Rotating flows
g of his paper he points out three possibilities:
quid is never steady.
dy, but our assumption that uI is small relative to the rotation
n near the object.
y and two dimensional.
Sir Geoffrey Ingram Taylor
possibility
is unlikely, since it must settle down eventually ! The
(7 March 1886 – 27 June 1975)
2) and (3). His paper, which can be downloaded from the course
actually what happens is possibility (3). This is really rather
tes) because there is only one way that it can really happen: An
st move atop the object.
Downloaded from rspa.royalsocietypublishing.org on April 16, 2014
ing flows
ations of a rotating fluid assuming that the rotation frequency
stic hydrodynamic flows in the problem. In other words, if ⌦
ion frequency, L is a horizontal lengthscale, and U is a typical
ame, we assumed that
U
Ro =
⌧ 1.
⌦L
(5)
Taylor - Proudman
http://ocw.mit.edu/
http://ocw.mit.edu/
Taylor - column
Ekman-Layer
http://ocw.mit.edu/
A viscous force between the Ekman layer and the stationary fluid causes fluid adjacent to the
boundary layer to accelerate and a Coriolis force propels it outward.
!
The result is a flow circulating along the boundary layers and downward along the center of
the cylinder.
only important on large scales.
⇢ @y
there
will
be
boundary
layers.
Let’s
divide
the
flow
into
three
regions:
(1)
A
boundary
viscid flow is two dimensional (and so no gradient in ⌦ across the cylinder
layer at the top plate; (2) a boundary layer at the bottom plate; and (3) a central inviscid1 @pI
0 small,
=
.
We therefore
anticipate that even though the Rossby number is
region.
⇢ @z
22
The
Ekman
layer
the inviscid
region
we would
the solution
is (uI , v(1)
I , wI ),
ndary layers. In
Let’s
divide
the
flow expect
intothat
three
regions:
Awhere
boundary
thebottom
same way
as1 @pbefore,
weaexpect
the
pressure gradient of the ou
I
late; (2) a boundary layer atInthe
plate;
and
(3)
central
inviscid
2⌦vI =
,
(526a)
We would now boundary
like to return
our
co↵ee cup
problem,
to get the
the structure
right answer
@x rotating
layer to
at ⇢the
wall.
Let’s consider
of t
@p
Boundary
I
so, we shallInner
consider
the2⌦u
e↵ect
walls
on the
flow
the bottom
z1=
0., There
theinviscid
equations
arewe calculated in the
= of
(526b)
Iwall,
region we would expect that the solution is (u
,
v
,
w
),
where
⇢ @y
I I
I
lecture. For starters, lets consider
a
jar
with
the top moving at angular
velocity
1 @pI
2
0 =
.
(526c)
1=@p
@ uour inviscid
I then
⇢
@z
the bottom moving 1at@pangular
velocity
⌦
.
Clearly,
if
⌦
⌦
B
B+ ⌫
2⌦v = T
,
I
2
2⌦v
, the out
(526a)
⇢ the
@x
@z
I as =
Let’s
try and
figure
what
happens
when
⌦
becomes
di↵erent
from ⌦
Inapplies.
the same way
before,
we
expect
pressure
gradient
of
the
outer
flow
to
force
T
⇢ @x
boundary layer at the rotating wall. Let’s consider the structure of the boundary layer
at I
1 @p
@2v
2⌦u =
+ ⌫ 2,
@pequations
the bottom wall, z = 0. There1the
are
I
⇢ @y
@z
= deviation
,
(526b)
22.1 2⌦u
AI small
⇢ @y
1 @pI
@2u
1(527a)
@pI
@2w
2⌦v =
+⌫ 2,
⇢ @x
@z
0 way
= to satisfy
+ ⌫ the2 ,no slip cond
1 @p⌦IB = ⌦ +
Suppose ⌦0T =⌦
and
✏.
Now
there
is
no
⇢ @z
@z
1 @pI
@2v
=
.
