Symmetry of the dielectric tensor In the book Principles of Optics by Max Born, in chapter XIV, the rate of change in the electric energy density $w_{e}$ is generalised to
\begin{equation}
\frac{dw_{e}}{dt} = \frac{1}{4\pi}\sum_{kl}\,E_{k}\epsilon_{kl}\dot{E}_{l}
\tag 1
\end{equation}
in order to take into account anisotropic media. It is said, however, that the right side of the equation above cannot be interpreted as the rate of change in the electric energy density unless
\begin{equation}
\frac{dw_{e}}{dt} = \frac{1}{4\pi}\sum_{kl}\,E_{k}\epsilon_{kl}\dot{E}_{l}\, = \frac{1}{8\pi}\sum_{kl}\,\epsilon_{kl}( E_{k}\dot{E}_{l}+ \dot{E}_{k}E_{l}) 
\tag 2
\end{equation}
that is, unless
\begin{equation}
\sum_{kl}\,\epsilon_{kl}( E_{k}\dot{E}_{l} - \dot{E}_{k}E_{l}) = 0 
\tag 3
\end{equation}
which implies $\epsilon_{kl} = \epsilon_{lk}$, given that $k$ and $l$ are dummy indices.
Now, for isotropic media $\epsilon_{kl} = \epsilon\,\delta_{kl}$ and equation (1) is as expected. I don't understand, however, why this expression can only be identified with the change in the electric energy density if the requirement of equation (2) is satisfied. To me, it all seems like circular reasoning because you can only write the equation (2) if the tensor is symmetric to start with. Can you help me understand this reasoning?
 A: The requirement of permutation symmetry in couplings of this form is a pretty universal feature, and the core reason for it is that for the energy to be a well-defined function of the state variables, you need it to be path-independent.
This is easiest to see using a concrete example, so consider a 2D case in which the susceptibility tensor reads
$$
\epsilon 
= \begin{pmatrix} \epsilon_{xx} & \epsilon_{yx} \\ \epsilon_{xy} & \epsilon_{yy} \end{pmatrix} 
= \begin{pmatrix} 0 & 0 \\ \epsilon_{xy} & 0 \end{pmatrix},
$$
and consider two processes that take $(E_x,E_y)$ from $(0,0)$ to $(E_0,E_0)$, 


*

*via the leg $(E_x,E_y): (0,0) \to (0,E_0)\to (E_0,E_0)$, versus

*via the leg $(E_x,E_y): (0,0) \to (E_0,0)\to (E_0,E_0)$,


with each side of the square traversed uniformly over a time $T$.
In the first process, you have 
$$
\frac{dw_{e}}{dt} 
= \frac{1}{4\pi}\sum_{kl}\,E_{k}\epsilon_{kl}\dot{E}_{l}
= \frac{1}{4\pi}\,E_{x}\epsilon_{xy}\dot{E}_{y}
= 0
$$
on the first leg, because ${E}_x=0$, and on the second leg you have $\dot{E}_y=0$, so you also get
$$
\frac{dw_{e}}{dt} 
= \frac{1}{4\pi}\sum_{kl}\,E_{k}\epsilon_{kl}\dot{E}_{l}
= \frac{1}{4\pi}\,E_{x}\epsilon_{xy}\dot{E}_{y}
= 0,
$$
and you conclude that $\Delta w_e=0$.
On the other hand, in the second process, you also have $\dot{E}_y=0$, so you also have $\frac{dw_{e}}{dt} =0$, but the closing side of the square is different, since
$$
\frac{dw_{e}}{dt} 
= \frac{1}{4\pi}\sum_{kl}\,E_{k}\epsilon_{kl}\dot{E}_{l}
= \frac{1}{4\pi}\,E_{x}\epsilon_{xy}\dot{E}_{y}
= \frac{1}{4\pi}\,E_{0}\epsilon_{xy}\frac{E_0}{T}
= \frac{1}{4\pi}\,\frac{1}{T}\epsilon_{xy}E_{0}^2
= 0,
$$
and you conclude that $\Delta w_e = \frac{1}{4\pi}\epsilon_{xy}E_{0}^2\neq 0$.

As you can see, the coupling tensor I started with is inconsistent with $w_e$ being a function of the state variables. A slightly more formalized version of the same argument is enough to show that this property is feasible if and only if the coupling tensor is symmetric in each pair of indices.
