Force-free magnetic field

If the current density is identically zero, then the magnetic field is the gradient of a magnetic scalar potential ${\displaystyle \phi }$:

${\displaystyle \mathbf {B} =-\nabla \phi .}$

The substitution of this into ${\displaystyle \nabla \cdot \mathbf {B} =0}$ results in Laplace's equation, ${\displaystyle \nabla ^{2}\phi =0,}$ which can often be readily solved, depending on the precise boundary conditions. In this case, the field is referred to as a potential field or vacuum magnetic field.

If the current density is not zero, then it must be parallel to the magnetic field, i.e., ${\displaystyle \mu _{0}\mathbf {j} =\alpha \mathbf {B} }$ where ${\displaystyle \alpha }$ is a scalar function known as the force-free parameter or force-free function. This implies that

${\displaystyle \nabla \times \mathbf {B} =\alpha \mathbf {B} ,}$

${\displaystyle \mathbf {B} \cdot \nabla \alpha =0.}$

The force-free parameter can be a function of position but must be constant along field lines.

When the force-free parameter ${\displaystyle \alpha }$ is constant everywhere, the field is called a linear force-free field (LFFF). A constant ${\displaystyle \alpha }$ allows for the derivation of a vector Helmholtz equation

${\displaystyle \nabla ^{2}\mathbf {B} =-\alpha ^{2}\mathbf {B} }$

by taking the curl of the nonzero current density equations above.

When the force-free parameter ${\displaystyle \alpha }$ depends on position, the field is called a nonlinear force-free field (NLFFF). In this case, the equations do not possess a general solution, and usually must be solved numerically.[1][2][3]^: 50–54