Magnetostatic equilibrium

A magnetostatic equilibrium is a time-independent configuration in which magnetic, pressure, gravitational, and any other body-force densities balance.

For an axisymmetric magnetic field $\mathbf{B}=\psi {\mathbf{e}}_{\varphi }/(r\sin \theta )+\nabla \times [\chi {\mathbf{e}}_{\varphi }/(r\sin \theta )]$, the poloidal magnetic flux function gives $B_r={\chi }_{\theta }/(r^2\sin \theta )$ and $B_{\theta }=-{\chi }_r/(r\sin \theta )$. Force balance implies locally $p=p(\chi )$ and $\psi =\psi (\chi )$ on regular connected flux surfaces. Direct calculation gives $(\nabla \times \mathbf{B})\times \mathbf{B}=-({\Delta }_{\ast }\chi +\psi {\psi }^{\prime })\nabla \chi /(r^2{\sin }^2\theta )$, with ${\Delta }_{\ast }\chi ={\chi }_{rr}+\sin \theta {\partial }_{\theta }({\chi }_{\theta }/\sin \theta )/r^2$. Equating this to ${\mu }_0\nabla p$ proves the displayed magnetostatic equilibrium equation, including its sign and dimensions.

Set $\psi =\alpha \chi$, $p=p_0+P\chi$ and $\chi =F(r){\sin }^2\theta$. A regular family is $F=A[\cos (\alpha r)-\sin (\alpha r)/(\alpha r)]+c{r}^2$, where $P=-{\alpha }^{2}c/{\mu }_0$. A zero magnetic field at radius a requires $F(a)=F^{\prime }(a)=0$. For nonzero A this yields the displayed condition with $x=\alpha a$ and $c=-A[\cos x-\sin x/x]/a^2$. The boundary pressure is $p_0$, independent of angle. The trigonometric pair is tied by regularity and the radial differential equation; its coefficients are not independent.

A plane-parallel magnetostatic equilibrium with constant downward gravitational acceleration $g$, horizontal magnetic field, constant isothermal sound speed $c_s$, and constant Alfvén speed $v_a$ hasIndeed the magnetohydrodynamic total pressure is $(c_s^2+v_a^2/2)\rho$, and its vertical derivative is $-g\rho$. The magnetic pressure therefore increases the scale height above that of an unmagnetized atmosphere.

A cylindrical magnetostatic equilibrium is an axisymmetric vector field and pressure configuration independent of the axial coordinate. Its radial Lorentz force density balances the pressure gradient, as expressed by cylindrical magnetostatic pressure balance.

In a Z-pinch, an axial electric current produces a toroidal magnetic field whose inward magnetic tension confines a plasma. A cylindrical magnetostatic equilibrium balances the resulting Lorentz force density against a pressure gradient.

For an axisymmetric vector field independent of the axial coordinate and regular on the axis, Gauss's law for magnetism gives $B_R=0$. The magnetostatic Ampère-Maxwell equation then gives $j_R=0$, $j_{\varphi }=-c{B}_z^{\prime }/(4\pi )$, and $j_z=c(R{B}_{\varphi }{)}^{\prime }/(4\pi R)$ in Gaussian units. Radial magnetostatic equilibrium isFor enclosed axial electric current $I(R)$, the magnetostatic Ampère-Maxwell equation gives $B_{\varphi }=2I/(cR)$, henceThe first term contains pressure and axial magnetic pressure; the second combines toroidal magnetic pressure with the hoop force from magnetic tension.

For a regular cylindrical flux tube with $\mathbf{B}=(0,B_{\varphi }(s),B_z(s))$, vanishing field beyond radius $s_0$, and exterior pressure $p_e$, define $⟨h⟩=2s_0^{-2}{\int }_0^{s_0}sh(s)ds$ and $B_p^2=2{\mu }_0{p}_e$. Integrating cylindrical magnetostatic pressure balance against $s^2$ gives the displayed identity. The toroidal magnetic pressure and hoop magnetic tension cancel in this cross-sectional axial-field average. Nonnegative gas pressure gives $⟨B_z^2⟩\le B_p^2$, with strict inequality precisely when $⟨p⟩>0$.

For a flux tube with $B_{\varphi }=\gamma (s/s_0)B_z$, regularity at the axis and cylindrical magnetostatic pressure balance give $B_z(0{)}^2=B_p^2-2{\mu }_{0}p(0)+{\gamma }^2⟨B_z^2⟩$. The flux-tube axial field virial identity and $p\ge 0$ give the bound. A twisted toroidal magnetic field can support a central axial field above the untwisted magnetic pressure limit even though its cross-sectional mean obeys that limit.

For constant axial magnetic field and $p=p_0(1+R^2/a^2{)}^{-k}$ with $p_0>0$ and $k>0$, cylindrical magnetostatic pressure balance gives, with $x=R^2/a^2$,The nontrivial electric current is finite at infinity exactly when $k>1$, in which case $I^2(\infty )=2\pi c^2{p}_0{a}^2/(k-1)$. Either sign of $I$ is possible because the pressure balance fixes its square.

With no axial electric current, cylindrical magnetostatic pressure balance reduces to constant $p+B_z^2/(8\pi )$. For $p=p_0{e}^{-R^2/a^2}$ and $B_z(0)=0$, its radial solutions areAs $R\downarrow 0$, $B_z\sim \sigma \sqrt{8\pi p_0}R/a$ and $j_{\varphi }\to -\sigma c\sqrt{8\pi p_0}/(4\pi a)$. These finite scalar limits do not give a smooth function of Cartesian position: $R=\sqrt{x^2+y^2}$ has a cusp at the axis, and the azimuthal unit vector has no unique limit there. Thus these data describe radial equilibrium away from the axis, but cannot satisfy smooth vector-field regularity on the axis without modifying the data.