Force-free magnetic field

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A force-free magnetic field is a magnetic field that 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. For a force free field, the electric current density is either zero or parallel to the magnetic field. The name "force-free" comes from being able to neglect the force from the plasma.

Basic Equations

Neglecting the effects of gravity, the Navier-Stokes equation for a plasma, in steady state, reads

0=-\nabla p+\mathbf{j}\times\mathbf{B},

where p is the thermal pressure,  \mathbf{B} is the magnetic field and \mathbf{j} is the electric current. Assuming that the gas pressure p is small compared to the magnetic pressure, i.e.,

p\ll B^2/2\mu

then the pressure term can be neglected. Here \mu is the magnetic permeability of the plasma. Therefore

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

This 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 to be determined. Combining this equation with Maxwell's equations:

\nabla\times\mathbf{B}=\mu\mathbf{j}

\nabla\cdot\mathbf{B}=0

and the vector identity:

\nabla\cdot(\nabla\times\mathbf{B})=0

leads to a pair of equations for \alpha and \mathbf{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

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 \alpha 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.

See also

References