# Energy in Electromagnetic Waves

Looking at diagrams of Electromagnetic Waves, it would appear to me that at certain times the waves have zero amplitude, and consequently zero energy. Indeed, substituting in the sinusoidal terms into the Poynting Vector equation, It would seem that at certain times the energy disappears. Why is this not the case?

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At certain positions in the waves, the EM field is zero and thus zero energy is stored at those positions. But at other positions, the EM field is at a maximum, and those points are local maxima of energy. That pattern of oscillation between zero energy and maximum energy moves in the direction of propagation of the wave but never changes - in particular, the maximum value of the EM field (the amplitude) stays constant, and there is no time at which the EM field is zero everywhere.

As for your conclusion from the definition of the Poynting vector that the energy disappears at certain times: it's not correct, but I couldn't tell you why without seeing how you did it. What I can do is show the calculation for an electromagnetic plane wave, defined by

$$\vec{E}(z,t) = E_0\hat{x}\sin(kz - \omega t)$$

The corresponding magnetic field is

$$\vec{B}(z,t) = \frac{1}{c}\hat{k}\times\vec{E}(z,t) = \frac{E_0}{c}\hat{y}\sin(kz - \omega t)$$

since I'm setting the direction of propagation as $\hat{k} = \hat{z}$. Check that this satisfies Maxwell's equations if you want. The energy density is

\begin{align} u(z,t) &= \frac{\epsilon_0}{2}E(z,t)^2 + \frac{1}{2\mu_0}B(z,t)^2 \\ &= \frac{\epsilon_0}{2}\biggl(E_0\hat{x}\sin(kz - \omega t)\biggr)^2 + \frac{1}{2\mu_0}\biggl(\frac{E_0}{c}\hat{y}\sin(kz - \omega t)\biggr)^2 \\ &= \epsilon_0 E_0^2\sin^2(kz - \omega t) \end{align}

using $\frac{1}{c^2} = \epsilon_0\mu_0$. This energy density does vary from point to point, but at any fixed time, if you take the average over one cycle, a length $\frac{2\pi}{k}$, you get

$$k\int_0^{2\pi/k} u(z,t)\mathrm{d}z = k\int_0^{2\pi/k} \epsilon_0 E_0^2\sin^2(kz - \omega t)\mathrm{d}z = k\epsilon_0 E_0^2 \frac{\pi}{k} = \pi\epsilon_0 E_0^2$$

which does not depend on time. So the average energy density is constant, it does not ever go to zero.

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If the E and B both have sin components, the Poynting vector should have a sin^2, shouldn't it? This should be equal to zero at some point-shouldn't it? –  Anthony May 15 '13 at 2:08
As I said, it is equal to zero at some points, but it also reaches a maximum at other points. –  David Z May 15 '13 at 2:10
@Tony: try to get the avg. of $sin^2\theta$..... –  ABC May 15 '13 at 2:12
I think the misunderstanding here is that EM waves aren't just an instantaneous point in time (position on the wave). They contain more of the wave than that. –  Brandon Enright May 15 '13 at 2:14
Yeah-Thank you all. –  Anthony May 15 '13 at 2:15