Department of Physics and Astronomy
Arizona State University
Tempe, AZ 85287
While scattering from a perfectly conducting sphere is relevant to more physical problems, the mathematics of scattering from an infinite cylinder is simpler. An infinite cylinder corresponds to the physical situation where the cylinder is much longer than the other length scales in the system. That means it must be much longer than its radius and it must be much longer than a wavelength. For simplicity only normal incidence will be discussed. That is for the cylinder axis along z , a plane wave is incident along x . It is convenient to solve for polarization along y and along z separately. Other polarizations give fields that are linear combinations of these. The radius of the circular cylinder is a .
Assume an e - it time dependence and take the real part of all quantities to get the physical fields.
Maxwell's curl equations outside the cylinder are
|- = - iEz||- = iBz|
|- = - iEx||- = iBx|
|- = - iEy||- = iBy||(1)|
|- = iBz|
|= - Ex|
|- = - iEz|
|= - By .||(3)|
In the first set the transverse (i.e. x and y ) components of the electric
t can be derived from Bz , and the equations become
|z = 0|
|( x )Bz = it .||(4)|
|z = 0|
|( x )Ez = - it .||(5)|
When we covered static problems, we separated variables for the Helmholtz equation. A complete set of angular functions can be written as e im, or the corresponding sines and cosines. The equation becomes Bessel's equation, with solutions combinations of Jn(/c) and Nn(/c) .
The incident plane wave for polarization along y is
|Bz(i) = E0e ix||(6)|
|Ez(i) = E0e ix||(7)|
|E0e ix||=||E0e icos = E0i nJn2 - )cos(n)|
|=||i n2 - )cos(n) .||(8)|
|H (1)n(x) = Jn(x) + iNn(x)|
|H (2)n(x) = Jn(x) - iNn(x)||(9)|
|H (1)n(x)||=||e i(x - [n + 1/2]/2)|
|H (2)n(x)||=||e - i(x - [n + 1/2]/2)||(10)|
For scattering problems, the boundary condition at infinity is that the only incoming waves come from the plane wave. Outgoing waves can come from either the incident plane wave or the scattered field.
A perfect conductor requires that the fields are zero inside. The curl equations require that the tangential electric field be continuous and the discontinuity in the tangential magnetic field is given by 4/c where is the surface current. In this case, the electric field condition gives the remaining boundary condition, and the magnetic field condition can be used to solve for the surface current on the conductor.
Notice that the boundary conditions on a circular cylinder will not couple different angular components. Nor will it couple the Bz , Ex , Ey terms to Ez , Bx , By . Therefore by solving for polarizations which have Bz = 0 (incident electric field along z called the transverse magnetic, TM, case) or Ez = 0 (incident electric field along y called the transverse electric, TE, case), the equations simplify. In both cases, since only cosn terms are excited by the plane wave, only those terms survive.
For the TE case where Ez = 0 , the boundary condition is E( = a) = 0 , or
|=a = 0 .||(11)|
|Ez( = a) = 0||(12)|
The TE solution is
|Bz||=||(2 - )i ncos(n)|
|=||E0e ix + (2 - )i ncos(n)n(1)||(13)|
|= - =||(14)|
|- 1||=||2ie isin .||(16)|
The scattered power per unit length divided by the incident power per
unit area gives the cross section per unit length of the cylinder. At
large distances the asymptotic expansion of the Hankel function gives
|=||(2 - )| - 1|2|
|=||(2 - )||(18)|
The TM case goes through nearly identically with
The total cross section as a function of a/c is shown in fig. 1. The differential cross section is shown in polar form in figs. 2-4 for three values of a/c .