%% LyX 1.3 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[12pt,letterpaper,american]{book}
\usepackage[T1]{fontenc}
\usepackage[latin1]{inputenc}
\usepackage{graphicx}

\makeatletter

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% LyX specific LaTeX commands.
%% Bold symbol macro for standard LaTeX users
\providecommand{\boldsymbol}[1]{\mbox{\boldmath $#1$}}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
% make units in Roman type with space before
\newcommand{\unit}[1]{\mbox{ #1}}

\usepackage{babel}
\makeatother
\begin{document}

\chapter{Fluid equations}


\section{Stress}

Two kinds of forces act on molecules in a fluid, short-range inter-molecular
forces, and long-range forces such as gravity. The two types of forces
require very different treatment in the fluid dynamical governing
equations.

Let us consider the intermolecular forces acting across the faces
of a tetrahedron of fluid, as shown in figure \ref{cap:traction}.
Think of the tetrahedron as a corner cut off of a cube. We assume
that this chunk of fluid is embedded in a larger fluid body. The force
per unit area on the $y-z$ face of the tetrahedron is $\mathbf{t}_{x}$,
on the $x-z$ face is $\mathbf{t}_{y}$, and on the $x-y$ face is
$\mathbf{t}_{z}$. On the oblique face it is simply $\mathbf{t}$.
The corresponding areas of each face are $A_{x}$, $A_{y}$, $A_{z}$,
and $A$. Assuming that the tetrahedron is small, so that the force
per unit area doesn't vary much over each face, Newton's second law
in the static case with no other forces acting tells us that\begin{equation}
A_{x}\mathbf{t}_{x}+A_{y}\mathbf{t}_{y}+A_{z}\mathbf{t}_{z}+A\mathbf{t}=0.\label{eq:tractionbal}\end{equation}


Noting that $A_{x}$ is just the projection of $A$ on the $y-z$
plane, we find that $A_{x}/A=\hat{\mathbf{x}}\cdot\hat{\mathbf{n}}$,
where $\hat{x}$ is the unit vector in the $x$ direction and $\hat{\mathbf{n}}$
is the outward normal unit vector to the oblique surface. Similar
relations hold for the other surfaces, resulting in\begin{equation}
\mathbf{t}=-(\mathbf{t}_{x}\hat{\mathbf{x}}+\mathbf{t}_{y}\hat{\mathbf{y}}+\mathbf{t}_{z}\hat{\mathbf{z}})\cdot\hat{\mathbf{n}}\equiv\mathbf{T}\cdot\hat{\mathbf{n}}.\label{eq:traction}\end{equation}
The quantity $\mathbf{T}$ is called the \emph{stress tensor}.

Note a convention that we have implicitly established: the sense of
the force per unit area $\mathbf{t}$ across the oblique surface is
that the fluid penetrated by the unit normal $\hat{\mathbf{n}}$ acts
on the fluid on the other side of the surface. This force per unit
area is called the \emph{traction} across the surface. Note further
that $\mathbf{t}_{x}$ indicates the traction across the $y-z$ plane,
not the $x$ component of a traction. Representation of components
requires a second subscript: $t_{yx}$ is the $y$ component of the
traction across the $y-z$ face of the tetrahedron, whereas $t_{xy}$
is the $x$ component of the traction across the $x-z$ face. Thus,
the first subscript represents the vector component of the force and
the second subscript indicates the face on which the force is acting.

%
\begin{figure}
\begin{center}\includegraphics[%
  scale=0.6]{traction.eps}\end{center}


\caption{\label{cap:traction} Molecular forces on a tetrahedron of fluid.}
\end{figure}


We now demonstrate that the stress tensor is symmetric. In matrix
form the stress tensor is\begin{equation}
\mathbf{T}=-\left(\begin{array}{ccc}
t_{xx} & t_{xy} & t_{xz}\\
t_{yx} & t_{yy} & t_{yz}\\
t_{zx} & t_{zy} & t_{zz}\end{array}\right),\label{eq:stressmatrix}\end{equation}
given the above convention on indices. Imagine a cube of fluid small
enough that the stress tensor does not vary significantly over the
cube. The shear or face-parallel components of the tractions on the
$x-z$ and $y-z$ faces are shown in figure \ref{cap:torque}, the
normal components being omitted for clarity. Recall that we assume
a static situation with the tractions being the only forces applied
to the cube. Under these conditions the torque on the cube must be
zero. However, computing the torque about the center of the cube quickly
leads us to the conclusion that $T_{xy}=T_{yx}$, etc. (The normal
components of the traction do not contribute to the torque.) 

