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Wave Equation in Cylindrical Coordinates

As pointed out above, the divergence of ${\bf B}$ is zero, so the wave equation reduces to

\nabla\times{(\nabla\times{{\bf B}})}= -{1\over c^2}{\partial^2{\bf B}\over\partial t^2}
\end{displaymath} (8)

The particular geometry I am interested in is the initial condition of a toroidal magnetic flux loop, which is to say, a magnetic field loop situated on a plane, concentrated between a minor and major radius. The magnetic field is always toroidal (i.e., in the $\hat{\phi}$ direction). Because of this symmetry, I will use cylindrical coordinates ($r$,$\phi$,$z$). In this arrangement the flux loop is placed in the $xy$ plane, with uniform magnetic field in the $\hat{\phi}$ direction. This further simplifies the equations because, at least initially, $B_r$ and $B_z$ are zero. Also, we can assume that derivatives with respect to $\phi$ are zero because the field is azimuthally uniform.

Standard calculus and physics textbooks contain the formulae for differential operators expressed in cylindrical coordinates. The curl becomes

$\displaystyle \nabla\times{\bf F}$ $\textstyle =$ $\displaystyle \left({1\over r}{\partial F_z\over\partial \phi} - {\partial F_\phi\over\partial z}\right)\hat{r}$ (9)
    $\displaystyle + \left({\partial F_r\over\partial z} - {\partial F_z\over\partial r}\right)\hat{\phi}$  
    $\displaystyle + {1\over r} \left({\partial(r F_\phi)\over\partial r} - {\partial F_r\over\partial \phi}\right)\hat{z}$  

The evaluation of this formula appears at first glance to be very difficult, but because of the symmetry in the initial conditions, we can set all $B_r$ and $B_z$ components, and all $\partial/\partial\phi$ terms to zero. This leaves only two terms remaining after performing the curl operation twice:

\nabla^2{{\bf B}} = -\left[{\partial^2B_\phi\over\partial z^...
... r}{\partial(rB_\phi)\over\partial r}\right)\right]\hat{\phi}.
\end{displaymath} (10)

Substituting back into the modified wave equation (8), I arrive at this differential equation,

\left[{\partial^2B_\phi\over\partial z^2} + {\partial\over\p...
...\hat{\phi} = {1\over c^2}{\partial^2{\bf B}\over\partial t^2}.
\end{displaymath} (11)

This is a very fortuitous result because the direction of the gradient term is in the azimuthal $\hat{\phi}$ direction. I can conclude that the magnetic field will always be in the azimuthal direction, and I can ignore $B_r$ and $B_z$ for all times with no loss of information. From this point I will use $B$ to refer to $B_\phi$. Also, it is worth noting that the above equation is very similar to, but not the same as, the divergence equation for the electric field in cylindrical coordinates. The $r$'s appear in different places, so the solutions will not be exactly the same, however I will show that ultimately the radial portion of the equation is solved by a standard Bessel function.

next up previous
Next: Solution by Separation of Up: Maxwell's Equations and a Previous: Wave Equation from Maxwell's
Craig Markwardt 2001-07-30