Sorry if this is a naive question but I've been struggling in trying to proof this for a week. Consider an electromagnetic wave with wave vector $\vec{k}=k\hat{n}$, the Maxwell stress tensor can be written as $T_{ij}=-un_in_j$, where $u$ is the energy density of the electromagnetic wave and it's independent of the wave polarization.

For simplicity, let's consider a monochromatic wave with arbitrary polarization. In the orthonormal base ($\hat{e_1},\hat{e_2},\hat{n}$), the fields can be expressed as:

$\vec{E}=(E_1\hat{e_1}+E_2\hat{e_2})e^{i(\vec{k}\cdot\vec{r}-\omega t)}$

with $\vec{B}=(\hat{n}\times\vec{E})/c$, so:

$\vec{B}=\frac{1}{c}(E_1\hat{e_2}-E_2\hat{e_1})e^{i(\vec{k}\cdot\vec{r}-\omega t)}$

The average energy density is defined as:


Using the above relations, we get:


On the other hand, the Maxwell stress tensor is defined as:


However, I have no idea how to express the Maxwell stress tensor as required, since I can't get rid of the polarization or can't properly express the components in the correct base. Thus my main question is how to properly express an reduce the tensor components (one example will be enough to get the idea).

For example, let's consider the term

$T_{xy}=\epsilon_0(E_xE_y+c^2B_xB_y)=\frac{\epsilon_0 (E_1^2+E_2^2)}{2} e^{2i(\vec{k}\cdot\vec{r}-\omega t)}\left (\frac{(E_1\hat{e_1}(\hat{x})+E_2\hat{e_2}(\hat{x}))((E_1\hat{e_1}(\hat{y})+E_2\hat{e_2}(\hat{y}))}{(E_1^2+E_2^2)}-\frac{((E_1\hat{e_2}(\hat{x})-E_2\hat{e_1}(\hat{x}))((E_1\hat{e_2}(\hat{y})-E_2\hat{e_1}(\hat{y}))}{(E_1^2+E_2^2)}\right )$

where I don't understand how to express the polarization base in terms of the Cartesian components, or how to get rid of the exponential (since we don't have a complex conjugate product).

If I try the particular case where I align the ($\hat{e_1},\hat{e_2},\hat{n}$) to ($\hat{x},\hat{y},\hat{z}$), I can find the non-zero components as:

$T_{xx}=u\frac{(E_1^2-E_2^2)}{(E_1^2+E_2^2)} \ \ , \ \ $ $T_{xy}=u\frac{(E_1E_2)}{(E_1^2+E_2^2)} \ \ , \ \ $ $T_{yy}=u\frac{(E_2^2-E_1^2)}{(E_1^2+E_2^2)} \ \ , \ \ $ $T_{zz}=u$

where only the $T_{zz}$ satisfies the required answer, while the others don't, so I don't really know how else I can proceed.

  • $\begingroup$ $u=\frac{\epsilon_0}{4}(\vec{E}\cdot\vec{E}^*+c^2\vec{B}\cdot\vec{B}^*)$ should be $u=\frac{\epsilon_0}{2}(\vec{E}\cdot\vec{E}^*+c^2\vec{B}\cdot\vec{B}^*)$ $\endgroup$
    – my2cts
    Mar 3, 2019 at 12:20
  • $\begingroup$ @my2cts For the energy density, you're right, the $1/2$ factor comes from taking the average but it's pointless if I'm not taking the average too of the Maxwell tensor. On the other hand, why did you change the sign in the last term so it vanishes? I copied the formula for the Maxwell stress tensor from Jackson's and it is the one in my original question. $\endgroup$
    – Charlie
    Mar 4, 2019 at 3:51
  • $\begingroup$ Really? I checked other books on there's still a positive sign in the last term. For example, here it has the same expression: en.wikipedia.org/wiki/Maxwell_stress_tensor $\endgroup$
    – Charlie
    Mar 4, 2019 at 18:38
  • $\begingroup$ A second error is that E and B are complex valued. You start out with the correct notation but then continue, following Jackson I presume, as if the fields are real valued. The stress tensor is correct. $\endgroup$
    – my2cts
    Mar 5, 2019 at 0:26

1 Answer 1


four years later you might not be that interested in the answer, but better late than never.

The way you write your waves is right and convenient, so I'll work with that notation.

$$\vec{E} = (E_1\hat{e}_1+E_2\hat{e}_2)e^{i(k\hat{n}\cdot\vec{r}-\omega t)},\\ \vec{B} =\frac{1}{c}\hat{n}\times\vec{E}= \frac{1}{c}(E_1 \hat{e}_2-E_2\hat{e}_1)e^{i(k\hat{n}\cdot \vec{r}-\omega t)},$$where $\lbrace \hat{e}_1, \hat{e}_2, \hat{n}\rbrace$ is an orthogonal basis and the amplitudes $E_1, E_2$ are complex numbers. The time-averaged Maxwell stress tensor is given by $$2\langle T_{ij}\rangle=\textrm{Re}\left[\epsilon_0 \left(E_i E_j^\ast-\frac{1}{2}\delta_{ij}|E|^2 \right)+\frac{1}{\mu_0}\left(B_i B_j^\ast-\frac{1}{2}\delta_{ij}|B|^2\right)\right],$$where $|E|^2=E_1^2 + E_2^2$ and $|B^2|=\frac{1}{c^2}\left(E_1^2+E_2^2\right)$. If you wonder about the factor of two, it comes from the following theorem (check Section 1.6.3 of Zangwill's book, the proof is easy): Let $A(\vec{r},t)=a(\vec{r})e^{-i\omega t}$ and $B(\vec{r},t)= b(\vec{r})e^{-i\omega t}$. If $T=2\pi/\omega$, then $$\langle \textrm{Re}[A]\textrm{Re}[B] \rangle = \frac{1}{T}\int_0^T \textrm{Re}[A]\textrm{Re}[B] dt = \frac{1}{2}\textrm{Re}[a b^\ast].$$

