Lecture 23 Newcomb’s Formulation for the Diffuse Pinch

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Lecture 23
Newcomb’s Formulation for the Diffuse Pinch

Overview

The cleanest way to organize ideal-MHD pinch stability is to start with Newcomb’s reduction of the screw pinch to a one-dimensional variational problem. Once that structure is visible, the diffuse-pinch limits become much easier to read: the $\theta $ pinch isolates line bending and magnetic compression, while the $z$ pinch isolates curvature and pressure drive. Together they show exactly what later gets compressed into the coefficient functions $f(r)$ and $g(r)$.

Historical Perspective

Diffuse-pinches forced the subject to move beyond surface-current intuition and toward genuinely distributed current profiles, while Newcomb supplied the mathematical language that made the general problem tractable Newcomb (1960). In that transition one sees, in unusually clean algebra, the three stabilizing agents that recur throughout ideal MHD: field-line bending, plasma compression, and boundary control. The payoff is that internal and external modes can be treated in one common variational framework rather than as disconnected special cases.

We use the energy principle developed earlier in Lecture 15, especially the plasma contribution (15.3), repeated here

\delta W_P = \frac 12\int _P dV\Bigg [ \begin {gathered} \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}} \\[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}} \end {gathered} \Bigg ]. \tag{23.1}

Diffuse-pinch limits. With that logical sequence in hand, it is useful to return to the two diffuse-pinch limits. They show, in the cleanest possible algebra, how line bending, compression, and curvature populate $\delta W$ before addressing the general screw pinch.

Perturbations are taken in the normal-mode form

\bm {\xi }(r,\theta ,z) = \left [\xi _r(r)\,\vect {e}_r+\xi _\theta (r)\,\vect {e}_\theta +\xi _z(r)\,\vect {e}_z\right ] e^{i m\theta +i k z}. \tag{23.2}

23.1 The $\theta $ pinch

A $\theta $ pinch carries only axial field,

\B _0 = B_z(r)\,\vect {e}_z, \qquad \J = -\frac {1}{\muo }\,\dd {B_z}{r}\,\vect {e}_\theta , \qquad J_\parallel = 0, \qquad \bm {\kappa }=0.

Radial force balance is therefore the total-pressure statement

\dd {}{r}\left (p+\frac {B_z^2}{2\muo }\right )=0 \qquad \Longrightarrow \qquad p(r)+\frac {B_z^2(r)}{2\muo }=\text {const}. \tag{23.4}

This is the cylindrical version of the static balance already encountered in the introductory equations, Eq. (1.8).

Incompressibility constraint. From (23.2),

\divergence \bm {\xi } = \left [ \frac {1}{r}\dd {}{r}(r\xi _r)+\frac {i m}{r}\xi _\theta +i k \xi _z \right ]e^{i m\theta +i k z}.

If we minimize the compressional part of $\delta W$ by taking $\divergence \bm {\xi }=0$, then for $k\neq 0$,

\xi _\parallel \equiv \xi _z = \frac {i}{k r}\left [\dd {}{r}(r\xi _r)+i m \xi _\theta \right ], \tag{23.6}

and equivalently

\divergence \bm {\xi }_\perp = \frac {1}{r}\dd {}{r}(r\xi _r)+\frac {i m}{r}\xi _\theta = -ik\,\xi _\parallel . \tag{23.7}

\begin{aligned}\delta W_P = \frac 12\int _P dV\Bigg [ & \frac {|\vect {Q}_\perp |^2}{\muo } + \frac {B^2}{\muo } \left |\divergence \vect {\xi }_\perp + \cancelto {0}{2\,\vect {\xi }_\perp \cdot \vect {\kappa }}\right |^2 + \cancelto {0}{\gamma p\,|\divergence \vect {\xi }|^2} \nonumber \\[0.5em] & - \cancelto {0}{2\,(\vect {\xi }_\perp \cdot \grad p) (\vect {\xi }_\perp ^{*}\cdot \vect {\kappa })} - {\cancelto {0}{\frac {J_\parallel }{B} (\vect {\xi }_\perp ^{*}\times \B )\cdot \vect {Q}_\perp }} \Bigg ].\end{aligned} \tag{23.8}

Perturbed field. Because $\B _0=B_z\vect {e}_z$,

\bm {\xi }\times \B _0 = \xi _\theta B_z\,\vect {e}_r-\xi _r B_z\,\vect {e}_\theta ,

\begin{aligned}\bm {Q} &\equiv \curl (\bm {\xi }\times \B _0) \nonumber \\ &= \underbrace {ik B_z\,\xi _r\,\vect {e}_r+ik B_z\,\xi _\theta \,\vect {e}_\theta }_{\bm {Q}_\perp } + \underbrace {\left [ \frac {1}{r}\dd {}{r}(rB_z\xi _r)-\frac {i m}{r}B_z\xi _\theta \right ]}_{Q_\parallel }\vect {e}_z .\end{aligned} \tag{23.10}

Hence

\bm {Q}_\perp = ik B_z(\xi _r\vect {e}_r+\xi _\theta \vect {e}_\theta ).

