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\begin{center}
\noindent{\bf LECTURE 3: Fluid Statics}
\end{center}
\vspace*{.1in}
We begin by considering static fluid configurations, for which the
stress tensor reduces to the form ${\bf T} = -p {\bf I}$, so that
${\bf n} \cdot {\bf T} \cdot {\bf n} = p$, and the
normal stress balance assumes the form:
\begin{equation}
p - \hat{p} ~=~ \sigma ~\nabla \cdot {\bf n}
\end{equation}
The pressure jump across the interface is balanced by the curvature
force at the interface.
Now since ${\bf n} \cdot {\bf T} \cdot {\bf s} = 0$ for a static system , the tangential stress
balance equation indicates that: $0 = \nabla \sigma$. This leads us to the following
important conclusion:
{\it There cannot be a static system in the presence of surface tension gradients}.
While pressure jumps can sustain normal stress jumps across a fluid interface,
they do not contribute to the tangential stress jump. Consequently, tangential
surface stresses can only be balanced by viscous stresses associated with
fluid motion.
We proceed by applying equation (1) to a number of static situations.
\vspace*{.2in}
\noindent{\bf 3.1 Stationary bubble}
\vspace*{.1in}
We consider a spherical bubble of radius $R$ submerged in a static fluid.
What is the pressure drop across the bubble surface?
\vspace*{.2in}
The curvature of the spherical surface is simply computed:
\begin{equation}
\nabla \cdot {\bf n} ~=~ \nabla \cdot {\bf r} ~=~\frac{1}{r^2}\frac{\partial}{\partial r} (r^2) ~=~\frac{2}{R}
\end{equation}
so the normal stress jump (1) indicates that
\begin{equation}
\hat{p} - p ~=~ \frac{2 \sigma}{R} ~~~.
\end{equation}
The pressure within the bubble is higher than that outside by an amount proportional
to the surface tension, and inversely proportional to the bubble size. It is thus that small
bubbles are louder than large ones when they burst at a free surface: champagne is
louder than beer. We note that soap bubbles in air have two surfaces that define the
inner and outer surfaces of the soap film; consequently, the pressure differential is
twice that across a single interface.
\vspace*{.2in}
\noindent{\bf 3.2 Static meniscus}
\vspace*{.1in}
\begin{figure}
\begin{center}
\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
%\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
\includegraphics[width = 0.7\textwidth]{log2}
\end{center}
\caption{A definitional sketch of a planar meniscus at an air-water interface.
The free surface is defined by $z = \eta(x)$, varying from its maximum elevation at
its point of contact with the wall ($x=0$) to zero at large $x$. The shape is prescribed
by the Young-Laplace equation.}
\label{fig2}
\end{figure}
Consider a situation where the pressure within a static fluid varies owing to
the presence of a gravitational
field, $p = p_0 + \rho g z$, where $p_0$ is the constant ambient pressure,
and $\vec{g}=-g\hat{z}$ is the gravitational acceleration.
The normal stress balance thus requires that the interface satisfy the
{\it Young-Laplace Equation}:
\begin{equation}
\rho g z ~=~ \sigma ~\nabla \cdot {\bf n}~~.
\end{equation}
The vertical gradient in fluid pressure must be balanced by the curvature
pressure; as the gradient is constant, the curvature must likewise increase
linearly with z. Such a situation arises in the static meniscus (see Figure 1).
\vspace*{.2in}
The shape of the meniscus is prescribed by two factors: the
contact angle between the air-water interface and the log, and the
balance between hydrostatic pressure and curvature pressure. We treat
the contact angle, $\theta$, as given; it depends on the physics of the
log-water-air interaction. The normal force balance is expressed by the
Young-Laplace equation, where now $\rho = \rho_w - \rho_a \approx \rho_w$
is the density difference between water and air.
