en-academic.com

Maxwell stress tensor

Electromagnetism

Electricity · Magnetism
Electrostatics Electric charge · Coulomb's law · Electric field · Electric flux · Gauss's law · Electric potential energy · Electric potential · Electrostatic induction · Electric dipole moment · Polarization density
Magnetostatics Ampère's law · Electric current · Magnetic field · Magnetization · Magnetic flux · Biot–Savart law · Magnetic dipole moment · Gauss's law for magnetism
Electrodynamics 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
Electrical Network Electrical conduction · Electrical resistance · Capacitance · Inductance · Impedance · Resonant cavities · Waveguides
Covariant formulation Electromagnetic tensor · EM Stress-energy tensor · Four-current · Electromagnetic four-potential
Scientists Ampère · Coulomb · Faraday · Gauss · Heaviside · Henry · Hertz · Lorentz · Maxwell · Tesla · Volta · Weber · Ørsted
v · d · e

The Maxwell Stress Tensor (named after James Clerk Maxwell) is a mathematical object in physics, more precisely it 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.

Motivation

Maxwell's equations in SI units in vacuum
(for reference)

Name	Differential form
Gauss's law (in vacuum)	$\nabla \cdot \mathbf{E} = \frac {\rho} {\epsilon_0}$
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 (in vacuum) (with Maxwell's correction)	$\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \epsilon_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 Maxwell's 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 Ampère'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

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

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

This would be relatively straightforward to calculate if we only could get rid of those curls, which are more demanding. 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

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

This expression contains every aspect of electromagnetism and momentum and is relatively easy to compute. It can be written more compactly by introducing the Maxwell stress tensor,

$\sigma_{i j} \equiv \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0} \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)\,$ ,

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

$\mathbf{f} + \epsilon_0 \mu_0 \frac{\partial \mathbf{S}}{\partial t}\, = \nabla \cdot \mathbf{\sigma}$ ,

where we have finally introduced the Poynting vector,

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

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:

$\sigma_{i j} = \epsilon_0 E_i E_j + \frac{1} {{\mu _0 }}B_i B_j - \frac{1}{2}\bigl( {\epsilon_0 E^2 + \tfrac{1} {{\mu _0 }}B^2 } \bigr)\delta _{ij}$ ,

where $\epsilon_0$ is the electric constant and μ₀ is the magnetic constant, $\mathbf{E}$ is the electric field, $\mathbf{B}$ is the magnetic field and δ_ij is Kronecker's delta. In Gaussian (CGS) units, more commonly used in physics textbooks, it is given by:

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

where $\mathbf{H}$ is the magnetizing field.

An alternative way of expressing this tensor is:

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

where $\otimes$ 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:

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

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. B_r is the flux density in the radial direction, and B_t is the flux density in the tangential direction.

References

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.

Maxwell stress tensor

Motivation

Equation

Magnetism only

See also

References