## Underwater Sound Channel

My objective here is to, starting with the most fundamental equations, arrive at the pressure profile for an underwater sound channel. A lot of what I derive will be more general, but I ultimately want to end with the pressure profile for two homogenous layers with constant sound speed and density, assuming a pressure-release surface and rigid bottom. But getting here will take a while, so I will first decide what my "fundamental starting position" will be.

## Navier-Stokes Equation

\[ \rho \left( \frac{\partial \textbf{v}}{\partial t} + \textbf{v}\cdot\nabla \textbf{v}\right) = -\nabla p + \textbf{f} + \mu \nabla^2 \textbf{v} + \left(\xi + \frac{\mu}{3}\right)\nabla\left(\nabla\cdot \textbf{v}\right)\]

\[ \frac{\partial \rho}{\partial t} + \nabla\cdot\left(\rho \textbf{v}\right) = 0\]

This comes straight from the Wikipedia page on the Navier-Stokes equation, where, to be more precise, I am using the compressible Newtonian fluid equations and the continuity equation. These equations can be derived, and that Wikipedia page gives a derivation, though I hope at some point to type up my own derivation to these equations, since they physically come from some very basic assumptions of mass and momentum conservation (combined with constitutive relations like Newton's law of viscosity). But for now, we take these equations for granted.

Let's talk about the various terms. The left hand side is "equivalent" to the $m\textbf{a}$ term in $\textbf{F}=m\textbf{a}$, and basically only involves the density $\rho$ and the "material derivative" of the velocity $\textbf{v}$. The material derivative is just like a time derivative you usually encounter for acceleration, except it includes the effect of convection. I hope to elaborate on this point in a later page, but for now, I take this as given, also. On the right hand side, we see the pressure $p$, the viscosity $\mu$, a body force $\textbf{f}$ (like gravity, E&M, etc), and the volume viscosity $\xi$ which arises in compressible fluid mechanics. To be honest, I've never encountered $\xi$ before, so my first simplification will be eliminating this term. Additionally, since the only body force we see in an ocean environment is gravity, which is a conservative force, we can lump the effect of gravity into pressure ($p \rightarrow p + \rho g z$). So the body force disappears.

Next, I want to cross out all the friction terms, but I'll do so with some justification. George Gabriel Stokes, back in 1845, derived the Stokes' Law of Sound Attenuation, which gives the pressure drop over a given distance due solely to friction. To decide on it's importance, let's calculate the distance in which the pressure drops to half of the expected value as a result of friction.

\[d_{\text{half pressure}} = \frac{3\ln(2)}{2} \frac{ \rho c^3}{\mu \omega^2}\]

Since there is a frequency dependence, we must decide what frequency we want to use as a gauge. Typical underwater acoustic frequencies range from 10Hz to 1MHz, though for scale, let's talk about this range. Humans can hear 20Hz to 20kHz on a good day, and bats can apparently hear up to 200kHz. Underwater, porpoises can hear up to 160kHz, and herring can only hear up to 4kHz. Ultrasound (like that used in the medical field, or for nondestructive evaluation of concrete, metal, etc), is usually in the range of 2 to 18MHz, though apparently can go up to single digit GHz for specialized projects. But, in keeping with our underwater acoustics desire, let's do 1kHz, since this this is in the middle of the underwater acoustics range (in a log-scale sense - the linear average of 100kHz and 10Hz is obviously about 50kHz, whereas in a log-scale, the average is 1kHz). That said, for 1kHz sound (which is apparently about C3 / Soprano C / High C on a piano) in a typical sea water environment, the distance to half the pressure strictly through viscous effects is about 70,000 miles, or just under 3 earth circumferences. Even if we bump the frequency up to 100kHz, the pressure would be half what we expected at a distance of about 7 miles / 11 km, which for realistically loud sources, is just about the limit that we could detect theoretically without any friction at all.

Basically, we went through all this to justify our elimination of $\mu$ (and $\xi$) from the Navier-Stokes equation - which for the mediocre 1kHz frequency at the fairly substantial 10km range, the pressures we model will differ from the actual pressures by 0.003%. There are other assumptions we'll make later that create much more error, so this should be considered a pretty safe assumption for underwater acoustics.

\[ \frac{\partial \rho}{\partial t} + \nabla\cdot\left(\rho \textbf{v}\right) = 0\]

This comes straight from the Wikipedia page on the Navier-Stokes equation, where, to be more precise, I am using the compressible Newtonian fluid equations and the continuity equation. These equations can be derived, and that Wikipedia page gives a derivation, though I hope at some point to type up my own derivation to these equations, since they physically come from some very basic assumptions of mass and momentum conservation (combined with constitutive relations like Newton's law of viscosity). But for now, we take these equations for granted.