Reduced energy and completion of the square. With $J_\parallel =0$ and $\bm {\kappa }=0$, the general plasma functional (23.1) reduces, under incompressibility, to

\begin{aligned}\delta W &= \frac {1}{2\muo }\int _P dV \left ( |\bm {Q}_\perp |^2 + B_z^2\,|\divergence \bm {\xi }_\perp |^2 \right ) \nonumber \\ &= \frac {1}{2\muo }\int _P dV\,B_z^2 \left [ k^2\left (|\xi _r|^2+|\xi _\theta |^2\right ) + \left | \frac {1}{r}\dd {}{r}(r\xi _r)+\frac {i m}{r}\xi _\theta \right |^2 \right ].\end{aligned} \tag{23.12}

Expanding the last modulus gives

\begin{aligned}\delta W &= \frac {1}{2\muo }\int _P dV\,B_z^2 \Bigg [ k^2|\xi _r|^2 +\frac {|(r\xi _r)'|^2}{r^2} +\left (k^2+\frac {m^2}{r^2}\right )|\xi _\theta |^2 \nonumber \\ &\hspace {3.4cm} +\frac {i m}{r^2}(r\xi _r)'^{*}\xi _\theta -\frac {i m}{r^2}(r\xi _r)'\xi _\theta ^* \Bigg ].\end{aligned}

Define

k_0^2(r)\equiv k^2+\frac {m^2}{r^2}.

Then the square completion is

\begin{aligned}\delta W &= \frac {1}{2\muo }\int _P dV\,B_z^2 \Bigg [ k^2|\xi _r|^2 + \left (\frac {1}{r^2}-\frac {m^2}{k_0^2 r^4}\right )|(r\xi _r)'|^2 + \left | k_0\xi _\theta -\frac {i m}{k_0 r^2}(r\xi _r)' \right |^2 \Bigg ].\end{aligned} \tag{23.15}

The minimizing choice is therefore

\xi _\theta = \frac {i m}{k_0^2 r^2}\,\frac {1}{r}\dd {}{r}(r\xi _r) = \frac {i m}{k^2r^2+m^2}\,\frac {1}{r}\dd {}{r}(r\xi _r), \tag{23.16}

and substituting it back gives

\delta W_{\min } = \frac {1}{2\muo }\int _P dV\,B_z^2 \left [ k^2|\xi _r|^2 + \frac {k^2}{k_0^2 r^2}|(r\xi _r)'|^2 \right ]. \tag{23.17}

For every finite $k\neq 0$ the $\theta $ pinch is ideally stable: $\delta W_{\min }>0$. The limit $k\rightarrow 0$ is only marginal, reflecting the absence of line bending in a perfectly long-wavelength translation.

23.2 The $z$ pinch

The $z$ pinch reverses the geometry:

\B _0 = B_\theta (r)\,\vect {e}_\theta , \qquad \J = \frac {1}{\muo r}\dd {}{r}\!\left (rB_\theta \right )\vect {e}_z, \qquad J_\parallel = 0.

Now the field lines are circles and have curvature

\bm {\kappa }=(\bvec \cdot \grad )\bvec =-\frac {\vect {e}_r}{r}.

Radial equilibrium becomes

\dd {p}{r}+\frac {B_\theta }{\muo r}\dd {}{r}(rB_\theta )=0 \qquad \Longleftrightarrow \qquad \dd {p}{r} = -\frac {1}{2\muo r}\dd {}{r}(rB_\theta ^2). \tag{23.20}

Equation (23.20) is exactly the $Z$-pinch balance discussed in the anisotropic-equilibrium lecture, Eq. (11.30).

