- #1
Rick16
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- TL;DR Summary
- I am trying to find the divergence of the Maxwell stress tensor.
This is from Griffiths' Electrodynamics, 3rd edition, page 352.
I am trying to calculate the divergence of the Maxwell stress tensor. The tensor is given as ##T_{ij} =\epsilon_0 (E_iE_j-\frac 1 2 \delta_{ij} E^2)+\frac 1 {\mu_0}(B_iB_j-\frac 1 2 \delta_{ij} B^2)##. To make things easier, I just want to focus on the part with the electrical field, i.e. I want to find the divergence of ##E_{ij}=E_iE_j-\frac 1 2 \delta_{ij}E^2##.
In matrix form, this tensor should look like this:
$$E_{ij}=E_iE_j-\frac 1 2 \delta_{ij}E^2=\begin{pmatrix}\frac 1 2 (E_x^2 -E_y^2 -E_z^2) & E_xE_y & E_xE_z \\ E_yE_x & \frac 1 2 (E_y^2 - E_x^2 -E_z^2) & E_yE_z \\ E_zE_x &E_zE_y & \frac 1 2 (E_z^2 -E_x^2 -E_y^2)\end{pmatrix}$$
According to the book, its divergence should be ##(\nabla \cdot \vec E)\vec E+(\vec E \cdot \nabla)\vec E- \frac 1 2 \nabla E^2##. Here is already a confusing part: Is ##(\nabla \cdot \vec E)\vec E## different from ##(\vec E \cdot \nabla)\vec E##? The components of a dot product are interchangeable, so what could ##\vec E \cdot \nabla## be, other than the divergence of ##\vec E##?
In any case, I continue with the calculation of the divergence:
$$\nabla \cdot E_{ij}=\sum_i \nabla_i E_{ij}=\begin{pmatrix}
\frac \partial {\partial x}E_{xx} +\frac \partial {\partial y}E_{yx}+ \frac \partial {\partial z}E_{zx} \\
\frac \partial {\partial x}E_{xy} +\frac \partial {\partial y}E_{yy}+ \frac \partial {\partial z}E_{zy} \\
\frac \partial {\partial x}E_{xz} +\frac \partial {\partial y}E_{yz}+ \frac \partial {\partial z}E_{zz}
\end{pmatrix} =$$
$$\begin{pmatrix}
\frac \partial {\partial x}(\frac 1 2 (E_x^2 -E_y^2 -E_z^2)) +\frac \partial {\partial y}E_yE_x+ \frac \partial {\partial z}E_zE_x \\
\frac \partial {\partial x}E_xE_y +\frac \partial {\partial y}(\frac 1 2 (E_y^2 - E_x^2 -E_z^2))+ \frac \partial {\partial z}E_zE_y \\
\frac \partial {\partial x}E_xE_z +\frac \partial {\partial y}E_yE_z+ \frac \partial {\partial z}\frac 1 2 (E_z^2 -E_x^2 -E_y^2)
\end{pmatrix} = $$
$$\begin{pmatrix}
E_x \frac {\partial E_x} {\partial x} + E_x \frac {\partial E_y} {\partial y} + E_x \frac {\partial E_z} {\partial z} \\
E_y \frac {\partial E_x} {\partial x} + E_y \frac {\partial E_y} {\partial y} + E_y \frac {\partial E_z} {\partial z} \\
E_z \frac {\partial E_x} {\partial x} + E_z \frac {\partial E_y} {\partial y} + E_z \frac {\partial E_z} {\partial z}
\end{pmatrix} =
\begin{pmatrix}
E_x\nabla \cdot \vec E \\ E_y \nabla \cdot \vec E \\ E_z \nabla \cdot \vec E
\end{pmatrix} = \nabla \cdot \vec E \begin{pmatrix}E_x \\ E_y \\ E_z \end{pmatrix} = (\nabla \cdot \vec E)\vec E
$$
This is only part of the answer that I was hoping to get, namely ##(\nabla \cdot \vec E)\vec E+(\vec E \cdot \nabla)\vec E- \frac 1 2 \nabla E^2##. What is wrong with my calculation?
