Saturday, 27 January 2018

electromagnetic radiation - How does energy transfer between B and E in an EM standing wave?


I'm trying to understand how an electric field induces a magnetic field and vice versa, its associated energy, as well as relating it to my understanding of waves on a string.


Using a standing wave as an example, I came up with the equations


$\vec{E}=A\sin(\omega t)\sin(kx)\hat{y}$


$\vec{B}=\frac{Ak}{w}\cos(\omega t)\cos(kx)\hat{z}$


I checked them against Maxwell's equations, and they're self-consistent. At time 0, this reduces to:


$\vec{B}=\frac{Ak}{w}\cos(kx)\hat{z}$


Since the electric field is 0, based on the Poynting vector, there's no energy transfer at this time. At this time, at a node where $\vec{B}=0$, there's neither electric field nor magnetic field. If there's no energy transfer, and no energy stored in either field, then how can an electric field exist at this point at some time later? How is the energy stored, or transferred from elsewhere?



Answer



The energy conservation is written $\dfrac{\partial u}{\partial t} + div \vec S=0$, where $u$ is the energy density $\vec E^2+\vec B^2$, and $\vec S$ is the Poynting vector $\vec E \wedge \vec B$ (skipping irrelevant constant factors).



If you choose $x,t$ such as $\sin \omega t=0$ and $\cos k x=0$, both $E$, $B$, the energy density $u$ and the Poynting vector $S$ will be zero.


We have $\vec S \sim \sin 2k x \sin 2\omega t ~\hat{x}$. The divergence of the Poynting vector will be $div \vec S\sim \cos 2k x \sin 2\omega t $, so it it zero too (because of the $t$ dependence ), and so $\dfrac{\partial u}{\partial t} = 0$. However, the first time derivative of $div(\vec S)$is not zero : $\dfrac{\partial (div \vec S)}{\partial t}\sim \cos 2k x \cos 2\omega t $, so, from the energy conservation equation, the second derivative of the density energy $\dfrac{\partial^2 u}{\partial t^2} = - \dfrac{\partial (div \vec S)}{\partial t}$ is not zero.


So, you may write :


$u(x,t+\epsilon) = \frac {\epsilon^2}{2} \dfrac{\partial^2 u}{\partial t^2} + o(\epsilon^2)$


So, at infinitesimal times after $t$, the energy density is not zero.


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