Case $m\neq 0$: curvature enters $\delta W$. Here the parallel displacement is $\xi _\parallel =\xi _\theta $. If we again minimize the compressional energy by taking $\divergence \bm {\xi }=0$, then

\xi _\parallel \equiv \xi _\theta = \frac {i}{m}\left [\dd {}{r}(r\xi _r)+ik r\,\xi _z\right ], \qquad (m\neq 0). \tag{23.21}

The perpendicular perturbed field is

\bm {Q}_\perp = \curl \Big [(\xi _r\vect {e}_r+\xi _z\vect {e}_z)\times B_\theta \vect {e}_\theta \Big ] = \frac {i m B_\theta }{r}\left (\xi _r\vect {e}_r+\xi _z\vect {e}_z\right ). \tag{23.22}

Substituting $\bm {\kappa }=-\vect {e}_r/r$ and $J_\parallel =0$ into (23.1) gives

\delta W = \frac {1}{2\muo }\int _P dV \left [ |\bm {Q}_\perp |^2 + B_\theta ^2\left |\divergence \bm {\xi }_\perp -2\frac {\xi _r}{r}\right |^2 + \frac {2\muo }{r}p'(r)\,|\xi _r|^2 \right ]. \tag{23.23}

Because

\divergence \bm {\xi }_\perp -2\frac {\xi _r}{r} = r\left (\frac {\xi _r}{r}\right )'+ik\xi _z,

the functional becomes

\delta W = \frac {1}{2\muo }\int _P dV \left [ \frac {m^2B_\theta ^2}{r^2}\left (|\xi _r|^2+|\xi _z|^2\right ) + B_\theta ^2\left |r\left (\frac {\xi _r}{r}\right )'+ik\xi _z\right |^2 + \frac {2\muo }{r}p'(r)\,|\xi _r|^2 \right ]. \tag{23.25}

Now complete the square in $\xi _z$:

\xi _z = \frac {i k r^2}{m^2+k^2r^2}\, r\left (\frac {\xi _r}{r}\right )'. \tag{23.26}

Substituting (23.26) back into (23.25) gives the reduced radial functional

\delta W_{\min } = \frac {1}{2\muo }\int _P dV \left [ \left (2\muo r\,p'(r)+m^2B_\theta ^2\right )\frac {|\xi _r|^2}{r^2} + \frac {m^2r^2B_\theta ^2}{m^2+k^2r^2} \left | \left (\frac {\xi _r}{r}\right )' \right |^2 \right ]. \tag{23.27}

The second term is stabilizing line bending. A simple sufficient condition for stability of a given $m\neq 0$ family is therefore

2 r\,p'(r)+\frac {m^2B_\theta ^2}{\muo }>0 \qquad \text {for all } r, \tag{23.28}

or, using (23.20),

-\,\dd {}{r}(rB_\theta ^2)+(1+m^2)B_\theta ^2>0, \tag{23.29}

equivalently

\frac {1}{B_\theta ^2}\dd {}{r}(rB_\theta ^2)<m^2-1. \tag{23.30}

Near the axis $B_\theta \sim (\muo J_0/2)r$, so the left-hand side tends to $3$. Hence the $m=1$ kink necessarily violates (23.30) near the axis. That is the classic ideal $z$-pinch kink.

Case $m=0$: sausage and compressibility. For the sausage mode compressibility matters and one cannot simply impose $\divergence \bm {\xi }=0$. Eliminating $\xi _z$ from the full compressible functional gives

\xi _z = \frac {i}{\gamma p + B_\theta ^2/\muo } \left [ \frac {rB_\theta ^2}{\muo }\left (\frac {\xi _r}{r}\right )' + \frac {\gamma p}{r}(r\xi _r)' \right ], \tag{23.31}

and a reduced energy of the schematic form

\delta W_E = \int _0^a r\,dr \left [ \frac {4\gamma p\,B_\theta ^2/\muo }{\gamma p + B_\theta ^2/\muo } + 2 r\,p' \right ]\frac {|\xi _r|^2}{r^2}. \tag{23.32}

A convenient way to read this is

-\,\frac {r p'}{p} < \frac {2\gamma \,B_\theta ^2/\muo }{\gamma p + B_\theta ^2/\muo }, \qquad (m=0), \tag{23.33}

which makes the competition between pressure-gradient drive and compressional restoring forces explicit: this is simply the interchange instability manifested in the $z$ pinch geometry.

Takeaways

The $\theta $ pinch teaches that line bending and magnetic compression can make a high-$\beta $ equilibrium ideally stable. The $z$ pinch teaches the opposite lesson: once field lines are curved, pressure and current can feed kink and sausage modes. Those same ingredients are exactly what Newcomb’s coefficients organize in the general screw-pinch problem.

23.3 The Screw Pinch

The general screw pinch contains both limiting cases at once. The axial field allows field-line bending along $z$, the azimuthal field introduces curvature, and their mixture produces the resonant combination $F(r)$. Newcomb’s reduction is therefore not a new physical principle; it is the careful algebraic organization of the same energy principle used above.

