Fourier acoustics

Pressure source

Begin with the Helmholtz equation \begin{align}\label{Helmster}\tag{1} \Laplacian p_\omega + k^2 p_\omega = 0\,. \end{align} Given field in the source plane \(p_\omega(x,y,0)\), find \(p_\omega(x,y,z>0)\). Start by taking the 2D spatial Fourier transform of Eq. \eqref{Helmster}: \begin{align*} \FTxy \bigg\{\bigg(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\bigg) p_\omega + k^2 p_\omega\bigg\} %= \frac{d^2 P}{dz^2} + (k^2-k_x^2-k_y^2) = 0\,. \end{align*} Using the derivative property and calling \(P= P(k_x,k_y,z)\), the above becomes \begin{align*} \frac{d^2 P}{dz^2} + k_z^2 P = 0\,,\quad k_z = \sqrt{k^2 - k_x^2 -k_y^2}\,. \end{align*} For forward propagation, the above 2nd order ODE reads \begin{align*} P(k_x,k_y,z) = P(k_x,k_y,0) e^{ik_zz}\,. \end{align*} To return to physical space, the inverse 2D spatial Fourier transform is taken: \begin{align*} p_\omega(x,y,z) %&= \frac{1}{4\pi^2}\iint_{-\infty}^\infty P(k_x,k_y,z) e^{i(k_xx +k_yy)} dk_x\, dk_y\\ %&= \frac{1}{4\pi^2} \iint_{-\infty}^\infty P(k_x,k_y,0) e^{i(k_xx + k_yy + k_zz)} dk_x\, dk_y\\ = \IFTxy \{P(k_x,k_y,0)e^{ik_zz}\}\,. \end{align*} Since \(P(k_x,k_y,0) = \FTxy\{p(k_x,k_y,0)\}\), the entire field can be written in terms of the source condition: \begin{align}\label{presssource}\tag{2} \boxed{p_\omega(x,y,z) = \IFTxy \{\FTxy\{p_\omega(x,y,0)\}e^{ik_zz}\}\,,} \end{align} where the \(z\) component of the wavenumber is \begin{align*} k_z &= \sqrt{k^2-k_x^2-k_y^2}\\ &=\begin{cases} \text{ positive real}\,\quad & k_x^2 + k_y^2 < k^2 \\ \text{ positive imaginary}\, \quad &k_x^2 + k_y^2 > k^2 \,. \end{cases} \end{align*} Note that the entire angular spectrum (all \(k_x,k_y\)) is required to reproduce the exact field.

For axisymmetric sources, Eq. \eqref{presssource} becomes \begin{align*} p_\omega (r,z) = \IHT \{\HT \{p_\omega(\rho,0)\} e^{ik_zz}\} \\ k_z = \sqrt{k^2 - \kappa^2}\,,\quad \kappa^2 = k_x^2 +k_y^2\,. \end{align*}

Example: Bessel beams

This topic was originally presented after the Fraunhofer approximation/before the Fresnel approximation were discussed. However, it serves as a good example of Fourier acoustics used for an axisymmetric pressure source.

Bessel (nondiffracting) beams were "discovered" in optics in the 80s [Durnan, JOSA A, 4, 651 (1987)].

