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Electricity · Magnetism
Free space · Lorentz force law · emf · Electromagnetic induction · Faraday’s law · Lenz's law · Displacement current · Maxwell's equations · EM field · Electromagnetic radiation · Liénard-Wiechert Potential · Maxwell tensor · Eddy current

The Maxwell Stress Tensor is a second rank tensor used in classical electromagnetism to represent the interaction between electric/magnetic 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.



Maxwell's equations (for handy reference)
Name Differential form
Gauss's law \nabla \cdot \mathbf{E} = \frac {\rho} {\varepsilon_0}
Gauss's law for magnetism \nabla \cdot \mathbf{B} = 0
Maxwell–Faraday equation
(Faraday's law of induction)
\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}
Ampère's circuital law
(with Maxwell's correction)
\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t}\

If we start out with the Lorentz force law, and write down the force per unit volume for an unknown charge distribution, the motivation for the Maxwell stress tensor becomes apparent.

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

This is the differential form of the Lorentz force law, the usual \mathbf{F} = q(\mathbf{E} + \mathbf{v}\times\mathbf{B}) can be obtained by integrating over the volume. If we now use Maxwells equations we can replace ρ and \mathbf{J} so that only the fields \mathbf{E} and \mathbf{B} are used:

 \mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}\,

where we have used Gauss' law and Ampere's law. Now, using the product rule, we can rewrite the time derivative to something that can be interpreted physically, namely the Poynting vector. We observe that

\frac{\partial}{\partial t} (\mathbf{E}\times\mathbf{B}) = \frac{\partial\mathbf{E}}{\partial t}\times \mathbf{B} + \mathbf{E} \times \frac{\partial\mathbf{B}}{\partial t} = \frac{\partial\mathbf{E}}{\partial t}\times \mathbf{B} - \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E})\,

where we have used Faradays law, and we can now rewrite \mathbf{f} as

\mathbf{f} = \epsilon_0 \left(\boldsymbol{\nabla}\cdot \mathbf{E} \right)\mathbf{E} + \frac{1}{\mu_0} \left(\boldsymbol{\nabla}\times \mathbf{B} \right) \times \mathbf{B} - \epsilon_0 \frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right) - \epsilon_0 \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E})\,,

or collecting terms with \mathbf{E} and \mathbf{B} we have

\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} - \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E}) \right] + \frac{1}{\mu_0} \left[\;\;\;\;\;\;\;\;\;\;\; - \mathbf{B}\times\left(\boldsymbol{\nabla}\times \mathbf{B} \right) \right] + \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)\,.

Notice that a term seems to be "missing" from the symmetry. We can, however, add this term to achieve symmetry since Gauss' law for magnetism tells us that \nabla \cdot \mathbf{B} = 0. This leaves us with

\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} - \mathbf{E} \times (\boldsymbol{\nabla}\times \mathbf{E}) \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B} )\mathbf{B} - \mathbf{B}\times\left(\boldsymbol{\nabla}\times \mathbf{B} \right) \right] + \epsilon_0\frac{\partial}{\partial t}\left( \mathbf{E}\times \mathbf{B}\right)\,.

Now, if we only could get rid of those curls (no-one likes to compute curls) this would be relatively straightforward to calculate. Fortunately we have the curl identity,

\tfrac{1}{2} \boldsymbol{\nabla} A^2 = \mathbf{A} \times (\boldsymbol{\nabla} \times \mathbf{A}) + (\mathbf{A} \cdot \boldsymbol{\nabla}) \mathbf{A} ,

so we end up with

\mathbf{f} = \epsilon_0\left[ (\boldsymbol{\nabla}\cdot \mathbf{E} )\mathbf{E} + (\mathbf{E}\cdot\boldsymbol{\nabla}) \mathbf{E} \right] + \frac{1}{\mu_0} \left[(\boldsymbol{\nabla}\cdot \mathbf{B} )\mathbf{B} + (\mathbf{B}\cdot\boldsymbol{\nabla}) \mathbf{B} \right] - \frac{1}{2} \boldsymbol{\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)\,.

It may not be pretty, but it contains every aspect of electromagnetism and momentum. And it's relatively easy to compute. But can't we write this more compactly? Yes, we can! We introduce the Maxwell stress tensor,

T_{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)\,,

and notice that all but the last term of the above can be written as the gradient of this:

f_j = \nabla \cdot \mathbf{T} + \epsilon_0 \mu_0 \frac{\partial S}{\partial t}\,,

where we have finally introduced the Poynting vector,

\mathbf{S} = \frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}.


In physics, the Maxwell stress tensor is the stress tensor of an electromagnetic field. In cgs units, it is given by:

\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 E is the electric field, H is the magnetic field and δij is Kronecker's delta.

In SI units, it is given by:

\sigma _{ij} = \varepsilon_0 E_i E_j + \frac{1} {{\mu _0 }}B_i B_j - \frac{1}{2}\bigl( {\varepsilon_0 E^2 + \tfrac{1} {{\mu _0 }}B^2 } \bigr)\delta _{ij} ,

where ε0 is the electric constant and μ0 is the magnetic constant.

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:

\sigma_{i j} = \frac{1}{\mu_0} B_i B_j - \frac{1}{2 \mu_0} B^2 \delta_{i j} \,.

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

\sigma_{r t} = \frac{1}{\mu_0} B_r B_t - \frac{1}{2 \mu_0} B^2 \delta_{r t} \,.

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.

See also


  • David J. Griffiths,"Introduction to Electrodynamics" pp. 351-352, Benjamin Cummings Inc., 2008
  • John David Jackson,"Classical Electrodynamics, 3rd Ed.", John Wiley & Sons, Inc., 1999.
  • Richard Becker,"Electromagnetic Fields and Interactions",Dover Publications Inc., 1964.


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