with respect to \(B\) at fixed \(\psi \): \[B\left .\pp {}{B}\left (\frac {Q}{B}\right )\right |_{\psi } = \frac {B}{\muo } - B\left .\pp {^2 p_\parallel }{B^2}\right |_{\psi }. \tag{11.25}\]
Now differentiate \[ p_\perp + \frac {B^2}{2\muo } = p_\parallel - B\left .\pp {p_\parallel }{B}\right |_{\psi } + \frac {B^2}{2\muo } \] with respect to \(B\): \[\begin {aligned} \left .\pp {}{B}\left (p_\perp +\frac {B^2}{2\muo }\right )\right |_{\psi } &= \left .\pp {p_\parallel }{B}\right |_{\psi } - \left .\pp {p_\parallel }{B}\right |_{\psi } - B\left .\pp {^2 p_\parallel }{B^2}\right |_{\psi } + \frac {B}{\muo } \\ &= \frac {B}{\muo } - B\left .\pp {^2 p_\parallel }{B^2}\right |_{\psi }. \end {aligned}\]
Comparing with Eq. (11.25) gives \[\boxed { \left .\pp {}{B}\left (p_\perp +\frac {B^2}{2\muo }\right )\right |_{\psi }>0. } \tag{11.27}\]
Again, this is being used here as a static sign condition, not yet as a wave-growth calculation.
11.3 Simple limits and the paraxial mirror bridge
The isotropic perpendicular balance.
When \(p_\perp =p_\parallel =p\), Eq. (11.13) reduces to
\[\grad _\perp \left (p+\frac {B^2}{2\muo }\right ) = \frac {B^2}{\muo }\vect {\kappa }. \tag{11.28}\]
This is the form from which familiar one-dimensional equilibria follow.
Slab equilibrium.
Take \[ \B =B(x)\,\ez , \qquad \vect {\kappa }=0. \] Then Eq. (11.28) becomes
\[\frac {d}{dx}\left (p+\frac {B^2}{2\muo }\right )=0. \tag{11.29}\]
Thus total pressure is constant across the slab.
The Z-pinch.
For a cylindrical pinch with \[ \B =B_\theta (r)\,\etheta , \qquad \vect {\kappa }=-\frac {\er }{r}, \] Eq. (11.28) becomes
\[\boxed { \frac {d}{dr}\left (p+\frac {B_\theta ^2}{2\muo }\right ) + \frac {B_\theta ^2}{\muo r}=0. } \tag{11.30}\]
The second term is the inward hoop stress.
The screw pinch.
If both \(B_\theta (r)\) and \(B_z(r)\) are present, only the azimuthal field contributes curvature. The equilibrium equation
becomes
\[\boxed { \frac {d}{dr}\left (p+\frac {B_z^2+B_\theta ^2}{2\muo }\right ) + \frac {B_\theta ^2}{\muo r}=0. } \tag{11.31}\]
This cylindrical prototype is the natural warm-up for toroidal equilibrium.
Why the paraxial mirror limit is different.
A mirror field is not really one-dimensional because the flux tube expands and contracts along the axis.
But when the tube is slender, the geometry simplifies. Let \(a\) be the tube radius and \(Z\) the axial scale length,
with
\[\epsilon \equiv \frac {a}{Z} \ll 1. \tag{11.32}\]
Write the axisymmetric field as \[ \B =B_r(r,z)\,\er + B_z(r,z)\,\ez . \] From \(\divergence \B =0\) and the estimates \[ \pp {}{r}\sim \frac {1}{a}, \qquad \pp {}{z}\sim \frac {1}{Z}, \] one finds \[\frac {B_r}{B_z}=\mathcal O(\epsilon ). \tag{11.33}\]
If a field line is represented by \(r=a(z)\), its curvature in the meridional plane is \[\kappa = \frac {|a''(z)|}{\left [1+(a'(z))^2\right ]^{3/2}} \simeq |a''(z)|. \tag{11.34}\]
Since \(a'\sim a/Z=\epsilon \) and \(a''\sim a/Z^2=\epsilon ^2/a\), \[\kappa \sim \frac {\epsilon ^2}{a}. \tag{11.35}\]
So curvature is down by two powers of the slenderness parameter.
Leading-order paraxial balance.
Because the curvature term is small, Eq. (11.13) reduces at leading order to
\[\boxed { \grad _\perp \left (p_\perp +\frac {B^2}{2\muo }\right )=0. } \tag{11.36}\]
Thus, on each axial slice, the sum of perpendicular plasma pressure and magnetic pressure is
approximately constant across the tube.
Use the exterior field to reconstruct the tube.