Newcomb starts with the straight screw-pinch equilibrium

\B (r)=B_\theta (r)\,\vect {e}_\theta +B_z(r)\,\vect {e}_z, \qquad p=p(r), \tag{23.34}

with radial force balance

\dd {}{r}\left (p+\frac {B_\theta ^2+B_z^2}{2\muo }\right ) +\frac {B_\theta ^2}{\muo r}=0. \tag{23.35}

Equivalently,

\muo p' =-\frac 12(B^2)'-\frac {B_\theta ^2}{r}, \qquad B^2\equiv B_\theta ^2+B_z^2, \tag{23.36}

where a prime denotes $d/dr$. This identity is what later removes the explicit equilibrium-field derivatives from $g(r)$.

Mode decomposition. Take

\bm {\xi }(r,\theta ,z)=\bm {\xi }(r)e^{i(m\theta +kz)}, \qquad m\in \mathbb {Z}, \tag{23.37}

and introduce the field-aligned basis

\bvec \equiv \frac {\B }{B}, \qquad B\equiv \sqrt {B_\theta ^2+B_z^2}, \qquad \bm {e}_\eta \equiv -\vect {e}_r\times \bvec = \frac {B_z}{B}\vect {e}_\theta -\frac {B_\theta }{B}\vect {e}_z. \tag{23.38}

Decompose the displacement as

\bm {\xi } = \xi (r)\,\vect {e}_r+\eta (r)\,\bm {e}_\eta +\xi _\parallel (r)\,\bvec . \tag{23.39}

Thus

\xi =\xi _r, \qquad \xi _\parallel =\frac {\xi _\theta B_\theta +\xi _z B_z}{B}, \qquad \eta =\frac {\xi _\theta B_z-\xi _z B_\theta }{B}. \tag{23.40}

The useful inverse relations are

\xi _\theta =\frac {B_z}{B}\eta +\frac {B_\theta }{B}\xi _\parallel , \qquad \xi _z=-\frac {B_\theta }{B}\eta +\frac {B_z}{B}\xi _\parallel . \tag{23.41}

The two helical combinations $F$ and $G$. Two combinations of $(m,k)$ and the equilibrium field occur everywhere:

F(r)\equiv \frac {m B_\theta }{r}+k B_z, \qquad G(r)\equiv \frac {m B_z}{r}-k B_\theta , \qquad k_0^2(r)\equiv k^2+\frac {m^2}{r^2}. \tag{23.42}

They are simply the projections of the wave vector parallel and perpendicular to the equilibrium field:

\B \cdot \grad =iF, \qquad \bm {e}_\eta \cdot \grad =i\frac {G}{B}. \tag{23.43}

They also satisfy the two identities

F^2+G^2=B^2k_0^2, \qquad F B_z-G B_\theta =kB^2. \tag{23.44}

Both identities are used in the square completion below.

Incompressibility eliminates the parallel displacement. Away from a rational surface, the plasma-compression term in (23.1) is minimized by taking

\divergence \bm {\xi }=0. \tag{23.45}

Since the basis vectors depend only on $r$ and the field has no radial component,

\divergence \bm {\xi } = \bvec \cdot \grad \xi _\parallel + \divergence \bm {\xi }_\perp , \qquad \bm {\xi }_\perp =\xi \,\vect {e}_r+\eta \,\bm {e}_\eta .

Moreover,

\divergence \bm {\xi }_\perp = \frac {(r\xi )'}{r}+i\frac {G}{B}\eta . \tag{23.47}

Using $\bvec \cdot \grad =(\B \cdot \grad )/B=iF/B$, the incompressibility constraint becomes

0=\frac {iF}{B}\xi _\parallel + \frac {(r\xi )'}{r}+i\frac {G}{B}\eta ,

\xi _\parallel = \frac {iB}{F}\left [\frac {(r\xi )'}{r}+i\frac {G}{B}\eta \right ] = \frac {iB}{F}\,\divergence \bm {\xi }_\perp , \qquad (F\neq 0). \tag{23.49}

At a rational surface $r=r_s$ where $F(r_s)=0$, this algebraic elimination becomes singular. That singularity is the mathematical signal that the parallel field-line-bending restoring force has vanished for that helical perturbation.