We define the free surface by $z = \eta(x)$; equivalently, we define a
functional $f(x,z) = z - \eta(x)$ that vanishes on the surface. The
normal to the surface is thus
\begin{equation}
{\bf n} ~=~ \frac{\nabla f}{|\nabla f|} ~=~ \frac{\hat{z} - \eta'(x) \bf{x}}{(1 + \eta'(x)^2)^{1/2}}
\end{equation}
As deduced in Appendix A, the curvature of the free
surface, $\nabla \cdot \hat{n}$, may be expressed as
\begin{equation}
\nabla\cdot\hat{n} = \frac{\eta_{xx}}{(1+\eta_x^2)^{3/2}} \approx
\eta_{xx} .
\end{equation}
Assuming that the slope of the meniscus remains small, $\eta_x <<1$,
allows one to linearize equation (13), so that (11) assumes the form:
\begin{equation}
\rho g \eta ~= ~ \sigma \eta_{xx} ~~.
\end{equation}
Applying the boundary condition $\eta(\infty)=0$ and the contact condition
$\eta_x(0)=-\cot\theta$, and solving (16) thus yields:
\begin{equation}
\eta (x) = \ell_c \cot(\theta) e^{-x/\ell_c} ,
\end{equation}
where $\ell_c=\sqrt{\frac{\sigma}{\rho g}}$ is the capillary
length. The meniscus formed by a log in water is exponential, dying
off on the scale of $\ell_c$.
\newpage
\noindent{\bf 3.3 Radial force on a circular hydraulic jump}
\vspace*{.1in}
%\begin{figure}[b]
%\begin{center}
%\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
%\includegraphics[width = 0.8\textwidth]{2jumps.pdf}
%\end{center}
%\caption{Circular hydraulic jumps formed by the impact of a vertical jet on a horizontal reservoir
%of fluid. We here demonstrate how to calculate the influence of surface tension on the
%jump radius.}
%\end{figure}
Hydraulic jumps may be generated when a vertical jet strikes a flat plate. The
jet spreads radially, giving rise to a fluid layer that generally thins with radius
until reaching a critical radius at which it increases dramatically (see Figure 2). We here
calculate the total radial force acting on the jump surface owing to the curvature
of the jump between points A and B, located at radii $R_1$ and $R_2$,
respectively. Assume that the points A and B are the points nearest the jump,
respectively, upstream and downstream, at which the slope of the surface
vanished identically.
\begin{figure}
\begin{center}
\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
%\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
\includegraphics[width = 0.8\textwidth]{defn}
\end{center}
\caption{A schematic illustration of the geometry of the circular hydraulic jump.
A jet of radius $a$ impacts a reservoir of outer depth $H$ at a speed $U$. The
curvature force associated with the surface tension $\sigma$ depends only on the
geometry of the jump; specifically, on the radial distance $\Delta R = R_2 - R_1$
and arclength $s$ between the two nearest points up and downstream of the jump at
which the surface $z = h(r)$ has vanishing slope.}
\label{fig2}
\end{figure}
We know that the curvature force per unit area is $\sigma (\nabla \cdot {\bf n}) {\bf n}$.
We must integrate the radial component of this force over the jump surface:
\begin{equation}
F_c ~=~ \sigma \int \int_S ~\nabla \cdot {\bf n} ~({\bf n} \cdot \hat{{\bf r}})~ dS
\end{equation}
Now defining the surface as $z = h(r)$ allows us to express the area element as
\begin{equation}
dS ~=~ r ~d\theta (dr^2 + dz^2)^{1/2} ~=~ r ~d\theta ~(1 + h_r)^{1/2} ~dr~,
\end{equation}
so that our radial force assumes the form
\begin{equation}
F_c ~=~ \sigma \int_0^{2\pi} \int_{R_1}^{R_2} ~\nabla \cdot {\bf n} ~({\bf n} \cdot \hat{{\bf r}})~ r ~(1 + h_r)^{1/2} ~dr~d\theta ~.
