Kirchhoff integral theorem

The integral has the following form for a monochromatic wave:[2][3][4]

${\displaystyle U(\mathbf {r} )={\frac {1}{4\pi }}\int _{S}\left[U{\frac {\partial }{\partial {\hat {\mathbf {n} }}}}\left({\frac {e^{iks}}{s}}\right)-{\frac {e^{iks}}{s}}{\frac {\partial U}{\partial {\hat {\mathbf {n} }}}}\right]dS,}$

where the integration is performed over an arbitrary closed surface S enclosing the observation point ${\displaystyle \mathbf {r} }$, ${\displaystyle k}$ in ${\displaystyle e^{iks}}$ is the wavenumber, ${\displaystyle s}$ in ${\displaystyle {\frac {e^{iks}}{s}}}$ is the distance from an (infinitesimally small) integral surface element to the point ${\displaystyle \mathbf {r} }$, ${\displaystyle U}$ is the spatial part of the solution of the homogeneous scalar wave equation (i.e., ${\displaystyle V(\mathbf {r} ,t)=U(\mathbf {r} )e^{-i\omega t}}$ as the homogeneous scalar wave equation solution), ${\displaystyle {\hat {\mathbf {n} }}}$ is the unit vector inward from and normal to the integral surface element, i.e., the inward surface normal unit vector, and ${\displaystyle {\frac {\partial }{\partial {\hat {\mathbf {n} }}}}}$ denotes differentiation along the surface normal (i.e., a normal derivative) i.e., ${\displaystyle {\frac {\partial f}{\partial {\hat {\mathbf {n} }}}}=\nabla f\cdot {\hat {\mathbf {n} }}}$ for a scalar field ${\displaystyle f}$. Note that the surface normal is inward, i.e., it is toward the inside of the enclosed volume, in this integral; if the more usual outer-pointing normal is used, the integral will have the opposite sign.

This integral can be written in a more familiar form

${\displaystyle U(\mathbf {r} )={\frac {1}{4\pi }}\int _{S}\left(U\nabla \left({\frac {e^{iks}}{s}}\right)-{\frac {e^{iks}}{s}}\nabla U\right)\cdot d{\vec {S}},}$

where ${\displaystyle d{\vec {S}}=dS{\hat {\mathbf {n} }}}$.[3]

A more general form can be derived for non-monochromatic waves. The complex amplitude of the wave can be represented by a Fourier integral of the form

${\displaystyle V(r,t)={\frac {1}{\sqrt {2\pi }}}\int U_{\omega }(r)e^{-i\omega t}\,d\omega ,}$

where, by Fourier inversion, we have

${\displaystyle U_{\omega }(r)={\frac {1}{\sqrt {2\pi }}}\int V(r,t)e^{i\omega t}\,dt.}$

The integral theorem (above) is applied to each Fourier component ${\displaystyle U_{\omega }}$, and the following expression is obtained:[2]

${\displaystyle V(r,t)={\frac {1}{4\pi }}\int _{S}\left\{[V]{\frac {\partial }{\partial n}}\left({\frac {1}{s}}\right)-{\frac {1}{cs}}{\frac {\partial s}{\partial n}}\left[{\frac {\partial V}{\partial t}}\right]-{\frac {1}{s}}\left[{\frac {\partial V}{\partial n}}\right]\right\}dS,}$

where the square brackets on V terms denote retarded values, i.e. the values at time t − s/c.

Kirchhoff showed that the above equation can be approximated to a simpler form in many cases, known as the Kirchhoff, or Fresnel–Kirchhoff diffraction formula, which is equivalent to the Huygens–Fresnel equation, except that it provides the inclination factor, which is not defined in the Huygens–Fresnel equation. The diffraction integral can be applied to a wide range of problems in optics.

Here, the derivation of the Kirchhoff's integral theorem is introduced. First, the Green's second identity as the following is used.

$${\displaystyle \int _{V}\left(U_{1}\nabla ^{2}U_{2}-U_{2}\nabla ^{2}U_{1}\right)dV=\oint _{\partial V}\left(U_{2}{\partial U_{1} \over \partial {\hat {\mathbf {n} }}}-U_{1}{\partial U_{2} \over \partial {\hat {\mathbf {n} }}}\right)dS,}$$

where the integral surface normal unit vector ${\displaystyle {\hat {\mathbf {n} }}}$ here is toward the volume ${\displaystyle V}$ closed by an integral surface ${\displaystyle \partial V}$. Scalar field functions ${\displaystyle U_{1}}$ and ${\displaystyle U_{2}}$ are set as solutions of the Helmholtz equation, ${\displaystyle \nabla ^{2}U+k^{2}U=0}$ where ${\displaystyle k={\frac {2\pi }{\lambda }}}$ is the wavenumber (${\displaystyle \lambda }$ is the wavelength), that gives the spatial part of a complex-valued monochromatic (single frequency in time) wave expression. (The product between the spatial part and the temporal part of the wave expression is a solution of the scalar wave equation.) Then, the volume part of the Green's second identity is zero, so only the surface integral is remained as

$${\displaystyle \oint _{\partial V}\left(U_{2}{\partial U_{1} \over \partial {\hat {\mathbf {n} }}}-U_{1}{\partial U_{2} \over \partial {\hat {\mathbf {n} }}}\right)dS=0.}$$