%
\begin{figure}
\begin{center}\includegraphics[%
  scale=0.6]{torque.eps}\end{center}


\caption{\label{cap:torque} Shear tractions acting on a cube.}
\end{figure}


We now relax the assumptions of static equilibrium and no other forces.
In this case the mass of the tetrahedron in figure \ref{cap:traction}
times the acceleration equals the total traction force plus the long
range or body force on the tetrahedron. If $L$ is a typical dimension
of the tetrahedron, then the mass scales as $L^{3}$. The body force
should also scale as $L^{3}$, since it should have a magnitude proportional
to the total number of molecules in the tetrahedron. However, the
traction forces scale as $L^{2}$, since they are proportional to
the area of the tetrahedron faces. Therefore the ratio of the traction
forces to the mass times the acceleration or to the body force scales
as $L^{-1}$. If we now take the limit of very small $L$, it is clear
that the traction forces dominate over the other two terms. Thus,
for an infinitesimal tetrahedron, the analysis for static conditions
with no body forces carries over to the more general case.


\section{Newtonian fluids}

In fluid dynamics we generally take what is called the \emph{Eulerian}
point of view, in which we consider our dependent variables such as
fluid velocity and density to be a function of position rather than
being identified with the individual parcels making up the fluid.
Thus, the velocity at some point refers at different times to different
parcels of fluids. The alternative point of view, in which variables
are referenced to fluid parcels, is called the \emph{Lagrangian} point
of view. It has occasional uses in fluid dynamics, but is much less
commonly seen than the Eulerian representation, which we use here
exclusively.

We now investigate the ways in which the fluid velocity $\mathbf{v}(\mathbf{x})$
can vary in the vicinity of some test point $\mathbf{x}_{0}$. Representing
vectors in component notation, we make a Taylor series expansion in
$v_{i}$ about the point $x_{0i}$:\begin{equation}
v_{i}(x_{j})=v_{i}(x_{0j})+\frac{\partial v_{i}}{\partial x_{j}}\delta x_{j},\label{eq:taylorvelocity}\end{equation}
where $\delta x_{j}=x_{j}-x_{0j}$ and where we use the Einstein convention
that repeated indices are summed.

The quantity $\partial v_{i}/\partial x_{j}$ is called the \emph{deformation
rate tensor}. It can be split into symmetric and antisymmetric parts\begin{equation}
\frac{\partial v_{i}}{\partial x_{j}}=\frac{1}{2}\left(\frac{\partial v_{i}}{\partial x_{j}}+\frac{\partial v_{j}}{\partial x_{i}}\right)+\frac{1}{2}\left(\frac{\partial v_{i}}{\partial x_{j}}-\frac{\partial v_{j}}{\partial x_{i}}\right)\equiv D_{ij}+R_{ij},\label{eq:defrate}\end{equation}
where $D_{ij}$ is called the \emph{strain rate tensor} and $R_{ij}$
is the \emph{rotation tensor}. To understand the origin of the latter
name, suppose that the fluid is locally in rigid body rotation about
the point $\mathbf{x}_{0}$. In this case the fluid velocity field
will locally take the form $\mathbf{v}=\mathbf{v}_{0}+\delta\mathbf{x}\times\Omega$,
where $\Omega$ is the angular rotation rate vector. The cross product
can be written in component form as $\epsilon_{ijk}\delta x_{j}\Omega_{k}$
where $\epsilon_{123}=\epsilon_{231}=\epsilon_{312}=1$, $\epsilon_{132}=\epsilon_{321}=\epsilon_{213}=-1$,
and $\epsilon_{ijk}$ equals zero if any two indices are zero. The
deformation rate tensor thus becomes\begin{equation}
\frac{\partial v_{i}}{\partial x_{j}}=\epsilon_{ijk}\Omega_{k},\label{eq:purerot}\end{equation}
which is antisymmetric. The strain rate tensor is therefore zero in
this case, and the deformation rate tensor equals the rotation tensor.
Furthermore, $\Omega$ is the dual vector of the rotation tensor.