\end{equation}
Now we use the appropriate forms for ${\bf n}$ and $\nabla \cdot {\bf n}$, respectively,
equations (4.5) and (4.6):
\begin{equation}
{\bf n} ~=~ \frac{\hat{z} - h_r \hat{r}}{(1 + h_r^2)^{1/2}} ~~~,~~~ \nabla \cdot {\bf n} ~=~ -\frac{1}{r}
\frac{d}{dr} ~\frac{r h_r}{(1 + h_r^2)^{1/2}} ~~~~,
\end{equation}
so that
\begin{equation}
(\nabla \cdot {\bf n})({\bf n} \cdot {\hat{\bf r}}) ~=~ ~\frac{h_r}{(1 + h_r^2)^{1/2}}~ \frac{1}{r}
\frac{d}{dr} ~\frac{r h_r}{(1 + h_r^2)^{1/2}}
\end{equation}
and the radial curvature force becomes
\begin{equation}
F_c~=~2 \pi \sigma \int_{R_1}^{R_2}~\frac{h_r}{(1 + h_r^2)^{1/2}}~ \frac{1}{r} \frac{d}{dr} \left( \frac{r h_r}{(1 + h_r^2)^{1/2}} \right) ~ r ~(1 + h_r)^{1/2} ~dr ~
\end{equation}
\begin{equation}
=~2 \pi \sigma \int_{R_1}^{R_2}~h_r~ \frac{d}{dr} \left( \frac{r h_r}{(1 + h_r^2)^{1/2}} \right) ~dr ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation}
Integrating by parts, $\int_a^b~ u dv ~=~uv|_a^b~-~\int_a^b~v~du$ with $u = h_r$, $v = \frac{rh_r}{(1+h_r^2)^{1/2}}$, yields
\begin{equation}
F_c ~=~ 2 \pi \sigma ~\left[~rh_r^2 \frac{1}{(1 + h_r^2)^{1/2}} \large|_{R_1}^{R_2} ~-~ \int_{R_1}^{R_2}
\frac{r h_r h_rr}{(1 + h_r^2)^{1/2}}~dr ~\right]
\end{equation}
\begin{equation}
= - 2 \pi \sigma ~ \int_{R_1}^{R_2} ~ \frac{r h_r h_rr}{(1 + h_r^2)^{1/2}}~dr ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{equation}
Integrating by parts again, with $u = r$, $v = (1+h_r^2)^{1/2}$, yields
\begin{equation}
F_c~=~2 \pi \sigma ~\left[ ~r (1 + h_r^2)^{1/2} |_{R_1}^{R_2} ~-~ \int_{R_1}^{R_2} (1 + h_r^2)^{1/2}
dr ~\right]
\end{equation}
\begin{equation}
= ~-2\pi \sigma \left[ (R_2 - R_1) ~-~ \int_{R_1}^{R_2} (1 + h_r^2)^{1/2} ~dr ~\right]
\end{equation}
since $h_r = 0$ at $r= R_1, R_2$ by assumption. We note also that
\begin{equation}
\int_{R_1}^{R_2} (1 + h_r^2)^{1/2} ~dr~ =~ \int_A^B (dr^2 + dz^2)^{1/2} ~=~\int_A^B dl ~=~S~,
\end{equation}
where $S$ is defined as the total arclength of the surface between points A and B. We define
$\Delta R = R_1 - R_2$ in order to obtain the simple result:
\begin{equation}
F_c ~=~ 2 \pi \sigma (S - \Delta R)~~~.
\end{equation}
We note that this relation yields reasonable results in two limits of interest. First,
for a flat interface, $S = \Delta R$, so that $F_c = 0$. Second, for an abrupt jump in
height of $\Delta H$, $\Delta R = 0$ and $S = \Delta H$, so that $F_c = 2 \pi \sigma \Delta H$.
This result is commensurate with the total force exerted by a cylindrical annulus of radius
$R$ and height $\Delta H$, which is deduced from the product of $\sigma$, the area $2 \pi R \Delta H$ and the curvature $1/R$:
\begin{equation}
F_c ~=~\sigma~(2 \pi R \Delta H~\frac{1}{R}) ~=~ 2 \pi \sigma \Delta H
\end{equation}
The influence of this force was found to be significant for small hydraulic jumps.
