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Considering an electric and magnetic plane wave, $E(t)=E_p(\omega)e^{i\omega t}$ and $H(t)=H_p(\omega)e^{i\omega t}$, where $E_p(\omega)$ and $H_p(\omega)$ represent the frequency dependent peak amplitudes of the wave. The time-averaged Poynting vector is defined as $<S_x> = \frac{1}{2}\text{Re}\{E_p(\omega)H_p^*(\omega)\}$ with unit $\left[\frac{V}{m}\frac{A}{m} = \frac{W}{m^2}\right]$ (see: https://en.wikipedia.org/wiki/Poynting_vector#Time-averaged_Poynting_vector).

How to determine the total power if there is a whole radiation spectrum? For example, if you want to know the power carried by the solar spectrum? I would expect something like

$$ S_{tot} = \frac{1}{2} \text{Re}\{ \int\limits_0^\infty E_p(\omega) H_p^*(\omega) d\omega \} \,.$$

However, this expression has a wrong unit $\left[\frac{V}{m}\frac{A}{m}\frac{1}{s} = \frac{W}{m^2}\frac{1}{s}\right]$...

My question: what is the correct expression of the total power in function of the reciprocal electric and magnetic field?

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  • $\begingroup$ What is $E_p(f)$? It looks like it might be the Fourier transform of the time-domain E-field, in which case it won't have units of $\text{V/m}$. Same for $H_p(f)$. If you saw these equations and definitions somewhere, consider mentioning where and linking the source if applicable. $\endgroup$
    – Puk
    Commented Oct 27, 2020 at 19:44
  • $\begingroup$ Thank you for the comment. I made some edits in the post, I hope it is more clear now. $\endgroup$
    – Frederic
    Commented Oct 27, 2020 at 19:56

1 Answer 1

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Confusion is caused by misleading notation. $E_p (\omega), H_p (\omega)$ is the first and the second case, are slightly different things.

In the case of monochromatic wave - they are the whole electric and magnetic field. Because there is a wave with only one frequency I would drop the $\omega$ index, and denote them as $E, H$.

And in the second case - it is a density of electric(magnetic) field for a given frequency.

To see the connection between the first and the second case you may write: $$ E = \int E_p (\omega) d \omega^{'} = \int E \ \delta (\omega - \omega^{'}) d \omega^{'} $$ So the units of $E_p (\omega)$ are $\left[\frac{V s}{m} \right]$. Now perform the integration ($T \rightarrow \infty$): $$ S = \frac{1}{2T} \text{Re} \left[\int_{-T}^{T} dt \int d \omega \int d \omega^{'} E_p (\omega) H_p^{*} (\omega^{'}) e^{i (\omega - \omega^{'}) t / T} \right] = \frac{1}{2} \text{Re} \left[ \int d \omega \ E_p (\omega) H_p^{*} (\omega) \right] $$ Checking the dimensions of S: $$ S = \left[\frac{V s}{m} \frac{A s}{m} \frac{1}{s} \frac{1}{s} \frac{s}{s}\right] = \left[\frac{W}{m^2} \right] $$

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    $\begingroup$ @Frederic 1) By $E_p (\omega)$ one means, that there is not just a monochromatic wave, but a wave packet, compising waves with different frequencies, like the daylight is not a single red or green wave, but a complicated mixture 2) $T$ is a time, over which the averaging is computed. During the period, the magintude of $S$ changes, but the average can be computed. Here it is tacitly assumed, that $T$ is largen than the period of any wave in the packet. $\endgroup$ Commented Oct 28, 2020 at 9:26
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    $\begingroup$ @Frederic 3) I've substituted the expansions of form $\int E_p (\omega) d \omega$ (same for $H_p (\omega)$), and I've denoted it with the prime index $\omega^{'}$ in order to distinguish from the omega. Then I've performed integration over time $\int_{-T}^{T} dt e^{i (\omega - \omega^{'}) t} = \delta (\omega - \omega^{'})$ and eliminated by delta function $\omega^{'}$. Here I am inaccurate , but the general idea works as follows. $\endgroup$ Commented Oct 28, 2020 at 9:31
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    $\begingroup$ @Frederic here I've been inaccurate, the division by the infinite time cancels. The correct formula is $\frac{1}{T} \int_{-T}^{T} e^{i \omega t / T} dt = 2 \pi \delta (\omega)$, so the integration over $t$ will give a factor of $T$, which will cancel with the $T$ in the denominator. I'll edit the answer $\endgroup$ Commented Oct 28, 2020 at 12:40
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    $\begingroup$ @Frederic useful reference - fourier.eng.hmc.edu/e102/lectures/ExponentialDelta.pdf $\endgroup$ Commented Oct 28, 2020 at 12:43
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    $\begingroup$ @Frederic integration over $\omega, \omega^{'}$ still remains, thus the factor $[\frac{1}{s^2}]$ and $s / s$ from the $dt$ and $1/T$ cancel each other. With the finite time interval, frequencies are restricted to lie in range $[-\pi/T, \pi /T]$ - so $1/T$ is some normalization in the exponent $e^{i \omega t / T}$, which scales the $\omega$ to $[-\pi, \pi]$ $\endgroup$ Commented Oct 28, 2020 at 16:34

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