In the general case, the rotation tensor represents the rotational
part of the local deformation while the strain rate tensor represents
that part of the deformation which changes the shape of the local
fluid element.

A Newtonian fluid is one in which the components of the stress tensor
are related linearly to the components of the strain rate tensor.
This relationship cannot be arbitrary, but must be expressible in
terms of a tensor expression, which makes the relationship independent
of coordinate system, and thus physically viable. This condition puts
strong limits on the possibilities, and the most general linear relationship
is\begin{equation}
T_{ij}=A\delta_{ij}+BD_{kk}\delta_{ij}+CD_{ij},\label{eq:linearstressstrain}\end{equation}
where $A$, $B$, and $C$ are scalars which may be functions of fluid
properties.

In the static case, $\mathbf{D}=0$ and only the first term on the
right side of equation (\ref{eq:linearstressstrain}) is important.
A traction $\mathbf{t}=\mathbf{T}\cdot\hat{\mathbf{n}}=A\hat{\mathbf{n}}$
occurs in this case. Experimentally, an inward pressure force is generally
needed to keep a fluid parcel in place in this situation. Since $\mathbf{n}$
points outward, that means that $A$ must be negative. We therefore
write $A=-p$, where $p$ is the pressure, i. e., the force per unit
area needed to confine the fluid parcel. By convention (the reasons
for which we will explore later), we write $B=\eta-2\mu/3$ and $C=2\mu$,
where $\mu$ is the \emph{coefficient of viscosity} and $\eta$ is
the \emph{second coefficient of viscosity}. Thus, we rewrite equation
(\ref{eq:linearstressstrain}) as\begin{equation}
T_{ij}=-p\delta_{ij}+(\eta-2\mu/3)\delta_{ij}D_{kk}+2\mu D_{ij}.\label{eq:newtonfluid}\end{equation}



\section{Mass continuity}

%
\begin{figure}
\begin{center}\includegraphics[%
  scale=0.6]{fluxbox.eps}\end{center}


\caption{\label{cap:fluxbox} Side view of a stationary 3-D cube of side $h$
with fluid flowing through it.}
\end{figure}


Referring to figure \ref{cap:fluxbox}, the mass of fluid per unit
time flowing into the cube from the left side is $\rho(x)v_{x}(x)h^{2}$,
whereas the mass flowing out of the right side is $\rho(x+h)v_{x}(x+h)h^{2}$,
where $\rho$ is the mass density of the fluid and $v_{x}$ is the
$x$ component of the fluid velocity. Similar expressions can be developed
for flow of mass in and out of the other faces of the cube.

The net flow of mass per unit time out of the box from the left and
right faces is\[
\left(\frac{\rho(x+h)v_{x}(x+h)-\rho(x)v_{x}(x)}{h}\right)h^{3}\approx\frac{\partial\rho v_{x}}{\partial x}h^{3}.\]
Since the cube is fixed in volume and location, the time rate of change
of mass in the box is $(\partial\rho/\partial t)h^{3}$. Equating
this to minus the rate at which mass is flowing out of the cube through
all six faces results in\[
\frac{\partial\rho}{\partial t}h^{3}\approx-\frac{\partial\rho v_{x}}{\partial x}h^{3}-\frac{\partial\rho v_{y}}{\partial y}h^{3}-\frac{\partial\rho v_{z}}{\partial z}h^{3}.\]
The approximation becomes exact when $h$ is allowed to become very
small, resulting in the \emph{mass continuity equation} when $h^{3}$
is canceled:\begin{equation}
\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})=0.\label{eq:masscont}\end{equation}


An alternate form of this equation may be obtained by noting that
$\nabla\cdot(\rho\mathbf{v})=\mathbf{v}\cdot\nabla\rho+\rho\nabla\cdot\mathbf{v}$:\begin{equation}
\frac{d\rho}{dt}+\rho\nabla\cdot\mathbf{v}=0,\label{eq:masscont2}\end{equation}
where\begin{equation}
\frac{d\rho}{dt}\equiv\frac{\partial\rho}{\partial t}+\frac{d\mathbf{x}}{dt}\cdot\nabla\rho=\frac{\partial\rho}{\partial t}+\mathbf{v}\cdot\nabla\rho\label{eq:parcelderiv}\end{equation}
is the time derivative of the density following a parcel in its flow.
This is sometimes called the \emph{parcel derivative} or the \emph{material
derivative}. It can also be applied to quantities other than the density.


