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# Force-free magnetic field

A force-free magnetic field is a type of field which arises as a special case from the magnetostatic equation in plasmas. This special case arises when the plasma pressure is so small, relative to the magnetic pressure, that the plasma pressure may be ignored, and so only the magnetic pressure is considered. The name "force-free" comes from being able to neglect the force from the plasma.

## Basic Equations

Start with the simplified magnetostatic equations, in which the effects of gravity may be neglected:

$0=-\nabla\rho+\mathbf{j}\times\mathbf{B}.$

Supposing that the gas pressure is small compared to the magnetic pressure, i.e.,

ρ < < B2 / 2μ

then the pressure term can be neglected, and we have:

$\mathbf{j}\times\mathbf{B} = 0$.

From Maxwell's equations:

$\nabla\times\mathbf{B}=\mu_{0}\mathbf{j}$

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

The first equation implies that: $\mu_{0}\mathbf{j}=\alpha\mathbf{B}$. e.g. the current density is either zero or parallel to the magnetic field, and where alpha is a spatial-varying function which must be determined.. Combing this equation with Maxwell's equations, leads to a pair of equations for alpha and B:

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

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

## Physical Examples

In the corona of the sun, the ratio of the gas pressure to the magnetic pressure is ~0.004, and so there the magnetic field is force-free.

## Mathematical Limits

• If the current density is identically zero, then the magnetic field is potential, i.e. the gradient of a scalar magnetic potential.
In particular, if $\mathbf{j}=0$
then $\nabla\times\mathbf{B}=0$ which implies, that $\mathbf{B}=\nabla\phi$.
The substitution of this into one of Maxwell's Equations, $\nabla\cdot\mathbf{B}=0$, results in Laplace's equation,
$\nabla^2\phi=0$,
which can often be readily solved, depending on the precise boundary conditions.
This limit is usually referred to as the potential field case.
• If the current density is not zero, then it must be parallel to the magnetic field, i.e.,
$\mu\mathbf{j}=\alpha \mathbf{B}$ which implies, that $\nabla\times\mathbf{B}=\alpha \mathbf{B}$, where α is some scalar function.
then we have, from above,
$\mathbf{B}\cdot\nabla\alpha=0$
$\nabla\times\mathbf{B}=\alpha\mathbf{B}$ , which implies that
$\nabla\times(\nabla\times\mathbf{B})=\nabla\times(\alpha\mathbf{B})$
There are then two cases:
Case 1: The proportionality between the current density and the magnetic field is constant everywhere .
$\nabla\times(\alpha\mathbf{B})= \alpha(\nabla\times\mathbf{B})=\alpha^2 \mathbf{B}$
and also
$\nabla\times(\nabla\times\mathbf{B})=\nabla(\nabla\cdot\mathbf{B}) -\nabla^2\mathbf{B}=-\nabla^2\mathbf{B}$,
and so
$-\nabla^2\mathbf{B} =\alpha^2 \mathbf{B}$
This is a Helmholtz equation.
• Case 2: The proportionality between the current density and the magnetic field is a function of position.
$\nabla\times(\alpha\mathbf{B})= \alpha(\nabla\times\mathbf{B})+\nabla\alpha\times\mathbf{B}=\alpha^2 \mathbf{B} +\nabla\alpha\times\mathbf{B}$
and so the result is coupled equations:
$\nabla^2\mathbf{B}+\alpha^2\mathbf{B}= \mathbf{B}\times\nabla\alpha$

and

$\mathbf{B}\cdot\nabla\alpha= 0$
In this case, the equations do not possess a general solution, and usually must be solved numerically.