📂Electrodynamics

Maxwell's Stress Tensor

Definition¹

The tensor $\mathbf{T}$ below is called the Maxwell stress tensor.

$$ \mathbf{T}=\overleftrightarrow{\mathbf{T}}=\begin{pmatrix} T_{xx} & T_{xy} & T_{xz} \\ T_{yx} & T_{yy} & T_{yz} \\ T_{zx} & T_{zy} & T_{zz} \end{pmatrix} $$

$$ T_{ij}=\epsilon_{0} \left( E_{i}E_{j}-\dfrac{1}{2}\delta_{ij}E^2 \right) + \dfrac{1}{\mu_{0}}\left(B_{i}B_{j}-\dfrac{1}{2}\delta_{ij}B^2 \right) $$

Here, $\delta_{ij}$ is the Kronecker delta.

Description

$2$th order tensor is defined as above. It appears in the process of deriving the force experienced by a charge in a volume $\mathcal{V}$. Being a $2$th order tensor, it has 9 components. For vectors with one lower index, they can be represented as $\vec{A}$, similarly, $2$th order tensors with two lower indices can be represented as $\overleftrightarrow{\mathbf{A}}$. However, the author considers this representation to be untidy and thus will represent vectors in the same manner, simply in boldface throughout this article.

The inclusion of the Kronecker delta in the formula makes the expressions for $i=j$ and $i \ne j$ quite different.

In the case of $i=j$

$$ T_{xx}=\epsilon_{0} \left( {E_{x}}^2-\dfrac{1}{2}E^2\right) +\dfrac{1}{\mu_{0}}\left( {B_{x}}^2-\dfrac{1}{2}B^2 \right) $$

since $E^2={E_{x}}^2+{E_{y}}^2+{E_{z}}^2$

$$ T_{xx}=\dfrac{\epsilon_{0}}{2} \left( {E_{x}}^2-{E_{y}}^2-{E_{z}}^2\right) +\dfrac{1}{2\mu_{0}}\left( {B_{x}}^2-{B_{y}}^2 -{B_{z}}^2\right) $$

In the case of $i \ne j$

$$ T_{xy}=\epsilon_{0}(E_{x}E_{y})+\dfrac{1}{\mu_{0}}(B_{x}B_{y}) $$

Inner Product

The inner product of the Maxwell stress tensor $\mathbf{T}$ with any vector $\mathbf{a}$, since it has components with only one lower index, is a vector ($1$th order tensor) and is as follows.

$$ \begin{align*} \mathbf{a} \cdot \mathbf{T} &=\begin{pmatrix} a_{x} & a_{y} & a_{z} \end{pmatrix}\begin{pmatrix} T_{xx} & T_{xy} & T_{xz} \\ T_{yx} & T_{yy} & T_{yz} \\ T_{zx} & T_{zy} & T_{zz} \end{pmatrix} \\ &= (a_{x}T_{xx}+a_{y}T_{yx}+a_{z}T_{zx},\ a_{x}T_{xy} + a_{y}T_{yy} + a_{z}T_{zy},\ a_{x}T_{xz} + a_{y}T_{yz} + a_{z}T_{zz}) \end{align*} $$

$$ \left( \mathbf{a} \cdot \mathbf{T} \right)_{j}=\sum \limits_{i=x,y,z}a_{i}T_{ij} $$

Divergence

The divergence of $\mathbf{T}$ is also a vector. The $j$ component of $\nabla \cdot \mathbf{T}$ is

$$ \begin{align*} & \left( \nabla \cdot \mathbf{T} \right)_{j} \\ =&\ \epsilon_{0} \left[ \nabla_{i}(E_{i}E_{j}) -\dfrac{1}{2}\nabla_{i}\delta_{ij}E^2 \right] + \dfrac{1}{\mu_{0}}\left[ \nabla_{i}(B_{i}B_{j})-\dfrac{1}{2}\nabla_{i}\delta_{ij}B^2 \right] \\ =&\ \epsilon_{0} \left[ \nabla_{i}E_{i}E_{j} +E_{i}\nabla_{i}E_{j} -\dfrac{1}{2}\nabla_{j}E^2 \right] + \dfrac{1}{\mu_{0}}\left[ \nabla_{i}B_{i}B_{j}+B_{i}\nabla_{i}B_{j}-\dfrac{1}{2}\nabla_{j}B^2 \right] \\ =&\ \epsilon_{0} \left[ (\nabla \cdot \mathbf{E})E_{j} +(\mathbf{E} \cdot \nabla) E_{j} -\dfrac{1}{2}\nabla_{j}E^2 \right] + \dfrac{1}{\mu_{0}}\left[ (\nabla \cdot \mathbf{B}) B_{j}+(\mathbf{B} \cdot \nabla) B_{j}-\dfrac{1}{2}\nabla_{j}B^2 \right] \end{align*} $$