\section{Momentum}

The corresponding equation for momentum is more complex in two respects:
first, momentum is a vector, so the flux of momentum per unit area
per unit time $\rho\mathbf{vv}$ is a tensor; second, forces as well
as mass transport can change the amount of momentum in the cube. The
momentum balance in the cube is therefore governed by\begin{equation}
\left(\frac{\partial\rho\mathbf{v}}{\partial t}+\nabla\cdot(\rho\mathbf{vv})\right)h^{3}=\mbox{{force on cube}}.\label{eq:ns1}\end{equation}


The stress force on the left and right $x$ faces of the cube is\begin{equation}
[\mathbf{T}(x)\cdot(-\hat{\mathbf{x}})+\mathbf{T}(x+h)\cdot(\hat{\mathbf{x}})]h^{2}\approx\frac{\partial\mathbf{T}}{\partial x}\cdot\hat{\mathbf{x}}h^{3},\label{eq:ns2}\end{equation}
with similar expressions for the $y$ and $z$ faces. The total force
is therefore $\nabla\cdot\mathbf{T}h^{3}$. The earth exerts a gravitational
force on the mass in the cube. In addition, the earth is rotating,
resulting in a centrifugal force and a Coriolis force in the rotating
reference frame of the earth. We combine gravity and centrifugal force
per unit mass into an effective gravity $\mathbf{g}$. The Coriolis
force per unit mass is given by $-2\Omega\times\mathbf{v}$, where
$\Omega$ is the rotation vector for the earth, i. e., it has magnitude
equal to $2\pi$ divided by the earth's rotation period and direction
parallel to the rotation axis of the earth.

Combining all of these effects and using the stress for a Newtonian
fluid given in equation (\ref{eq:newtonfluid}) results in the following
\emph{momentum equation}:\begin{equation}
\frac{\partial\rho\mathbf{v}}{\partial t}+\nabla\cdot(\rho\mathbf{vv}-\mathbf{T})=\rho\mathbf{g}-2\rho\Omega\times\mathbf{v}.\label{eq:momentum}\end{equation}
To a very good approximation in atmospheric convection, $\eta$ and
$\mu$ can be considered constants. Furthermore, since atmospheric
flow is highly subsonic, it is nearly incompressible on small spatial
scales over which viscous effects are important, which means that
terms involving both the divergence of the velocity and viscous constants
can be ignored. Under these conditions the equations simplify to\begin{equation}
\frac{\partial\rho\mathbf{v}}{\partial t}+\nabla\cdot(\rho\mathbf{vv})+\nabla p-\rho\mathbf{g}+2\rho\Omega\times\mathbf{v}-\mu\nabla^{2}\mathbf{v}=0.\label{eq:momentum2}\end{equation}


An alternate form of this equation may be obtained by noting that\begin{equation}
\frac{\partial\rho\mathbf{v}}{\partial t}+\nabla\cdot(\rho\mathbf{vv})=\rho\left(\frac{\partial\mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}\right)+\left(\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})\right)\mathbf{v}.\label{eq:advexpand}\end{equation}
The second term on the right side is zero by virtue of the mass continuity
equation (\ref{eq:masscont}), which means that equation (\ref{eq:momentum2})
can be rewritten\begin{equation}
\frac{d\mathbf{v}}{dt}+\frac{1}{\rho}\nabla p-\mathbf{g}+2\Omega\times\mathbf{v}-\nu\nabla^{2}\mathbf{v}=0,\label{eq:momentum3}\end{equation}
where $\nu\equiv\mu/\rho$ is the \emph{kinematic viscosity}.


\section{Incompressible fluid without diffusion}

We now consider an incompressible fluid, which nevertheless may vary
in density with position. An example of such a fluid is salty water
with the value of the salinity varying from parcel to parcel. Ignoring
diffusion for now, the density obeys the equation\begin{equation}
\frac{d\rho}{dt}=0,\label{eq:incomp}\end{equation}
which simply indicates that the density of a parcel doesn't change.
Combining this with the mass continuity equation (\ref{eq:masscont2})
yields the condition\begin{equation}
\nabla\cdot\mathbf{v}=0.\label{eq:incomp2}\end{equation}