Consider the pressure source \begin{align*} p_\omega(x,y,0) = p_\omega(\rho) = p_0J_0(\alpha \rho)\,, \end{align*} where \(\alpha\) is a constant, not a direction cosine. The solution is found by taking \begin{align*} p(\rho,z) = \IHT\{\HT\{p_0(\rho)\}e^{ik_zz}\}\,, \end{align*} where \(k_z = \sqrt{k^2 - \kappa^2}\). The Hankel transform is found by looking at the delta function handout: \begin{align*} \HT\{p_0(\rho)\} &= p_0 \int_{0}^{\infty} J_0(\alpha\rho)J_0(\kappa\rho) \rho\,d\rho\\ &= \frac{p_0}{\alpha} \delta(\kappa-\alpha)\,. \end{align*} Thus the pressure field is \begin{align*} p_0(\rho,z) &= \frac{p_0}{\alpha} \int_{0}^{\infty} \delta(\kappa-\alpha) e^{i(k^2-\kappa^2)^{1/2}} J_0(\kappa \rho) \kappa\, d\kappa \\ &= p_0 J_0(\alpha \rho) e^{i(k^2 - \alpha^2)^{1/2}z} \\ & \propto J_0(\alpha\rho) \text{ for all } z\,, \end{align*} i.e., there is no diffraction in the field. Note that the beam propagates if \(\alpha\) is less than \(k\). The phase speed is \begin{align*} c_\text{ph} = \frac{\omega}{k_z} = \frac{\omega}{\sqrt{k^2-\alpha^2}} = \frac{c_0}{\sqrt{1 - (\alpha/k)^2}}\,. \end{align*} The glaring issue with the realization of such a non-diffracting beam is that \begin{align*} J_0(\alpha\rho) \propto \frac{1}{\sqrt{\rho}}\,,\quad \alpha\rho \gg 1\,, \end{align*} so the source power is \begin{align*} W = \oint \vec{I} \cdot d\vec{S} = \frac{p_0^2}{2\rho_0c_0}\int_{0}^{\infty} J_0^2(\alpha\rho) \pi \rho \,d\rho \to \infty\,. \end{align*} That is to say, an infinite amount of energy is required to generate such a beam! For practical (local) realization of a Bessel beam, the source needs to be truncated before \(a\) and \(0\) outside, i.e., \begin{align*} p_0(\rho) = p_0\, J_0(\alpha \rho)\,,\quad \rho \leq a\,. \end{align*}

Velocity source

For a velocity source (which is more practical), we should recall the linearized Newton's law: \begin{align*} \rho_0 \frac{\partial \vec{u}}{\partial t} + \grad p = 0\,, \end{align*} or for time-harmonic waves, \(-i\omega\rho_0 \vec{u}_\omega + \grad p_\omega = 0\). Solving for \(\vec{u}\) gives \begin{equation*} \vec{u}_\omega = \frac{\grad p_\omega}{ik\rho_0c_0}\,. \end{equation*} The 2D spatial Fourier transform of the left-hand side of the above is simply \begin{align*} \FTxy \{\vec{u}_\omega (x,y,z)\} = \vec{U}(k_x,k_y,z). \end{align*} Meanwhile, for the right-hand side, note that by the derivative property of the Fourier transform, \begin{align*} \FTxy\{\grad p_\omega(x,y,z)\} &= \FTxy\bigg\{\bigg(\ex \frac{\partial}{\partial x} + \ey \frac{\partial}{\partial y} + \ez \frac{\partial }{\partial z}\bigg)p_\omega\bigg\}\\ &= i(k_x \ex + k_y \ey)P(k_x,k_y,z) +\ez \FTxy\bigg\{\frac{\partial p_\omega}{\partial z}\bigg\}\,. \end{align*} But since \(P(k_x,k_y,z) = P(k_x,k_y,0) e^{ik_zz}\), the last term of the above, \(\ez \FTxy\{\partial p_\omega/\partial z\}\), is \begin{align*} \ez \FTxy\bigg\{\frac{\partial p_\omega}{\partial z} \bigg\} &= \ez \FTxy \bigg\{\frac{\partial}{\partial z}\IFTxy [P(k_x,k_y,0) e^{ik_zz}] \bigg\}\\ &=\ez \FTxy \{\IFTxy ik_z [P(k_x,k_y,0) e^{ik_zz}] \}\\ &=\ez ik_z P(k_x,k_y,0) e^{ik_zz}\\ &=\ez ik_z P(k_x,k_y,z)\,. \end{align*} So \begin{align*} \FTxy\{ \grad p_\omega(x,y,z)\} &= i(k_x \ex + k_y \ey)P(k_x,k_y,z) + \ez ik_z P(k_x,k_y,z) \\ &=i\vec{k} P(k_x,k_y,z)\,, \end{align*} where \begin{align*} \vec{k} &= k_x\ex + k_y\ey + (k^2 - k_x^2 - k_y^2)^{1/2}\ez = \vec{k} (k_x,k_y)\,. \end{align*} Since \(\vec{u}_\omega = \frac{\grad p_\omega}{ik\rho_0c_0}\), its 2D spatial Fourier transform is \begin{align}\label{U}\tag{3} \vec{U}(k_x,k_y,z) &= \frac{1}{\rho_0c_0} \frac{\vec{k}}{k} P(k_x,k_y,z)\,. \end{align} Assuming that \(\vec{u}_\omega (x,y,0) = u_0(x,y)\ez\), Eq. \eqref{U} evaluated in the source plane \(z=0\) is \begin{align*} \vec{U}_0(k_x,k_y) &= \FTxy \{u_0(x,y)\} \ez = U_0 (k_x,k_y)\ez\\ &= \frac{1}{\rho_0c_0} \frac{k_z \ez}{k} P_0(k_x,k_y) \end{align*} Solving for the 2D spatial Fourier transform of the source pressure gives \[P_0(k_x,k_y) = \rho_0 c_0 \frac{k}{k_z} U(k_x,k_y)\,.\] Since \(p_\omega(x,y,z) = \IFTxy \{P_0(k_x,k_y)e^{ik_zz}\}\), the full field due to a velocity source is \begin{align*} \boxed{p_\omega (x,y,z) = \rho_0c_0\IFTxy \{U_0(k_x,k_y)\frac{k}{k_z} e^{-ikz}\}\,,} \end{align*} where \(P_0(k_x,k_y) = \FTxy\{p_0(x,y)\}\). Note that there is a singularity at \(k_z = 0\), corresponding to the perimeter of the radiation circle.

