Rayleigh integrals

The only merit of which I personally am conscious was that of having pleased myself by my studies, and any results that may be due to my researches were owing to the fact that it has been a pleasure for me to become a physicist.

–John Strutt, 3rd Baron Rayleigh

Fourier acoustics is now used to derive Rayleigh's famous diffraction integrals. The plane wave decomposition (angular spectrum) of a spherical wave is first needed to derive these integrals.

Contents:

• Angular spectrum of a spherical wave
• Derivation of the first Rayleigh integral using Fourier acoustics
• Derivation of the second Rayleigh integral using Fourier acoustics

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Angular specturm of a spherical wave

Let \(f(x,y,z) = e^{ikr}/r\) be the functional dependence of the radiation due to a point source at \(r=0\) (i.e., the amplitude factor has been neglected), where \(r = \sqrt{x^2+y^2+z^2}\) and \(\rho = \sqrt{x^2 + y^2}\). To obtain the plane wave decomposition (angular spectrum) of the spherical wave in all space, first calculate the 2D spatial Fourier transform of the wave at the source \(f(x,y,0)\): \begin{align*} F_0(\kappa) = F_0(k_x,k_y) &= \FTxy \{f(x,y,0)\} = \iint_{-\infty}^\infty \frac{e^{ik\rho}}{\rho} e^{-i(k_xx + k_yy)}\, dx\, dy\\ &= \FTxy \{f(x,y,0)\} = \iint_{-\infty}^\infty \frac{e^{ik\rho}}{\rho} e^{-i\vec{\kappa}\cdot \vec{\rho}} \rho \, d\rho \, d\phi \end{align*} As before, due to axisymmetry, take \(\vec{\kappa} = \kappa \hat{e}_x\), giving \begin{align*} \vec{\kappa} \cdot \vec{\rho} = \kappa\rho \cos\phi\,. \end{align*} So \begin{align*} F_0(\kappa)= \int_{0}^{2\pi} d\phi \int_{0}^{\infty} e^{i(k-\kappa \cos \phi)\rho}d\rho\,. \end{align*} This is a tricky integral to take, since the integrand of the \(\rho\) integral is oscillatory. To evaluate the integral, let \(k = k + i\epsilon\) and then take \(\epsilon \to 0\). \begin{align*} F_0(\kappa) &= \int_{0}^{2\pi} d\phi \frac{e^{i(k-\kappa\cos\phi)} e^{-\epsilon\rho}}{i(k-\kappa \cos\phi)-\epsilon}\bigg\rvert_{\rho=0}^{\rho = \infty}\\ &= -\int_{0}^{2\pi} \frac{d\phi}{i(k-\kappa \cos\phi)\rho-\epsilon}\\ &= i\int_{0}^{2\pi} \frac{d\phi}{k-\kappa \cos\phi} = i \frac{2\pi}{\sqrt{k^2- \kappa^2}} = \frac{i2\pi}{k_z}\,. \end{align*} Note that \(e^{ikr}/r\) is symmetric about the plane \(z=0\), as shown below:

With the Fourier transform of the source condition taken, the full field can therefore be calculated for all \(z\) (both positive and negative) using the standard Fourier acoustics procedure: \begin{align*} f(x,y,z) = e^{ikr}/r &= \IFTxy\{F_0(\kappa) e^{ik_z|z|}\}\\ &= \IFTxy\bigg\{\frac{i2\pi}{k_z} e^{ik_z|z|}\bigg\}\,. \end{align*} Taking the 2D spatial Fourier transform of both sides above gives the plane wave decomposition of a spherical wave: \begin{align*} \boxed{\FTxy \{e^{ikr}/r\} = i2\pi \frac{e^{ik_z|z|}}{k_z}\,.} \end{align*} Since \(F(\kappa) = 2\pi F_H(\kappa)\), the above can conveniently be written in terms of the Hankel transform \(\HT_1\), \begin{align*} \HT_1 \{e^{ikr}/r\} = i\frac{e^{ik_z|z|}}{k_z}\,. \end{align*}

