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

3,160

## 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

• loltospoon about 2 years

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:

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

• loltospoon about 5 years
I'm sorry to say this but we haven't really learned about phasors, so I don't really understand any of this.
• ZeroTheHero about 5 years
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).
• loltospoon about 5 years
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?
• loltospoon about 5 years
Furthermore, do I use $\vec{E_I}$ or $\vec{E_R}$ when computing the cross product? What about for $\vec{B}$?
• Christina about 5 years
@loltospoon There is no time dependence on theta and you should look at your book for what "complex fields" mean in the first place. The time average of the whole cosine squared term is just going to be 1/2. You really should try understanding why the equations are what they are and why complex numbers are used, instead of just plugging things into equations you are given!