The curvature and the compression factor. For the straight screw pinch,

\bm {\kappa }=(\bvec \cdot \grad )\bvec = -\frac {B_\theta ^2}{B^2}\,\frac {\vect {e}_r}{r}. \tag{23.50}

Therefore

\begin{aligned}\divergence \bm {\xi }_\perp +2\bm {\kappa }\cdot \bm {\xi }_\perp &= \frac {(r\xi )'}{r}+i\frac {G}{B}\eta -\frac {2B_\theta ^2}{rB^2}\xi .\end{aligned} \tag{23.51}

It is convenient to name the radial derivative that appears here:

S\equiv \frac {(r\xi )'}{r}=\xi '+\frac {\xi }{r}. \tag{23.52}

Then the magnetic-compression factor is

\divergence \bm {\xi }_\perp +2\bm {\kappa }\cdot \bm {\xi }_\perp = S-\frac {2B_\theta ^2}{rB^2}\xi +i\frac {G}{B}\eta . \tag{23.53}

The perpendicular perturbed magnetic field. The part of $\bm {\xi }\times \B $ that matters is independent of $\xi _\parallel $:

\bm {\xi }\times \B =B\eta \,\vect {e}_r-B\xi \,\bm {e}_\eta =B\eta \,\vect {e}_r-B_z\xi \,\vect {e}_\theta +B_\theta \xi \,\vect {e}_z. \tag{23.54}

Taking the curl component by component gives

\begin{aligned}Q_r &= \frac {1}{r}\frac {\partial }{\partial \theta }(B_\theta \xi e^{i(m\theta +kz)}) -\frac {\partial }{\partial z}(-B_z\xi e^{i(m\theta +kz)}) = iF\xi , \\[0.4em] Q_\theta &= ikB\eta -(B_\theta \xi )', \\[0.4em] Q_z &= -\frac {1}{r}(rB_z\xi )'-i\frac {m}{r}B\eta .\end{aligned} \tag{23.55}

Projecting onto $\bm {e}_\eta =(B_z\vect {e}_\theta -B_\theta \vect {e}_z)/B$ gives

\begin{aligned}Q_\eta &= \frac {B_zQ_\theta -B_\theta Q_z}{B} \nonumber \\ &= iF\eta + \frac {\xi }{B} \left [ B_\theta B_z'-B_zB_\theta '+\frac {B_\theta B_z}{r} \right ].\end{aligned} \tag{23.58}

Define the bracketed equilibrium combination

\mathcal {A} \equiv B_\theta B_z'-B_zB_\theta '+\frac {B_\theta B_z}{r}. \tag{23.59}

Then

\bm {Q}_\perp =Q_r\vect {e}_r+Q_\eta \bm {e}_\eta = iF\xi \,\vect {e}_r+ \left (iF\eta +\frac {\mathcal {A}}{B}\xi \right )\bm {e}_\eta . \tag{23.60}

The parallel current is

\muo J_\parallel = \frac {B_z}{rB}(rB_\theta )' - \frac {B_\theta }{B}B_z', \tag{23.61}

and the combination $\mathcal {A}$ is related to it by

\mathcal {A}+\muo J_\parallel B = \frac {2B_\theta B_z}{r}. \tag{23.62}

This identity is one of the main cancellations in the derivation.

Write the energy density before eliminating $\eta $. With $\divergence \bm {\xi }=0$, the plasma-compression term is gone. Multiplying the remaining integrand in (23.1) by $\muo $, write

\delta W = \frac {1}{2\muo }\int dV\,\mathcal {L}, \tag{23.63}

where

\begin{aligned}\mathcal {L} =&\;|\bm {Q}_\perp |^2 +B^2\left |S-\frac {2B_\theta ^2}{rB^2}\xi +i\frac {G}{B}\eta \right |^2 +\frac {2\muo p'B_\theta ^2}{rB^2}|\xi |^2 \nonumber \\[0.4em] &\;- \muo J_\parallel \left [ \frac {1}{B}(\bm {\xi }_\perp ^*\times \B )\cdot \bm {Q}_\perp \right ].\end{aligned} \tag{23.64}

The pressure–curvature term has become

-2(\bm {\xi }_\perp \cdot \grad p)(\bm {\xi }_\perp ^*\cdot \bm {\kappa }) = \frac {2p'B_\theta ^2}{rB^2}|\xi |^2. \tag{23.65}

For the current term, use

\bm {\xi }_\perp ^*\times \B =B\eta ^*\vect {e}_r-B\xi ^*\bm {e}_\eta ,

so that

\frac {1}{B}(\bm {\xi }_\perp ^*\times \B )\cdot \bm {Q}_\perp = iF(\eta ^*\xi -\xi ^*\eta )-\frac {\mathcal {A}}{B}|\xi |^2. \tag{23.67}

Collect only the terms that contain $\eta $. The coefficient of $|\eta |^2$ comes from $|Q_\eta |^2$ and magnetic compression:

