A Z-Pinch That Would Work, If It Were Stable

The problem

Design a Z-pinch to reach fusion-relevant temperature, check its equilibrium with the Bennett relation — then compute how long it survives.

Step 1: the equilibrium

The Bennett relation follows from radial force balance and is independent of profile shape:

\[\frac{\mu_0I^2}{8\pi} = Nk_B(T_e+T_i)\]

Take \(I = 1\) MA and a line density \(N = 10^{19}\) ions per metre:

\[k_B(T_e+T_i) = \frac{\mu_0I^2}{8\pi N} = \frac{(4\pi\times10^{-7})(10^{6})^2}{8\pi\times10^{19}} = \frac{4\pi\times10^{5}}{8\pi\times10^{19}} = 5\times10^{-15}\ \text{J}\]

Converting: \(5\times10^{-15}/1.602\times10^{-19} = 3.1\times10^{4}\) eV, so

\[T_e = T_i \approx 16\ \text{keV}\]

Fusion-relevant temperature from one megaamp and no external field whatsoever. In 1952 this calculation looked like a shortcut to a power station.

Step 2: the supporting numbers

Take a pinch radius \(a = 1\) cm. Then:

\[n = \frac{N}{\pi a^2} = \frac{10^{19}}{\pi(10^{-2})^2} = 3.2\times10^{22}\ \text{m}^{-3}\]

\[B_\theta(a) = \frac{\mu_0I}{2\pi a} = \frac{(4\pi\times10^{-7})(10^6)}{2\pi(10^{-2})} = 20\ \text{T}\]

Check the triple product against the Lawson criterion: \(nT = 3.2\times10^{22}\times16 = 5.1\times10^{23}\) keV·m⁻³, so ignition needs only

\[\tau_E \gtrsim \frac{3\times10^{21}}{5.1\times10^{23}} = 6\ \text{ms}\]

Six milliseconds. Not obviously unreasonable.

Step 3: the Alfvén time

\[\rho = nm_i = 3.2\times10^{22}\times1.67\times10^{-27} = 5.3\times10^{-5}\ \text{kg/m}^3\]

\[v_A = \frac{B}{\sqrt{\mu_0\rho}} = \frac{20}{\sqrt{(4\pi\times10^{-7})(5.3\times10^{-5})}} = \frac{20}{8.2\times10^{-6}} = 2.4\times10^{6}\ \text{m/s}\]

\[\tau_A = \frac{a}{v_A} = \frac{10^{-2}}{2.4\times10^6} = 4.1\times10^{-9}\ \text{s}\]

Step 4: the verdict

Ideal MHD instabilities grow on the Alfvén timescale. The \(m=0\) sausage and \(m=1\) kink modes have growth rates \(\gamma \sim v_A/a\), so the column distorts in a few \(\tau_A\):

\[\tau_{\rm instability} \sim 10\,\tau_A \approx 41\ \text{ns}\]

Compare with what is needed:

\[\frac{\tau_E^{\rm required}}{\tau_{\rm instability}} = \frac{6\times10^{-3}}{4.1\times10^{-8}} \approx 1.5\times10^{5}\]

The pinch must survive well over a hundred thousand times longer than it does. Not a factor to engineer away. Every early pinch experiment saw exactly this: beautiful compression, then sausages and kinks, then a spray of plasma into the wall, all inside a microsecond. The neutrons those machines produced turned out to come from beam-target reactions in the disrupting necks, not from thermonuclear burn — a distinction that took years and some embarrassment to establish.

Step 5: what fixes it

Add axial \(B_z\). Compressing or bending the column now costs magnetic energy. Stability needs roughly \(B_z^2 > B_\theta^2/2\), converting the Z-pinch into a screw pinch — and, closed into a torus with the right safety factor, into a tokamak. This is, historically, how the tokamak was arrived at.
Sheared axial flow. Velocity shear across the column can suppress both modes; this is still an active line of fusion research.
Or: give up on stability and go fast. If the pinch only needs to live 100 ns, use it as a pulsed X-ray source rather than a reactor. That is exactly what Sandia's Z machine does — wire-array Z-pinches producing the most intense laboratory X-ray source on Earth, used to drive inertial-confinement capsules.

The lesson

Equilibrium is easy; stability is the whole subject. The Bennett relation is exact, profile-independent, and completely silent on whether the configuration survives being nudged — which is why the energy principle, and not force balance, is the tool that decides whether a confinement scheme is viable.

Pinch equilibria · Kink instability · Energy principle · Magnetic stress tensor · Lawson criterion · Virial theorem