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There is a problem I'm trying to solve for some time now and is about the standard (?) approximation that it is made when one tries to solve the Helmholtz equation in inhomogeneous media, that is

\begin{align} \nabla^2\vec{E} + k^2\vec{E}=-\nabla(\vec{E}\cdot\frac{\nabla\epsilon}{\epsilon})\approx0 \end{align}

where \begin{equation} k^2=\omega^2\mu\epsilon, \qquad \vert\frac{\nabla\epsilon}{\epsilon_0}\rvert\lambda<<1 ,\qquad \lambda =wavelength\end{equation} and only epsilon is a function of position. I have searched all the literature I could find and there is not one that explains why this approximation is valid even though it is widely accepted. I expanded the first equation for the $x$ component of the electric field, namely

\begin{equation} \frac{\partial^2 E_x}{\partial^2 x^2}+\frac{\partial^2 E_x}{\partial y^2}+\frac{\partial^2 E_x}{\partial z^2} +k^2 E_x=\frac{\partial E_x}{\partial x}\frac{\partial\epsilon}{\partial x}+\frac{\partial E_y}{\partial x}\frac{\partial\epsilon}{\partial y}+\frac{\partial E_z}{\partial x}\frac{\partial\epsilon}{\partial z}+E_x\frac{\partial^2 \epsilon}{\partial x^2}+E_y\frac{\partial^2 \epsilon}{\partial x\partial y}+E_z\frac{\partial^2 \epsilon}{\partial x\partial z}-\frac{E_x \frac{\partial \epsilon}{\partial x}+E_y \frac{\partial \epsilon}{\partial y}+E_z \frac{\partial \epsilon}{\partial z}}{\epsilon^2}\frac{\partial \epsilon}{\partial x}\end{equation}

It's not at all obvious how the small variation over a wavelength assumption can be used to make the right hand side zero. There are even second order derivatives over epsilon appearing about which we known nothing.

This question was asked here before but the answer seems to me to be completely nonsensical. For example, it makes the approximation \begin{equation} \lvert \nabla (\vec{E} \cdot \frac{\nabla \epsilon}{\epsilon})\rvert\approx kE\frac{\nabla \epsilon}{\epsilon} \end{equation} which to me makes zero sense. He just ignored the higher derivatives of epsilon and just assumed without giving any reason that the electric field is of the form \begin{equation} \vec{E}=E e^{jkr} \end{equation} with $k$ being a constant, which is not correct by assumption.

Anyone else who stumbled across this problem?

Qmechanic
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user1524841
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1 Answers1

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As you say, the RHS of the Helmholtz equation is approximately zero if $(\nabla \ln \epsilon) \lambda \ll 1$ or $\nabla \ln \epsilon \ll k$. In the answer you refer to, I did indeed claim that \begin{equation} \lvert \nabla (\vec{E} \cdot \frac{\nabla \epsilon}{\epsilon})\rvert\approx kE\frac{\nabla \epsilon}{\epsilon} + E\frac{\nabla^2 \epsilon}{\epsilon} \end{equation} but where I implicitly neglected the second term and assumed the electric field is represented by $E_0 \exp(i\vec{k}\cdot \vec{r})$.

This will be justified if we can use the slowly varying envelope approximation, which says that the properties of the waves are changing slowly in space compared to a wavelength. In those circumstances we can say that $$\nabla \epsilon \simeq \alpha\frac{\epsilon}{\lambda} \simeq \alpha\ k\ \epsilon$$ with $\alpha \ll 1$ and that any second derivatives, are given by $$\nabla^2 \epsilon \simeq \alpha\ k \ \nabla \epsilon \simeq \alpha^2\ k^2 \epsilon$$

If I now use these latter relations in the original equation above $$ \lvert \nabla (\vec{E} \cdot \frac{\nabla \epsilon}{\epsilon})\rvert\approx \alpha k^2 E + \alpha^2 k^2 E\ . $$ And if $\alpha \ll 1$, then $\alpha^2 \ll \alpha$ and the second term can be neglected.

That is then the justification for ignoring second derivatives and if indeed $\alpha \ll 1$, then the RHS of the Helmholtz equation is $\simeq 0$ and we are justified in assuming that a wave solution of the form $\vec{E} = \vec{E_0} \exp(i\vec{k}\cdot \vec{r})$ will be approximately correct.

Note that this cannot be used if $\epsilon$ or $k$ change significantly over a wavelength.

ProfRob
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