F^2|\eta |^2+G^2|\eta |^2=(F^2+G^2)|\eta |^2=B^2k_0^2|\eta |^2. \tag{23.68}

The cross term between $\eta $ and $S$ comes only from magnetic compression:

iBG(\eta S^*-\eta ^*S). \tag{23.69}

The cross term between $\eta $ and $\xi $ is where the cancellation is easiest to miss. From $|Q_\eta |^2$ and the current term one obtains

\begin{aligned}& iF\left (\frac {\mathcal {A}}{B}+\muo J_\parallel \right ) (\eta \xi ^*-\eta ^*\xi ) \nonumber \\ &\hspace {2cm}= \frac {2iF B_\theta B_z}{rB} (\eta \xi ^*-\eta ^*\xi ),\end{aligned} \tag{23.70}

where (23.62) was used. From the curvature shift inside the magnetic-compression square one obtains

-\frac {2iG B_\theta ^2}{rB} (\eta \xi ^*-\eta ^*\xi ). \tag{23.71}

Adding (23.70) and (23.71), and using $F B_z-G B_\theta =kB^2$, gives

\frac {2ikBB_\theta }{r}(\eta \xi ^*-\eta ^*\xi ). \tag{23.72}

Therefore the full $\eta $-dependent part of $\mathcal {L}$ is

\begin{aligned}\mathcal {L}_\eta =&\; B^2k_0^2|\eta |^2 +iBG(\eta S^*-\eta ^*S) +\frac {2ikBB_\theta }{r}(\eta \xi ^*-\eta ^*\xi ) \nonumber \\ =&\; B^2k_0^2|\eta |^2 +iB\left [\eta \left (GS+\frac {2kB_\theta }{r}\xi \right )^* -\eta ^*\left (GS+\frac {2kB_\theta }{r}\xi \right )\right ].\end{aligned} \tag{23.73}

Now the square completion is explicit:

\begin{aligned}\mathcal {L}_\eta =&\; B^2k_0^2 \left | \eta - \frac {i}{Bk_0^2}\left (GS+\frac {2kB_\theta }{r}\xi \right ) \right |^2 \nonumber \\ &\;- \frac {1}{k_0^2} \left | GS+\frac {2kB_\theta }{r}\xi \right |^2.\end{aligned} \tag{23.74}

Thus the minimizing value of the binormal displacement is

\eta = \frac {i}{Bk_0^2}\left (GS+\frac {2kB_\theta }{r}\xi \right ) = \frac {i}{r\,k_0^2\,B} \left [G(r\xi )'+2kB_\theta \xi \right ]. \tag{23.75}

This is the precise point at which $\eta $ disappears from Newcomb’s problem. The square itself is non-negative, so minimizing sets it to zero; the negative last term in (23.74) must still be kept in the reduced radial functional.

The reduced integrand before integration by parts. After substituting (23.75), the reduced integrand can be written in the compact form

\mathcal {L}_{\rm red} =a|S|^2+b(S\xi ^*+S^*\xi )+c|\xi |^2, \tag{23.76}

where

a=\frac {F^2}{k_0^2}. \tag{23.77}

This coefficient comes from

B^2|S|^2-\frac {G^2}{k_0^2}|S|^2 =\frac {B^2k_0^2-G^2}{k_0^2}|S|^2 =\frac {F^2}{k_0^2}|S|^2.

The mixed coefficient is

b =-\frac {2B_\theta ^2}{r} -\frac {2kB_\theta G}{rk_0^2}. \tag{23.79}

The first term in $b$ comes from the curvature shift in $B^2|S-2B_\theta ^2\xi /(rB^2)|^2$; the second comes from the negative square term in (23.74). Finally,

\begin{aligned}c =&\; \underbrace {F^2}_{|Q_r|^2} + \underbrace {\frac {2B_\theta B_z}{rB^2}\mathcal {A}}_{|Q_\eta |^2\;{\rm plus\;current}} + \underbrace {\frac {4B_\theta ^4}{r^2B^2}}_{{\rm curvature\;shift}} + \underbrace {\frac {2\muo p'B_\theta ^2}{rB^2}}_{{\rm pressure\text {--}curvature}} - \underbrace {\frac {4k^2B_\theta ^2}{r^2k_0^2}}_{{\rm eliminated}\;\eta \;{\rm square}} .\end{aligned} \tag{23.80}

This equation is often the most useful bookkeeping step: it shows exactly where the pieces of the final $g(r)$ originate before the equilibrium relation has hidden the derivatives of $B_\theta $ and $B_z$.