Let's talk about the various terms. The left hand side is "equivalent" to the $m\textbf{a}$ term in $\textbf{F}=m\textbf{a}$, and basically only involves the density $\rho$ and the "material derivative" of the velocity $\textbf{v}$. The material derivative is just like a time derivative you usually encounter for acceleration, except it includes the effect of convection. I hope to elaborate on this point in a later page, but for now, I take this as given, also. On the right hand side, we see the pressure $p$, the viscosity $\mu$, a body force $\textbf{f}$ (like gravity, E&M, etc), and the volume viscosity $\xi$ which arises in compressible fluid mechanics. To be honest, I've never encountered $\xi$ before, so my first simplification will be eliminating this term. Additionally, since the only body force we see in an ocean environment is gravity, which is a conservative force, we can lump the effect of gravity into pressure ($p \rightarrow p + \rho g z$). So the body force disappears.

Next, I want to cross out all the friction terms, but I'll do so with some justification. George Gabriel Stokes, back in 1845, derived the Stokes' Law of Sound Attenuation, which gives the pressure drop over a given distance due solely to friction. To decide on it's importance, let's calculate the distance in which the pressure drops to half of the expected value as a result of friction.

\[d_{\text{half pressure}} = \frac{3\ln(2)}{2} \frac{ \rho c^3}{\mu \omega^2}\]

Since there is a frequency dependence, we must decide what frequency we want to use as a gauge. Typical underwater acoustic frequencies range from 10Hz to 1MHz, though for scale, let's talk about this range. Humans can hear 20Hz to 20kHz on a good day, and bats can apparently hear up to 200kHz. Underwater, porpoises can hear up to 160kHz, and herring can only hear up to 4kHz. Ultrasound (like that used in the medical field, or for nondestructive evaluation of concrete, metal, etc), is usually in the range of 2 to 18MHz, though apparently can go up to single digit GHz for specialized projects. But, in keeping with our underwater acoustics desire, let's do 1kHz, since this this is in the middle of the underwater acoustics range (in a log-scale sense - the linear average of 100kHz and 10Hz is obviously about 50kHz, whereas in a log-scale, the average is 1kHz). That said, for 1kHz sound (which is apparently about C3 / Soprano C / High C on a piano) in a typical sea water environment, the distance to half the pressure strictly through viscous effects is about 70,000 miles, or just under 3 earth circumferences. Even if we bump the frequency up to 100kHz, the pressure would be half what we expected at a distance of about 7 miles / 11 km, which for realistically loud sources, is just about the limit that we could detect theoretically without any friction at all.

Basically, we went through all this to justify our elimination of $\mu$ (and $\xi$) from the Navier-Stokes equation - which for the mediocre 1kHz frequency at the fairly substantial 10km range, the pressures we model will differ from the actual pressures by 0.003%. There are other assumptions we'll make later that create much more error, so this should be considered a pretty safe assumption for underwater acoustics.

## Linearize Navier-Stokes Equation

After crossing out lots of terms from the last section, we arrive out the simplified Navier-Stokes and continuity equations: \[ \rho \left( \frac{\partial \textbf{v}}{\partial t} + \textbf{v}\cdot\nabla \textbf{v}\right) = -\nabla p \]\[ \frac{\partial \rho}{\partial t} + \nabla\cdot\left(\rho \textbf{v}\right) = 0\]These simplifications to Navier-Stokes are called the Euler equations. Notice the nonlinear terms in this coupled nonlinear partial differential equation (basically a worst case scenario in terms of possible differential equations). This equation is, in general, very difficult to solve, and analytic answers may not even be possible (hence the field of CFD - computational fluid dynamics). However, acoustics usually involves very small pressure, density, and velocity fluctuations. For this reason, we can linearize the Navier-Stokes equation to arrive at a more convenient form. Let's make the assumption:

\[ p(\textbf{r},t) = p_0(\textbf{r}) + p_1(\textbf{r},t) \] \[\textbf{v}(\textbf{r},t) = \textbf{v}_0 (\textbf{r}) + \textbf{v}_1(\textbf{r},t)\] \[\rho(\textbf{r},t) = \rho_0 (\textbf{r}) + \rho_1(\textbf{r},t)\]

