Pressure

We denote the pressure field on surface 1 as $p^{(1)}(x,y)$ and the pressure field on surface 2 as $p^{(2)}(x,y)$. In plane wave propagation we saw that the pressure is taken to be the same on both sides of the discontinuity. In the multimodal case the pressure field is matched at either side. For the case shown where $S_2>S_1$ the pressure is matched on the air they share, $S_1$.

$$p^{(1)} = p^{(2)} \hspace{0.5cm} {\mathrm on} \hspace{0.5cm} S_1, \hspace{0.5cm} S_1<S_2$$ (B.1)

Now the concept of orthogonality of the modes is used. Recalling our orthogonality relation from equation (2.47):

$$\int\limits_S \psi_n \psi_m dS = S \delta_{nm}$$ (B.2)

it follows that the integration of the mode profile $\psi_n$ with a general pressure field, $p$ will extract the component of the $n$th mode in the pressure field. Using the fact that the pressure may be expressed as a sum of the modes from equation 2.26 (ignoring the ${\mathrm exp}(i \omega t)$ time factor):

$$\int\limits_S \psi_n p dS = \int\limits_S \psi_n \sum\limits_{m=0}^{\infty} \psi_m P_m dS = P_n S.$$ (B.3)

We will use $P_n^{(1)}$ and $P_n^{(2)}$ to denote the complex mode amplitudes on surfaces 1 and 2 respectively. $\psi_n^{(1)}$ and $\psi_n^{(2)}$ are the corresponding mode profiles on surfaces 1 and 2. Combining equations (B.1) and B.3) gives

$$P_n^{(1)} = \frac{1}{S_1} \int\limits_{S_1} \psi_n^{(1)} p^{... ..._n^{(1)} \sum\limits_{m=0}^{\infty} \psi_m^{(2)} P_m^{(2)} dS$$ (B.4)

which may be written as

$$P_n^{(1)} = \sum\limits_{m=0}^{\infty} F_{nm} P_m^{(2)}$$ (B.5)

where

$$F_{nm} = \frac{1}{S_1}\int\limits_{S_1}\psi_{n}^{(1)}\psi_{m}^{(2)}dS.$$ (B.6)

Using matrix notation, ${\mathbf P}^{(2)}$ is a column vector whose entries are given by $P_m^{(2)}$ and

$${\mathbf P}^{(1)}= F {\mathbf P}^{(2)}, \mbox{\hspace{1cm}} S_1 < S_2$$ (B.7)

where F is a matrix with elements $F_{nm}$. We have now proved equations (2.79) and (2.80) from chapter 2.

We now have a formula giving the pressure on a smaller cross-section at a discontinuity from the pressure on a larger cross-section on the other side. Since $S_1$ and $S_2$ are assumed to not be separated by any distance along the $z$ axis the formula holds whatever side the largest cross-section is on. Consider if $S_1>S_2$. Now section 1 is the larger cross-section meaning that we just have to interchange the labels 1 and 2 in equation (B.7):

$${\mathbf P}^{(2)}= V {\mathbf P}^{(1)}, \mbox{\hspace{1cm}} S_1 > S_2$$ (B.8)

where $V$ is a matrix with the elements defined by

$$V_{nm} = \frac{1}{S_2}\int\limits_{S_2}\psi_{n}^{(2)}\psi_{m}^{(1)}dS$$ (B.9)