$\hspace{5mm}$Considering that $E_i = \vec{E}\cdot\hat{x}_i = E_1 e_1^i + E_2 e_2^i,$ where $\hat{x}_i$ is the $i$-th cartesian unit vector, we have $$2\langle T_{ij}\rangle=\textrm{Re}\left[\epsilon_0 \left(|E_1|^2 e_1^i e_1^j + |E_2|^2 e_2^i e_2^j + E_1 E_2^* e_1^i e_2^j + E_2 E_1^* e_2^i e_1^j - \frac{1}{2}\delta_{ij}|E|^2 \right) + \frac{1}{\mu_0 c^2}\left(|E_1|^2 e_2^i e_2^j +|E_2|^2 e_1^i e_1^j - E_1 E_2^* e_2^i e_1^j - E_1^* E_2 e_1^i e_2^j - \frac{1}{2}\delta_{ij}|E|^2\right)\right].$$I don't respect the index position because for this problem it doesn't really matter. Note that $1/(\mu_0 c^2) = \epsilon_0 \mu_0/\mu_0 = \epsilon_0$. Thus, we can factorize several terms $$2\langle T_{ij}\rangle = \textrm{Re}\bigg\lbrace\epsilon_0 \left[ |E|^2 \left( e_1^i e_1^j + e_2^i e_2^j\right) + (E_1 E_2^\ast - E_1^\ast E_2)(e_1^i e_2^j - e_2^i e_1^j)-\delta_{ij}|E|^2 \right]\bigg\rbrace.$$ Let us analyze the first term in parentheses $$e_1^i e_1^j + e_2^i e_2^j$$by substituting $e_2^i = (\hat{n}\times \hat{e}_1)^i = \epsilon^{ilm}n^l e_1^m$ and $e_2^j=\epsilon^{jpq}n^p e_1^q$ (repeated indices sum). Then $$e_2^i e_2^j = \epsilon^{ilm}\epsilon^{jpq}e_1^m e_1^q n^l n^p.$$ Next we use the following identity, which you can find in Wikipedia (https://en.wikipedia.org/wiki/Levi-Civita_symbol), \begin{align*}\epsilon^{ilm}\epsilon^{jpq} = \begin{vmatrix} \delta^{ij} & \delta^{ip} & \delta^{iq} \\ \delta^{lj} & \delta^{lp} & \delta^{lq} \\ \delta^{mj} & \delta^{mp} & \delta^{mq}\end{vmatrix} &= \delta^{ij}(\delta^{lp}\delta^{mq}-\delta^{lq}\delta^{mp})- \delta^{ip}(\delta^{lj}\delta^{mq} - \delta^{lq}\delta^{mj})\\ &+\delta^{iq}(\delta^{lj}\delta^{mp}-\delta^{lp}\delta^{mj}).\end{align*} Then \begin{align*} e_2^i e_2^j &= \delta^{ij}(n^l n^l e_1^m e_1^m - n^l e_1^l e_1^m n^m)-n^i(n^j e_1^m e_1^m-e_1^j n^l e_1^l)\\ &+e_1^i(n^j e_1^m n^m - n^p n^p e_1^j).\end{align*}Since $\hat{n}\cdot \hat{e}_1 = 0$ and $\hat{n}\cdot \hat{n} = \hat{e}_1\cdot\hat{e}_1 = 1$, the latter reduces to $$e_2^i e_2^j = \delta^{ij}-n^i n^j-e_1^i e_1^j.$$

If we substitute in the stress tensor we obtain $$2\langle T_{ij}\rangle =\textrm{Re} \left[-\epsilon_0 |E|^2 n^i n^j + \epsilon_0 (E_1 E_2^\ast - E_1^\ast E_2)(e^i_1 e_2^j - e_2^i e_1^j)\right].$$ The last term is zero for a wave with linear polarization, because for that case the amplitudes $$E_1 = |E_1|e^{i\delta_1}, \quad E_2 = |E_2|e^{i \delta_2},$$satisfy $\delta_2-\delta_1 = m\pi$, with $m\in\mathbb{Z}.$ For other types of polarization the term will still vanish when you take the real part, because $E_1 E_2^\ast - E_1^\ast E_2 = 2 i \textrm{Im}(E_1 E_2^\ast)$, $\textit{i.e.}$ it's purely imaginary. This leaves us with $$2\langle T_{ij}\rangle = -\epsilon_0 |E|^2n^i n^j.$$

$\hspace{5mm}$On the other hand, the time-averaged energy density is $$2\langle u\rangle = \frac{\epsilon_0}{2}\vec{E}\cdot\vec{E}^\ast + \frac{1}{2\mu_0} \vec{B}\cdot \vec{B}^\ast = \frac{\epsilon_0}{2}|E|^2 + \frac{1}{2\mu_0} \frac{|E|^2}{c^2} = \epsilon_0|E|^2 .$$Therefore $$\langle T_{ij} \rangle = -\langle u \rangle n^i n^j.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.