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Up: PHY531-PHY532 Classical Electrodynamics

Scattering at Normal Incidence from an Infinite Perfectly Conducting Circular Cylinder

K.E. Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ 85287

1 Introduction

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 - i$\scriptstyle\omega$t time dependence and take the real part of all quantities to get the physical fields.

Maxwell's curl equations outside the cylinder are

$\displaystyle{\frac{\partial B_y}{\partial x}}$ - $\displaystyle{\frac{\partial B_x}{\partial y}}$ = - i$\displaystyle{\frac{\omega}{c}}$Ez    $\displaystyle{\frac{\partial E_y}{\partial x}}$ - $\displaystyle{\frac{\partial E_x}{\partial y}}$ = i$\displaystyle{\frac{\omega}{c}}$Bz   
$\displaystyle{\frac{\partial B_z}{\partial y}}$ - $\displaystyle{\frac{\partial B_y}{\partial z}}$ = - i$\displaystyle{\frac{\omega}{c}}$Ex    $\displaystyle{\frac{\partial E_z}{\partial y}}$ - $\displaystyle{\frac{\partial E_y}{\partial z}}$ = i$\displaystyle{\frac{\omega}{c}}$Bx   
$\displaystyle{\frac{\partial B_x}{\partial z}}$ - $\displaystyle{\frac{\partial B_z}{\partial x}}$ = - i$\displaystyle{\frac{\omega}{c}}$Ey    $\displaystyle{\frac{\partial E_x}{\partial z}}$ - $\displaystyle{\frac{\partial E_z}{\partial x}}$ = i$\displaystyle{\frac{\omega}{c}}$By (1)
The system is invariant along z so the z derivatives in Maxwell's equation give zero. Notice that this makes Maxwell's equations break up into two uncoupled sets:
$\displaystyle{\frac{\partial E_y}{\partial x}}$ - $\displaystyle{\frac{\partial E_x}{\partial y}}$ = i$\displaystyle{\frac{\omega}{c}}$Bz        
$\displaystyle{\frac{\partial B_z}{\partial y}}$ = - $\displaystyle{\frac{i \omega}{c}}$Ex        
$\displaystyle{\frac{\partial B_z}{\partial x}}$ = $\displaystyle{\frac{i \omega}{c}}$Ey      (2)
and
$\displaystyle{\frac{\partial B_y}{\partial x}}$ - $\displaystyle{\frac{\partial B_x}{\partial y}}$ = - i$\displaystyle{\frac{\omega}{c}}$Ez        
$\displaystyle{\frac{\partial E_z}{\partial y}}$ = $\displaystyle{\frac{i \omega}{c}}$Bx        
$\displaystyle{\frac{\partial E_z}{\partial x}}$ = - $\displaystyle{\frac{i \omega}{c}}$By .      (3)

In the first set the transverse (i.e. x and y ) components of the electric field $\mbox{\boldmath$\space {E} $}$t can be derived from Bz , and the equations become

$\displaystyle\left [ \frac{\partial^2}{\partial x^2} +\frac{\partial^2 }{\partial y^2}
+\frac{\omega^2}{c^2} \right]B$z = 0        
($\displaystyle\mbox{\boldmath$ {\hat {z}} $}$ x $\displaystyle\mbox{\boldmath$ {\nabla} $}$)Bz = i$\displaystyle{\frac{\omega}{c}}$$\displaystyle\mbox{\boldmath$ {E} $}$t .      (4)
and in the second set $\mbox{\boldmath$\space {B} $}$t can be derived from Ez
$\displaystyle\left [ \frac{\partial^2}{\partial x^2} +\frac{\partial^2 }{\partial y^2}
+\frac{\omega^2}{c^2} \right]E$z = 0        
($\displaystyle\mbox{\boldmath$ {\hat {z}} $}$ x $\displaystyle\mbox{\boldmath$ {\nabla} $}$)Ez = - i$\displaystyle{\frac{\omega}{c}}$$\displaystyle\mbox{\boldmath$ {B} $}$t .      (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$\scriptstyle\phi$, or the corresponding sines and cosines. The $\rho$ equation becomes Bessel's equation, with solutions combinations of Jn($\omega$$\rho$/c) and Nn($\omega$$\rho$/c) .