We now have all the ingredients needed to compute the evolution of
an incompressible fluid. However, in many cases the density variations
in a fluid are fractionally small, which allows certain simplifying
approximations to be made. Let us assume that $\rho=\rho_{0}+\rho'$,
where $\rho_{0}$ is a constant reference density and $\rho'$ is
small in magnitude compared to $\rho_{0}$. The incompressibility
equation therefore becomes\begin{equation}
\frac{d\rho'}{dt}=0.\label{eq:incompprime}\end{equation}


The application of this approximation to the momentum equation is
slightly more complex. In an equilibrium situation with a fluid at
rest, equation (\ref{eq:momentum3}) reduces to hydrostatic balance:\begin{equation}
\frac{1}{\rho}\nabla p+g\hat{\mathbf{z}}=0.\label{eq:hydrostatic}\end{equation}
The horizontal component of this equation tells us that the horizontal
gradient of pressure is zero, i. e., pressure is a function only of
$z$. The vertical component obeys\begin{equation}
\frac{1}{\rho}\frac{dp}{dz}+g=0,\label{eq:hydrostatic2}\end{equation}
where the conversion to a total $z$ derivative reflects the lack
of dependence of $p$ on $x$ and $y$.

We now obtain the pressure distribution $p_{0}(z)$ hydrostatically
consistent with our constant reference density $\rho_{0}$:\begin{equation}
p_{0}(z)=g\rho_{0}(z_{0}-z)\label{eq:hydrostatic3}\end{equation}
where $z_{0}$ is the elevation at which $p_{0}=0$. Assuming that
$p=p_{0}(z)+p'$, where the magnitude of $p'$ is taken to be much
less than $p_{0}$, then the pressure gradient and gravity terms can
be approximated by\begin{equation}
\frac{1}{\rho_{0}+\rho'}\nabla(p_{0}+p')-\mathbf{g}\approx\frac{1}{\rho_{0}}\nabla p_{0}-\mathbf{g}+\frac{1}{\rho_{0}}\nabla p'-\frac{\rho'}{\rho_{0}^{2}}\nabla p_{0}=\nabla\pi-b\hat{\mathbf{z}},\label{eq:hydrostatic4}\end{equation}
where we have dropped nonlinear terms in primed quantities, and where
the \emph{kinematic pressure} is defined $\pi\equiv p'/\rho_{0}$
and where the \emph{buoyancy} is given by $b=-g\rho'/\rho_{0}$. The
momentum equation thus becomes\begin{equation}
\frac{d\mathbf{v}}{dt}+\nabla\pi-b\hat{\mathbf{z}}+2\Omega\times\mathbf{v}=0\label{eq:momentum4}\end{equation}
where we have ignored viscosity as well as diffusion. Equation (\ref{eq:incompprime})
can be rewritten in terms of the buoyancy:\begin{equation}
\frac{db}{dt}=0.\label{eq:buoyancygov}\end{equation}


The approximations used in equations (\ref{eq:incomp2}), (\ref{eq:momentum4}),
and (\ref{eq:buoyancygov}) are jointly called the \emph{Boussinesq
approximation}.


\section{Ideal gas without diffusion}

For the subsonic, non-turbulent flow of an ideal gas with no radiative
processes or latent heat release, the specific entropy $s$ written
in terms of temperature $T$ and pressure $p$\begin{equation}
s=C_{p}\ln(T/T_{0})-R\ln(p/p_{0})\label{eq:dryentropy}\end{equation}
is conserved to a high degree of accuracy. The constants $C_{p}$
and $R$ are the specific heat of the gas at constant pressure and
the universal gas constant divided by the molecular weight of the
gas and $T_{0}$ and $p_{0}$are constant reference values of temperature
and pressure. Again ignoring diffusive processes, this implies\begin{equation}
\frac{ds}{dt}=0.\label{eq:entgov}\end{equation}


A related variable is the \emph{potential temperature}\begin{equation}
\theta\equiv T_{0}\exp(s/C_{p})=T(p_{0}/p)^{\kappa},\label{eq:pottemp}\end{equation}
where $\kappa=R/C_{p}$. The potential temperature is also conserved
by parcels under these conditions,\begin{equation}
\frac{d\theta}{dt}=0,\label{eq:pottempgov}\end{equation}
and it can be thought of as a surrogate for the entropy. By using
the ideal gas law $p/\rho=RT$, we can rewrite the pressure term in
the momentum equation:\begin{equation}
\frac{1}{\rho}\nabla p=\frac{RT}{p}\nabla p=\frac{R\theta}{p}(p/p_{0})^{\kappa}\nabla p=\theta C_{p}\nabla(p/p_{0})^{\kappa}\equiv\theta\nabla\Pi.\label{eq:exnerdef}\end{equation}
The quantity $\Pi=C_{p}(p/p_{0})^{\kappa}$ is called the \emph{Exner
function}.

