# Maxwell stress tensor

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The Maxwell stress tensor (named after James Clerk Maxwell) is a second rank tensor used in classical electromagnetism to represent the interaction between electromagnetic forces and mechanical momentum. In simple situations, such as a point charge moving freely in a homogeneous magnetic field, it is easy to calculate the forces on the charge from the Lorentz force law. When the situation becomes more complicated, this ordinary procedure can become impossibly difficult, with equations spanning multiple lines. It is therefore convenient to collect many of these terms in the Maxwell stress tensor, and to use tensor arithmetic to find the answer to the problem at hand.

## Motivation

Lorentz force (per unit 3-volume) f on a continuous charge distribution (charge density ρ) in motion. The 3-current density J corresponds to the motion of the charge element dq in volume element dV and varies throughout the continuum.

As outlined below, the electromagnetic force is written in terms of E and B, using vector calculus and Maxwell's equations symmetry in the terms containing E and B are sought for, and introducing the Maxwell stress-tensor simplifies the result.

Maxwell's equations in SI units in vacuum
(for reference)
Name Differential form
Gauss's law (in vacuum) ${\displaystyle \nabla \cdot {\mathbf {E} }={\frac {\rho }{\epsilon _{0}}}}$
Gauss's law for magnetism ${\displaystyle \nabla \cdot \mathbf {B} =0}$
${\displaystyle \nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}}$
Ampère's circuital law (in vacuum)
(with Maxwell's correction)
${\displaystyle \nabla \times \mathbf {B} =\mu _{0}\mathbf {J} +\mu _{0}\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\ }$

in the above relation for conservation of momentum, ${\displaystyle \nabla \cdot {\mathbf {\sigma } }}$ is the momentum flux density and plays a role similar to ${\displaystyle \mathbf {S} }$ in Poynting's theorem.

## Equation

In physics, the Maxwell stress tensor is the stress tensor of an electromagnetic field. As derived above in SI units, it is given by:

${\displaystyle \sigma _{ij}=\epsilon _{0}E_{i}E_{j}+{\frac {1}{\mu _{0}}}B_{i}B_{j}-{\frac {1}{2}}{\bigl (}{\epsilon _{0}E^{2}+{\tfrac {1}{\mu _{0}}}B^{2}}{\bigr )}\delta _{ij}}$,

where ε0 is the electric constant and μ0 is the magnetic constant, E is the electric field, B is the magnetic field and δij is Kronecker's delta. In Gaussian cgs unit, it is given by:

${\displaystyle \sigma _{ij}={\frac {1}{4\pi }}\left(E_{i}E_{j}+H_{i}H_{j}-{\frac {1}{2}}(E^{2}+H^{2})\delta _{ij}\right)}$,

where H is the magnetizing field.

An alternative way of expressing this tensor is:

${\displaystyle {\overset {\leftrightarrow }{\mathbf {\sigma } }}={\frac {1}{4\pi }}\left[\mathbf {E} \otimes \mathbf {E} +\mathbf {H} \otimes \mathbf {H} -{\frac {E^{2}+H^{2}}{2}}(\mathbf {\hat {x}} \otimes \mathbf {\hat {x}} +\mathbf {\hat {y}} \otimes \mathbf {\hat {y}} +\mathbf {\hat {z}} \otimes \mathbf {\hat {z}} )\right]}$,

where ⊗ is the dyadic product.

The element ij of the Maxwell stress tensor has units of momentum per unit of area times time and gives the flux of momentum parallel to the ith axis crossing a surface normal to the jth axis (in the negative direction) per unit of time.

These units can also be seen as units of force per unit of area (negative pressure), and the ij element of the tensor can also be interpreted as the force parallel to the ith axis suffered by a surface normal to the jth axis per unit of area. Indeed the diagonal elements give the tension (pulling) acting on a differential area element normal to the corresponding axis. Unlike forces due to the pressure of an ideal gas, an area element in the electromagnetic field also feels a force in a direction that is not normal to the element. This shear is given by the off-diagonal elements of the stress tensor.

## Magnetism only

If the field is only magnetic (which is largely true in motors, for instance), some of the terms drop out, and the equation in SI units becomes:

${\displaystyle \sigma _{ij}={\frac {1}{\mu _{0}}}B_{i}B_{j}-{\frac {1}{2\mu _{0}}}B^{2}\delta _{ij}\,.}$

For cylindrical objects, such as the rotor of a motor, this is further simplified to:

${\displaystyle \sigma _{rt}={\frac {1}{\mu _{0}}}B_{r}B_{t}-{\frac {1}{2\mu _{0}}}B^{2}\delta _{rt}\,.}$

where r is the shear in the radial (outward from the cylinder) direction, and t is the shear in the tangential (around the cylinder) direction. It is the tangential force which spins the motor. Br is the flux density in the radial direction, and Bt is the flux density in the tangential direction.