Where each of the "1" terms are infinitesimal compared to the corresponding "0" terms. Also, each of the "1" terms have a mean (in time) of zero, whereas the "0" terms are constant in time. Spatially, all the "0" terms should be slowly varying (with respect to a wavelength that hasn't been defined yet), whereas the "1" terms can vary however they like. Lastly, we assume the fluid is, on average, stationary, or at least approximately so, allowing us to eliminate $\textbf{v}_0$. With these assumptions in mind, plugging them into the Navier-Stokes and continuity equations, we get:

\[ (\rho_0+\rho_1) \left( \frac{\partial \textbf{v}_1}{\partial t} + \textbf{v}_1\cdot\nabla \textbf{v}_1\right) = -\nabla (p_0+p_1) \] \[ \frac{\partial }{\partial t}(\rho_0 + \rho_1) + \nabla\cdot\left((\rho_0+\rho_1) \textbf{v}_1\right) = 0\]

Because of the infinitesimal nature of the "1" terms, we can eliminate anything quadratic in "1" terms (hence why this procedure is called linearization). Considering all the assumptions made before, we arrive at:

\[ \rho_0 \frac{\partial \textbf{v}_1}{\partial t} +\nabla p_1=0 \] \[ \frac{\partial \rho_1}{\partial t} + \nabla\cdot\left(\rho_0 \textbf{v}_1\right) = 0\]

Note that at this point, I have two equations to solve for three variables: $\textbf{v}_1$, $\rho_1$, and $p_1$. I need a third equation, which will come in the form of a constitutive relation relating $\rho$ to $p$. If I expand $\rho$ as a function of $p$ in a Taylor series, and, in keeping with the linearity, keep only the first term, I'd get:

\[ \rho(p) = \rho_0 + \left(\frac{d\rho}{d p}\right)_{p=p_0}(p-p_0) + O((p-p_0)^2) \]

I won't derive it here, but I could argue through a dimensional analysis that $\left(\frac{d\rho}{dp}\right)_{S} = \frac{1}{c^2}$. Therefore, just keeping the first order, I get: $p_1 = c^2 \rho_1$. This is my last equation that will allow me to rearrange and solve.

\[ p(\textbf{r},t) = p_0(\textbf{r}) + p_1(\textbf{r},t) \] \[\textbf{v}(\textbf{r},t) = \textbf{v}_0 (\textbf{r}) + \textbf{v}_1(\textbf{r},t)\] \[\rho(\textbf{r},t) = \rho_0 (\textbf{r}) + \rho_1(\textbf{r},t)\]

Where each of the "1" terms are infinitesimal compared to the corresponding "0" terms. Also, each of the "1" terms have a mean (in time) of zero, whereas the "0" terms are constant in time. Spatially, all the "0" terms should be slowly varying (with respect to a wavelength that hasn't been defined yet), whereas the "1" terms can vary however they like. Lastly, we assume the fluid is, on average, stationary, or at least approximately so, allowing us to eliminate $\textbf{v}_0$. With these assumptions in mind, plugging them into the Navier-Stokes and continuity equations, we get:

\[ (\rho_0+\rho_1) \left( \frac{\partial \textbf{v}_1}{\partial t} + \textbf{v}_1\cdot\nabla \textbf{v}_1\right) = -\nabla (p_0+p_1) \] \[ \frac{\partial }{\partial t}(\rho_0 + \rho_1) + \nabla\cdot\left((\rho_0+\rho_1) \textbf{v}_1\right) = 0\]

Because of the infinitesimal nature of the "1" terms, we can eliminate anything quadratic in "1" terms (hence why this procedure is called linearization). Considering all the assumptions made before, we arrive at:

\[ \rho_0 \frac{\partial \textbf{v}_1}{\partial t} +\nabla p_1=0 \] \[ \frac{\partial \rho_1}{\partial t} + \nabla\cdot\left(\rho_0 \textbf{v}_1\right) = 0\]

Note that at this point, I have two equations to solve for three variables: $\textbf{v}_1$, $\rho_1$, and $p_1$. I need a third equation, which will come in the form of a constitutive relation relating $\rho$ to $p$. If I expand $\rho$ as a function of $p$ in a Taylor series, and, in keeping with the linearity, keep only the first term, I'd get:

\[ \rho(p) = \rho_0 + \left(\frac{d\rho}{d p}\right)_{p=p_0}(p-p_0) + O((p-p_0)^2) \]

I won't derive it here, but I could argue through a dimensional analysis that $\left(\frac{d\rho}{dp}\right)_{S} = \frac{1}{c^2}$. Therefore, just keeping the first order, I get: $p_1 = c^2 \rho_1$. This is my last equation that will allow me to rearrange and solve.

UNDER CONSTRUCTION!!!