(526c)
=
+ ⌫ the
, whole flow spin (527b)
⇢ @z2⌦u while
both the top and bottom
⇢ having
@y
@z 2
r · u = 0.at angular velocity ⌦. Le
1 @ptry
@ 2 wcompute the secondary flow that is induced.
I
first to the rotating frame,
and
to
0 =
+ the
⌫ 2 ,outer flow to force(527c)
as before, we expect the pressure
gradient
of
the
⇢
@z
@z boundary layer approximation
Here we have made the
that @/@z
@
rthe
·u =
0.
(527d)
t the rotating wall. Let’s consider
structure
of equation
the boundary
layer
at w is much smaller th
From
the
continuity
we deduce
that
have made theare
boundary layer
approximation
@/@z101@/@x,
@/@y. @p /@z = 0, and the equatio
z = 0. ThereHere
theweequations
ponents
parallel
to the that
boundary
so that
I
From the continuity equation we deduce that w is much smaller than the velocity components parallel to the boundary so that
2 @pI /@z = 0, and the equations become
1 @pI
@ u
@2u
2⌦(v v(527a)
,
2⌦v =
+⌫ 2,
I) = ⌫
@2u
2
⇢ @x 2⌦(v@z vI ) = ⌫ 2 ,
(528a) @z
@z
2v
@
2
2
1 @pI 2⌦(u
@ vuI ) = ⌫ @ v .
2⌦(u (527b)
uI(528b)
) = ⌫ 2.
2
2⌦u =
+ ⌫ 2,
@z
@z
⇢ @y
@z
These are the equations we must solve. Acheson has a good trick. Multiplying the second
These
are
the equations we must solve. Acheson has a good trick. M
equation by i and adding
the
1 @p
@2w
I two yields
0 =
+ ⌫ by2 ,i2 and adding the two yields (527c)
equation
⇢ @z
@z⌫ @ f = 2⌦if,
(529a)
@z 2
@2f
r · u = 0.
(527d)
Here we have made the boundary layer approximation that @/@z
@/@x, @/@y.
From the continuity equation we deduce that w is much smaller than the velocity components parallel to the boundary so that @pI /@z = 0, and the equations become
Solution
2⌦(v
2⌦(u
@2u
vI ) = ⌫ 2 ,
@z
@2v
uI ) = ⌫ 2 .
@z
(528a)
(528b)
These are the equations we must solve. Acheson has a good trick. Multiplying the second
equation by i and adding the two yields
@2f
⌫ 2 = 2⌦if,
@z
(529a)
where
f =u
uI + i(v
vI ).
The solution is obtained by guessing f ⇠e↵z , which yields ↵2 = 2⌦i/⌫. Hence,
p
(1+i)z ⇤
(1+i)z ⇤
⇤
f = Ae
+ Be
,
z = z ⌦/⌫.
(529b)
(530)
102
We require that as z ⇤ ! 1, f ! 0. This implies that A = 0. We are in the frame
of reference moving with the bottom plate, so the no slip boundary condition at z = 0
requires that f (z = 0) = uI ivI . Splitting f into its real an imaginary parts implies
u = uI
e
z⇤
v = vI
e
z⇤
(uI cos(z/ ) + vI sin(z/ )),
(vI cos(z/ )
uI sin(z/ )).
This is the velocity profile in the boundary layer.
What about the z-component? From the divergence free condition, we have
✓ ◆1/2
✓
◆
(531)
(532)
u = uI
e
z⇤
v = vI
e
z⇤
(uI cos(z/ ) + vI sin(z/ )),
(vI cos(z/ )
(531)
uI sin(z/ )).
(532)
This is the velocity profile in the boundary layer.
What about the z-component? From the divergence free condition, we have
✓ ◆1/2
✓
◆
⌦
@w
@w
@u @v
@vI
@uI
z⇤
⇤
=
=
=
e
sin
z
.
⌫
@z ⇤
@z
@x @y
@x
@y
Integrating from z ⇤ = 0 to 1 gives
✓ ◆
1 ⌦
w=
2 ⌫
1/2 ✓
@vI
@x
@uI
@y
◆
=
!