Convert from $S=(r\xi )'/r$ to $\xi '$. The volume element supplies a factor of $r$, so the radial integrand is $r\mathcal {L}_{\rm red}$. Since $S=\xi '+\xi /r$,

\begin{aligned}r\mathcal {L}_{\rm red} =&\; r a|\xi '|^2 + (a+rb)(\xi '\xi ^*+\xi '^*\xi ) + \left (\frac {a}{r}+2b+rc\right )|\xi |^2.\end{aligned} \tag{23.81}

The middle term is a total derivative plus a leftover radial coefficient:

\int _0^a (a+rb)(\xi '\xi ^*+\xi '^*\xi )\,dr = \left [(a+rb)|\xi |^2\right ]_0^a - \int _0^a (a+rb)'|\xi |^2\,dr . \tag{23.82}

Regularity removes the axis contribution. Therefore

f(r)=ra=\frac {rF^2}{k_0^2}, \tag{23.83}

and, before simplification,

g(r)=\frac {a}{r}+2b+rc-(a+rb)'. \tag{23.84}

The boundary coefficient is also explicit. Directly from (23.79),

\begin{aligned}a+rb &= \frac {F^2}{k_0^2}-2B_\theta ^2-\frac {2kB_\theta G}{k_0^2} \nonumber \\ &= \frac {F^2-2kB_\theta G-2B_\theta ^2k_0^2}{k_0^2} = \frac {F F^\dagger }{k_0^2},\end{aligned} \tag{23.85}

where

F^\dagger (r)\equiv kB_z-\frac {mB_\theta }{r}. \tag{23.86}

Thus the surface term left by integration by parts is

\left [\frac {F F^\dagger }{k_0^2}\right ]_{r=a}|\xi _a|^2, \qquad \xi _a\equiv \xi (a). \tag{23.87}

Simplify $g(r)$. Substitute (23.77)–(23.80) into (23.84). Before using force balance, the coefficient of $|\xi |^2$ is

\begin{aligned}\widetilde g =&\; \underbrace {\frac {F^2}{rk_0^2}}_{a/r} - \underbrace {\frac {4B_\theta ^2}{r} +\frac {4kB_\theta G}{rk_0^2}}_{2b} + \underbrace {rF^2 +\frac {2B_\theta B_z}{B^2}\mathcal {A} +\frac {4B_\theta ^4}{rB^2} +\frac {2\muo p'B_\theta ^2}{B^2} -\frac {4k^2B_\theta ^2}{rk_0^2}}_{rc} - \underbrace {\left (\frac {F F^\dagger }{k_0^2}\right )'}_{(a+rb)'} .\end{aligned} \tag{23.88}

This long expression is the raw collection of terms. To reduce it, insert

\mathcal {A}=B_\theta B_z'-B_zB_\theta '+\frac {B_\theta B_z}{r}, \qquad F^\dagger =kB_z-\frac {mB_\theta }{r},

and collect powers of $k$ and $m/r$. The result can be written as

\begin{aligned}\widetilde g =&\; \frac {2k^2}{k_0^2}\,\muo p' + \left (\frac {k_0^2r^2-1}{k_0^2r^2}\right )rF^2 + \frac {2k^2}{r k_0^4} \left (kB_z-\frac {mB_\theta }{r}\right )F \nonumber \\ &\;- \frac {2F F^\dagger }{B^2k_0^2} \left [ \muo p'+\frac 12(B^2)'+\frac {B_\theta ^2}{r} \right ].\end{aligned} \tag{23.90}

The bracket in the second line is exactly the radial force-balance condition (23.36). Hence the last line vanishes, leaving

g(r) = \underbrace {\frac {2k^2}{k_0^2}\,\muo p'}_{\text {pressure drive}} + \underbrace {\left (\frac {k_0^2r^2-1}{k_0^2r^2}\right )rF^2}_{\text {line bending / geometry}} + \underbrace {\frac {2k^2}{r k_0^4} \left (kB_z-\frac {mB_\theta }{r}\right )F}_{\text {helical coupling}} . \tag{23.91}

The origin of the three terms is now visible. The pressure-drive term is the part of the pressure–curvature contribution that survives the equilibrium simplification. The line-bending/geometric term comes from expanding $S=\xi '+\xi /r$ and integrating the mixed term by parts. The final coupling term comes from the same integration-by-parts coefficient $a+rb$, together with the subtraction produced by the completed $\eta $ square.

Newcomb’s one-dimensional functional. Combining the radial integral with the angular and axial integrals gives

\frac {\delta W}{2\pi ^2 R_0/\muo } = \int _0^{a} \left [ f(r)\,|\xi '|^2+g(r)\,|\xi |^2 \right ]dr + \left [ \frac {F F^\dagger }{k_0^2} \right ]_{r=a} |\xi _a|^2. \tag{23.92}

The key point is that all of the dangerous local behavior is concentrated where $f\propto F^2$ becomes small. At a rational surface, $F(r_s)=0$, the leading coefficient of the radial derivative vanishes, and the Newcomb equation becomes singular.