Intensity

How does one calculate intensity using Fourier acoustics? Recall that \begin{align*} \vec{I} &= \langle p\vec{u}\rangle\\ &= \langle \Re \{p_\omega e^{-i\omega t}\} \Re \{\vec{u}_\omega e^{-i\omega t}\}\rangle\\ &= \frac{1}{2} \Re \{p_\omega^* \vec{u}_\omega\} = \frac{1}{2} \Re \{p_\omega \vec{u}_\omega^*\}\,, \end{align*} where \begin{align*} p_\omega (x,y,z) &= \IFTxy\{P(k_x,k_y,z)\}\\ \text{and}\quad \vec{u}_\omega (x,y,z) &= \IFTxy\{\vec{U}(k_x,k_y,z)\}=\IFTxy\{U_x\hat{e}_x + U_y\hat{e}_y + U_z\hat{e}_z\}\,. \end{align*}

Code

The whole code is nondimensionalized. The \(z\) axis is nondimensionalized by \(z_0\), the Rayleigh distance, sometimes called the "collumnation distance" or "diffraction length" \(ka^2/2 \sim S/\lambda\) (surface area to wavelength). The transverse axis is nondimensionalized by the source radius \(a\) (\(R = r/a\)), as is the wavenumber (\(K = ka\)). The code is set up to handle a velocity source, but it can easily be modified to handle pressure sources by implementing the discussion above.

The code itself contains all of these definitions. This sheet contains a full explanation and discussion of the code.

Radiation from plate due to 1D bending wave

The bending wave has the form \begin{align*} \vec{u}(x,y,0,t) &= u_0 e^{i(k_bx-\omega t)} \ez\, \quad k_b = \omega/c_b\\ &= u_0(x) e^{-i\omega t} \end{align*} where the normal particle velocity on the surface is \begin{align*} u_0(x) = u_0 e^{ik_b}\,. \end{align*}