If the exterior vacuum field is \(B_{\rm ext}(z)\), then Eq. (11.36) implies
\[p_\perp (\psi ,z)+\frac {B^2(\psi ,z)}{2\muo } = \frac {B_{\rm ext}^2(z)}{2\muo }. \tag{11.37}\]
Hence \[B(\psi ,z) = \sqrt {B_{\rm ext}^2(z)-2\muo p_\perp (\psi ,z)}. \tag{11.38}\]
For an axisymmetric flux function, \[r^2(\psi ,z) = 2\int _0^{\psi }\frac {d\psi '}{B(\psi ',z)}. \tag{11.39}\]
This is the natural mirror-machine analogue of solving for flux surfaces in toroidal geometry.
11.4 Anisotropic pressure equilibrium
Now allow gyrotropic anisotropy,
\[\tens {P} = p_\perp \tens {I} + (p_\parallel - p_\perp ) {\vect {b}} {\vect {b}}, \qquad {\vect {b}} = \frac {\B }{B}.\]
Force balance becomes
\[\boxed { \J \times \B = \divergence \tens {P}. } \tag{11.41}\]
Parallel constraint
Projecting along \(\vect {b}\) gives
\[\boxed { \frac {\partial p_\parallel }{\partial \ell } + \frac {p_\perp - p_\parallel }{B} \frac {\partial B}{\partial \ell } = 0, } \tag{11.42}\]
where \(\partial /\partial \ell = {\vect {b}} \cdot \grad \).
Assuming \(p_{\perp ,\parallel } = p_{\perp ,\parallel }(\psi ,B)\),
\[\boxed { \pp {p_\parallel }{B} + \frac {p_\perp - p_\parallel }{B} = 0,\qquad \mbox { or } \qquad p_\perp = p_\parallel - B \frac {\partial p_\parallel }{\partial B} } \tag{11.43}\]
Perpendicular balance and cancellation algebra
Define
\[\sigma = \frac {1}{\muo } + \frac {p_\perp - p_\parallel }{B^2}.\]
Consider the following slight variation on \(\vect {J}\times \vect {B}\):
\[\B \times (\curl \sigma \B ) = B^2 \grad _\perp \sigma + \sigma \B \times (\curl \B ),\]
and expanding explicitly, \[\begin {aligned} B^2 \grad _\perp \sigma &= \grad _\perp p_\perp - \grad _\perp p_\parallel - \frac {p_\perp - p_\parallel }{B^2} \grad _\perp B^2 , \\ \sigma \B \times (\curl \B ) &= \left ( 1 + \frac {\muo (p_\perp - p_\parallel )}{B^2} \right ) \left (\grad _\perp \frac {B^2}{2 \muo } - B^2 \sigma \vect {\kappa }\right ). \end {aligned}\]
Adding terms, the \(\grad _\perp B\) contributions cancel using (11.43), leaving
\[\begin {aligned} B^2 \grad _\perp \sigma + \sigma \B \times (\curl \B ) &= \grad _\perp p_\perp - \grad _\perp p_\parallel - \frac {p_\perp - p_\parallel }{B^2}\grad _\perp B^2 \\ &\quad + \left ( 1 + \frac {\muo (p_\perp - p_\parallel )}{B^2} \right ) \left (\grad _\perp \frac {B^2}{2 \muo } - B^2 \sigma \vect {\kappa }\right ) \\ &= - \nabla _\perp p_\parallel + \pp {p_\parallel }{B}\grad _\perp B \\ &= \pp {p_\parallel }{\psi }\grad _\perp \psi . \end {aligned}\]
Carrying through the axisymmetric representation yields the anisotropic Grad equation:
This proof now implies
\[\begin {aligned} (\nabla \times \vect {B})\times \vect {B} &= \frac {1}{\sigma }\left .\pp {p_\parallel }{\psi }\right |_B\,\nabla _\perp \psi + B^2 \nabla _\perp \ln \sigma \end {aligned}\]
From Eq. (11.50) we also note that
\[(\curl \vect {B})\times \vect {B} = -\frac {\Delta ^\star \psi }{R^2}\nabla _\perp \psi\]
Carrying this through with the axisymmetric representation yields and taking the dot product with \(\nabla \psi \)
\[\boxed { \Delta ^\star \psi = - \frac {R^2}{\sigma }\,\pp {p_\parallel }{\psi } \;-\; \, \nabla _\perp \psi \cdot \nabla _\perp \ln \sigma . } \tag{11.50}\]
For practical purposes it may be more useful to consider the dimensionless parameter \[\begin {aligned} \bar {\sigma } & = \mu _0 \sigma \\ & = 1 + \frac {\beta _\perp - \beta _\parallel }{2} \\ & = 1 + \mu _0 \left (\frac {p_\perp - p_\parallel }{B^2}\right ) \\ & = 1 + \mu _0 \left (\frac {p_\parallel - B dp_\parallel /dB - p_\parallel }{B^2}\right ) \\ & = 1 - \frac {\mu _0}{B} \dd {p_\parallel }{B}. \end {aligned}\]
\[\boxed { \Delta ^\star \psi = - \mu _0 \frac {R^2}{\bar {\sigma }}\,\pp {p_\parallel }{\psi } \;-\; \, B^2 R^2 \frac { \partial \ln \bar {\sigma }}{\partial \psi } - \nabla _\perp \psi \cdot \nabla B \left . \frac { \partial \ln \bar {\sigma }}{\partial B}\right |_\psi . }\]
Equivalent Form.