Now ${\displaystyle U_{2}}$ is set as the solution of the Helmholtz equation to find and ${\displaystyle U_{1}}$ is set as the spatial part of a complex-valued monochromatic spherical wave ${\displaystyle U_{1}={\frac {e^{iks}}{s}}}$ where ${\displaystyle s}$ is the distance from an observation point ${\displaystyle P}$ in the closed volume ${\displaystyle V}$. Since there is a singularity for ${\displaystyle U_{1}={\frac {e^{iks}}{s}}}$ at ${\displaystyle P}$ where ${\displaystyle s=0}$ (the value of ${\displaystyle {\frac {e^{iks}}{s}}}$ not defined at ${\displaystyle s=0}$), the integral surface must not include ${\displaystyle P}$. (Otherwise, the zero volume integral above is not justified.) A suggested integral surface is an inner sphere ${\displaystyle S_{1}}$ centered at ${\displaystyle P}$ with the radius of ${\displaystyle s_{1}}$ and an outer arbitrary closed surface ${\displaystyle S_{2}}$.

Then the surface integral becomes

$${\displaystyle \oint _{S_{1}}\left(U_{2}{\partial \over \partial {\hat {\mathbf {n} }}}{\frac {e^{iks}}{s}}-{\frac {e^{iks}}{s}}{\partial \over \partial {\hat {\mathbf {n} }}}U_{2}\right)dS+\oint _{S_{2}}\left(U_{2}{\partial \over \partial {\hat {\mathbf {n} }}}{\frac {e^{iks}}{s}}-{\frac {e^{iks}}{s}}{\partial \over \partial {\hat {\mathbf {n} }}}U_{2}\right)dS=0.}$$

For the integral on the inner sphere ${\displaystyle S_{1}}$,

$${\displaystyle {\frac {\partial }{\partial {\hat {\mathbf {n} }}}}{\frac {e^{iks}}{s}}=\nabla {\frac {e^{iks}}{s}}\cdot {\hat {\mathbf {n} }}=\left({\frac {ik}{s}}-{\frac {1}{s^{2}}}\right)e^{iks},}$$

and by introducing the solid angle ${\displaystyle d\Omega }$ in ${\displaystyle dS=s^{2}d\Omega }$,

$${\displaystyle \oint _{S_{1}}\left(U_{2}{\partial \over \partial {\hat {\mathbf {n} }}}{\frac {e^{iks}}{s}}-{\frac {e^{iks}}{s}}{\partial \over \partial {\hat {\mathbf {n} }}}U_{2}\right)dS=\oint _{S_{1}}\left(U_{2}\left({\frac {ik}{s}}-{\frac {1}{s^{2}}}\right)e^{iks}-{\frac {e^{iks}}{s}}{\partial \over \partial {\hat {\mathbf {n} }}}U_{2}\right)s^{2}d\Omega =\oint _{S_{1}}\left(iksU_{2}-U_{2}-s{\frac {\partial }{\partial {\hat {\mathbf {n} }}}}U_{2}\right)e^{iks}d\Omega }$$

due to ${\displaystyle {\frac {\partial }{\partial {\hat {\mathbf {n} }}}}U_{2}=\nabla U_{2}\cdot {\hat {\mathbf {n} }}={\frac {\partial }{\partial s}}U_{2}}$. (The spherical coordinate system which origin is at ${\displaystyle P}$ can be used to derive this equality.)

By shrinking the sphere ${\displaystyle S_{1}}$ toward the zero radius (but never touching ${\displaystyle P}$ to avoid the singularity), ${\displaystyle e^{iks}\to 1}$ and the first and last terms in the ${\displaystyle S_{1}}$ surface integral becomes zero, so the integral becomes ${\displaystyle -4\pi U_{2}}$. As a result, denoting ${\displaystyle U_{2}}$, the location of ${\displaystyle P}$, and ${\displaystyle S_{2}}$ by ${\displaystyle U}$, the position vector ${\displaystyle \mathbf {r} }$, and ${\displaystyle S}$ respectively,

$${\displaystyle U(\mathbf {r} )={\frac {1}{4\pi }}\oint _{S}\left(U{\partial \over \partial {\hat {\mathbf {n} }}}{\frac {e^{iks}}{s}}-{\frac {e^{iks}}{s}}{\partial \over \partial {\hat {\mathbf {n} }}}U\right)dS.}$$