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I have come across an issue with the use of complex electric and magnetic fields that I just cannot quite figure out. I will lay out my thought-process and I would like to know if and why it is wrong.

The intensity is defined as the time-averaged norm of the Poynting vector $\vec{P}$ ($P$ without arrow indicates the norm).

$$\vec{P}=\frac{\vec{E}\times \vec{B} }{\mu_0 c} \qquad I(t)=\langle P \rangle _t$$

In the case of a plane wave, it can be shown from Maxwell's equations that $P = \frac{||\vec{E}||^2}{\mu_0}$. And so for the following expression for the electric field: $$E = ||\vec{E}|| = E_0 \cos(\omega t - kz)$$ We obtain the following expression for the intensity: $$I = \frac{1}{2\mu_0 c} E_0^2$$

However, it can be quite useful to instead use the complex electric field $\tilde E$, defined such that $E = Re(\tilde E)$. Since only the real part is physically relevant, it is my understanding we should only apply linear operations to $\tilde E$ so as to not "mix" the real and imaginary parts.

For our single plane wave, the maths works out such that the intensity is proportional to the module squared of the complex electric field: \begin{equation} I = \frac{1}{2\mu_0 c}|\tilde E|^2 \end{equation}

To me this is just a nice coincidence but not a general expression for arbitrary combination of plane waves since $|\cdot|^2$ is not linear. However, I often see this same expression being used to determine the intensity of non-monochromatic light.

Say we consider a two plane waves of frequency $\omega_1$ and $\omega_2$:

$$E_{tot} = E_0[\cos(\omega_1 t - k_1 z) + \cos(\omega_2 t - k_2 z)]$$ $$\tilde E_{tot} = E_0 [e^{\omega_1 t - k_1 z} + e^{\omega_2 t - k_2 z}]$$

My algebra tells me that $\langle E_{tot}^2 \rangle_t \neq \frac{1}{2}|\tilde E_{tot}|^2$, so what have I missed? This expression with the module squared truly seems to appear everywhere in textbooks and optics literature. Yet, I am convinced that it is incorrect.

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First thing to notice is that the Poynting Vector is "redefine" when dealing with complex functions. This is something usually overlooked by books, Jackson's for exemple doesn't even bother changing notation to call attention to that, so if you don't read the text you can get confused.

So, if you have a eletromagnetic wave (in the vacuum): $$ \begin{align} \vec{E} (\vec{r},t) = \vec{E}_{0} \, \cos(\omega\ t - \vec{k} \cdot \vec{r}) \tag{1}\label{eq1} \\ \vec{H} (\vec{r},t) = \hat{k} \times \frac{\vec{E}}{Z_{0}} \,, \tag{2}\label{eq2} \end{align} $$

the Poynting Vector is: $$ \vec{S} (\vec{r},t) = \vec{E} (\vec{r},t) \times \vec{H} (\vec{r},t) \ . \tag{3}\label{eq3} $$

Now, if we have the complex form of the fields: $$ \begin{align} \vec{E} (\vec{r},t) &= \Re \ [ \ \vec{\tilde{E}} (\vec{r},t) \ ] = \Re \ [ \ \vec{E}_{0} \ e^{ \ i \ ( \ \omega\ t \ - \ \vec{k} \ \cdot \ \vec{r} \ )} \ ] = \Re \ [ \ \vec{E}_{0} \ e^{-\ i \ \vec{k} \ \cdot \ \vec{r} } \ e^{ i \ \omega\ t } \ ] \\ &= \Re \ [ \ \vec{\tilde{E}}(\vec{r}) \ e^{ i \ \omega\ t } \ ] = \frac{1}{2} \ [ \ \vec{\tilde{E}}(\vec{r}) \ e^{ i \ \omega\ t } + \ \vec{\tilde{E}}^{\, \bf*}(\vec{r}) \ e^{ -i \ \omega\ t } \ ] \,, \end{align} \tag{4}\label{eq4} $$

where $\vec{\tilde{E}}(\vec{r})$ is the phasor (a complex function with phase and magnitude that change with position; it is time independent), we get: $$ \begin{align} \vec{S} (\vec{r}, t) &= \frac{1}{2} [ \vec{\tilde{E}}(\vec{r}) e^{ i \ \omega \ t } + \vec{\tilde{E}}^{\ \bf*}(\vec{r}) e^{ -i \ \omega \ t } ] \times \frac{1}{2} [ \vec{\tilde{H}}(\vec{r}) e^{ i \ \omega \ t } + \vec{\tilde{H}}^{\ \bf*}(\vec{r}) e^{ -i \ \omega \ t } ] \\ &= \frac{1}{2} \Re [ \vec{\tilde{E}}(\vec{r}) \times \vec{\tilde{H}}^{\ \bf*}(\vec{r}) ] + \frac{1}{2} \Re [ \vec{\tilde{E}}(\vec{r}) \times \vec{\tilde{H}}(\vec{r}) e^{ 2 \ i \ \omega \ t } ] \,, \end{align} \tag{5}\label{eq5} $$

whos average in time is: $$ \langle \ \vec{S} \ \rangle_{t} = \frac{1}{T} \int_{0}^{T} \vec{S}(\vec{r}, t) \ dt = \frac{1}{2} \ \Re \ [ \ \vec{\tilde{E}}(\vec{r}) \times \vec{\tilde{H}}^{\, \bf*}(\vec{r}) \ ] \,, \tag{6}\label{eq6} $$

that way you can define a "complex" Poynting Vector: $$ \vec{S}_{c} \equiv \frac{1}{2} \ \vec{\tilde{E}}(\vec{r}) \times \vec{\tilde{H}}^{\, \bf*}(\vec{r}) \,, \tag{7}\label{eq7} $$

and the average of the Poyting Vector is the real part of this quantity: $$ \langle \ \vec{S}(\vec{r},t) \ \rangle_{t} = \Re \ [ \ \vec{S}_{c} (\vec{r}) \ ] \ . \tag{8}\label{eq8} $$

