Magnetic tension

In MHD, the magnetic tension force can be derived from the Cauchy momentum equation:

$${\displaystyle \rho \left({\frac {\partial }{\partial t}}+\mathbf {V} \cdot \nabla \right)\mathbf {V} =\mathbf {J} \times \mathbf {B} -\nabla p.}$$

The first term on the right hand side of the above equation represents the Lorentz force and the second term represents pressure gradient forces. Using Ampère's law, ${\displaystyle \mu _{0}\mathbf {J} =\nabla \times \mathbf {B} }$, and the vector identity

$${\displaystyle \nabla (\mathbf {a} \cdot \mathbf {b} )=(\mathbf {a} \cdot \nabla )\mathbf {b} +(\mathbf {b} \cdot \nabla )\mathbf {a} +\mathbf {a} \times (\nabla \times \mathbf {b} )+\mathbf {b} \times (\nabla \times \mathbf {a} ),}$$

we obtain the following equation:

$${\displaystyle \rho \left({\frac {\partial }{\partial t}}+\mathbf {V} \cdot \nabla \right)\mathbf {V} =-\nabla \left({\frac {B^{2}}{2\mu _{0}}}\right)+{(\mathbf {B} \cdot \nabla )\mathbf {B} \over \mu _{0}}-\nabla p.}$$

The first and last gradient terms are associated with the total pressure which is the sum of the magnetic and thermal pressures; ${\displaystyle p+B^{2}/2\mu _{0}}$. The second term represents the magnetic tension.

We can separate the force due to changes in the magnitude of ${\displaystyle \mathbf {B} }$ and its direction by writing ${\displaystyle \mathbf {B} =B\mathbf {b} }$ with ${\displaystyle B=|\mathbf {B} |}$ and ${\displaystyle \mathbf {b} }$ a unit vector. Some vector identities give

$${\displaystyle -\nabla \left({\frac {B^{2}}{2\mu _{0}}}\right)+{(\mathbf {B} \cdot \nabla )\mathbf {B} \over \mu _{0}}=-(1-\mathbf {b} \mathbf {b} )\cdot \nabla \left({\frac {B^{2}}{2\mu _{0}}}\right)+\left({\frac {B^{2}}{\mu _{0}}}\right)(\mathbf {b} \cdot \nabla )\mathbf {b} }$$

The first term is the magnetic pressure due solely to changes in ${\displaystyle B}$ in directions perpendicular to ${\displaystyle \mathbf {B} }$, while the second term is the "tension" due solely to changes in the direction of ${\displaystyle \mathbf {B} }$ (or curvature of magnetic field lines).

A more rigorous way to look at this is through Maxwell stress tensor. The Lorentz force equation

$${\displaystyle \mathbf {F} =q(\mathbf {E} +\mathbf {v} \times \mathbf {B} )}$$

gives the force per unit volume:

$${\displaystyle \mathbf {f} =\rho \mathbf {E} +\mathbf {J} \times \mathbf {B} }$$

This, after some algebra and using Maxwell's equations to replace the current, leads to

$${\displaystyle \mathbf {f} =\epsilon _{0}\left[(\nabla \cdot \mathbf {E} )\mathbf {E} +(\mathbf {E} \cdot \nabla )\mathbf {E} \right]+{\frac {1}{\mu _{0}}}\left[(\nabla \cdot \mathbf {B} )\mathbf {B} +(\mathbf {B} \cdot \nabla )\mathbf {B} \right]-{\frac {1}{2}}\nabla \left(\epsilon _{0}E^{2}+{\frac {1}{\mu _{0}}}B^{2}\right)-\epsilon _{0}{\frac {\partial }{\partial t}}\left(\mathbf {E} \times \mathbf {B} \right).}$$

This result can be re-written more compactly by introducing the Maxwell stress tensor,

$${\displaystyle \sigma _{ij}\equiv \epsilon _{0}\left(E_{i}E_{j}-{\frac {1}{2}}\delta _{ij}E^{2}\right)+{\frac {1}{\mu _{0}}}\left(B_{i}B_{j}-{\frac {1}{2}}\delta _{ij}B^{2}\right).}$$

All but the last term of the above expression for the force density, ${\displaystyle \mathbf {f} }$, can be written as the divergence of the Maxwell tensor:

$${\displaystyle \mathbf {f} +\epsilon _{0}\mu _{0}{\frac {\partial \mathbf {S} }{\partial t}}=\nabla \cdot \mathbf {\sigma } ,}$$

which gives the electromagnetic force density in terms of Maxwell stress tensor, ${\displaystyle \sigma _{ij}}$, and the Poynting vector, ${\displaystyle \mathbf {S} =\mathbf {E} \times \mathbf {B} /\mu _{0}}$. Now, the magnetic tension is implicitly included inside ${\displaystyle \sigma _{ij}}$. The implication of the above relation is the conservation of momentum. Here, ${\displaystyle \nabla \cdot \mathbf {\sigma } }$ is the momentum flux density and plays a role similar to ${\displaystyle \mathbf {S} }$ in Poynting's theorem.