# Diffraction Transfer Matrix for Infinite Depth

## Introduction

This is an extension of the Diffraction Transfer Matrix (which only applied to finite depth) to infinite depth. This is based upon the results in Peter and Meylan 2004.

## Calculation of the diffraction transfer matrix for bodies of arbitrary geometry

To calculate the diffraction transfer matrix in infinite depth, we require the representation of the Infinite Depth, Free-Surface Green Function in cylindrical eigenfunctions,

$G(r,\theta,z;s,\varphi,c) = \frac{\mathrm{i}\alpha}{2} \, \mathrm{e}^{\alpha (z+c)} \sum_{\nu=-\infty}^{\infty} H_\nu^{(1)}(\alpha r) J_\nu(\alpha s) \mathrm{e}^{\mathrm{i}\nu (\theta - \varphi)}$
$+ \frac{1}{\pi^2} \int\limits_0^{\infty} \psi(z,\eta) \frac{\eta^2}{\eta^2+\alpha^2} \psi(c,\eta) \sum_{\nu=-\infty}^{\infty} K_\nu(\eta r) I_\nu(\eta s) \mathrm{e}^{\mathrm{i}\nu (\theta - \varphi)} \mathrm{d}\eta,$

$r \gt s$, given by Peter and Meylan 2004b where

$\psi(z,\eta) = \cos \eta z + \frac{\alpha}{\eta} \sin \eta z.$

We assume that we have represented the scattered potential in terms of the source strength distribution $\varsigma^j$ so that the scattered potential can be written as

$\phi_j^\mathrm{S}(\mathbf{y}) = \int\limits_{\Gamma_j} G (\mathbf{y},\mathbf{\zeta}) \, \varsigma^j (\mathbf{\zeta}) \mathrm{d}\sigma_\mathbf{\zeta}, \quad \mathbf{y} \in D,$

where $D$ is the volume occupied by the water and $\Gamma_j$ is the immersed surface of body $\Delta_j$. The source strength distribution function $\varsigma^j$ can be found by solving an integral equation. The integral equation is described in Wehausen and Laitone 1960. Substituting the eigenfunction expansion of the Green's function (green_inf) into the above integral equation, the scattered potential can be written as

$\phi_j^\mathrm{S}(r_j,\theta_j,z) = \mathrm{e}^{\alpha z} \sum_{\nu = - \infty}^{\infty} \bigg[ \frac{\mathrm{i}\alpha}{2} \int\limits_{\Gamma_j} \mathrm{e}^{\alpha c} J_\nu(\alpha s) \mathrm{e}^{-\mathrm{i}\nu \varphi} \varsigma^j(\mathbf{\zeta}) \mathrm{d}\sigma_\mathbf{\zeta} \bigg] H_\nu^{(1)} (\alpha r_j) \mathrm{e}^{\mathrm{i}\nu \theta_j}$
$+ \int\limits_{0}^{\infty} \psi(z,\eta) \sum_{\nu = - \infty}^{\infty} \bigg[ \frac{1}{\pi^2} \frac{\eta^2 }{\eta^2 + \alpha^2} \int\limits_{\Gamma_j} \psi(c,\eta) I_\nu(\eta s) \mathrm{e}^{-\mathrm{i}\nu \varphi} \varsigma^j({\mathbf{\zeta}}) \mathrm{d}\sigma_{\mathbf{\zeta}} \bigg] K_\nu (\eta r_j) \mathrm{e}^{\mathrm{i}\nu \theta_j} \mathrm{d}\eta,$

where $\mathbf{\zeta}=(s,\varphi,c)$ and $r\gts$. This restriction implies that the eigenfunction expansion is only valid outside the escribed cylinder of the body.

