- #1
Chris L T521
Gold Member
MHB
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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:
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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.