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The general ideal magnetic-energy balance has outward-surface contribution ${\mu }_0^{-1}\oint [(u\cdot B)B-B^{2}u]\cdot dS$ and volume contribution $-\int u\cdot (J\times B)dV$. The latter vanishes for a force-free magnetic field. For an axisymmetric boundary rotating tangentially, $u\cdot B=\Omega f(\psi )$ and the directed flux of a thin tube is $2\pi d\psi$. Pair its exit and entry endpoints to obtain the displayed energy injection, counting each boundary endpoint once. Equal endpoint angular velocities inject no magnetic energy.

Let $K$ be the kinetic energy, $M$ the magnetic energy and $E=K+M$:Dot the magnetohydrodynamic momentum equation with $\mathbf{v}$ and integrate. Incompressibility and the zero boundary velocity remove the advective and pressure work. The viscous term integrates to $-\rho \nu {\int }_D|\nabla \mathbf{v}{|}^2$. ThusThe ideal magnetohydrodynamic induction equation gives, with no boundary contribution,The Lorentz force density therefore transfers energy between the field and the fluid without creating total energy. Since $\nabla \cdot \mathbf{v}=0$ and $\mathbf{v}=0$ on the boundary, another integration by parts gives $\int |\nabla \mathbf{v}{|}^2=\int |{\mathbf{\omega }}_v{|}^2$, where ${\mathbf{\omega }}_v=\nabla \times \mathbf{v}$ is vorticity. The viscous magnetic-relaxation energy identity isThe conserved nonzero magnetic helicity prevents all the field energy from disappearing. Fix a bounded inverse-curl convention for the magnetic vector potential, so that $\Vert \mathbf{A}{\Vert }_2\le C_D\Vert \mathbf{B}{\Vert }_2$. The Cauchy-Schwarz inequality gives the magnetic-energy lower bound from helicity,Consequently $E$ decreases to a finite positive limit andUnder the intended regular long-time interpretation of smooth vorticity, the dissipation cannot keep producing peaks of fixed height and arbitrarily short duration. More precisely, if $d(t)=\Vert {\mathbf{\omega }}_v(t){\Vert }_2^2$ is uniformly continuous, uniformly continuous integrable dissipation tends to zero proves $d(t)\to 0$. The Poincaré inequality and the identity above then imply $\Vert \mathbf{v}(t){\Vert }_2\to 0$.

To identify the asymptotic state, suppose the regular relaxation has a limit, or work on a compact invariant limiting set on which the energy is continuous. On such a set the energy is constant, so the displayed identity forces ${\mathbf{\omega }}_v=0$, and the zero boundary velocity forces $\mathbf{v}=0$. The induction equation is then stationary, and the momentum equation becomesThis is magnetostatic equilibrium, with nonzero field allowed and required here by conserved magnetic helicity. It is not necessarily a force-free magnetic field: a pressure gradient can balance the magnetic force. If the field develops a current sheet, equilibrium is understood with the corresponding interface force balance.

There is a mathematical qualification to the smoothness premise. Smoothness at each finite time alone does not prove uniform long-time bounds, uniform continuity of dissipation or convergence to one limiting field. The energy calculation proves finite total dissipation unconditionally for a smooth solution; the settling conclusion is the regular-relaxation argument just given. A rigorous global convergence theorem requires that additional long-time control. In particular, smooth vorticity does not require smooth electric current density in the limiting state.

Put $I=R{B}_{\varphi }$ and define ${\Delta }_{\ast }\psi ={\psi }_{RR}-R^{-1}{\psi }_R+{\psi }_{zz}$. Direct cylindrical differentiation givesThe azimuthal component of the Lorentz force is proportional to $I_z{\psi }_R-I_R{\psi }_z$. A force-free magnetic field therefore has parallel meridional gradients of $I$ and $\psi$, implying $I=f(\psi )$ on connected regular flux surfaces. ConsequentlyWith this dependence the poloidal part of $\nabla \times B$ is $f^{\prime }(\psi )B_p$. Its full cross product with $B$ isVanishing force therefore requires ${\Delta }_{\ast }\psi +f{f}^{\prime }=0$ wherever the flux gradient is nonzero, extending by regularity through ordinary isolated nulls. Finally, evaluating the three-dimensional cylindrical divergence gives $R^2\nabla \cdot (R^{-2}\nabla \psi )={\Delta }_{\ast }\psi$, soThis derives the Grad-Shafranov equation for a force-free magnetic field. The function $f$ is set by boundary/current information; it is not an additional fixed constant. Its single-valued global form presumes the usual connected flux-surface labeling. The equation also makes $\nabla \times B=f^{\prime }(\psi )B$, explicitly showing that current is parallel to the magnetic field.

A magnetostatic force-free magnetic field is in equilibrium when its electric current exerts no Lorentz force density:Thus $\mathbf{J}$ is pointwise parallel to $\mathbf{B}$, so for some scalar $\alpha (\mathbf{r})$,Using the magnetostatic Ampère's law, $\nabla \times \mathbf{B}=(4\pi /c)\mathbf{J}$, givesTaking the divergence and using $\nabla \cdot \mathbf{B}=0$ yieldsThe force-free parameter is therefore constant along each magnetic field line.

Define the accumulated rotation angleThe coupled first-order equations describe a rotation of $(B_y,B_z)$ and have the general solutionDirect differentiation verifies both force-free equations, and $B_y^2+B_z^2=C_1^2+C_2^2$ is constant. Thus a one-dimensional force-free magnetic field rotates without changing its magnitude.

In a magnetostatic equilibrium, inertia is absent and all forces balance. A force-free magnetic field is the limiting case in which the Lorentz force density vanishes:Thus the current density is parallel to the magnetic field, so for some scalar force-free parameter $\alpha (\mathbf{r})$,Using Ampère's law in magnetostatics givesTaking the divergence and using both the divergence of a curl is zero and $\nabla \cdot \mathbf{B}=0$ yieldssoThe force-free parameter is therefore constant along every magnetic field line.

A force-free magnetic field obeysequivalently $\mathbf{j}\times \mathbf{B}=0$: its current is parallel to its magnetic field. In a very low-density exterior, material pressure and inertia are too small to balance a finite Lorentz force. Quasistatic force balance therefore drives the magnetic field toward a force-free configuration.

The magnetic energy isUsing the ideal-MHD induction equation and integrating the identity $\mathbf{B}\cdot \nabla \times (\mathbf{u}\times \mathbf{B})$ by parts givesThe volume term vanishes for a force-free magnetic field, leaving