To calculate the field in the fluid medium, first take the 2D spatial Fourier transform of the source condition, \begin{align*} U_0(k_x,k_y) &= \FTxy\{u_0(x)\} \\ &= u_0 \iint_{-\infty}^\infty e^{-i(k_x-k_b)x -ik_yy} dx\, dy\\ &= 4\pi^2 u_0 \delta(k_x-k_b)\delta(k_y)\,. \end{align*} Now convert to a pressure source using the result from the section on velocity sources. \begin{align*} P(k_x,k_y) &= \rho_0c_0 \frac{k}{k_z} U_0(k_x,k_y)\\ &= 4\pi^2 \rho_0c_0 u_0 \frac{k}{k_z} \delta(k_x-k_b)\delta(k_y)\,. \end{align*} The pressure field is found by the standard Fourier acoustics procedure: \begin{align*} p_\omega(x,y,z) &= \IFTxy \{P_0(k_x,k_y) e^{ik_zz}\}\\ &= 4\pi^2 \rho_0c_0 u_0 k \, \IFTxy \{\delta(k_x-k_b) \delta(k_y) {e^{ik_zz}}/{k_z}\}\\ &= \rho_0c_0 u_0 k \iint_{-\infty}^\infty \delta(k_x-k_b)\delta(k_y) \frac{e^{i(k^2-k_x^2 -k_y^2)^{1/2}}}{\sqrt{k^2-k_x^2-k_y^2}} e^{ik_xx + ik_yy} dk_xdk_y\\ &= \frac{\rho_0c_0 u_0 k}{\sqrt{k^2-k_b^2}} e^{ik_bx + i(k^2-k_b^2)^{1/2}z}\,. \end{align*} If \(k > k_b\) (or \(c_b > c_0\)), then the wave propagates into the fluid, while if \(k < k_b\) (or \(c_b < c_0\)), then the wave is evanescent in the fluid.

For additional insight, convert pressure to the particle velocity: \begin{align*} \vec{u}_\omega (x,y,z) &= \grad p_\omega/ik\rho_0c_0\\ &= u_0 \frac{k_b\ex }{\sqrt{k^2-k_b^2}} e^{ik_bx + i\sqrt{k^2 - k_b^2}z}\,. \end{align*} The intensity can then be calculated: \begin{align*} \vec{I} &= \frac{1}{2}\Re \{p_\omega \vec{u}_\omega^*\}\\ &= \frac{\rho_0c_0u_0^2}{2(1-k_b^2/k^2)} \bigg[\frac{k_b}{k} \ex + \sqrt{1-k_b^2/k^2}\ez\bigg]\,\quad k_b \,{<}\, k\\ &= \frac{\rho_0c_0u_0^2}{2(1-k_b^2/k^2)} \frac{k}{k_b} e^{-2(k_b^2 - k^2)^{1/2}z} \quad k_b>k\,. \end{align*} In the second case, there is no power radiated in the \(z\) direction.

As a sanity check, what if \(c_b \to \infty\)? Then one should obtain a plane wave in the \(z\) direction. Indeed, for \(k_b\to 0\), \begin{align*} p_\omega(x,y,z) &= \rho_0c_0 u_0 e^{ikz}\,. \end{align*}

Generalization to elastic waves in isotropic solids

Let \(\vec{\xi}\) be the particle displacement. It can be separated into a irrotational and rotational components, \begin{align*} \vec{\xi} = \vec{\xi}_l + \vec{\xi}_t\,, \end{align*} where \(\vec{\xi}_l\) is the longitudinal (compressional) wave displacement (\(\curl \vec{\xi}_l = 0\)), and where \(\vec{\xi}_t\) is the shear (transverse) wave displacement (\(\divergence \vec{\xi}_t = 0\)). These two displacements above satisfy wave equations of their own. This was shown without proof, but the derivation is worked out from first principles here.

