Download Introduction to Soft Matter Physics- Lecture 5

Introduction to Soft Matter Physics- Lecture 5
Submitted by: Ilya Osherov
In this lecture we are going to discuss two topics:
1. Fluid dynamics
2. Surface waves
1. Fluid Dynamics
1.1 Simple Shear
Figure 1: Simple Shear
Force acting on a surface in a viscous fluid is decomposed into components normal to the surface, i.e. pressure, and tangential to the surface,
called shear stress σ. In general, the mechanical stress is a tensor, whose
diagonal components characterize pressure and non-diagonal components
characterize shear stress. For common fluids, the shear stress is proportional to the gradient of velocity field, with viscosity defined as a coefficient
of proportionality, such as in Fig. 1.
f
∂vx
v
= σxy = η
=η
S
∂y
h
Viscosity causes energy losses, which are calculated per unit time and
per unit volume are proportional to viscosity and to the square of velocity
gradient. E.g. in Fig. 1:
v 2
fv
∂vx 2
∆E
=−
= −η
= −η
∆t∆V
Sh
h
∂y
We can think of viscous phenomena in terms of momentum transport.
The shear stress applied to the upper plate is in fact the momentum flux
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Jp injected into the system by the force on the boundary. We can easily see
this in the following way.
First we write:
F
F ∆t
∆p
σxy =
=
=
= Jp
S
S∆t
S∆t
Then
∂vx
J = −η
(1)
∂y
is a typical relation from the theory of transport phenomena: velocity gradient leads to the associated flux of momentum. The momentum is getting
transported from layer to layer by shear forces in between the layers, or, on
the molecular level, by molecules jumping from one layer to another and
transporting the moment with them.
It is easy to see without solving any equations that in stationary conditions (after long enough time passed since the beginning of motion) the
velocity gradient is constant across the liquid depth in Fig. 1: in the stationary conditions the velocities of the layers do not change. This means
that the momentum is transported from the upper plate downwards without
”sticking” anywhere, i.e. the momentum flux is constant across the liquid
and therefore via Eq. 1 velocity gradient is constant too.
We can estimate the characteristic time scales of establishing the gradients by introducing kinematic viscosity ν = η/ρ. Kinematic viscosity has
units of m2 /s, same units as diffusion coefficient. In fact, for gases it is equal
to the diffusion coefficient of molecules. Shear stress thereby propagates in
a diffusion-like manner with a kinetic rate determined by ν. E.g. within
time t the stress propagates at the distace:
√
y ∼ νt
Thus for the system in Fig. 1 the characteristic time to reach the stationary state can be estimated from:
tH =
H2
v
Lets see what happens to the system as it is reaching the steady state.
At the beginning the momentum starts propagating in the liquid. Only
the upper part of the pool knows about the motion. The gradient is not
constant. It equals 0 at the bottom and to some value at the top. The
plate is already in motion but the information about the motion has not yet
propagated through the whole pool. Lets focus on some layer at the center
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of the pool. Momentum enters this layer from above. The momentum that
will leave this layer downwards will be smaller (since the gradient on the
upper side of the layer is larger than that on the lower side), thus some
momentum will accumulate in the layer and the layer will accelerate. At
this time all the layers are accelerating, the motion is propagating downwards from the upper plate and the velocity gradient is getting distributed
over a progressively wider range. After some time the velocity gradient will
become constant over the depth of the liquid, and the system will stabilize.
The layers will no longer accelerate. All the momentum that will come from
above will move through the whole system until it reaches the bottom.
1.2 Navier-Stokes eq.
The main equation of fluid dynamics is the Navier-Stokes eq.
∂~v
ρ
+ (~v ∇)~v = −∇p + η∇2~v + sometimes volume forces, e.g ρ~g
∂t
Were ρ is fluid density and p is pressure.
It is essentially just an application of the second law of Newton to the unit
volume portion of fluid.
Navier-Stokes equation has to be complemented by the continuity equation,
which for incompressible fluid is just
div ~v = 0
For compressible liquids
div(ρ~v ) = 0
Where ρ is water density. Notice that the amount of the liquid in total is
constant (the amount of mass does not change).
Navier-Stokes equations are differential equations and have to be supplemented by boundary conditions, which are typically the non-slip conditions,
i.e. the velocity of the liquid layer adjacent to the solid surface is equal to
the velocity of that surface.
We also define a kinetic viscosity which has the units of diffusion coefficient
and for gases (not liquids) is in fact equal to the diffusion coefficient (check
your Thermo II notes).
η
ν=
ρ
Reynolds number is an estimate of the relative importance of the nonlinear term in the Navier-Stokes equation:
Re =
v 2 /l
vl
vlρ
v∇v
≈
=
,
=
ν∇2 v
νv/l2
ν
η
3
(2)
where v and l are respectively the characteristic velocities and characteristic
sizes of the objects.
