# Maxwell stress tensor | Wikipedia audio article

The Maxwell stress tensor (named after James

Clerk Maxwell) is a symmetric second-order 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.

In the relativistic formulation of electromagnetism, the Maxwell’s tensor appears as a part of

the electromagnetic stress–energy tensor which is the electromagnetic component of

the total stress–energy tensor. The latter describes the density and flux of energy and

momentum in spacetime.==Motivation==As outlined below, the electromagnetic force

is written in terms of E and B. Using vector calculus and Maxwell’s equations, symmetry

is sought for in the terms containing E and B, and introducing the Maxwell stress tensor

simplifies the result. in the above relation for conservation of

momentum, ∇

⋅ σ {\displaystyle \nabla \cdot \mathbf {\sigma

} } is the momentum flux density and plays a role

similar to S {\displaystyle \mathbf {S} }

in Poynting’s theorem. The above derivation assumes complete knowledge

of both ρ and J (both free and bounded charges and currents). For the case of nonlinear materials

(such as magnetic iron with a BH-curve), the nonlinear Maxwell stress tensor must be used.==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: σ i

j=ϵ E i E j + 1 μ B i B j − 1

2 ( ϵ E 2 + 1 μ B 2 ) δ i

j {\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: σ i

j=1 4

π ( E i E j + H i H j − 1

2 ( E 2 + H 2 ) δ i

j ) {\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: σ ↔=1 4

π [ E ⊗ E + H ⊗ H − E 2 + H 2 2 I ] {\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}}\mathbb {I} \right]}

where ⊗ is the dyadic product, and the last tensor is the unit dyad: I ≡ ( 1 1 1 )=

( x

^ ⊗ x

^ + y

^ ⊗ y

^ + z

^ ⊗ z

^ ) {\displaystyle \mathbb {I} \equiv \left({\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}}\right)=(\mathbf

{\hat {x}} \otimes \mathbf {\hat {x}} +\mathbf {\hat {y}} \otimes \mathbf {\hat {y}} +\mathbf

{\hat {z}} \otimes \mathbf {\hat {z}} )} The element ij of the Maxwell stress tensor

has units of momentum per unit of area per unit 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: σ i

j=1 μ B i B j − 1 2 μ B 2 δ i

j . {\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: σ r

t=1 μ B r B t − 1 2 μ B 2 δ r

t . {\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.==In Electrostatics==

In electrostatics the effects of magnetism are not present. In this case the magnetic

field vanishes, B={\displaystyle \mathbf {B}=\mathbf {0} }

, and we obtain the electrostatic Maxwell stress tensor. It is given in component form

by σ i

j=ε E i E j − 1

2 ε E 2 δ i

j {\displaystyle \sigma _{ij}=\varepsilon _{0}E_{i}E_{j}-{\frac

{1}{2}}\varepsilon _{0}E^{2}\delta _{ij}} and in symbolic form by σ=ε E ⊗ E − 1

2 ε ( E ⋅ E ) I {\displaystyle {\boldsymbol {\sigma }}=\varepsilon

_{0}\mathbf {E} \otimes \mathbf {E} -{\frac {1}{2}}\varepsilon _{0}(\mathbf {E} \cdot

\mathbf {E} )\mathbf {I} } where I {\displaystyle \mathbf {I} }

is the appropriate identity tensor (usually 3

× 3 {\displaystyle 3\times 3}

).==Eigenvalue==

The eigenvalues of the Maxwell stress tensor are given by: {

λ }

={ − ϵ E 2 + B 2 / μ 2 , ± ( ϵ E 2 − B 2 / μ 2 ) 2 + ϵ μ ( E ⋅ B ) 2 } {\displaystyle \{\lambda \}=\left\{-{\frac

{\epsilon _{0}E^{2}+B^{2}/\mu _{0}}{2}},~\pm {\sqrt {\left({\frac {\epsilon _{0}E^{2}-B^{2}/\mu

_{0}}{2}}\right)^{2}+{\frac {\epsilon _{0}}{\mu _{0}}}\left({\boldsymbol {E}}\cdot {\boldsymbol

{B}}\right)^{2}}}\right\}} These eigenvalues are obtained by iteratively

applying the Matrix Determinant Lemma, in conjunction with the Sherman-Morrison Formula.