The columns of the diffraction transfer matrix are the coefficients of the eigenfunction expansion of the scattered wavefield due to the different incident modes of unit-amplitude. The elements of the diffraction transfer matrix of a body of arbitrary shape are therefore given by

$({\mathbf B}_j)_{pq} = \frac{\mathrm{i}\alpha}{2} \int\limits_{\Gamma_j} \mathrm{e}^{\alpha c} J_p(\alpha s) \mathrm{e}^{-\mathrm{i}p \varphi} \varsigma_q^j(\mathbf{\zeta}) \mathrm{d}\sigma_\mathbf{\zeta}$

and

$({\mathbf B}_j)_{pq} = \frac{1}{\pi^2} \frac{\eta^2}{\eta^2 + \alpha^2} \int\limits_{\Gamma_j} \psi(c,\eta) I_p(\eta s) \mathrm{e}^{-\mathrm{i}p \varphi} \varsigma_q^j(\mathbf{\zeta}) \mathrm{d}\sigma_\mathbf{\zeta}$

for the propagating and the decaying modes respectively, where $\varsigma_q^j(\mathbf{\zeta})$ is the source strength distribution due to an incident potential of mode $q$ of the form

$\phi_q^{\mathrm{I}}(s,\varphi,c) = \mathrm{e}^{\alpha c} H_q^{(1)} (\alpha s) \mathrm{e}^{\mathrm{i}q \varphi}$

for the propagating modes, and

$\phi_q^{\mathrm{I}}(s,\varphi,c) = \psi(c,\eta) K_q (\eta s) \mathrm{e}^{\mathrm{i}q \varphi}$

for the decaying modes.

## The diffraction transfer matrix of rotated bodies

For a non-axisymmetric body, a rotation about the mean centre position in the $(x,y)$-plane will result in a different diffraction transfer matrix. We will show how the diffraction transfer matrix of a body rotated by an angle $\beta$ can be easily calculated from the diffraction transfer matrix of the non-rotated body. The rotation of the body influences the form of the elements of the diffraction transfer matrices in two ways. Firstly, the angular dependence in the integral over the immersed surface of the body is altered and, secondly, the source strength distribution function is different if the body is rotated. However, the source strength distribution function of the rotated body can be obtained by calculating the response of the non-rotated body due to rotated incident potentials. It will be shown that the additional angular dependence can be easily factored out of the elements of the diffraction transfer matrix.

The additional angular dependence caused by the rotation of the incident potential can be factored out of the normal derivative of the incident potential such that

$\frac{\partial \phi_{q\beta}^{\mathrm{I}}}{\partial n} = \frac{\partial \phi_{q}^{\mathrm{I}}}{\partial n} \mathrm{e}^{\mathrm{i}q \beta},$

where $\phi_{q\beta}^{\mathrm{I}}$ is the rotated incident potential. Since the integral equation for the determination of the source strength distribution function is linear, the source strength distribution function due to the rotated incident potential is thus just given by

$\varsigma_{q\beta}^j = \varsigma_q^j \, \mathrm{e}^{\mathrm{i}q \beta}.$

This is also the source strength distribution function of the rotated body due to the standard incident modes.

The elements of the diffraction transfer matrix $\mathbf{B}_j$ are given by equations in the previous section. Keeping in mind that the body is rotated by the angle $\beta$, the elements of the diffraction transfer matrix of the rotated body are given by

$(\mathbf{B}_j^\beta)_{pq} = \frac{\mathrm{i}\alpha}{2} \int\limits_{\Gamma_j} \mathrm{e}^{\alpha c} J_p(\alpha s) \mathrm{e}^{-\mathrm{i}p (\varphi+\beta)} \varsigma_{q\beta}^j(\mathbf{\zeta}) \mathrm{d}\sigma_\mathbf{\zeta},$

and

$(\mathbf{B}_j^\beta)_{pq} = \frac{1}{\pi^2} \frac{\eta^2}{\eta^2 + \alpha^2} \int\limits_{\Gamma_j} \psi(c,\eta) I_p(\eta s) \mathrm{e}^{-\mathrm{i}p (\varphi+\beta)} \varsigma_{q\beta}^j(\mathbf{\zeta}) \mathrm{d}\sigma_\mathbf{\zeta},$

for the propagating and decaying modes respectively. Thus the additional angular dependence caused by the rotation of the body can be factored out of the elements of the diffraction transfer matrix. The elements of the diffraction transfer matrix corresponding to the body rotated by the angle $\beta$, $\mathbf{B}_j^\beta$, are given by

$(\mathbf{B}_j^\beta)_{pq} = (\mathbf{B}_j)_{pq} \, \mathrm{e}^{\mathrm{i}(q-p) \beta}.$

As before, $(\mathbf{B})_{pq}$ is understood to be the element of $\mathbf{B}$ which corresponds to the coefficient of the $p$th scattered mode due to a unit-amplitude incident wave of mode $q$. This equation applies to propagating and decaying modes likewise.