In colloidal physics, the objects are so small (typically < 1µm), that
Re 1 and nonlinear term in the Navier-Stokes equation is neglected in
most of the cases.
Solving field equations is not an easy task even for small Reynolds Numbers for which we can neglect the non linear part.
(~v ∇)~v
So far there is no general solution for Navier-Stokes differential equations
and there is a one milion dollar award prize to be given to the one who will
solve the Millennium Problem. However there are solutions in some private
cases and it is possible to get a good intuition without actually solving these
equations. We are going to discuss a few of the problems in order to develop
such an intuition. We can learn many things about the system if we understand a few basic principles in fluid dynamics.
The main principles are:
• The motion of an object in fluid leads to gradients in velocity, pressure
and shear stress (see below) at the distances of about the object size.
• Pressure field propagates with the speed of sound: infinitely fast for
colloidal-type problems
• Shear stress propagates in a diffusion-like manner with a ”diffusion”
coefficient of ν (see Eq. 1.34), so that within time t the shear stress
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propagates to the distances of ∼ (νt) 2
1.3 Basic problem- Periodic Pure Shear
Now we demonstrate the use of the above principles by looking at a classical
problem of fluid dynamics. The system is similar to the one we dealt with
in 1.1 (Fig 1). We apply tangential harmonic force f ∼ sin(ωt) on a surface
of unit area in a viscous fluid. The plate moves in one direction of half of
the period T , during that time the motion propagates downwards, then the
plate changes the direction canceling the effect of prior motion. Thus from
the third principal we infer that the shear stress in this system propagates
and the motion is confined within the distance
r
p
ν
δ ∼ νT /2 ∼
ω
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1.4 Basic problem - Drag force on a moving sphere, Stokes law
A sphere of radius R moves through the liquid with constant velocity
v. Liquid exerts a viscous (drag) force acting on the sphere. What is this
force? The exact calculation gives F = 6πηRv (Stoke’s law). Let’s see how
to obtain this law (up to a numerical coefficient) from qualitative arguments.
We understand that shear stress acts over the surface of the sphere, that
the shear stress is proportional to velocity gradient, and, the main point,
that velocity in the liquid drops from ∼ v at the surface of the sphere to
∼ 0 over the distances of the order of radius R. Thus the velocity gradient
can be estimated as v/R. Then
F ∼ σS ∼ η∇vS = η
v
4πR2 ∼ 4πηRv
R
The exact solution is
F = 6πηRν
1.5 Basic problem- Velocity field around a moving Sphere
Figure 2: One sphere in a viscous liquid
A sphere moves with low and constant velocity in a liquid. We are
interested in how the velocity decays away from the sphere. The exact
solution is rather tedious. Thus would like to solve the problem qualitatively.
It is intuitive to think that the further the point in liquid is from the sphere
the less it feels the effect of the motion.
We assume that the velocity field of the sphere decays as a power law:
v(r) ∼ r−n . What power n is equal to?
Why does liquid move at all? Lets first consider mass conservation in
the system. The sphere moves, thus it has to push liquid from the forward
direction and get some liquid filled on its back. In order to estimate the
power law, let’s think first of a system of an expanding non moving sphere.
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Its volume grows, the liquid is pushed outwards and according to the conservation of mass law, same mass has to pass through all the closed surfaces
around the sphere. Thus we get that that the flux of the velocity is constant
v(r)4πr2 = const → v(r) ∼ r−2
The motion of the sphere can be imagined as an expansion in the front and
shrinking at the back of the sphere and because of linearity of equations for
(Re << 1), well get a field similar to electrical dipole
v(r) ∼ r−3
This in fact was a valid consideration. However, the real dependence is
v(r) ∝ r−1 . We didnt get the correct result because we ignored viscosity and
only considered mass conservation. Another reason for the motion of liquid
is viscosity. To take that into account we should consider momentum conservation. Going into the reference frame of the sphere we see that the motion
is stationary and therefore the momentum transferred from the sphere to
the neighboring layer should pass without changes to system boundaries far
away:
Jp 4πr2 = Const ⇒ η4πr2
dv
dv
1
1
4πr2 = Const ⇒
∝ 2 → v(r) ∝
dr
dr
r
r
1.5 Basic problem- Two spheres problem (interactions)
Figure 3: Two shperes in a viscous liquid
A force acting on the second sphere will move it and produce the motion
of the liquid around it decaying ∝ 1/r with the distance from the second
sphere. The first sphere will be dragged with this motion of the liquid.
