# Electromagnetic fields for empty space

• May 2nd 2008, 04:08 PM
nairbdm
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• May 3rd 2008, 12:19 AM
Isomorphism
Quote:

Originally Posted by nairbdm
Show that in empty space, you can get electromagnetic waves. Start by writing down Maxwell's equations for empty space, where there are no currents or charges (p=0,j=0). Then show that you can derive wave equations from these empty space Maxwell's equations. That is, you should get:
(del)^2E=(1/(c^2))(2nd derivative with respect to t of E)
and
(del)^2B=(1/(c^2))(2nd derivative with respect to t of B)

Find out what c is in terms of Î knot and m knot.
Hint: to begin, operate on one of Maxwell’s equations with either del, del (dot), or del X. Then use the table of vector identities involving del.
Del = upside-down triangle (operator)

Thanks.

$\nabla\cdot \textbf{D} = 0$

$\nabla\cdot \textbf{B} = 0$

$\nabla \times \textbf{E} = -\frac{\partial \textbf{B}}{\partial t}$

$\nabla \times \textbf{H} = \frac{\partial \textbf{D}}{\partial t}$

$\nabla \times (\nabla \times \textbf{E}) = \nabla(\nabla\cdot \textbf{E}) - \nabla^2(\textbf{E})$

Now $\textbf{D} = \epsilon \textbf{E}$ since empty space is an isotropic medium.

$\nabla\cdot \textbf{D} = 0 \Rightarrow \nabla\cdot \textbf{E} = 0$

So
$\nabla \times (\nabla \times \textbf{E}) = - \nabla^2(\textbf{E})$

But $- \nabla^2(\textbf{E}) = \nabla \times (\nabla \times \textbf{E}) = -\frac{\partial(\nabla \times\textbf{B})}{\partial t} = -\mu \frac{\partial(\nabla \times\textbf{H})}{\partial t} = -\mu\epsilon \frac{\partial^2 \textbf{E}}{\partial t^2}$

Now use $\frac1{c^2} = \mu\epsilon$ to obtain $\nabla^2(\textbf{E}) = \frac1{c^2}\frac{\partial^2 \textbf{E}}{\partial t^2}$
• May 3rd 2008, 09:14 PM
Isomorphism
Quote:

Originally Posted by nairbdm
thanks thats a big help, but how would you prove that (del)^2B=(1/(c^2))(2nd derivative with respect to t of B) in empty space.

Note that the equations are symmetric in E and -B. You can derive it similarly.