Noting that the characteristic equation matrix, σ

↔ −

λ I {\displaystyle {\overleftrightarrow {\sigma

}}-\lambda \mathbf {\mathbb {I} } } , can be written as σ

↔ −

λ I=

− ( λ

+ V ) I + ϵ E E T + B B T μ {\displaystyle {\overleftrightarrow {\sigma

}}-\lambda \mathbf {\mathbb {I} }=-\left(\lambda +V\right)\mathbf {\mathbb {I} } +\epsilon

_{0}\mathbf {E} \mathbf {E} ^{T}+{\frac {\mathbf {B} \mathbf {B} ^{T}}{\mu _{0}}}} where V

=1

2 ( ϵ E 2 + B 2 / μ ) {\displaystyle V={\frac {1}{2}}\left(\epsilon

_{0}E^{2}+B^{2}/\mu _{0}\right)} we set U=

− ( λ

+ V ) I + ϵ E E T {\displaystyle \mathbf {U}=-\left(\lambda

+V\right)\mathbf {\mathbb {I} } +\epsilon _{0}\mathbf {E} \mathbf {E} ^{T}} Applying the Matrix Determinant Lemma once,

this gives us det ( σ

↔ −

λ I )=( 1

+ μ −

1 B T U −

1 B ) det ( U ) {\displaystyle \det {\left({\overleftrightarrow

{\sigma }}-\lambda \mathbf {\mathbb {I} } \right)}=\left(1+\mu _{0}^{-1}\mathbf {B} ^{T}\mathbf {U} ^{-1}\mathbf

{B} \right)\det {\left(\mathbf {U} \right)}} Applying it again yields, det ( σ

↔ −

λ I )=( 1

+ μ −

1 B T U −

1 B ) ( 1

− ϵ E T E ( λ

+ V ) ) ( −

λ −

V ) 3 {\displaystyle \det {\left({\overleftrightarrow

{\sigma }}-\lambda \mathbf {\mathbb {I} } \right)}=\left(1+\mu _{0}^{-1}\mathbf {B} ^{T}\mathbf {U} ^{-1}\mathbf

{B} \right)\left(1-{\frac {\epsilon _{0}\mathbf {E} ^{T}\mathbf {E} }{\left(\lambda +V\right)}}\right)\left(-\lambda

-V\right)^{3}} From the last multiplicand on the RHS, we

immediately see that λ

=−

V {\displaystyle \lambda=-V}

is one of the eigenvalues. To find the inverse of U {\displaystyle \mathbf {U} }

, we use the Sherman-Morrison formula: U −

1=

− ( λ

+ V ) −

1 − ϵ E E T ( λ

+ V ) 2 − ( λ

+ V ) ( ϵ E T E ) {\displaystyle \mathbf {U} ^{-1}=-\left(\lambda

+V\right)^{-1}-{\frac {\epsilon _{0}\mathbf {E} \mathbf {E} ^{T}}{\left(\lambda +V\right)^{2}-\left(\lambda

+V\right)\left(\epsilon _{0}\mathbf {E} ^{T}\mathbf {E} \right)}}} Factoring out a ( −

λ −

V ) {\displaystyle \left(-\lambda -V\right)}

term in the determinant, we are left with finding the zeros of the rational function: ( − ( λ

+ V ) − ϵ ( E ⋅ B ) 2 μ ( − ( λ

+ V ) + ϵ E T E ) ) ( − ( λ

+ V ) + ϵ E T E ) {\displaystyle \left(-\left(\lambda +V\right)-{\frac

{\epsilon _{0}\left(\mathbf {E} \cdot \mathbf {B} \right)^{2}}{\mu _{0}\left(-\left(\lambda

+V\right)+\epsilon _{0}\mathbf {E} ^{T}\mathbf {E} \right)}}\right)\left(-\left(\lambda +V\right)+\epsilon

_{0}\mathbf {E} ^{T}\mathbf {E} \right)} Thus, once we solve − ( λ

+ V ) ( − ( λ

+ V ) + ϵ E 2 ) − ϵ ( E ⋅ B ) 2 μ=

0 {\displaystyle -\left(\lambda +V\right)\left(-\left(\lambda

+V\right)+\epsilon _{0}E^{2}\right)-{\frac {\epsilon _{0}\left(\mathbf {E} \cdot \mathbf

{B} \right)^{2}}{\mu _{0}}}=0} we obtain the other two eigenvalues.==See also==

Ricci calculus Energy density of electric and magnetic fields

Poynting vector Electromagnetic stress–energy tensor

Magnetic pressure Magnetic tension force