This type of interactions between the motions of two objects mediated by
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the liquid are called Hydrodynamic Interaction. Notice that they decay extremely slowly, like ∝ r−1 , much slower than any other interaction. That’s
why they usually cannot be neglected in the calculations of dynamics of the
system (however, they are totally irrelevant for the statistical properties of
a system).
A rather symmetric way of writing down these interactions is given below
(valid under certain approximation):
v1 =
F~1
F~2
+
6πηR1 6πηr12
i.e. the velocity of a bead is caused by the forces acting on this bead
directly as well as by other forces applied to other objects which act on the
bead indirectly, via hydrodynamic interactions .
2. Surface waves (Water waves)
2.1 Introduction
Water waves are one of the oldest topics in fluid mechanics. We are going
to discuss two different types of surface waves Gravitational waves and Capillary waves. Each type got its name because of the origin of its potential
energy. Gravitational waves are connected to the earth gravitation potential
where is capillary waves are originated in surface energy of the water.
Generally all the waves in nature are partly gravitational and cappillar. For
shorter wavelengthes the capilary part is more important and for long wavelengthes the gravitational part is what determines the behavior. We are
interested to find out what is the dispersion relation and also to understand
the reason for fading of the waves.
2.2 Gravitational waves
One has to consider gravitational potential energy, kinetic energy of the fluid
and viscous losses.
2.2.1 Deep water (wavelength λ << H depth):
Motion is confned to the surface layer of λ in depth; all of the viscous losses
happen there; the characteristic velocity of the fluid is similar to surface
elevation depreciation speed ∼ ḣ.
Like in all oscillators the total energy of the system is the sum of kinetic
and potential energies which are functions of distance from the equilibrium
point. For now lets assume a regular harmonic oscillator and ignore energy
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losses. We also assume λ wavelength.
We will try to construct an equation similar to harmonic oscillator. Although
it is possible to calculate it precisely we will try to get qualitative answer
up to pre-factors.
First we assume a closed system, a very deep pool. Now lets say that there is
a distortion with characteristic height- h on the surface. The characteristic
velocity is ḣ.
We want to find the gravitational potential energy. The general formula is
Mgh. The mass of the moving fluid is
M = ρV = ρhLx Ly
And we get that the potential energy
U = M gh = (ρhLx Ly ) gh = (ρLx Ly )gh2
Kinetic energy is proportional to the square of velocity
1
1
K = M ḣ2 = (ρLx Ly λ) ḣ
2
2
So the total energy E = U + K = Const after differentiation with time
U̇ + K̇ = 2 (ρLx Ly ) ghḣ +
1
(ρLx Ly λ) ḧ
2
And if we ignore the factors we get an eq. of the form
ḧ +
g
h=0
λ
And this is a harmonic oscillator with dispersion relation
r
g
√
ω=
= gq
λ
The group velocity is
vg =
√
1
dw
∼√ ∼ λ
dq
q
From this we see that that group velocity grows as function of wavelength.
The only parameter we have in the problem that tells us how much the wave
penetrates the water is wavelength λ. There are many ways to explain this.
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One of which is: even waves with very low amplitude crate motion of the
water to the depth because water has to pass very long distances (meters).
Wavelength can be up to several meters and thats what will determine how
much the disturbance will advance in the liquid. Now well deal with energy
losses (viscosity). Well get damped harmonic oscillator.
K̇ + U̇ + Ė = 0
Ė = −η(∇v)2 V ≈ −η
ḣ
λ
!2
λLx Ly
And from these two eq. We can get the damping coefficient
Ė η
Γ=
= 2 = νq 2
K
λ ρ
2.2.2 Shallow water (h H λ): Motion occurs throughout the
λ
depth of the liquid; the characteristic velocity is much larger (by ∼ H
) than
surface elevation/depreciation velocity due to continuity; the largest velocity
1
gradient and hence the main losses occur in a thin (∼ (v/w) 2 ) layer near
the bottom. In this case the calculation of potential energy is same as in
2.2.1 but kinetic energy changes.
1
1
K = M ν 2 = (ρLx Ly H)ν 2
2
2
This time we have to move the same amount of mass but through a
thinner layer (H instead of λ). From the continuity Eq. the fluid velocity
will be much higher than that of surface elevation/depreciation.
ḣλ = vH → v =
ḣλ
H
And we get
!2
1
1
ḣλ
K = M v 2 = (ρLx Ly H)
2
2
H
r
p
k
1p
ω=
=
qH = q gH
m
λ
This is linear dispersion relation. Now we add energy losses. It is natural
to deal with them in same way as in 2.1.1 and to get
!2
ḣ
V2
Ė = −η
HLx Ly ≈ −ηLx Ly
λ
H
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But this is small contribution compared to the friction losses in the thin
layer δ at the bottom of the pool where the problem is similar to that of
periodic motion of a plate:
ḣ
Ė = −η
δ
!2
δLx Ly ≈ −ηLx Ly
Where
V2
δ
r
V
ω
Now we can the damping coffisient and we will get that for shallow water
Ė 2ηH V 2
=
Γ=
K
δρ ḣλ
δ=
2.3 Capillary waves
The main factor in the potential energy is surface tension energy. When we
create a distortion in the pool, we create a new surface and this costs us
energy.