ˆI
2
r
⌫
,
⌦
(533)
(534)
where !
ˆ I is the vorticity in the inviscid flow. Thus, if !
ˆ I > 0 (i.e., the bottom boundary is
moving slower than the main body of fluid) then there is flow from the boundary layer into
the fluid.
22.2
Matching
Now we have these Ekman layers at the top and the bottom. What we just assumed was
that the boundary is moving at frequency ⌦. If it is not, but instead moving at an angular
frequency ⌦B relative to the rotating frame, then we need to change the boundary conditions
a little in the rotating frame. In this case
◆
⇣ ⌫ ⌘1/2 ✓ !
ˆI
w=
⌦B .
(535a)
⌦B
2
We could derive this, but it is intuitive since (ˆ
!I 2⌦B ) is the vorticity of the interior flow
relative to the moving lower boundary. Similarly, if ⌦T denotes the angular velocity of the
rigid upper boundary relative to the rotating frame, then there is a small z-component of
velocity up into the boundary layer
◆
⇣ ⌫ ⌘1/2 ✓
!
ˆI
w=
⌦T
.
(535b)
⌦T
2
1 ⌦
@vI
@uI I
!I ⌫
w
=
=
,
(534)
moving slower than the main body
2 of
⌫ fluid) then
@x there
@y is flow2from
⌦ the boundary layer into
the fluid.
where !I is the vorticity in the inviscid flow. Thus, if !I > 0 (i.e., the bottom boundary is
moving slower than the main body of fluid) then there is flow from the boundary layer into
22.2 Matching
the fluid.
Now we have these Ekman layers at the top and the bottom. What we just assumed was
that22.2
the boundary
is moving at frequency ⌦. If it is not, but instead moving at an angular
Matching
frequency ⌦B relative to the rotating frame, then we need to change the boundary conditions
Now we have these Ekman layers at the top and the bottom. What we just assumed was
a little in the rotating frame. In this case
that the boundary is moving at frequency ⌦.✓ If it is not,
◆ but instead moving at an angular
⇣
⌘
1/2 then
frequency ⌦B relative to the rotating⌫frame,
!
ˆ I we need to change the boundary conditions
w=
⌦B .
(535a)
a little in the rotating frame. In this⌦case
2
B
⇣ ⌫ ⌘1/2 ⇣ !
⌘
I
We could derive this, but it is intuitive
!I 2⌦B⌦
) is the
vorticity of the interior(535a)
flow
w = since (ˆ
B .
⌦B
relative to the moving lower boundary. Similarly,
if2 ⌦T denotes the angular velocity of the
rigid
relative
to intuitive
the rotating
then
there is a small z-component of
Weupper
couldboundary
derive this,
but it is
since frame,
(!I 2⌦
B ) is the vorticity of the interior flow
velocity
up to
into
boundary
relative
thethe
moving
lower layer
boundary. Similarly, if ⌦T denotes the angular velocity of the
◆ there is a small z-component of
rigid upper boundary relative to the
then
⇣ ⌫ rotating
⌘1/2 ✓ frame,!
ˆI
=
⌦T
.
(535b)
velocity up into the boundarywlayer
⌦T
2
⇣ ⌫ ⌘1/2 ⇣
!I ⌘
Now in our container both are whappening.
Since
of z
=
⌦T uI , vI and
. wI are all independent
(535b)
⌦T
2
then so is !
ˆ I . Thus, the only way the experiment could work is if the induced value of !
ˆI
Nowcases
in ourmatches.
containerThis
bothimplies
are happening.
Since uI , vI and wI are all independent of z
from both
that
then so is !I . Thus, the only way the experiment could work is if the induced value of !I
!
ˆI =
⌦T + ⌦B .
(536)
from both cases matches. This implies
that
I
With ⌦B = 0 and ⌦T = ✏ we have that!I!