The incident plane wave for polarization along y is

Bz(i) = E0e i$\scriptstyle{\frac{\omega}{c}}$x (6)

and for polarization along z

Ez(i) = E0e i$\scriptstyle{\frac{\omega}{c}}$x (7)

where E0 is the amplitude of the electric field. These must be linear combinations of the general solutions. A simple way to get the coefficients is to multiply them by e - in$\scriptstyle\phi$ and integrate over $\phi$ . The integral is given in Jackson problem 3.16 with the result
E0e i$\scriptstyle{\frac{\omega}{c}}$x = E0e i$\scriptstyle{\frac{\omega}{c}}$$\scriptstyle\rho$cos $\scriptstyle\phi$ = E0$\displaystyle\sum_{n=0}^{\infty}$i nJn$\displaystyle\left (\frac{\omega}{c}\rho \right)($2 - $\displaystyle\delta_{n0}^{}$)cos(n$\displaystyle\phi$)   
  = $\displaystyle{\frac{E_0}{2}}$$\displaystyle\sum_{n=0}^{\infty}$i n$\displaystyle\left [
H^{(1)}_n\left (\frac{\omega}{c}\rho \right )
+H^{(2)}_n\left (\frac{\omega}{c}\rho \right ) \right]($2 - $\displaystyle\delta_{n0}^{}$)cos(n$\displaystyle\phi$) . (8)
The second form is written in terms of the Hankel functions
H (1)n(x) = Jn(x) + iNn(x)        
H (2)n(x) = Jn(x) - iNn(x)      (9)
which for large argument become
H (1)n(x) = $\displaystyle\sqrt{\frac{2}{\pi x}}$e i(x - [n + 1/2]$\scriptstyle\pi$/2)   
H (2)n(x) = $\displaystyle\sqrt{\frac{2}{\pi x}}$e - i(x - [n + 1/2]$\scriptstyle\pi$/2) (10)
which shows that H (1) corresponds to outgoing cylindrical waves and H (2) corresponds to incoming cylindrical waves with e - i$\scriptstyle\omega$t time dependence.

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$\pi$$\mbox{\boldmath$\space {K} $}$/c where $\mbox{\boldmath$\space {K} $}$ 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$\phi$ terms are excited by the plane wave, only those terms survive.

For the TE case where Ez = 0 , the boundary condition is E$\scriptstyle\phi$($\rho$ = a) = 0 , or

$\displaystyle\left .\frac{\partial B_z}{\partial \rho} \right\vert _$$\displaystyle\rho$=a = 0 . (11)

For the TM case where Bz = 0 , the boundary condition is

Ez($\displaystyle\rho$ = a) = 0 (12)

The TE solution is

Bz = $\displaystyle{\frac{E_0}{2}}$$\displaystyle\sum_{n=0}^{\infty}$(2 - $\displaystyle\delta_{n0}^{}$)i ncos(n$\displaystyle\phi$)$\displaystyle\left [ H^{(2)}_n\left (\frac{\omega}{c}\rho \right )
+ \alpha^{(TE)}_n H_n^{(1)} \left (\frac{\omega}{c}\rho \right ) \right]$   
  = E0e i$\scriptstyle{\frac{\omega}{c}}$x + $\displaystyle{\frac{E_0}{2}}$$\displaystyle\sum_{n=0}^{\infty}$(2 - $\displaystyle\delta_{n0}^{}$)i ncos(n$\displaystyle\phi$)$\displaystyle\left [ \alpha^{(TE)}_n-1 \right]H$n(1)$\displaystyle\left (\frac{\omega}{c}\rho \right)$ (13)
and the cylinder boundary conditions makes