Its importance relative to the hydrostatic pressure in containing the jump is given by
\begin{equation}
\frac{\hbox{Curvature}}{\hbox{Gravity}} ~=~ \frac{\sigma/R}{\rho g \Delta H} ~=~ \frac{H}{R} B_0^{-1}
\end{equation}
where the Bond number is defined here as
\begin{equation}
B_0 ~=~\frac{\rho g H^2}{\sigma} ~~~.
\end{equation}
The simple result (19) was derived by Bush \& Aristoff (2003), who also presented the
results of an experimental study of circular hydraulic jumps. Their experiments indicated
that the curvature force becomes appreciable for jumps with characteristic radius of
less than 3 cm.
While the influence of surface tension on the radius of the circular hydraulic jump is
generally small, it may have a qualitative influence on the shape of the jump. In particular,
it may prompt the axisymmetry-breaking instability responsible for the polygonal
hydraulic jump structures discovered by Elegaard {\it et al.} ({\it Nature}, 1998),
and more recently studied by Bush, Hosoi \& Aristoff (2004).
See www-math.mit.edu/~jumps.html .
\vspace*{.2in}
\noindent{\bf 3.4 Floating Bodies}
\vspace*{.1in}
Floating bodies must be supported by some combination of buoyancy and
curvature forces. Specifically, since the fluid pressure beneath the interface
is related to the atmospheric pressure $P_0$ above the interface by
\[ p ~= ~P_0 ~+~ \rho g z ~+ ~ \sigma \nabla \cdot {\bf n} ~~~,\]
\noindent one may express the vertical force balance as
\begin{equation}
Mg ~=~ {\bf z} \cdot \int_c -p ~{\bf n} ~d\ell ~=~ F_b ~+~F_c~~~.
\end{equation}
\noindent The buoyancy force
\begin{equation}
F_b ~=~ {\bf z} \cdot \int_C \rho g ~z {\bf n} ~d\ell ~= ~ \rho g V_c
\end{equation}
\noindent is thus simply expressible in terms of the volume $V_c$ of fluid
displaced above the object and insight the line of tangency (Figure 3).
A simple expression for the curvature force may be deduced using the
first of the Frenet-Serret equations (see Appendix B).
\begin{equation}
F_c ~=~ {\bf z} \cdot \int_C \sigma (\nabla \cdot {\bf n}) {\bf n} ~d\ell ~= ~ \sigma {\bf z} \cdot ({\bf t_1} - {\bf t_2}) ~=~ 2 \sigma~ \hbox{sin} \theta
\end{equation}
\noindent At the interface, the buoyancy and curvature forces must balance precisely,
so the Young-Laplace relation is satisfied:
\begin{equation}
0 ~=~ \rho g z ~+ ~ \sigma \nabla \cdot {\bf n}
\end{equation}
Integrating the fluid pressure over the meniscus yields the vertical force balance:
\begin{equation}
F_b^m ~+ ~F_c^m ~= ~0~~~.
\end{equation}
\noindent where
\begin{equation}
F_b^m ~=~ {\bf z} \cdot \int_{C_m} \rho g z {\bf n} ~d\ell ~= ~ \rho g V_m
\end{equation}
\begin{equation}
F_c^m ~=~ {\bf z} \cdot \int_{C_m} \sigma (\nabla \cdot {\bf n}) {\bf n} ~d\ell ~= ~ \sigma {\bf z} \cdot ({\bf t_x} - {\bf t_2}) ~=~ - 2 \sigma~ \hbox{sin} \theta
\end{equation}
where we have again used the Frenet-Serret equations to evaluate the curvature force.
\begin{figure}
\begin{center}
\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
%\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
\includegraphics[width = 0.9\textwidth]{floater}
\end{center}
\caption{A non-wetting two-dimensional body of radius $r$ and mass $M$ floats on a free surface with
surface tension $\sigma$. In general, its weight $Mg$ must be supported by some combination of
curvature and buoyancy forces. $V_c$ and $V_m$ denote
the fluid volumes displaced, respectively, inside and outside the line of tangency. }
\label{fig2}
\end{figure}
Equations (27)-(31) thus indicate that the curvature force acting on the floating
body is expressible in terms of the fluid volume displaced {\it outside} the line
of tangency:
\begin{equation}
F_c ~=~ \rho g V_m ~~~~~.