ˆ I==⌦T✏.+Thus,
a
⌦B . the flow in the inner region has
(536)
velocity which is entirely set by the boundary layers. Note that there is no viscosity in this
With ⌦
0 and ⌦plays
have
that !I = ✏. the
Thus,
theWe
flowhave
in the
inner region
has a
B =
T = ✏awe
formula,
but
viscosity
role
in determining
flow.
completely
di↵erent
velocityfor
which
set by
the⌫ boundary
layers. Note that there is no viscosity in this
behaviour
⌫ =is0 entirely
and in the
limit
! 0.
formula, but viscosity plays a role in determining the flow. We have completely di↵erent
behaviour for ⌫ = 0 and in the limit ⌫ ! 0.
102
103
22.2
Matching
Now we have these Ekman layers at the top and the bottom. What we just assumed was
Spin-down of this apparatus
that the boundary is moving at frequency ⌦. If it is not, but instead moving at an angular
frequency
⌦B to
relative
the rotating
frame, of
then
needcup.
to change
boundary
We
now want
finallytosolve
the spin-down
ourwe
co↵ee
To dothe
so we
assumeconditions
the co↵ee
a little
in athe
rotatingwith
frame.
In and
this acase
cup
to be
cylinder
a top
bottom both rotating with angular velocity ⌦ + ✏.
✓ is reduced
◆ to ⌦. How long does it take to
At t = 0 the angular velocity of the ⇣boundaries
⌫ ⌘1/2 !
ˆI
w=
⌦B .
(535a)
reach a steady state?
⌦B
2
We use the time-dependent formula
We could derive this, but it is intuitive since (ˆ
!I 2⌦B ) is the vorticity of the interior flow
@uI
1 @pI
relative to the moving lower boundary.2⌦v
Similarly,
if ⌦T, denotes the angular velocity (537a)
of the
=
I
⇢ @x then there is a small z-component of
rigid upper boundary relative to@tthe rotating frame,
@vI
1 @pI
velocity up into the boundary layer
+ 2⌦uI =
.
(537b)
@t ⇣
⇢
@y
✓
◆
⌫ ⌘1/2
!
ˆI
w = with respect
⌦T to y and
. the second with respect(535b)
Then di↵erentiate the first equation
to x.
⌦T
2
Subtracting the latter from the former, and using the continuity equation, we obtain the
Now equation
in our container both are happening. Since uI , vI and wI are all independent of z
vorticity
then so is !
ˆ I . Thus, the only way the@!
experiment
work is if the induced value of !
ˆI
@wcould
I
I
(538)
from both cases matches. This implies@t
that= 2⌦ @z .
Now !I is independent of z, so
!
ˆ =⌦ +⌦ .
(536)
22.3
I
T
B
Z L
@!
I
With ⌦B = 0 and ⌦T = ✏ we@!
have
ˆ II ==✏.2⌦[w(L)
Thus, thew(0)].
flow in the inner region has
a
dz that
=L !
(539)
@tthe boundary
@t layers. Note that there is no viscosity in this
velocity which is entirely set
0 by
formula, but viscosity plays a role in determining the flow. We have completely di↵erent
The
velocity
andinopposite
at⌫ the
two boundaries (flow is leaving both boundary
behaviour
foris⌫ equal
= 0 and
the limit
!
0.
layers), and has magnitude (⌫/⌦)1/2 !I /2. Thus
p
102
@!I
2 ⌦⌫
=
!I ,
(540)
@t
L
p
implying that vorticity is decreasing in the interior with a characteristic decay time L/2 ⌦⌫.
Atmospheric general circulation
23.1
23.1
23
Water waves
Deep water
water waves
waves
Deep
The flow
flow is
is assumed
assumed
to be
be inviscid,
inviscid,
andthe
as it
itRiver
is initially
initially
irrotational
must remain
remain
so.
