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Suppose the electric and magnetic fields $e(x)$ and $h(x)$ with units $\left[\frac{V}{m}\right]$ and $\left[\frac{A}{m}\right]$ and their Fourier transforms $E(k)$ and $H(k)$ with units $\left[V\right]$ and $\left[A\right]$. Let's assume that the electric and magnetic field are perpendicular to each other, then the Poynting vector becomes $S(x)=e(x)h(x)$. The spatial average Poynting vector $S$ is then defined as the spatial avaverage of $e(x)h(x)$ and can be written in function of the Fourier transforms $E(k)$ and $H(k)$ as \begin{align} S &= \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X} e(x)h(x) dx \\ &= \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X} \int\limits_{-\infty}^{\infty} E(k)e^{ikx} dk \int\limits_{-\infty}^{\infty} H(k')e^{ik'x} dk' dx \\ &= \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}E(k) H(k')\left[ \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X}e^{i(k+k')x} dx \right] dk dk' \\ &= 2 \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}E(k) H(k')\delta(k'+k) dk dk' \\ &= 2 \int\limits_{-\infty}^{\infty} E(k) H(-k)dk \\ \end{align}

The question: the unit of the Poynting vector $S$ should be $\left[\frac{VA}{m^2}\right]$ whereas the unit of the final result is $\left[\frac{VA}{m}\right]$. How can this be and how to physically interpret it?

Example: if you know the Fourier transforms $E(k)$ and $H(k)$, and you want to calculate the average Poynting vector $S$, then you will use $S=2 \int\limits_{-\infty}^{\infty} E(k) H(-k)dk$ but this will give another unit as expected?! So, then you have the wrong result?

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    $\begingroup$ I'd say that $\lim_{X\rightarrow\infty} \int_{-X}^X \exp\left[i (k+k')x\right] dx = 2 \pi \delta(k+k')$. So I don't know why are you assuming that $\lim_{X\rightarrow\infty} \frac{1}{X}\int_{-X}^X \exp\left[i (k+k')x\right] dx$ produces also $\delta(k+k')$. They have different units, to begin with. $\endgroup$
    – secavara
    Oct 29, 2020 at 16:12
  • $\begingroup$ I second this. It is just the error in the claculations. One can see from the definition that the dimensions of the average are the same as those of the Poynting vetcor. $\endgroup$
    – Roger V.
    Mar 9, 2022 at 5:14

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You haven't written the time average of $S$, you've written the spatial average of $S$, but that doesn't change your question. In calculating the average, you are interpreting $e(x)h(x)$ as a density. Poynting per meter. The Fourier transform converts this into spatial frequency space, so it becomes a density in spatial frequency space. That is, the F.T. is Poynting per wave number, the "amount" of $S$ at each wave number.

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  • $\begingroup$ Okay, this I understand. But how does it work with the equations? Then $e(x)$ and $h(x)$ should have units $[\frac{V}{m^{3/2}}]$ and $[\frac{A}{m^{3/2}}]$ to make $e(x)h(x)$ with unit $[\frac{VA}{m^2}\frac{1}{m}]$ which is Poynting per meter? $\endgroup$
    – Frederic
    Oct 29, 2020 at 11:06
  • $\begingroup$ The point made is correct, but does not refer to the context in the OP. There it is an error in the calculation. Properly defined average (and it is properly defined in the OP) has the same dimensions as the quantity being averaged. Using spectral density would be probably more appropriate here. $\endgroup$
    – Roger V.
    Mar 9, 2022 at 5:17

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