Surface tension is a necessary feature accompanying phase separation.
If not for the surface tension materials would never part from each other. If
the surface energy is smaller then 0 it means that materials like to mix. Now
well try to understand the source of surface energy by focusing on phase
separation between liquid and gas. Any phase is seeking to minimize its free
energy, which has two contributions, energy and entropic terms. Gas phase
goes for maximal entropy, liquid state compromises on entropy in order
to reduce interaction energy due to attraction forces between molecules.
Molecules on the surface do not really belong to any of the phases and
have an extra free energy associated with them. E.g. as compared to the
molecules in the bulk of liquid, they have smaller amount of neighbors and
thus lack some negative attractive interactions. This extra free energy is
surface tension. In thermodynamics we defined surface tension α as free
energy per unit surface:
α = fS /∆S
If we create a wave in the water the surface area becomes (by means of
differential geometry)
s
2 2
Z
dh
dh
S = dxdy 1 +
+
dx
dy
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How much spare surface have we created?
s
2 2 Z
Z
dh
dh
∆S = dxdy 1 +
+
− dxdy
dx
dy
So far we were accurate, now we come back to physics and make an
approximation. Lets say all the waves are small (little distortions), from
this follows that we have small angles in the problem.
2
dh
<< 1
dx
2
dh
<< 1
dy
Now we can use Taylor expansion:
!
Z
1 dh 2 1 dh 2
∆S = dxdy 1 +
+
−1
2 dx
2 dy
!
dh 2
dxdy
+
dy
2
Z
1
h
2
=
dxdy∇ h ≈ Lx Ly
2
λ
1
=
2
Z
dh
dx
2
Thermal fluctuations create small waves in the liquid. These waves can
be measured by means of laser light reflection. They are the size of a few
nanometers. Now we make an approximation of contribution of surface
energy to the total potential energy of the wave.
2
h
Fα → U = αLx Ly
λ
U ≈ αLx Ly q 2 h2
The total dispersion relation for capillary gravitational waves is
ω = gq +
α 3
q
ρ
For qg > αρ q 2 the waves are mostly gravitational
→ λ2 >
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α
ρg
Figure 4: Wave Speed as Function of Wave Number
We define capillary length
r
lc =
α
ρg
For lc < λ the gravitational part is more important, otherwise the capillary part is stronger.
√
For gravitational waves: ω ∼ q → u ∼ √1q
√
√
For capillary waves: ω ∼ q q → u ∼ q
For weak perturbations of the surface h(x; y) = h(r⊥ ) the change in
surface energy is:
" 2 #
Z
Z
α
∂h 2
∂h
α
∂h 2
∆Fs ≈
dxdy
+
=
d~rn
2
∂x
∂y
2
∂~rn
Otherwise the treatment is similar to gravitational waves. In general,
both capillary and gravitational effects have to be taken into account.
Lets draw the speed of the waves as function of q We can create surface
waves only from certain minimal speed u0 . As long as we create slower
distortion we will not see capillary waves. There is a minimal surface wave
velocity under which waves do not begin to from. For example if we move an
object through a pool of water at a very small velocity and star accelerating
itll take some time until well see triangle waves behind the object. This
effect is similar to Cherenkov Effect in solid states physics. Cherenkov
radiation (also spelled Cerenkov or sometimes ?erenkov) is electromagnetic
radiation emitted when a charged particle passes through an insulator at a
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speed greater than the speed of light in the medium. The characteristic
”blue glow” of nuclear reactors is due to Cherenkov radiation. It is named
after Pavel Alekseyevich Cherenkov, the 1958 Nobel Prize winner who was
the first to rigorously characterize it.
2.4. The connection between soap and waves
What will happed if we spill a barrel of soap molecules on a water surface?
There is no significant change in surface energy but Surfactant molecules
change the conditions on the surface to (almost) sticky boundary conditions.
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Hence, the viscous losses occur in a thin layer near the surface ∼ (v/ω) 2 .
This will cause significant chance in the behavior of waves. This happens
because soap molecules at large enough concentration create flat layer of
molecules that are strongly connected to each other and their interaction
swallows the waves.
2.5 Sources
http://en.wikipedia.org/wiki/Cherenkov effect
http://www.fluidmech.net/jscalc/waves.htm
http://www.galleryoffluidmechanics.com/waves/plageb.htm
http://www.claymath.org/millennium/Navier-Stokes Equations
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