Fluid
If you look
out onto
Charles,
theititwaves
thatso.
are
immediately
The
to
and
as
is
irrotational
must
Fluid
motion is
is therefore
therefore
described
by
the velocity
velocity
potentialthere
(u,v)
v) =
=
r ,many
, and
and satisfies
satisfies
Laplace’s
motion
the
potential
(u,
r
Laplace’s
waves described
on the by
water.
However,
are
di↵erent
types of wav
equation (incompressibility
(incompressibility condition)
condition)
equation
oceans, which have profound e↵ects on our surroundings. The most dr
22 = 0.
r
(541a)
r =
0. generated by earthquakes
(541a)
Tsunami, which is a wave
train
and volcanoes
these,
however,
let’s begin by considering the motion of a disturbance on
The momentum
momentum
equation
becomes
The
equation
becomes
@r
@r
11 r(r )22 =
+
+ 2r(r ) =
@t
@t
2
11 rp
rp
⇢
⇢
r ,,
r
(541b)
(541b)
where is
is the
the gravitational
gravitational potential
potential such
such that
that gg =
= r
r .. This
This can
can be
be integrated
integrated to
to give
give
where
the unsteady
unsteady Bernoulli
Bernoulli relation
relation
@@ + 11 (r )22 + pp + = C(t).
+ 2(r ) + ⇢ + = C(t).
@t
@t
2
⇢
104
(542)
(542)
Here, C(t) is
is aa time
time dependent
dependent constant
constant that
that does
does not
not a↵ect
a↵ect the
the flow,
flow, which
which isis related
related to
to
only through
through spatial
spatial gradients.
gradients. The
The surface
surface is
is h(x,
h(x,t)
t) and
and we
we have
have the
the kinematic
kinematic condition
condition
@h
@h
@h
@h
+
u
= vv
+u
=
@t
@x
@t
@x
(543)
(543)
on y = h(x,
h(x, t).
t). This
This simply
simply states
states that
that ifif you
you choose
choose an
an element
element of
of fluid
fluid on
on the
the surface,
surface, the
the
rate at which
which that
that part
part of
of the
the surface
surface rises
rises or
or falls
falls is,
is, by
by definition,
definition, the
the vertical
vertical velocity.
velocity.
Finally, we require
require that
that the
the pressure
pressure be
be atmospheric,
atmospheric, pp00 at
at the
the surface.
surface. From
From the
the unsteady
unsteady
Bernoulli relation
relation we
we get
get
on y = h(x, t). This simply states that if you choose an element of fluid on the surface, the
Here,
is athat
timepart
dependent
that does
not is,
a↵ect
flow, which
relatedvelocity.
to
rate
at C(t)
which
of the constant
surface rises
or falls
by the
definition,
the isvertical
only through
spatial
gradients.
The surface
is h(x, t) and
havesurface.
the kinematic
condition
Finally,
we require
that
the pressure
be atmospheric,
p0 we
at the
From the
unsteady
Bernoulli relation we get
@h
@h
(543)
@
1 +2u 2= v
+ @t(u +@xv ) + gh = 0
(544)
@t
2
y = h(x, t). This simply states that if you choose an element of fluid on the surface, the
ononh(x,
t), where we have chosen the constant C(t) appropriately to simplify things.
rate at which that part of the surface rises or falls is, by definition, the vertical velocity.
The equations we have derived so far take account of the e↵ect of gravity on the free
Finally, we require that the pressure be atmospheric, p0 at the surface. From the unsteady
surface. We have ignored one important factor, however, which is surface tension. It costs
Bernoulli relation we get
energy to create waves, as they@have1greater
surface area than a flat surface. From our
2
2
+ (u
+ v ) + gh = 0
(544)
earlier work we know that a pressure
@t
2 jump exists across a distorted interface. If p0 is
atmospheric
pressure,
thenchosen
the pressure
at theC(t)
fluid
surface is to simplify things.
on h(x, t), where
we have
the constant
appropriately
Include
surface
tension
The equations we have derived so far take account of the e↵ect of gravity on the free
@ 2 h(x, t)
surface. We have ignored one important
however,
p = pfactor,
. which is surface tension. It costs
(545)
0
2
@x
energy to create waves, as they have greater surface area than a flat surface. From our
earlier work
we know
that
a pressure
jump
exists across
distorted
Including
surface
tension
in our
pressure
condition
at the asurface,
we interface.
