which expresses the observation that particle have a mass that will resist a change in speed under the influence of an applied force (inertial reaction).
planewave
Assume a plane wave that propagates in the x-direction
(35)\[\begin{equation}
\hat{p}=p_+\exp{i(\omega t - kx)}
\end{equation}\]
Euler equation
(36)\[\begin{equation}
\nabla p +\rho_0\frac{\partial u}{\partial t}=0
\end{equation}\]
in x-direction
(37)\[\begin{equation}
\frac{\partial p}{\partial x} +\rho_0\frac{\partial u}{\partial t}=0
\end{equation}\]
results in particle velocity
(38)\[\begin{equation}
\hat{u}_x = -\frac{1}{i \omega\rho}\frac{\partial\hat{p}}{\partial x}=\frac{k}{\omega\rho}p_+\exp{i(\omega t - kx)} =
\frac{p_+}{\rho c}\exp{i(\omega t - kx)}=\frac{\hat{p}}{\rho c}
\end{equation}\]
The paricle velocity is obtained by integration and using the Euler approximation to the pressure gradient one gets
(39)\[\begin{equation}
u(t)=u(0)-\frac{1}{\rho}\int_0^t{\nabla p(\tau) d\tau}
\end{equation}\]
Using \(u(0)=0\) one ontains fot the sound intensity
(40)\[\begin{equation}
I=-\frac{1}{T}\frac{1}{\rho} \int_0^Tp(t)\Biggl(\int_0^t{\nabla p(\tau) d\tau}\Biggr) dt
\end{equation}\]
Using next the Fourier synthesis
(41)\[\begin{equation}
p(t)=\frac{1}{2\pi}\int_{-\pi}^{\pi}P(\omega)\exp(-i\omega t)d\omega
\end{equation}\]
and with
(42)\[\begin{equation}
\int_{0}^{t}\exp(-i\omega \tau)d\tau=\frac{i}{\omega}\Biggl(\exp(-i\omega t)-1\Biggr)
\end{equation}\]
and therefore
(43)\[\begin{equation}
\int_0^t \nabla p(\tau)d\tau=
\frac{i}{2\pi}\int_{-\pi}^{\pi}\Biggl(\frac{\nabla P(\omega)}{\omega}\Biggr)\Biggl(\exp(-i\omega t)-1\Biggr)d\omega
\end{equation}\]
we obtain
(44)\[\begin{equation}
I=-\frac{i}{\rho}\frac{1}{2\pi}\frac{1}{2\pi}
\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} P(\omega')
\Biggl(\frac{\nabla P(\omega)}{\omega}\Biggr) \Biggl(\frac{1}{T}\int_0^T \exp(-i\omega' t)\Biggl(\exp(-i\omega t)-1\Biggr) dt \Biggr) d\omega d\omega'
\end{equation}\]
Assume the temporal ingration
(45)\[\begin{equation}
\frac{1}{T}\int_0^T \exp(-i\omega' t)\Biggl(\exp(-i\omega t)-1\Biggr) dt = \delta (\omega' + \omega = 0)
\end{equation}\]
evaluates to one if and only if \(\omega'+\omega=0\) and to zero otherwise, so that
(46)\[\begin{equation}
I=-\frac{i}{\rho}\frac{1}{2\pi}\frac{1}{2\pi}
\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} P(-\omega)
\Biggl(\frac{\nabla P(\omega)}{\omega}\Biggr) d\omega d\omega'
\end{equation}\]
or
(47)\[\begin{equation}
I=-\frac{i}{\rho}\frac{1}{2\pi}
\int_{-\pi}^{\pi} P^*(\omega) \Biggl(\frac{\nabla P(\omega)}{\omega}\Biggr) d\omega
\end{equation}\]
Approximating the sound pressure and pressure gradient by pressure measurements of 2 sensors labeled (1,2)
(48)\[\begin{equation}
P(\omega)=\frac{P_2(\omega)+P_1(\omega)}{2}
\end{equation}\]
(49)\[\begin{equation}
\nabla P(\omega)=\frac{P_2(\omega)-P_1(\omega)}{d_{2,1}}
\end{equation}\]
(50)\[\begin{equation}
P^*(\omega)\nabla P(\omega) = \frac{1}{2d_{2,1}}\{(P_2^*(\omega)+P_1^*(\omega)) (P_2(\omega)-P_1(\omega))\}
\end{equation}\]
(51)\[\begin{equation}
P^*(\omega)\nabla P(\omega) = \frac{1}{2d_{2,1}}\{ |P_2(\omega)|^2 - |P_1(\omega)|^2 +P_2(\omega)P_1^*(\omega)-P_2^*(\omega)P_1(\omega)\}
\end{equation}\]
With \(|P_2(\omega)|^2 = |P_1(\omega)|^2\)
(52)\[\begin{equation}
P^*(\omega)\nabla P(\omega) = \frac{1}{2d_{2,1}}\{ P_2(\omega)P_1^*(\omega)-P_2^*(\omega)P_1(\omega)\}
\end{equation}\]
or
(53)\[\begin{equation}
P^*(\omega)\nabla P(\omega) = \frac{i}{d_{2,1}}\text{Im}\{ P_2(\omega)P_1^*(\omega)\}
\end{equation}\]
and finally the spectral sound intensity along sensor pair (1,2) is approximated by
(54)\[\begin{equation}
I_{2,1}(\omega)=-\frac{i}{\rho d_{2,1}} \frac{ P^*(\omega)\Delta P(\omega)}{\omega}
=-\frac{1}{\rho d_{2,1}} \frac{\text{Im}\{P_2(\omega) P_1^* (\omega)\}}{\omega}
\end{equation}\]