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Problem of the Week #76 - September 9th, 2013

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Chris L T521

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Jan 26, 2012
995
Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Background Info: Maxwell's equations relating the electric field $\mathbf{E}$ and magnetic field $\mathbf{H}$ as they vary with time in a region containing no charge and no current can be stated as follows:

\[\begin{array}{ccc} \begin{aligned}\mathrm{div}\,\mathbf{E} &= 0 \\ \mathrm{curl}\,\mathbf{E} &= -\frac{1}{c}\frac{\partial \mathbf{H}}{\partial t} \end{aligned} & & \begin{aligned}\mathrm{div}\,\mathbf{H} &= 0 \\ \mathrm{curl}\,\mathbf{H} &= \frac{1}{c}\frac{\partial\mathbf{E}}{\partial t}\end{aligned}\end{array}\]

where $c$ is the speed of light.

Problem: Use the above equations to prove the following:

(a) $\displaystyle \nabla\times (\nabla\times \mathbf{E}) = -\frac{1}{c^2}\frac{\partial^2 \mathbf{E}}{\partial t^2}$

(b) $\displaystyle \nabla\times (\nabla\times \mathbf{H}) = -\frac{1}{c^2}\frac{\partial^2 \mathbf{H}}{\partial t^2}$

(c) $\displaystyle\nabla^2\mathbf{E} = \frac{1}{c^2}\frac{\partial^2 \mathbf{E}}{\partial t^2}$

(d) $\displaystyle\nabla^2\mathbf{H} = \frac{1}{c^2}\frac{\partial^2 \mathbf{H}}{\partial t^2}$

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Note: $\nabla \times\mathbf{F}$ and $\nabla\cdot\mathbf{F}$ also denotes the curl and divergence of a vector field $\mathbf{F}$ respectively.

Hint:
For (c) & (d), the indentity $\mathrm{curl}\,(\mathrm{curl}\,(\mathbf{F}) = \mathrm{grad}\,(\mathrm{div}\,(\mathbf{F})) - \nabla^2\mathbf{F}$ may come in handy.

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by Ackbach. You can find his solution below.

For part a): We start with Faraday's Law, invoke Clairaut's Theorem (the electric and magnetic fields will have continuous second-order mixed partial derivatives, because they are physical quantities), and then substitute in Ampere's Law:

\begin{align*}
\nabla \times \vec{E}&=- \frac{1}{c} \, \frac{\partial \vec{H}}{ \partial t}\\
\nabla \times (\nabla \times \vec{E})&= \nabla \times \left(- \frac{1}{c} \, \frac{\partial \vec{H}}{ \partial t} \right)\\
&= - \frac{1}{c} \, \frac{\partial}{ \partial t} \left(\nabla \times \vec{H}\right)\\
&= - \frac{1}{c} \, \frac{\partial}{ \partial t} \left( \frac{1}{c} \frac{ \partial \vec{E}}{ \partial t} \right)\\
&= - \frac{1}{c^{2}} \frac{ \partial^{2} \vec{E}}{ \partial t^{2}}.
\end{align*}

Part b) is exactly analogous with part a), except that you start with Ampere's Law and substitute in Faraday's Law.

For part c): We start with the result of part a), and note the vector identity $\nabla \times (\nabla \times \vec{E})= \nabla( \nabla \cdot \vec{E})- \nabla^{2} \vec{E}$. The first term on the RHS vanishes because of Gauss's Law for Electric Fields, and you are left with
$$-\nabla^{2} \vec{E}=- \frac{1}{c^{2}} \frac{ \partial^{2} \vec{E}}{ \partial t^{2}},$$
from which the result follows.

Part d), again, is exactly analogous to part c), except that we will need to invoke Gauss's Law for Magnetic Fields instead.
 
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