have that If p0 is
atmospheric pressure, then the pressure at the fluid surface is
@
1 2
@2h
2
+ (u + v ) @+2 h(x,
gh t)
=0
@t
2p = p0
⇢. @x2
2
@x
(546)
(545)
atIncluding
y = h(x,surface
t).
tension in our pressure condition at the surface, we have that
@
1
+ (u2 + v 2 ) + gh
@t
2
at y = h(x, t).
105
105
@2h
=0
⇢ @x2
(546)
Linearize
We now follow the same procedure as in the last lecture and assume all the variables to
We now follow the same procedure as in the last lecture and assume all the variables to
be small,
so that we can linearise the equations. The linearised system of equations consists
be small, so that we can linearise the equations. The linearised system of equations consists
of Laplace’s equation and the boundary conditions at y = 0:
of Laplace’s equation and the boundary conditions at y = 0:
We now follow the same procedure
to
r22 as=in 0the last lecture and assume all the variables(547a)
= 0 The linearised system of equations consists
(547a)
be small, so that we can linearise r
the
@hequations.
@
(547b)
@h =conditions
@ (x, 0,
of Laplace’s equation and the boundary
att),
y = 0:
@t = @y (x, 0, t),
(547b)
@t
@y
2
@
@22 h
r
= 0
(547a)
=
gh + @ h2 ,
(547c)
@
@h
@
@t =
gh + ⇢ @x2 ,
(547c)
=
(x, 0,⇢t),@x
(547b)
@t
@t
@y
These conditions arise because we have Taylor expanded
terms such as
2
@
@ h terms such as
These conditions arise because we have
Taylor expanded
=
gh +
,
(547c)
2
⇢ y@x
v(x, h, t) @t
= v(x, 0, t) + hv
(x, 0, t),
(548)
v(x, h, t) = v(x, 0, t) + hvy (x, 0, t),
(548)
These conditions arise because we have Taylor expanded terms such as
and then ignored nonlinear terms. We guess solutions of the form
and then ignored nonlinear terms. We guess solutions of the form
(548)
kyv(x, h, t) = v(x, 0, t) + hvy (x,
ky0, t),
= Aeky sin(kx !t) ,
h = ✏eky cos(kx !t),
(549)
= Ae sin(kx !t) ,
h = ✏e cos(kx !t),
(549)
and then ignored nonlinear terms. We guess solutions of the form
ky , as
knowing that these satisfy Laplace’s equation (we have ignored terms of the form e ky
ky
knowing that these satisfy
Laplace’s
equation
have
ignored!t),
terms of the form (549)
e , as
Aeky
sin(kx
!t)terms
,
h(we
=
✏e
cos(kx
the surface is at y = 0=and
we
need all
to
disappear
as y ! 1). Putting these
into
the surface is at y = 0 and we need all terms to disappear as y ! 1). Putting these into
the surface boundary conditions (8) and (9) gives
these satisfy
Laplace’s
equation
(we have ignored terms of the form e ky , as
theknowing
surface that
boundary
conditions
(8) and
(9) gives
the surface is at y = 0 and we need all
to disappear as y ! 1). Putting these into
!✏ terms
= Ak,
(550a)
Ak,
(550a)
the surface boundary conditions (8)!✏
and =
(9) gives
2
k ✏
!A = g✏ + k 2 ✏ .
(550b)
!✏
=
Ak,
(550a)
!A
g✏ + ⇢ .
