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Fresnel-Kirchhoff Integral Theorem
Adrian Down
April 17, 2006
1 Derivation
1.1 Green’s Theorem
This version of Green’s Theorem follows from the Divergence Theorem for a vector field G, ∫ ∇∇∇ · G dτ =
G · da =
G · ˆn da (1)
where ˆn ≡ daˆ. Substitute G = V ∇∇∇U into (1), where U and V are scalar fields. Use the identity for the divergence of a scalar field,
∇∇∇ · (f A) = ∇∇∇f · A + f∇∇∇ · A ⇒ ∇∇∇ · G = ∇∇∇V · ∇∇∇U + V ∇^2 U (2)
Also rewrite the directional derivative of U ,
G · nˆ = V ∇∇∇U · nˆ = V
∂U
∂n
Substituting (2) and (3) into (1), ∫ (∇∇∇V · ∇∇∇U ) dτ +
V ∇^2 U dτ =
V
∂U
∂n
da (4)
Now, interchange U and V and perform the same steps to obtain, ∫ (∇∇∇U · ∇∇∇V ) dτ +
U ∇^2 V dτ =
U
∂V
∂n
da (5)
Taking the difference of (4) and (5) gives Green’s Theorem,
V ∇^2 U − U ∇^2 V
dτ =
V
∂U
∂n
− U
∂V
∂n
da
1.2 Wave equation
Now apply Green’s Theorem to solutions to the wave equation. Recall the wave equation for an arbitrary scalar function V (not the electrostatic po- tential in this case),
∇^2 Vphys −
v^2
∂^2 Vphys ∂t^2
= 0 v =
c n
where v is the phase velocity if the wave. Suppose that Vphys depends on time as e±ıωt. We can then write the physical function Vphys in the familiar way as the real part of a complex function,
Vphys(r, t) = <
V (r)e±ıωt
With this assumption, V satisfies the Scalar Helmholtz equation, ( ∇^2 + k^2
V = 0 (6)
Applying the same assumptions to U , ( ∇^2 + k^2
U = 0 (7)
Substituting (6) and (7) into Green’s theorem, we obtain Green’s theorem for solution to the scalar Helmholtz equation,
V
∂U
∂n
− U
∂V
∂n
da
1.3 Spherical waves
1.3.1 Complex representation
We now apply the previous results to the case of spherical waves. The com- plex component of a spherical wave is independent of time, and so the wave can be represented,
Vphys = <
V 0 e−ıωt
the term in (^) R^12 dominates. Keeping only this term and substituting (9) into the second term of (8),
∮
r=R
−U
R^2
eık(0)R^2 dΩ = − 4 πU (0) (10)
where U (0) is the function U evaluated at the origin. Substituting (10) into (8), yields the Kirchhoff Integral Theorem (KIT),
U (0) =
4 π
outer
eıkr r
∂U
∂n
− U
∂n
eıkr r
da
Note. This result could apply equally well to any point P contained within the outer surface provided the vector r points from the point P to a point on the surface.
2 Application of the KIT to diffraction
2.1 Motivation
The typical diffracting device consists of a baffle containing an aperture. Consider a source of radiation on one side of the baffle, with vector r′^ from the source to the aperture. Our goal is to obtain the characteristics of the radiation at some point P on the other side of the baffle, given the behavior of the source and the characteristics of the aperture. We apply the KIT by taking the outer surface to contain the baffle. The baffle is assumed to be large enough that the only radiation reaching the point of observation passes through the aperture of the baffle. Thus we can extend the outer surface to close around the point P without introducing any additional contributions to the integral.
2.2 Optical disturbance
2.2.1 Definition
Applying the KIT, the scalar function U is chosen to represent the amplitude of the emitted radiation, so that |U |^2 ∝ I. U is known as the optical distur- bance. Because U must be a scalar function, this is the best description of the vector EM fields that we can hope to achieve. In this approximation, we loose
all information about the polarization of the emitted radiation. Describing vector fields by a scalar field is a gross approximation, but it is necessary to make the mathematics manageable. Due to the opaqueness of the aperture, the only radiation reaching the observation point is transmitted through the aperture in the baffle. Thus U is effectively 0 over the surface of integration except over the area of the aperture, and the integration need only be taken over this region,
U (0) =
4 π
aperture
eıkr r
∂U
∂n
− U
∂n
eıkr r
da (11)
Note. U and ∂U∂n must be finite on the area of the aperture.
2.2.2 Explicit form
Assume that the source is point-like, so that the radiation emitted there- from can be described by outward-propagating spherical waves. Take the hereto unspecified function U to be optical disturbance corresponding to the radiation emitted by this source,
U = U 0
eıkr ′
r′^
where U is the time-independent part of the outward-propagating spherical wave Uphys, as in the case of V.
Uphys = <
U e−ıωt
Note. The sign of the exponent of Uphys is opposite that of Vphys because U represents an outward propagating wave.
Substituting (12) into (11),
U (P ) =
4 π
ap
eıkr r
∂n
eıkr ′
r′
eıkr ′
r′
∂n
eıkr r
da
As before, we can rewrite the directional derivatives. The derivative of U can be taken with respect to r′, instead of r as in the case of V ,
∂ ∂n
eıkr ′
r′^
eıkr ′
r′^
· ˆn
ık −
r′
eıkr
′
r′^
ˆr′^ · nˆ
Note. The same arguments can be applied to (^) ∂n∂ e ıkr r.