Finishing the calculation we get: $$ \begin{align} \langle \ \vec{S}(\vec{r},t) \ \rangle_{t} &= \frac{1}{2} \ \Re \ [ \ \vec{\tilde{E}}(\vec{r}) \times \vec{\tilde{H}}^{\, \bf*}(\vec{r}) \ ] = \frac{1}{2Z_{0}} \ \Re \ [ \ \vec{\tilde{E}}(\vec{r}) \times \hat{k} \times \vec{\tilde{E}}^{\, \bf*}(\vec{r}) \ ] \\ &= \frac{1}{2Z_{0}} \ \Re \ \{ \ [ \ \vec{\tilde{E}}(\vec{r}) \cdot \vec{\tilde{E}}^{\, \bf*}(\vec{r}) \ ] \hat{k} - [ \ \vec{\tilde{E}}(\vec{r}) \cdot \hat{k} \ ] \ \vec{\tilde{E}}^{\, \bf*}(\vec{r}) \ \} \\ &= \frac{1}{2Z_{0}} \ | \vec{\tilde{E}}(\vec{r}) |^{2} \ \hat{k} = \frac{1}{2Z_{0}} \ | \vec{E}_{0} |^{2} \ \hat{k} \,, \end{align} \tag{9}\label{eq9} $$ as expected.

For a dichromatic wave: $$ \begin{align} \vec{\tilde{E}}_{tot} (\vec{r},t) &= \vec{E}_{0} [\ e^{i \ ( \ \omega_{1} t \ - \ \vec{k}_{1} \ \cdot \ \vec{r} \ )} + e^{i \ ( \ \omega_{2} t \ - \ \vec{k}_{2} \ \cdot \ \vec{r} \ )} \ ] \\ &= \vec{\tilde{E}}_{1}(\vec{r}) \ e^{i \ \omega_{1}\ t} + \vec{\tilde{E}}_{2}(\vec{r}) \ e^{i \ \omega_{2}\ t} \,, \end{align} \tag{10}\label{eq10} $$ you can see by yourself that the time average operation will return zeros for the combinations of the time periodic exponentials, just like before. In the end: $$ \begin{align} \langle \ \vec{S}(\vec{r},t) \ \rangle_{t} &= \frac{1}{2Z_{0}} \ \Re \ \{ \ [ \ \vec{\tilde{E}}_{1}(\vec{r}) \cdot \vec{\tilde{E}}^{*}_{1}(\vec{r}) \ ] \hat{k}_{1} + \ [ \ \vec{\tilde{E}}_{2}(\vec{r}) \cdot \vec{\tilde{E}}^{*}_{2}(\vec{r}) \ ] \hat{k}_{2} \ \} \\ &= \frac{1}{2Z_{0}} \ [ \ | \vec{\tilde{E}}_{1} |^{2} \hat{k}_{1} \ + \ | \vec{\tilde{E}}_{2} |^{2} \hat{k}_{2} \ ] = \frac{1}{2Z_{0}} \ | \vec{E}_{0} |^{2} \ ( \hat{k}_{1} + \hat{k}_{2} ) \,, \end{align} \tag{11}\label{eq11} $$

if the waves travels in the same direction $\hat{k}$: $$ \frac{1}{Z_{0}} \ | \vec{E}_{0} |^{2} \ \hat{k} \ . \tag{12}\label{eq12} $$

Notes:

$\boldsymbol{\cdot}$ The direction of wave propagation and the direction of the wave vector can be different if the medium is not (lossless) isotropic [1].

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  • $\begingroup$ I see, the complex pointing vector is already time averaged. I did not know about phasor notation. $\endgroup$ Commented Jun 4 at 13:30
  • $\begingroup$ Is there a convention for a complex Poynting vector that is not time averaged? In optics, when phase modulating optical signals we can end up with oscillating terms with low enough frequencies such that they aren't averaged out by the photodetector. $\endgroup$ Commented Jun 4 at 13:40
  • $\begingroup$ I would not phrase it that way, but, yes, for the monochromatic case the time average of the Poynting vector equals the real part of that "complex" Poynting vector, defined that way. $\endgroup$
    – koy
    Commented Jun 5 at 18:07
  • $\begingroup$ If your signals don't average to zero, do the manipulation up to equation $(5)$, then, when the temporal average is measured, the time exponentials will not go to zero in the time interval considered. You can do that with the real version of the fields too. The same goes if you have more than one frequency, but the expression will have various terms (which I omitted in my answer). $\endgroup$
    – koy
    Commented Jun 5 at 18:20
  • $\begingroup$ Phasors are a excellent tool when dealing with waves/signals. See this for more information: on phasors; about the mathematical description of signals using phasors; visualization of two phasors with different frequencies being added to form the combined signal (related to your question); visualization of three phasors with different frequencies being added to form the combined signal. $\endgroup$
    – koy
    Commented Jun 5 at 18:49

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