Compressional waves

The compressional wave equation is \begin{align*} \Laplacian \vec{\xi}_l &= \frac{1}{c_l^2} \frac{\partial \vec{\xi}_l}{\partial t^2} \end{align*} where the longitudinal sound speed is \begin{align*} c_l &= \sqrt{\frac{K + 4\mu/3}{\rho}}\,, \end{align*} where \(K\) is the bulk modulus and \(\mu\) is the shear modulus. By the Helmholtz decomposition theorem, \(\vec{\xi}_\ell\) is defined as the gradient of a scalar potential \(\phi\), \begin{align*} \vec{\xi}_l = \grad \phi\,, \end{align*} giving \begin{align*} \Laplacian \phi = \frac{1}{c_l} \frac{\partial^2\phi}{\partial t^2}\,. \end{align*} Assuming \(\phi\) is time harmonic results in the Helmholtz equation, which is the same as for waves in fluids: \begin{align*} \Laplacian \phi + k^2\phi = 0\,. \end{align*} Thus we can swap out \(p\) for \(\phi\) in Fourier theory presented earlier, and one can obtain the full compressional wave field: \begin{align*} \phi_\omega(x,y,z) &= \IFTxy \{\FTxy[\phi_\omega(x,y,0)] e^{ik_zz}\}\,. \end{align*}

Shear waves

The wave equation for shear waves is vectorial, i.e., \begin{align}\label{vectorwave}\tag{1} \Laplacian \vec{\xi}_t = \frac{1}{c_t^2}\frac{\partial \vec{\xi}_t}{\partial t^2}\, \quad c_t = \sqrt{\mu/\rho} \end{align} The other condition noted above is that \(\divergence \vec{\xi}_t = 0\), which is often utilized to to defined terms of the vector potential \(\vec{\xi}_t = \curl \vec{\psi}\) (analogous to how the longitudinal displacement was defined in terms of a scalar potential above). However, even with the vector potential, we would still have three components, so let us follow Landau and Lifshitz and simply use the Cartesian components of \(\vec{\xi}_t\), \begin{align*} \vec{\xi}_t = \xi_x\ex + \xi_y\ey + \xi_z\ez\,, \end{align*} such that Eq. \eqref{vectorwave} becomes \begin{align} \Laplacian \xi_x + k_t^2 \xi_x &= 0\label{Helmers1} \tag{2}\\ \Laplacian \xi_y + k_t^2 \xi_y &= 0 \label{Helmers2} \tag{3}\\ \Laplacian \xi_z + k_t^2 \xi_z &= 0\,. \end{align} Note that in the equations above, only two components are independent, because the third is related by \(\divergence \vec{\xi}_t = 0\), which in Cartesian coordinates reads \begin{align*} \frac{\partial \xi_x}{\partial x} + \frac{\partial \xi_y}{\partial y} + \frac{\partial \xi_z}{\partial z} = 0\,. \end{align*} Eqs. \eqref{Helmers1} and \eqref{Helmers2} are solved separately using Fourier acoustics: \begin{align*} \xi_x(x,y,z) &= \IFTxy\{\FTxy\{\xi_x(x,y,0)\}e^{ik_zz}\}\,. \end{align*}

Example: Axisymmetric shear waves

Torsional waves

In torsional waves, the transverse displacement is given by \begin{align*} \vec{\xi}_t(\rho,\phi,z=0) &= \xi_\phi (\rho) \hat{e}_\phi\,, \end{align*} where the coordinates and unit vectors are shown below:

Then \begin{align*} \xi_x &= -\xi_\phi(\rho) \sin\phi\\ \xi_y &= \xi_\phi(\rho) \cos\phi\\ \xi_z &= 0 \end{align*} and \begin{align*} \divergence \vec{\xi}_t &= \frac{1}{\rho} \frac{\partial (\rho \xi_\rho)}{\partial \rho} + \frac{1}{\rho} \frac{\partial \xi_\phi}{\partial \phi} + \frac{\partial \xi_z}{\partial z} \end{align*} Since \(\vec{\xi}_t(\rho,\phi,z=0) = \xi_\phi (\rho) \hat{e}_\phi\), the above equation reduces to \(\divergence \vec\psi = 0\). Meanwhile, the Laplacian in cylindrical coordinates is \begin{align*} \Laplacian &= \frac{\partial^2}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial}{\partial \rho} + \frac{1}{\rho^2}\frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial z^2}\,. \end{align*} Thus the Laplacian of the wave variable in Eqs. \eqref{Helmers1} and \eqref{Helmers2} read \begin{align*} \Laplacian \xi_x &= \bigg( \frac{\partial^2}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial}{\partial \rho} + \frac{1}{\rho^2}\frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial z^2} \bigg) (-\xi_\phi \sin\phi)\\ \Laplacian \xi_y &= \bigg( \frac{\partial^2}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial}{\partial \rho} + \frac{1}{\rho^2}\frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial z^2} \bigg) (\xi_\phi \cos\phi)\,, \end{align*} and both Eqs. \eqref{Helmers1} and \eqref{Helmers2} yield \begin{align*} \bigg( \frac{\partial^2}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial}{\partial \rho} + \frac{1}{\rho^2}\frac{\partial^2}{\partial \phi^2} \bigg) (\xi_\phi \cos\phi) \end{align*} \begin{align}\label{Hankster}\tag{5} \bigg(\frac{\partial^2}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial}{\partial \rho} - \frac{1}{\rho^2}\bigg) \xi_\phi + \frac{\partial^2\xi_\phi}{\partial z^2} + k^2_t \xi_\phi \end{align} The above equation is now solved using Hankel transforms. Define the \(n\)th order Hankel transform pair: \begin{align*} F_H^{n}(\kappa) &= \HT^n \{f(\rho)\} = \int_{0}^{\infty} f(\rho) J_n (\kappa \rho )\rho d\rho\\ f(\rho) &= \IHT_n \{F_H^{n}(\kappa)\} = \int_{0}^{\infty} F_H^{(n)}(\kappa) J_n (\kappa \rho )\kappa d\kappa\,. \end{align*} Because of the identity \begin{align*} \int_{0}^{\infty} J_n(\kappa\rho) J_n(\kappa\rho')\kappa d\kappa = \frac{\delta(\rho-\rho')}{\rho} \end{align*} and \begin{align*} \IHT_n\bigg\{\bigg(\frac{\partial^2}{\partial\rho^2} + \frac{1}{\rho}\frac{\partial}{\partial\rho} - \frac{n^2}{\rho^2}\bigg)f(\rho) \bigg\} = -k^2 {F_H}^n(\kappa) \end{align*} So Eq. \eqref{Hankster} becomes \begin{align*} \frac{d^2 \hat{\xi}_\phi}{dz^2} + (k_t^2 - \kappa^2) \hat{\xi}_\phi = 0 \end{align*} Thus the solution for the torsional wave field is \begin{align*} \xi_\phi (\rho,z) = \IHT_1\{\HT_1 [\xi_\phi(\rho,0)] e^{ik_zz}\} \end{align*}

Example: Uniform circular motion:

\begin{align*} \xi_\phi(\rho) &= \xi_0 \frac{\rho}{a} \circfn(\rho/a)\\ \hat{\xi}(\kappa) &= \HT_1 \{\xi_\phi(\rho)\}\\ &= \frac{\xi_0}{a} \int_{0}^{a} \rho J_1 (\kappa \rho) \rho d\rho,\quad t = \kappa \rho\\ &= \frac{\xi_0}{\kappa^3a } \int_{0}^{\kappa a}J_1(t) t^2 dt \end{align*} Note that \begin{align*} \int_{0}^{x}J_{n-1}(t) t^n dt = x^n J_n(x) \end{align*} so \begin{align*} \hat{\xi}_\phi (\kappa) &= \xi_0 \frac{a}{\kappa} J_2(\kappa a)\\ &=0\,\quad \kappa -0, \end{align*} because \(J_2(x) = x^2/8 + \Order(x^4)\) The directivity is thus proportional to \begin{align*} \bigg|\frac{J_2(\kappa a)}{\kappa a}\bigg|\,, \end{align*} and thus there is a zero field on the propagation axis.

