The Poynting vector (in Gaussian units) is given by $\mathbf{S} = \frac{c}{8\pi}\left(\mathbf{E}\times\mathbf{H}^*\right)$, and the magnetic field is given by $\mathbf{H} = \frac{c}{\mu\omega}\left(\mathbf{k}\times\mathbf{E}\right)$ where $\mathbf{k}$ is the propagation vector.

The average energy flux through a surface of normal unit vector $\mathbf{\hat{n}}$ is $$\Re\left\{\mathbf{S}\cdot\mathbf{\hat{n}}\right\} = \frac{c^2}{8\pi\omega\mu}\Re\left\{\mathbf{\hat{n}}\cdot\left[\mathbf{E}\times\left(\mathbf{k}^*\times\mathbf{E}^*\right)\right]\right\}$$

Using the vector triple product expansion I get: $$ = \frac{c^2}{8\pi\omega\mu}\Re\left\{\left(\mathbf{\hat{n}}\cdot\mathbf{k}^*\right)\left|\mathbf{E}\right|^2 - \left(\mathbf{\hat{n}}\cdot\mathbf{E}^*\right)\left(\mathbf{E}\cdot\mathbf{k}^*\right)\right\}$$

Now for an evanescent wave, the propagation vector $\mathbf{k}$ has a purely imaginary normal component, so $$\Re\left\{\left(\mathbf{\hat{n}}\cdot\mathbf{k}^*\right)\left|\mathbf{E}\right|^2\right\} = 0$$

Therefore I am left with: $$\Re\left\{\mathbf{S}\cdot\mathbf{\hat{n}}\right\} = -\frac{c^2}{8\pi\omega\mu}\Re\left\{\left(\mathbf{\hat{n}}\cdot\mathbf{E}^*\right)\left(\mathbf{E}\cdot\mathbf{k}^*\right)\right\}$$

In theory the evanescent wave is supposed to have zero energy flux across the surface, so this expression should equal zero. But I can't see this mathematically.

I know that $\mathbf{E}\cdot\mathbf{k} = 0$, but here we have $\mathbf{k}^*$ instead. We can use the relation $\mathbf{k}^* = \mathbf{k} - 2i\Im\left\{\mathbf{k}\right\}$ to write it as:

$$\Re\left\{\mathbf{S}\cdot\mathbf{\hat{n}}\right\} = \frac{c^2}{4\pi\omega\mu}\Re\left\{i\left(\mathbf{\hat{n}}\cdot\mathbf{E}^*\right)\left(\mathbf{E}\cdot\Im\left\{\mathbf{k}\right\}\right)\right\}$$

So this would be zero if $\left(\mathbf{\hat{n}}\cdot\mathbf{E}^*\right)\left(\mathbf{E}\cdot\Im\left\{\mathbf{k}\right\}\right)$ is purely real. But this does not look obvious to me, given that $\mathbf{E}$ is in general complex-valued as well (its components might have phase shifts, for example).

How can I show that the average of the normal Poynting vector is zero for an evanescent wave?


2 Answers 2


I realized soon after writing the question that the quantity is indeed real. That is because for evanescent waves, the normal component of $\mathbf{k}$ is purely imaginary but the transverse component is purely real.

So $\Im\left\{\mathbf{k}\right\} = k_n\mathbf{\hat{n}}$ where $k_n$ is a real scalar.

Then the product is $k_n\left(\mathbf{\hat{n}}\cdot\mathbf{E}^*\right)\left(\mathbf{\hat{n}}\cdot\mathbf{E}\right)$.

We separate $\mathbf{E}$ into real and imaginary components to obtain: $$k_n\left[\mathbf{\hat{n}}\cdot\left(\Re\left\{\mathbf{E}\right\}-i\Im\left\{\mathbf{E}\right\}\right)\right]\left[\mathbf{\hat{n}}\cdot\left(\Re\left\{\mathbf{E}\right\}+i\Im\left\{\mathbf{E}\right\}\right)\right]$$

$$ = k_n\left[\left(\mathbf{\hat{n}}\cdot\Re\left\{\mathbf{E}\right\}\right)^2 + \left(\mathbf{\hat{n}\cdot\Im\left\{\mathbf{E}\right\}}\right)^2\right]$$

Since $\mathbf{\hat{n}}$ is a real vector, this quantity is purely real, so the result follows.


The above answer is incorrect.

The expression $(\vec{k^*}.\vec{n})(\vec{E}.\vec{E^*})=(\vec{k^*}.\vec{n})|E|^2$ is not equal to $k_n(\vec{n}.\vec{E^*})(\vec{n}.\vec{E})$

  • $\begingroup$ I'm not sure which part you are referring to. What I ended up doing was to write $\Im\left\{\mathbf{k}\right\} = k_n\hat{\mathbf{n}}$ where $k_n$ is some real scalar. Then $\left(\mathbf{\hat{n}}\cdot\mathbf{E}^*\right)\left(\mathbf{E}\cdot\Im\left\{\mathbf{k}\right\}\right) = \left(\mathbf{\hat{n}}\cdot\mathbf{E}^*\right)\left(\mathbf{E}\cdot\left(k_n\hat{\mathbf{n}}\right)\right) = k_n\left(\mathbf{\hat{n}}\cdot\mathbf{E}^*\right)\left(\mathbf{E}\cdot\hat{\mathbf{n}}\right) = k_n\left(\mathbf{\hat{n}}\cdot\mathbf{E}^*\right)\left(\hat{\mathbf{n}}\cdot\mathbf{E}\right)$ $\endgroup$
    – Tob Ernack
    Dec 16, 2022 at 19:33
  • $\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$ Dec 16, 2022 at 19:44

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