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Following these notes: http://www.colorado.edu/physics/phys4510/phys4510_fa05/Chapter6.pdf

This is likely just an artifact of me getting confused by notation and not understanding vectors or how to parse $<|\vec{E}(x,t)|>$, where $<...>$ is a time average. Here is the context:

Suppose we have a EM wave given by $$\vec{E}(x,t) = \vec{E_0}e^{i(kx-\omega t+\phi)} $$

The intensity/irradiance is given by (this is going to be wrong at some point, but I'm not sure where) \begin{align} I(x) =& \frac{1}{\eta}<|\vec{E}(x,t)|^2> \\ =& \frac{1}{\eta}<\vec{E}(x,t) \vec{E}(x,t)^*> \\ =& \frac{1}{\eta}<\vec{E_0}e^{i(kx-\omega t+\phi)}\vec{E_0}e^{-i(kx-\omega t+\phi)}> \\ =& \frac{1}{\eta}<\vec{E_0}\vec{E_0}> \\ \end{align}

At which point we've eliminated the time dependence and $<...>$ should be irrelevant, but we also know $$I(x) = \frac{E_0^2}{2}$$ since we're time-averaging.

Clearly something goes wrong somewhere. My question is, where does it go wrong, and how can I reconcile the phasor notation result with what we would get if we just stuck with writing the field as a cosine?

A more direct question is, what does the absolute value imply in the first line? Is it the magnitude of the phasor? The vector?

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  • $\begingroup$ The mistake is on the first line: the actual electric field is $\text{Re}(\mathbf{E})$, so the intensity is proportional to $\langle \text{Re}(\mathbf{E})^2 \rangle$. This is not equal to $\langle \mathbf{E}^2 \rangle$, which is larger by a factor of two since it includes both the real and imaginary parts. $\endgroup$ – knzhou Jan 16 '17 at 0:50
  • $\begingroup$ If the resource you were using wrote $|\mathbf{E}|$, they meant the magnitude of the physical electric field. That is, take the real part of your phasor to get a vector, then take the magnitude of that vector. You're right that the notation is a bit vague; it would be better to mark phasors with something else, like a tilde over them, e.g. $\widetilde{\mathbf{E}}$. $\endgroup$ – knzhou Jan 16 '17 at 0:51
  • $\begingroup$ Thanks for the reply! I just added a link to the notes I'm reading. If $|\textbf{E}|$ implies the magnitude of the field, what does it mean to take the time average? $\endgroup$ – dkv Jan 16 '17 at 0:57
  • $\begingroup$ I'm not totally sure what you're asking -- the time average of a function $f(t)$ is $\lim_{T \to \infty} \int_0^T f(t) \, dt\, /T$. Did you mean to ask something else? $\endgroup$ – knzhou Jan 16 '17 at 0:59
  • $\begingroup$ Oops, sorry, mixed up "magnitude" and "amplitude." That makes sense. But then how does the author then go from $|\textbf{E}|^2$ to $\textbf{E} \cdot \textbf{E}^*$? i.e. at which point do we take the real part? $\endgroup$ – dkv Jan 16 '17 at 1:18

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