I Maxwell stress tensor

[Rick16]

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?

 

[renormalize]

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 $\vec E$?

No, the two expressions $\nabla\cdot\vec{E}$ and $\vec{E}\cdot\nabla$ are very different. The first is simply the scalar quantity $\nabla\cdot\vec{E}=\frac{\partial E_{x}}{\partial x}+\frac{\partial E_{y}}{\partial y}+\frac{\partial E_{z}}{\partial z}$, whereas the second is the differential operator $\vec{E}\cdot\nabla=E_{x}\frac{\partial}{\partial x}+E_{y}\frac{\partial}{\partial y}+E_{z}\frac{\partial}{\partial z}$ that acts on whatever is to its immediate right.

 

[JimWhoKnew]

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?

In addition to #2, you'll recover more missing terms by using brackets and carrying out the derivatives properly (without shortcuts). For example$$\frac{\partial}{\partial y}E_{yx}=\frac{\partial}{\partial y}(E_y E_x)=\dots$$

 

[Rick16]

Thank you for your comments. I obviously made two wrong assumptions.

I did not understand $\vec E \cdot \nabla$, and when calculating derivatives like $\frac \partial {\partial y} (E_yE_x)$ I assumed that $\frac {\partial E_x} {\partial y}=0$. I reasoned that $E_x$ is the x-component of the electric field and thus has no y-dependence. Apparently things are a little more complicated with tensors. Griffiths writes that "$T_{ij}$ is the force (per unit area) in the ith direction acting on an element of surface oriented in the jth direction". So, when I calculate the derivative $\frac \partial {\partial y} (E_yE_x)$ and I look at $\frac {\partial E_x} {\partial y}$, I have to interpret this as the field being oriented in the x-direction but having a force acting on it in the y-direction and therefore $\frac {\partial E_x} {\partial y} \neq 0$. Is this correct?

With this new understanding of the differential operator and the tensor elements I have made great progress, but I am stuck again, close to the finish line. Here is what I have now:

$$\nabla \cdot E_{ij}=(\nabla \cdot \vec E)\vec E+(\vec E \cdot \nabla)\vec E+1/2
\begin{pmatrix}
\frac \partial {\partial x}(E_y^2 +E_z^2 -E_x^2)\\
\frac \partial {\partial y}(E_x^2 +E_z^2 -E_y^2)\\
\frac \partial {\partial z}(E_x^2 +E_y^2 -E_z^2)
\end{pmatrix}$$

What I need is
$$\nabla \cdot E_{ij}=(\nabla \cdot \vec E)\vec E+(\vec E \cdot \nabla)\vec E-1/2
\begin{pmatrix}
\frac \partial {\partial x}(E_y^2 +E_z^2 +E_x^2)\\
\frac \partial {\partial y}(E_x^2 +E_z^2 +E_y^2)\\
\frac \partial {\partial z}(E_x^2 +E_y^2 +E_z^2)
\end{pmatrix}$$

Is it possible that $\begin{pmatrix} \frac \partial {\partial x}(E_y^2 +E_z^2 -E_x^2)\\ \frac \partial {\partial y}(E_x^2 +E_z^2 -E_y^2)\\ \frac \partial {\partial z}(E_x^2 +E_y^2 -E_z^2) \end{pmatrix}$ is the same as $-\begin{pmatrix} \frac \partial {\partial x}(E_y^2 +E_z^2 +E_x^2)\\ \frac \partial {\partial y}(E_x^2 +E_z^2 +E_y^2)\\ \frac \partial {\partial z}(E_x^2 +E_y^2 +E_z^2) \end{pmatrix}$?

 

[JimWhoKnew]

and therefore $\frac {\partial E_x} {\partial y} \neq 0$. Is this correct?

To improve your understanding of $~\frac {\partial E_x} {\partial y}~$, consider the simple example of the electrostatic field of a small spherical charge (which can be regarded as almost pointlike) at the origin.
What is $~E_x(1,0,0)$ ?
What is $~E_x(1,1,0)$ ?
What is $~\left. \frac {\partial E_x} {\partial y}\right|_{(1,1,0)}~$ ?
The lesson: when you encounter terms that you don't understand (like $~\frac {\partial E_x} {\partial y}~$ in the present case), always try to see what happens in some simple examples. That way, you may gain some intuition about the physical meaning, rather than just a sequence of mathematical manipulations.

With this new understanding of the differential operator and the tensor elements I have made great progress, but I am stuck again, close to the finish line.

If you'll post a detailed calculation, as you did in OP, it will be easier to spot possible errors.

Edit: corrected equations.

 

[TSny]

The last term on the right is wrong. This term should represent the divergence of just the part of $E_{ij}$ given by $-\frac 1 2 \delta_{ij}E^2$.

 

[Rick16]

That way, you may gain some intuition about the physical meaning, rather than just a sequence of mathematical manipulations.

I see you see right through me. I am indeed the type that likes to run away with mathematical manipulations, forgetting about everything else.

If you'll post a detailed calculation, as you did in OP, it will be easier to spot possible errors.

Here is it (I did it in Word, which is much easier to do):

 

[JimWhoKnew]

Please use LaTeX. The calculations are displayed blurry and hard to read on my laptop.

In passage from the second to the third line of your calculation, where did all the minus signs go?

 

[Rick16]

In passage from the second to the third line of your calculation, where did all the minus signs go?

Problem solved! Thanks a lot. I feel really stupid now. I constantly have trouble with these pesky minus signs, but this time I have really outdone myself.

Now, that I finally have the solution, I wonder if this is the only way to do it. Isn't there an easier way? This is such a cumbersome process. And what about higher rank tensors, which cannot be written in matrix form at all?

 

[JimWhoKnew]

Isn't there an easier way?

There is. If you practice Einstein's summation convention a little, where repeated indices imply summation:$$A_iB_i:=\sum_{i=x,y,z} A_i B_i \quad ,$$you get immediately\begin{align}
\partial_i E_{ij} & =\partial_i(E_i E_j)-\frac12\partial_i(\delta_{ij}E^2) \nonumber\\
& =(\partial_iE_i)E_j+E_i\partial_i E_j-\frac12\partial_j E^2\nonumber\\
& =\left[(\nabla\cdot\vec{E})\vec{E}+(\vec{E}\cdot\nabla)\vec{E}-\frac12\nabla E^2\right]_j\nonumber
\end{align}

 

[Rick16]

Great, thanks a lot!