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Math Help - Electromagnetic fields for empty space

  1. #1
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    Last edited by nairbdm; May 8th 2008 at 02:17 PM.
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    Quote Originally Posted by nairbdm View Post
    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 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}
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  3. #3
    Lord of certain Rings
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    Quote Originally Posted by nairbdm View Post
    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.
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