Derivation of first Rayleigh integral using Fourier acoustics

Given an arbitrary normal velocity distribution \begin{align*} \vec{u} &= (x,y,0,t) = u_0(x,y) e^{-i\omega t} \ez \,. \end{align*} Follow the recipe for the calculation of the field. The 2D spatial Fourier transform of the source condition, and its mapping to a pressure source, is \begin{align*} U_0(k_x,k_y,0) &= \FTxy \{u_0(x,y)\}\,,\\ P_0(k_x,k_y,0) &= \rho_0c_0 \frac{k}{k_z} U(k_x,k_y,0)\,. \end{align*} Then the solution to the Helmholtz equation is \begin{align*} p_\omega(x,y,z) &= \IFTxy\{P_0(k_x,k_y) e^{ik_zz}\}\\ &= \rho_0c_0 k\, \IFTxy \{U_0 (k_x,k_y) e^{ik_zz}/k_z\}\\ &= \rho_0c_0 k\, \IFTxy \{U_0 (k_x,k_y)\} ** \IFTxy\{ e^{ik_zz}/k_z\}\\ &= \rho_0c_0 k\, u(x,y) ** \frac{e^{ikr}}{i2\pi r}\,, \end{align*} where \(r= \sqrt{x^2 + y^2 + z^2}\), and where the result from the previous section, \(\FTxy \{e^{ikr}/r\} = i2\pi e^{ik_z|z|}/k_z\), has been used to evaluate the third line above. Writing the convolution explicitly and denoting \[R = \sqrt{(x-x_0)^2 + (y-y_0)^2 + z^2}\] results in the so-called Rayleigh integral of the first kind of acoustics: \begin{align*} \boxed{p_\omega (x,y,z) = -\frac{ik\rho_0c_0}{2\pi} \iint_{-\infty}^\infty u_0(x_0,y_0) \frac{e^{ikR}}{R} dx_0 dy_0\,.} \end{align*}

Derivation of second Rayleigh integral using Fourier acoustics

Note that this is called the first Rayleigh integral of optics, because it is more common in optics to calculate the electric field beyond the source in terms of the electric field at the source.

Now consider a pressure source \begin{align*} p(x,y,0,t) = p_0(x,y) e^{-i\omega t}\,. \end{align*} Then take its 2D spatial Fourier transform: \begin{align*} P_0(k_x,k_y) = \FTxy\{p_0(x,y)\}\,. \end{align*} The full field is given by \begin{align*} p_\omega(x,y,z) &= \IFTxy \{P_0(k_x,k_y) e^{ik_zz}\}\\ &= p_0(x,y) ** \IFTxy\{e^{ik_zz}\} \end{align*} Note that \begin{align*} \IFTxy\{e^{ik_zz}\} &= \IFTxy\{ik_z e^{ik_zz}/ik_z\}\\ &= \frac{\partial}{\partial z} (-e^{ikr}/2\pi r) \end{align*} where the first line holds by multiplication and division by \(1\), and where second line holds because \(\FTxy\{e^{ikr}/r\} = (i2\pi/k_z) e^{ik_z|z|}\) and thus \(-e^{ikr}/2\pi r = -\IFTxy \{(i/k_z) e^{ik_z|z|}\} = \IFTxy \big\{ \frac{1}{ik_z} e^{ik_z|z|} \big \}\), so \begin{align*} \IFTxy \{ e^{ik_z|z|}\} = \IFTxy \bigg\{ \frac{\partial}{\partial z} \bigg[\frac{1}{ik_z} e^{ik_z|z|} \bigg]\bigg \} &= \frac{\partial}{\partial z} \IFTxy \bigg\{\bigg[\frac{1}{ik_z} e^{ik_z|z|} \bigg]\bigg \}\\ &= \frac{\partial}{\partial z}(-e^{ikr}/2\pi r) \,. \end{align*} So the pressure field can be written as a convolution: \begin{align*} p_\omega(x,y,z) &= p_0(x,y) ** \frac{\partial}{\partial z} (-e^{ikr}/2\pi r) \,. \end{align*} Writing the convolution explicitly results in \begin{align*} p_\omega(x,y,z) = -\frac{1}{2\pi} \iint_{-\infty}^\infty p_0(x_0,y_0) \frac{\partial}{\partial z} \frac{e^{ikR}}{R} dx_0 dy_0 \,. \end{align*} Noting that \begin{align*} \frac{\partial }{\partial z} (e^{ikR}/R) = (ik- 1/R) \frac{e^{ikR}}{R} \frac{\partial R}{\partial z} \end{align*} and \begin{align*} \frac{\partial R}{\partial z} &= \frac{\partial}{\partial z }\sqrt{(x-x_0)^2 + (y-y_0)^2 + z^2}\\ &= \frac{2z}{2R} = z/R\,, \end{align*} the partial derivative in the above integral becomes \begin{align*} \frac{\partial }{\partial z} (e^{ikR}/R) &= (ik- 1/R) \frac{e^{ikR}}{R} \frac{z}{R}\\ &= ikz (1- 1/ikR) \frac{e^{ikR}}{R^2}\,. \end{align*} The integral above then becomes the second Rayleigh integral of acoustics, \begin{align*} \boxed{p_\omega(x,y,z) = -\frac{ikz}{2\pi} \iint_{-\infty}^\infty p_0(x_0,y_0) (1- 1/ikR) \frac{e^{ikR}}{R^2} dx_0 dy_0\,.} \end{align*}