The Boussinesq approximation can be made in the case of an ideal gas.
Density variations in the earth's atmosphere are related primarily
to the hydrostatic variation in pressure with height and with sound
waves. Subsonic flows and waves with vertical scales small compared
to the vertical scale of hydrostatic pressure changes still exhibit
density variations, but these variations are small. Thus, we can ignore
$d\rho/dt$ compared to the individual terms of $\rho\nabla\cdot\mathbf{v}$,
allowing us to use equation (\ref{eq:incomp2}).

For the momentum equation we assume that $\theta=\theta_{0}+\theta'$,
where $\theta_{0}$ is a constant reference potential temperature,
and $\Pi=\Pi_{0}+\Pi'$, where $\Pi_{0}(z)$ is hydrostatically related
to $\theta_{0}$, i. e.,\begin{equation}
\theta_{0}\nabla\Pi_{0}+g\hat{\mathbf{z}}=0.\label{eq:hydrostatic5}\end{equation}
Following the same procedure used for the incompressible fluid case,
we come up with the following momentum equation:\begin{equation}
\frac{d\mathbf{v}}{dt}+\nabla(\theta_{0}\Pi')-\frac{g\theta'}{\theta_{0}}+2\Omega\times\mathbf{v}=0.\label{eq:momentum5}\end{equation}


If we define $\pi=\theta_{0}\Pi'$ and $b=g\theta'/\theta_{0}$, then
equation (\ref{eq:momentum5}) takes precisely the same form as equation
(\ref{eq:momentum4}). Starting from equation (\ref{eq:pottempgov}),
it is easy to show that the buoyancy as defined here also satisfies
equation (\ref{eq:buoyancygov}). Thus, under the conditions employed
in this section, an ideal gas obeys the same set of equations as an
incompressible fluid under the Boussinesq approximation. This demonstrates
that the ocean and the atmosphere exhibit essentially similar behavior
in many circumstances.


\subsection*{Problems}

\begin{enumerate}
\item Show by a scale analysis that the argument for the symmetry of the
stress tensor carries over to the non-static case with body forces
when the cube in figure \ref{cap:torque} is of infinitesimal size.
\item Given a fluid velocity field $\mathbf{v}=Sy\hat{\mathbf{x}}$ where
$S$ is a constant:

\begin{enumerate}
\item Compute the strain rate and rotation rate tensors.
\item Find the rotation vector associated with the rotation rate tensor.
\item Find the principal axes and eigenvalues associated with the strain
rate tensor.
\item Find the viscous part of the stress tensor assuming a Newtonian fluid.
\end{enumerate}
\item Appealing to standard classical mechanics, write separately the gravitational
and inertial components of $\mathbf{g}$ in equation (\ref{eq:momentum})
in terms of the actual gravitational field, the rotation vector of
the earth, and the radius vector $\mathbf{r}$ relative to the center
of the earth. Use this to show that if the rotation rate of the earth
changes, then the definition of {}``horizontal'' changes as well.
\item Suppose for a fluid $\partial v_{x}/\partial t=0$, but $v_{x}=Cx$,
where $C$ is a constant. Find the $x$ component of the total force
per unit mass on the fluid. Assuming further that $v_{y}=v_{z}=0$,
find $d\ln\rho/dt$.
\item Suppose an ideal gas is at rest in a uniform gravitational field.

\begin{enumerate}
\item If the gas has constant entropy with height, determine how the pressure
and temperature vary with height.
\item Determine how the pressure and entropy vary with height if the temperature
is constant with height.
\end{enumerate}
\item Suppose instead that the gravitational field obeys $g=g_{0}(r_{0}/r)^{2}$,
where $g_{0}$ is the gravitational field at radius $r_{0}$. Determine
the distribution of pressure with radius $r$ for an ideal gas atmosphere
of constant temperature.
\end{enumerate}

\end{document}