\end{equation}
The relative magnitude of the buoyancy and curvature forces supporting a floating,
non-wetting body is thus prescribed by the relative magnitudes of the volumes of the
fluid displaced inside and outside the line of tangency:
\begin{equation}
\frac{F_b}{F_c} ~=~ \frac{V_c}{V_m} ~~~~.
\end{equation}
We note that this result has been generalized to 3D floating, non-wetting bodies
by J. Keller (Phys. Fluids, 1998). For 2D bodies, we note that since the meniscus
will have a length comparable to the capillary length, $l_c = (\sigma/(\rho g))^{1/2}$,
the relative magnitudes of the buoyancy and curvature forces,
\begin{equation}
\frac{F_b}{F_c} ~ \approx ~ \frac{r}{l_c} ~~~~,
\end{equation}
is prescribed by the relative magnitudes of the body size and capillary length.
Very small floating objects ($r \ll l_c$) are supported principally by curvature rather than
buoyancy forces. This result has been extended to three-dimensional floating
objects by Keller (1998, {\it Phys. Fluids}, {\bf 10}, 3009-3010).
\vspace*{.2in}
\noindent{\bf Water-walking Insects}
\vspace*{.1in}
\begin{figure}
\begin{center}
\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
%\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
\includegraphics[width = 1.\textwidth]{front}
\end{center}
\caption{Water-walking insects deflect the free surface, thus generating
curvature forces that bear their weight. The water strider has characteristic
length 1cm and weight 1-10 mg. }
\label{fig2}
\end{figure}
\begin{figure}
\begin{center}
\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
%\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
\includegraphics[width = 0.7\textwidth]{natfig1b.pdf}
\end{center}
\caption{The water strider legs are covered with hair, rendering them effectively
non-wetting. The tarsal segment of its legs rest on the free surface. The free surface
makes an angle $\theta$ with the horizontal, resulting in an upward curvature force
per unit length $\sigma$ sin$\theta$ that bears the insect's weight. }
\label{fig2}
\end{figure}
Small objects such as paper clips, pins or insects
may reside at rest on a free surface provided the curvature
force induced by their deflection of the free surface is sufficient to bear their weight.
For example, for a body of contact length $L$ and total mass $M$,
static equilibrium on the free surface requires that:
\begin{equation}
\frac{Mg}{2 \sigma L \hbox{sin} \theta} ~<~ 1
\end{equation}
where $\theta$ is the angle of tangency of the floating body.
This simple criterion is an important geometric constraint on water-walking
insects. Figure 4 indicates the dependence of contact length on body weight
for 250 species of water-striders, the most common water walking insect.
Note that the solid line corresponds to the requirement (35) for static
equilibrium. Smaller insects maintain a considerable margin of safety,
while the larger striders live close to the edge. The size of water-walking
insects is limited by the constraint (35).
\begin{figure}
\begin{center}
\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
%\centerline{$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$}
\includegraphics[width = 1.\textwidth]{mega}
\end{center}
\caption{The relation between the maximum curvature force $F_s = \sigma P$ and
body weight $F_g = Mg$ for 342 species of water striders. $P= 2(L_1 + L_2 + L_3)$
is the combined lengths of the tarsal segments (see strider B).
From Hu, Chan \& Bush ({\it Nature}, {\bf 424}, 2003). }
\label{fig2}
\end{figure}
If body proportions were independent of size $L$, one would expect the
body weight to scale as $L^3$ and the curvature force as $L$. Isometry
would thus suggest a dependence of the form $F_c \sim F_g^{1/3}$,
represented as the dashed line in Figure 5. The fact that the best fit
line has a slope considerably larger than $1/3$ indicates a variance from
isometry: the legs of large water striders are proportionally longer.