\[\vect {J}\times \vect {B} = \nabla _\perp p_\perp + (p_\parallel - p_\perp ) \vect {\kappa }\]
that leads directly to all of the following equivalent formes for \(J_\phi \): \[\begin {aligned} \mu _0 J_\phi & = -\frac {\Delta ^\star \psi }{ R} \\ J_\phi & = \frac {\vect {B}}{B^2} \times \left ( \nabla _\perp p_\perp + (p_\parallel - p_\perp ) \vect {\kappa } \right ) \\ \mu _0 J_\phi & = \frac {\vect {B}\times \nabla _\perp p_\perp }{B^2} + \frac { \beta _\parallel - \beta _\perp }{2} \vect {B}\times \vect {\kappa } \\ \mu _0 J_\phi & = - \mu _0 \frac {R}{\bar {\sigma }}\,\pp {p_\parallel }{\psi } \;-\; \, B^2 R \frac { \partial \ln \bar {\sigma }}{\partial \psi } - \frac {\nabla _\perp \psi \cdot \nabla B }{R} \left . \frac { \partial \ln \bar {\sigma }}{\partial B}\right |_\psi . \end {aligned}\]
Checks.
- If \(p_\perp =p_\parallel =p(\psi )\) then \(\bar {\sigma } \to 1 \) and \(\nabla \ln \bar {\sigma }\to 0\), and reduces to the isotropic GS structure (with the usual \(FF'\) term from the
toroidal field absent).
- The \(\nabla \psi \cdot \nabla \ln \sigma \) term encodes the fact that anisotropy introduces additional \((R,Z)\) dependence through \(B(R,Z)\).
- One must always make sure that the mirror condition (Eq. (11.27)) and firehose condition
(Eq. (11.20)) are satisfied.
Bridge to the next lecture.
The paraxial mirror problem is the cleanest anisotropic example because it stays close to one-dimensional
intuition. The next lecture goes in the opposite direction: it specializes back to isotropic pressure, keeps
full axisymmetric geometry, and derives the Grad–Shafranov equation for truly two-dimensional toroidal
equilibria.
Experimental perspective.
Equilibrium is where kinetic ideas first become impossible to hide from the experimentalist. In mirror
devices, the dependence of pressure on \(B\) is built directly into confinement geometry. In toroidal devices, the
same stress-balance logic survives but is reorganized by flux surfaces, current closure, and shaping coils.
The point of doing the algebra carefully is that the diagnostic questions are always the same: what is
balancing what, where does the tension sit, and which stress term is being inferred from the
measurements?
Takeaways
- Static anisotropic equilibrium begins with the gyrotropic tensor, not with a specific
closure law.
- Parallel balance gives Eq. (11.11); perpendicular balance gives Eq. (11.13).
- If \(p_{\perp ,\parallel }=p_{\perp ,\parallel }(\psi ,B)\), then Eq. (11.16) follows. In the isotropic limit this collapses to \(p=p(\psi )\).
- Slab, pinch, mirror, and toroidal equilibria are all different geometric faces of the
same underlying stress-balance problem.
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Problems
Problem 11.1. Derive the anisotropic equilibrium equations carefully
Part A.
Starting from \(\tens {P}=p_\perp \tens {I}+(p_\parallel -p_\perp )\vect {b}\vect {b}\), reproduce Eq. (11.8). Do not skip the use of \(\divergence (B\vect {b})=0\).
Part B.
Combine Eq. (11.8) with Eq. (11.9) and derive Eqs. (11.11) and (11.13).
Part C.
Assume \(p_{\perp ,\parallel }=p_{\perp ,\parallel }(\psi ,B)\) and derive Eq. (11.16) step by step. Then show that the isotropic limit implies \(p=p(\psi )\).
Part D.
Starting from Eq. (11.28), derive the slab, Z-pinch, and screw-pinch balances, Eqs. (11.29), (11.30), and
(11.31).
Part E.
In the paraxial ordering, show explicitly that the curvature term in Eq. (11.13) is smaller than the radial
pressure-gradient term by \(\mathcal O(\epsilon ^2)\).