$\displaystyle\alpha_{n}^{(TE)}$ = - $\displaystyle{\frac{H_n'^{(2)} \left (\frac{\omega}{c}a \right )}{H'^{(1)}_n \left (\frac{\omega}{c}a \right )}}$ = $\displaystyle{\frac{N_n'\left (\frac{\omega}{c}a \right )
+iJ_n'^{(2)} \left (\...
 ... \left (\frac{\omega}{c}a \right )
-i J'_n \left (\frac{\omega}{c}a \right ) }}$ (14)

where a prime on the Bessel function indicates the derivative with respect to argument. Notice that the numerator of the $\alpha_{n}^{(TE)}$ expression is the complex conjugate of the denominator. This shows that $\alpha_{n}^{(TE)}$ has unit magnitude and is therefore only a phase. Physically this comes about because the cylinder is rotationally invariant and has zero resistance so no energy is lost and energy in a particular incoming multipole is not scattered out of that multipole. The magnitude of the outgoing field component must remain the same, only the phase can change. Sometimes this is shown explicitly by defining scattering phase shifts,

e 2i$\scriptstyle\delta_{n}^{(TE)}$ $\displaystyle\equiv$ $\displaystyle\alpha_{n}^{(TE)}$ (15)

which makes
tan $\displaystyle\delta_{n}^{(TE)}$ = $\displaystyle{\frac{J_n'^{(2)} \left (\frac{\omega}{c}a \right ) }{ N'_n \left (\frac{\omega}{c}a \right )}}$   
$\displaystyle\alpha_{n}^{(TE)}$ - 1 = 2ie i$\scriptstyle\delta_{n}^{(TE)}$sin $\displaystyle\delta_{n}^{(TE)}$ . (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

$\displaystyle{\frac{d\sigma^{(TE)}}{d\phi}}$ = $\displaystyle{\frac{c}{2\pi \omega}}$$\displaystyle\left \vert\sum_{n=0}^\infty (2-\delta_{n0}) [\alpha_n^{(TE)}-1]
\cos (n\phi) \right\vert^$2   
  = $\displaystyle{\frac{2 c}{\pi \omega}}$$\displaystyle\left \vert\sum_{n=0}^\infty (2-\delta_{n0}) e^{i\delta_n^{(TE)}}
\sin \delta_n^{(TE)} \cos (n\phi) \right\vert^$2 (17)
and the total cross section becomes
$\displaystyle\sigma^{(TE)}_{}$ = $\displaystyle{\frac{c}{\omega}}$$\displaystyle\sum_{n=0}^{\infty}$(2 - $\displaystyle\delta_{n0}^{}$)|$\displaystyle\alpha_{n}^{(TE)}$ - 1|2   
  = $\displaystyle{\frac{4 c}{\omega}}$$\displaystyle\sum_{n=0}^{\infty}$(2 - $\displaystyle\delta_{n0}^{}$)$\displaystyle\sin^{2}_{}$$\displaystyle\delta_{n}^{(TE)}$ (18)

The TM case goes through nearly identically with

$\displaystyle\alpha_{n}^{(TM)}$ = - $\displaystyle{\frac{H_n^{(2)}\left (\frac{\omega}{c} a\right )}{H_n^{(1)}\left (\frac{\omega}{c} a \right )}}$   
tan $\displaystyle\delta_{n}^{(TM)}$ = $\displaystyle{\frac{J_n\left (\frac{\omega}{c} a\right )}{N_n \left (\frac{\omega}{c} a \right )}}$ (19)
and the cross section results are identical with TE replaced by TM .

The total cross section as a function of $\omega$a/c is shown in fig. 1. The differential cross section is shown in polar form in figs. 2-4 for three values of $\omega$a/c .


  
Figure 1: The total cross section of as a function of $\omega$a/c for the two polarizations. Notice that both are going to twice the geometric cross section as expected.
\begin{figure}
\rotatebox {270}{
\includegraphics[width=8cm]{cyltot.ps}}\end{figure}


  
Figure 2: Polar plot of the differential cross section divided by a as a function of angle for $\omega$a/c equal to 1 for both polarizations.
\begin{figure}
\rotatebox {270}{
\includegraphics[width=8cm]{cylang1.ps}}\end{figure}


  
Figure 3: Polar plot of the differential cross section divided by a as a function of angle for $\omega$a/c equal to 5 for both polarizations.
\begin{figure}
\rotatebox {270}{
\includegraphics[width=8cm]{cylang5.ps}}\end{figure}


  
Figure 4: Polar plot of the differential cross section divided by a as a function of angle for $\omega$a/c equal to 10 for both polarizations.
\begin{figure}
\rotatebox {270}{
\includegraphics[width=8cm]{cylang10.ps}}\end{figure}


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