\vspace*{.5in}
\begin{center}
{\bf Appendix A: Computing curvatures}
\end{center}
\vspace*{.1in}
We see the appearance of the divergence of the surface normal, $\nabla \cdot {\bf n}$,
in the normal stress balance. We proceed by briefly reviewing how to formulate this
curvature term in two common geometries.
\vspace*{.1in}
In cartesian coordinates $(x,y,z)$, we consider a surface defined by $z=h(x,y)$.
We define a functional $f(x,y,z) = z - h(x,y)$ that necessarily vanishes on the
surface. The normal to the surface is defined by
\begin{equation}
{\bf n} ~=~ \frac{\nabla f}{|\nabla f|} ~=~ \frac{\hat{z} - h_x \hat{x} - h_y \hat{y}}{(1 + h_x^2 + h_y^2)^{1/2}}
\end{equation}
and the local curvature may thus be computed:
\begin{equation}
\nabla \cdot {\bf n} ~=~ \frac{-(h_{xx} + h_{yy}) - (h_{xx}h_y^2 + h_{yy}h_x^2) + 2 h_x h_y h_{xy}}{(1+ h_x^2 + h_y^2)^{3/2}}
\end{equation}
In the simple case of a two-dimensional interface, $z = h(x)$, these results assume the simple
forms:
\begin{equation}
{\bf n} ~=~ \frac{\hat{z} - h_x \hat{x}}{(1 + h_x^2)^{1/2}}~~~,~~~\nabla \cdot {\bf n} ~=~ \frac{-h_{xx}}{(1+ h_x^2)^{3/2}} ~~~~.
\end{equation}
Note that ${\bf n}$ is dimensionless, while $\nabla \cdot {\bf n}$ has the units of $1/L$.
\vspace*{.1in}
In 3D polar coordinates $(r, \theta, z)$, we consider a surface defined by $z= h(r, \theta)$.
We define a functional $g(r, \theta, z) = z - h(r, \theta)$ that vanishes on the surface, and
compute the normal:
\begin{equation}
{\bf n} ~=~ \frac{\nabla g}{|\nabla g|} ~=~ \frac{\hat{z} - h_r \hat{r} - \frac{1}{r} h_{\theta} \hat{\theta}}{(1 + h_r^2 + \frac{1}{r^2} h_{\theta}^2)^{1/2}} ~~~,
\end{equation}
from which the local curvature is computed:
\begin{equation}
\nabla \cdot {\bf n} ~=~ \frac{-h_{\theta \theta} - h_r^2 h_{\theta \theta} + h_r h_{\theta} - r h_r -
\frac{2}{r} h_r h_{\theta}^2 - r^2 h_{rr} - h_{rr} h_{\theta}^2 + h_r h_{\theta} h_{r \theta}}{r^2(1 + h_r^2 + \frac{1}{r^2} h_{\theta}^2)^{3/2}} ~~~~.
\end{equation}
In the case of an exisymmetric interface, $z = h(r)$, these results reduce to:
\begin{equation}
{\bf n} ~=~ \frac{\hat{z} - h_r \hat{r}}{(1 + h_r^2)} ~~~,~~~ \nabla \cdot {\bf n} ~=~ \frac{- r h_r - r^2 h_{rr}}{r^2(1 + h_r^2)^{3/2}} ~~~~.
\end{equation}
\vspace*{.4in}
\begin{center}
\noindent{\bf Appendix B: Frenet-Serret Equations}
\end{center}
\vspace*{.2in}
Differential geometry yields relations that are often useful in computing
curvature forces on 2D interfaces.
\begin{equation}
(\nabla \cdot {\bf n})~ {\bf n} ~=~ \frac{d{\bf t}}{d\ell}
\end{equation}
\begin{equation}
- (\nabla \cdot {\bf n})~ {\bf t} ~=~ \frac{d{\bf n}}{d\ell}
\end{equation}
\vspace*{.2in}
\noindent Note that the LHS of the first of (42) is proportional
to the curvature force acting on an interface.
\end{document}