# Taking the Curl of Maxwell's Equations to show that E and B obey the wave equation

#### knowlewj01

1. The problem statement, all variables and given/known data

This question is closely related to physics but it's in a maths assignment paper i have so here it is:

By taking curls of the following equations:

$\nabla \times \bf{E} = -\frac{1}{c}\frac{\partial\bf{B}}{\partial t}$

$\nabla \times \bf{B} = \frac{1}{c}\frac{\partial\bf{E}}{\partial t}$

show that both E and B obey the wave equation with speed c, that is:

$\nabla^2\bf{E} = \frac{1}{c^2}\frac{\partial^2\bf{E}}{\partial t^2}$

and

$\nabla^2\bf{B} = \frac{1}{c^2}\frac{\partial^2\bf{B}}{\partial t^2}$

2. Relevant equations

$\nabla\cdot\bf{E} = 0$
$\nabla\cdot\bf{B} = 0$
$\nabla \times \bf{E} = -\frac{1}{c}\frac{\partial\bf{B}}{\partial t}$
$\nabla \times \bf{B} = \frac{1}{c}\frac{\partial\bf{E}}{\partial t}$

3. The attempt at a solution

I dont really know how to start this, so i took a wild guess. I started by looking at the first equation with $\nabla \times \bf{E}$ in it. and did the cross product, needless to say i got some horrible expression.

given that the divergence of both vector fields are 0, we should be able to write these vector fields as the curl of another vector field, such that:

$\bf{E} = \nabla \times \bf{A}$

iff

$\nabla\cdot\bf{E} = 0$

i think this is the starting point for the question. But i hit a brick wall at this point, anyone know what is next? thanks

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#### knowlewj01

Re: Taking the Curl of Maxwell's Equations to show that E and B obey the wave equatio

i have got a bit further with this, not sure how to finish it off though.

take curls of the equations:

$\nabla \times \left(\nabla\times\bf{E}\right) = -\frac{1}{c}\frac{\partial}{\partial t}\left(\nabla\times\bf{B}\right)$ [1]

$\nabla \times \left(\nabla\times\bf{B}\right) = \frac{1}{c}\frac{\partial}{\partial t}\left(\nabla\times\bf{E}\right)$ [2]

Use the identity: $\nabla\times\left(\nabla\times A\right) = \nabla\left(\nabla\cdot A\right) - \nabla^2 A$

and since $\nabla\cdot \bf{E} = \nabla\cdot \bf{B} = 0$

eqn's [1] and [2] become:

$\frac{1}{c}\frac{\partial}{\partial t}\left(\nabla\times\bf{B}\right) = \nabla^2\bf{E}$ [3]
$\frac{1}{c}\frac{\partial}{\partial t}\left(\nabla\times\bf{E}\right) = -\nabla^2\bf{B}$ [4]

is this correct? i dont know wether i should curl something that is already operated by a d/dt also, where should the second time derivative come into it?

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