Nearfield acoustical holography (NAH)

This topic was originally presented after focused sources were discussed. See Williams pgs. 89-92.

Consider two planes, \(z_1 = \) constant and \(z_2\) = constant, where \(z_2 > z_1\). The purpose of nearfield acoustical holography (NAH) is to calculate the field near the source given the field far away. NAH can be used to investigate structural integrity of a vehicle, for example.

Given \(p_\omega (x,y,z_1)\), then the field at \(z_2\) is \begin{align}\tag{1}\label{1holo} p_\omega(x,y,z_2) = \IFTxy\{\FTxy{p_\omega(x,y,z_1)} e^{ik_z(z_2-z_1)}\}\,, \end{align} where the projection of the wavenumber in the \(z\) direction is, for propagating and evanescent waves, respectively, \begin{align} k_z &= \sqrt{k^2 - k_x^2 -k_y^2}\,,\quad k_x^2 + k_y^2 \leq k^2 \tag{2a}\label{2aholo}\\ &= i\sqrt{k_x^2 + k_y^2- k^2}\,,\quad k_x^2 + k_y^2 > k^2 \tag{2b}\label{2bholo} \end{align} Eq. \eqref{1holo} is inverted to calculate the pressure field in the plane \(z_1 =\) constant: \begin{align}\label{3holo}\tag{3} p_\omega(x,y,z_1) &= \IFTxy\{\FTxy[p_\omega(x,y,z_2)] e^{-ik_z(z_2-z_1)}\}\,, \end{align} Thus as \(z_2-z_1\) increases (\(z_1\) moved farther to the left with \(z_2\) fixed), the evanescent waves grow. Thus Eq. \eqref{3holo} is not a physical solution to the wave equation. Normally, \(z_2\) and \(z_1\) are defined as \begin{alignat*}{2} z_2 &= z_H &&= \text{ hologram plane}\\ z_1 &= z_I &&= \text{ image plane} \,, \end{alignat*} and the \(z\) component of \(\vec{u}\) is normally desired at \(z= z_I\). Let the normal component of the particle velocity in the image plane be \begin{align*} u_0(x,y,z_I) = \vec{u}(x,y,z_I) \cdot \ez\,. \end{align*} Then the Fourier transform of the normal velocity field in the image plane is \begin{align*} U_0 (k_x,k_y,z_I) &= \frac{1}{\rho_0c_0}\frac{k_z}{k} P_0(k_x,k_y,z_I)\,. \end{align*} Eq. \eqref{3holo} can thus converted to give the partical velocity in the image plane, \(z_1 = \) constant: \begin{align*} u_0 (x,y,z_I) &= \IFTxy\{\FTxy [p_\omega(x,y,z_H)] G_H(k_x,k_y,z_H-z_I)\}\,, \end{align*} where the propagator above is \begin{align*} G_H(k_x,k_y,z_H-z_I) &= \frac{1}{\rho_0c_0}\frac{k_z}{k} \begin{cases} e^{-i(k^2 -k_x^2 -k_y^2)^{1/2} (z_H - z_I)} \,, \quad k_x^2 +k_y^2 \leq k^2\\ e^{(k_x^2 + k_y^2 - k^2)^{1/2}(z_H - z_I)} \,, \quad k_x^2 +k_y^2 > k^2 \end{cases} \end{align*}

The engineering challenge is to have \(|z_H- z_I| \lesssim \lambda \) to measure the evanescent waves, without which resolution is limited to \(\sim \lambda\), because of the restriction that \(k_x^2 + k_y^2 \leq k^2\) for propagating waves, or \begin{align*} \frac{1}{\lambda_{x,y}} \leq \frac{1}{\lambda^2} \implies \lambda_{x,y}^2 \geq \lambda^2\,. \end{align*} Thus \(\lambda\) is the minimum resolvable spatial scale if only propagating waves are measured in the hologram plane \(z_H =\) constant.