Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(lu) #1

APPLICATIONS OF COMPLEX VARIABLES


x-dependence ofn(x), the solution would be (as usual)


y(x)=Aexp(−ik 0 nx), (25.43)

with bothAandnconstant. The quantityk 0 nxwould be real and would be called


the ‘phase’ of the wave; it increases linearly withx.


As a first variation on this simple picture, we may allowf(x) to be complex,

though, for the moment, still constant. Thenn(x) is still a constant, albeit a


complex one:


n=μ+iν.

The solution is formally the same as before; however, whilst it still exhibits


oscillatory behaviour, the amplitude of the oscillations either grows or declines,


depending upon the sign ofν:


y(x)=Aexp(−ik 0 nx)=Aexp[−ik 0 (μ+iν)x]=Aexp(k 0 νx) exp(−ik 0 μx).

This solution withνnegative is the appropriate description for a wave travelling


in a uniform absorbing medium. The quantityk 0 (μ+iν)xis usually called the


complex phaseof the wave.


We now allowf(x), and hencen(x), to be both complex and varying with

position, though, as we have noted earlier, there will be restrictions on how


rapidlyf(x) may vary if valid solutions are to be obtained. The obvious extension


of solution (25.43) to the present case would be


y(x)=Aexp[−ik 0 n(x)x], (25.44)

but direct substitution of this into equation (25.42) gives


y′′+k 02 n^2 y=−k^20 (n′^2 x^2 +2nn′x)y−ik 0 (n′′x+2n′)y.

Clearly the RHS can only be zero, as is required by the equation, ifn′andn′′are


both very small, or if some unlikely relationship exists between them.


To try to improve on this situation, we consider how the phaseφof the solution

changes as the wave passes through an infinitesimal thicknessdxof the medium.


The infinitesimal (complex) phase changedφfor this is clearlyk 0 n(x)dx,and


therefore will be


∆φ=k 0

∫x

0

n(u)du

for a finite thicknessxof the medium. This suggests that an improvement on


(25.44) might be


y(x)=Aexp

(
−ik 0

∫x

0

n(u)du

)

. (25.45)


This is still not an exact solution as now


y′′(x)+k^20 n^2 (x)y(x)=−ik 0 n′(x)y(x).
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