I am trying to calculate the divergence of the Maxwell stress tensor. The tensor is given as ##T_{ij} =\epsilon_0 (E_iE_j-\frac 1 2 \delta_{ij} E^2)+\frac 1 {\mu_0}(B_iB_j-\frac 1 2 \delta_{ij} B^2)##. To make things easier, I just want to focus on the part with the electrical field, i.e. I want to find the divergence of ##E_{ij}=E_iE_j-\frac 1 2 \delta_{ij}E^2##.
In matrix form, this tensor should look like this:
$$E_{ij}=E_iE_j-\frac 1 2 \delta_{ij}E^2=\begin{pmatrix}\frac 1 2 (E_x^2 -E_y^2 -E_z^2) & E_xE_y & E_xE_z \\ E_yE_x & \frac 1 2 (E_y^2 - E_x^2 -E_z^2) & E_yE_z \\ E_zE_x &E_zE_y & \frac 1 2 (E_z^2 -E_x^2 -E_y^2)\end{pmatrix}$$
According to the book, its divergence should be ##(\nabla \cdot \vec E)\vec E+(\vec E \cdot \nabla)\vec E- \frac 1 2 \nabla E^2##. Here is already a confusing part: Is ##(\nabla \cdot \vec E)\vec E## different from ##(\vec E \cdot \nabla)\vec E##? The components of a dot product are interchangeable, so what could ##\vec E \cdot \nabla## be, other than the divergence of ##\vec E##?
In any case, I continue with the calculation of the divergence:
$$\nabla \cdot E_{ij}=\sum_i \nabla_i E_{ij}=\begin{pmatrix}
\frac \partial {\partial x}E_{xx} +\frac \partial {\partial y}E_{yx}+ \frac \partial {\partial z}E_{zx} \\
\frac \partial {\partial x}E_{xy} +\frac \partial {\partial y}E_{yy}+ \frac \partial {\partial z}E_{zy} \\
\frac \partial {\partial x}E_{xz} +\frac \partial {\partial y}E_{yz}+ \frac \partial {\partial z}E_{zz}
\end{pmatrix} =$$
$$\begin{pmatrix}
\frac \partial {\partial x}(\frac 1 2 (E_x^2 -E_y^2 -E_z^2)) +\frac \partial {\partial y}E_yE_x+ \frac \partial {\partial z}E_zE_x \\
\frac \partial {\partial x}E_xE_y +\frac \partial {\partial y}(\frac 1 2 (E_y^2 - E_x^2 -E_z^2))+ \frac \partial {\partial z}E_zE_y \\
\frac \partial {\partial x}E_xE_z +\frac \partial {\partial y}E_yE_z+ \frac \partial {\partial z}\frac 1 2 (E_z^2 -E_x^2 -E_y^2)
\end{pmatrix} = $$
$$\begin{pmatrix}
E_x \frac {\partial E_x} {\partial x} + E_x \frac {\partial E_y} {\partial y} + E_x \frac {\partial E_z} {\partial z} \\
E_y \frac {\partial E_x} {\partial x} + E_y \frac {\partial E_y} {\partial y} + E_y \frac {\partial E_z} {\partial z} \\
E_z \frac {\partial E_x} {\partial x} + E_z \frac {\partial E_y} {\partial y} + E_z \frac {\partial E_z} {\partial z}
\end{pmatrix} =
\begin{pmatrix}
E_x\nabla \cdot \vec E \\ E_y \nabla \cdot \vec E \\ E_z \nabla \cdot \vec E
\end{pmatrix} = \nabla \cdot \vec E \begin{pmatrix}E_x \\ E_y \\ E_z \end{pmatrix} = (\nabla \cdot \vec E)\vec E
$$
This is only part of the answer that I was hoping to get, namely ##(\nabla \cdot \vec E)\vec E+(\vec E \cdot \nabla)\vec E- \frac 1 2 \nabla E^2##. What is wrong with my calculation?