23.4 How the Euler–Lagrange equation emerges

Tutorial

Start with a one-dimensional functional. After the angular and parallel structure have been eliminated, the reduced energy has the generic form
$\mathcal {W}[\xi ] = \int _0^a \left (f|\xi '|^2+g|\xi |^2\right )dr +\mathcal {B}\,|\xi _a|^2, \tag{23.93}$ where $\xi _a\equiv \xi (a)$ and $\mathcal {B}$ represents whatever boundary term is left after the vacuum or wall has been integrated out.

Take a variation. Write
$\xi (r)\rightarrow \xi (r)+\epsilon \,\eta (r), \qquad \eta (0)\ \text {regular},$ and differentiate with respect to $\epsilon $ at $\epsilon =0$. For a real trial function, $\begin{aligned}\frac {1}{2}\,\delta \mathcal {W} &= \int _0^a \left (f\xi '\eta '+g\xi \eta \right )dr +\mathcal {B}\,\xi _a\eta _a .\end{aligned} \tag{23.95}$
If one keeps $\xi $ complex, the same result follows by varying $\xi $ and $\xi ^\ast $ independently and then taking the real part at the end.

Do the integration by parts explicitly. The derivative on $\eta $ is the awkward term. Move it onto $\xi $:
$\begin{aligned}\int _0^a f\xi '\eta '\,dr &= \left [f\xi '\eta \right ]_0^a-\int _0^a (f\xi ')'\eta \,dr.\end{aligned}$
Therefore
$\begin{aligned}\frac {1}{2}\,\delta \mathcal {W} &= \left [f\xi '\eta \right ]_0^a -\int _0^a \left [(f\xi ')'-g\xi \right ]\eta \,dr +\mathcal {B}\,\xi _a\eta _a .\end{aligned} \tag{23.97}$
This is the key step: the entire ODE comes from moving one derivative off the test function.

Interior Euler–Lagrange equation. If the edge displacement is fixed, then $\eta (a)=0$. Regularity also removes the axis term, and the remaining interior variation is
$\delta \mathcal {W} = -2\int _0^a \left [(f\xi ')'-g\xi \right ]\eta \,dr .$ Since $\eta (r)$ is otherwise arbitrary, the minimizing displacement must satisfy $\dd {}{r}\left (f(r)\,\dd {\xi }{r}\right )-g(r)\,\xi =0. \tag{23.99}$ This is the Newcomb equation.

Natural boundary condition. If the boundary is free, then $\eta (a)$ is arbitrary as well. The coefficient of $\eta (a)$ in (23.97) must therefore vanish:
$f(a)\,\xi '(a)+\mathcal {B}\,\xi (a)=0. \tag{23.100}$ The vacuum or wall changes only $\mathcal {B}$; it does not change the interior Euler–Lagrange operator.

Why this matters in practice. The minimizing perturbation is not guessed. It is the solution of (23.99) that is regular at the axis and satisfies either a fixed or natural boundary condition at the edge. This is why one can turn a variational stability problem into an ordinary differential equation without losing the energy-principle content.

23.5 How Newcomb’s logic should be applied

First restrict to internal perturbations. The first question is whether the plasma can lower its energy without moving the boundary. So one first minimizes the reduced functional over displacements that are regular at the axis and satisfy

\xi _a \equiv \xi (a)=0. \tag{23.101}

This is the fixed-boundary problem. Any negative $\delta W$ found in this restricted class is an internal instability.

Then check the singular local limits. Within that fixed-boundary class, the most delicate radii are the rational surfaces $F(r_s)=0$, where field-line bending disappears. The localized pressure-driven tests developed in the next lecture—Suydam in a cylinder and Mercier in a torus—are therefore part of the internal problem, not a separate external calculation.

Then solve the global Newcomb equation. If the local tests are passed, one integrates the regular solution of (23.99) from the axis outward and checks for nodes. A node signals that a lower-energy trial function exists. A node-free solution in the fixed-boundary class is the candidate internally stable solution.

Only after the fixed-boundary problem is stable does one release the edge. The external problem is the enlarged variational problem in which the boundary is allowed to move, so $\xi _a$ is no longer constrained to vanish. In that sector one must add the vacuum and wall energy. A nonzero $\xi _a$ means the plasma boundary participates in the perturbation; that is what makes the mode external.

Caution

It is worth being explicit about notation. For internal modes one imposes $\xi _a=0$. For external modes one allows $\xi _a\neq 0$. The sign of $\xi _a$ is just an amplitude convention, so writing $\xi _a>0$ does not by itself prove instability. Instability means that the minimized total energy is negative. What $\xi _a\neq 0$ tells you is that the boundary has moved and the mode belongs to the external sector.

Bibliography

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.