# Help finding an alternative to the time average of the Poynting vector

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## Solution 1

You seem to have two misconceptions:

1. If $\vec{E}$ is defined to be the real part of the complex field $\vec{E}_c$, where $\vec{E}_c$ is a mathematical tool and not physical. The formula for the Poynting vector involves $\vec{E}\times \vec{B}$, which is very much NOT the real part of $\tilde{E}_c\times \tilde{B}_c$. That is to say, you WANT to calculate $\mathrm{Re}(\vec{E}_c)\times \mathrm{Re}(\vec{B}_c)$, not $\mathrm{Re}(\vec{E}_c\times\vec{B}_c)$ (which is what you have calculated). If you proceeded with the cross product you have in your post, you would in fact find the time average of $\vec{E}_c\times\vec{B}_c$ to be zero!
2. For computing a time average, you don't need the duration of the event. The time average as $T\to \infty$ and the time average for $T$ equal to one period (which would be $T=2\pi/\omega$) are equal, so you can compute either. For example, $$\frac{1}{2\pi} \int_0^{2\pi} \sin^2(\theta)d\theta=\lim_{T\to \infty} \frac{1}{T} \int_0^{T}\sin^2(\theta)d\theta=\frac{1}{2}$$(the graph of sine squared is centered around 1/2, so this agrees with intuition)

To solve the problem, recalculate the cross product correctly as $\mathrm{Re}(\vec{E}_c)\times \mathrm{Re}(\vec{B}_c)$, and then plug in $T=2\pi/\omega$.

## Solution 2

The canonical expression, valid in the far field, is to use the (complex) phasor form $\vec E_s$ and $\vec B_s$. In this form the Poynting vector is simply $$\langle\vec S\rangle =\frac{1}{2\mu}\hbox{Re}\left[\vec E_s\times \vec B_s^*\right]\, . \tag{1}$$ You can find a derivation of this result on this wiki page. The Poynting vector is usually calculated in terms of $\vec H$, rather than $\vec B$.

Edit: for propagation in a lossless medium the result is just $$\langle \vec S\rangle =\frac{1}{2}\frac{\vert \vec E\vert^2}{\eta_0}\hat k$$ where $\eta_0\approx 120\pi$ is the impedance of vacuum, and $\hat k$ is a unit vector in the direction of propagation: $\hat k= \vec k/\vert \vec k\vert$.

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### loltospoon

Updated on April 26, 2020

The time averaged Poynting vector is:

$$\langle \vec{S} \rangle = \frac{1}{T}\int_0^T\frac{1}{\mu_0}(\vec{E} \times \vec{B}) dt$$

But I am not given the total time of the event. Is there another way to find the time average of the Poynting vector here?

The context:

A plane wave hits a conducting surface at an angle $\theta_I$ and I need to find the radiation pressure given by:

$$P = \frac{2\langle S \rangle}{c}\cos ^2(\theta_I)$$

Based on an answer given that said that I should use the real parts the electric and magnetic fields, I've found $\vec{E} \times \vec{B}$ to be:

$$\vec{E} \times \vec{B} = (E_0)^2 \left [\cos(zk_I\sin(\theta)+xk_I\cos(\theta)-wt)\right ]^2 \left [\frac{\sin \theta_I}{c}\hat x + \frac{\cos \theta_I}{c}\hat z \right ]$$

Write your $\vec E$ and $\vec B$ in terms of exponentials and strip off the time-dependent factor. This should give you the phasor, i.e. something like $$\vec E\to \vec E_s: = \hat n_\perp E_{0}e^{i k (x\sin\theta_i-z\cos\theta_i)}.$$ Here, $\hat n_\perp$ is a unit vector in the direction of $\vec E$ that is perpendicular to the direction of propagation. Do same with $\vec B$. Proceed as per (1).
Ok so I've edited my question to use the real parts. But now something doesn't make sense - I can't really do the time average here because all the $\theta \to \theta_I$. These are fixed angles and therefore fixed values for $\sin \theta_I$ and $\cos \theta_I$. Does that make sense?
Furthermore, do I use $\vec{E_I}$ or $\vec{E_R}$ when computing the cross product? What about for $\vec{B}$?