# 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