(550b)
⇢2
k ✏
These conditions arise because we have Taylor expanded terms such as
ky
ky
knowing that these satisfy
Laplace’s
terms!t),
of the form e ky , as(549)
= Ae
sin(kxequation
!t) , (we hhave
= ✏eignored
cos(kx
the surface is at y = 0 and wev(x,
need
all=terms
into
h, t)
v(x, 0,to
t) disappear
+ hvy (x, 0,as
t), y ! 1). Putting these
(548)
ky , as
that theseconditions
satisfy Laplace’s
equation
the knowing
surface boundary
(8) and (9)
gives (we have ignored terms of the form e
andsurface
then ignored
Weall
guess
solutions
of the form
the
is at ynonlinear
= 0 and terms.
we need
terms
to disappear
as y ! 1). Putting these into
= Ak,
(550a)
the surface boundary
conditions !✏
(8) and
(9)hgives
ky
= Aeky sin(kx !t) ,
= ✏e
cos(kx
!t),
(549)
2
k ✏
!A !✏
= g✏
+
.
(550b)
= (we
Ak,⇢have
knowing that these satisfy Laplace’s equation
ignored terms of the form e ky , as(550a)
k 2 ✏ as y ! 1). Putting these into
the surface is at y = 0 and we need all terms to disappear
!A = g✏ +
.
(550b)
Eliminating
A we
get theconditions
dispersion(8)relation
the surface
boundary
and (9) gives
⇢
3
k
2 relation
Ak, .
(550a)
Eliminating A we get the dispersion
!!✏
==
gk +
(551)
⇢ k2 ✏
3
!A =
(550b)
2 g✏ + ⇢ k .
! =
+ simplest
.
What are the consequences of this relation?
Ongkthe
level we know that the phase(551)
⇢
speed,
c,
of
a
disturbance
is
given
by
the
relation
c
=
!/k.
Thus
Eliminating A we get the dispersion relation
What are the consequences of this relation? On the simplest level we know that the phase
k k3
22 g
speed, c, of a disturbance is given by
relation
c! the
== gk
+
. c = !/k. Thus
(552)
(551)
k +⇢ ⇢ .
g
k
2 gravity
The What
relative
importance
of
surface
tension
and
in. determining
given(552)
c
=
+
are the consequences of this relation? On the simplest
level we wave
knowmotion
that theisphase
⇢ we have gravity waves, for which
by the Bond number B = k 2 /⇢g. If B < k1 then
Insert
speed, c, of a disturbance
is given by the relation
c = !/k. Thus
o
o
longer
travel faster.
If Bo tension
> 1 thenand
we gravity
have capillary
waves, for
which
shorter
Thewavelengths
relative importance
of surface
in determining
wave
motion
is given
g
k
wavelengths
travelnumber
faster. BFor
water,
thec2Bond
becomes
unity
for wavelengths
=
by the Bond
k 2 /⇢g.
If B+
< .1 then
we have
gravity
waves, (552)
for ofwhich
o =
o number
k
⇢
about
2
cm,
and
this
accounts
for
the
di↵erent
ring
can observe
a stone
longer wavelengths travel faster. If Bo > 1 then patterns
we have you
capillary
waves, when
for which
shorter
andwavelengths
a raindrop
intofaster.
water.
The
relative fall
importance
of surface
tension
gravity
in determining
is given
travel
For water,
theand
Bond
number
becomes wave
unitymotion
for wavelengths
of
2
by the2 Bond
number
Bo = k for
/⇢g.theIfdi↵erent
Bo < 1 ring
thenpatterns
we have you
gravity
which
about
cm, and
this accounts
can waves,
observeforwhen
a stone
longer
wavelengths
travel
faster.
If Bo > relation
1 then we have capillary waves, for which shorter
23.2
Properties
thewater.
dispersion
and
a raindrop
fallof
into
wavelengths travel faster. For water, the Bond number becomes unity for wavelengths of
In describing
anand
arbitrary
wave disturbance,
we consider
the integral
about 2 cm,
this accounts
for the di↵erent
ring patterns
you can observe when a stone
23.2
Properties
the dispersion
Z +1 relation
and a raindrop
fall intoofwater.
ikx
h(x, disturbance,
t) =
ĥkwe
(t)econsider
dk, the integral
In describing an arbitrary wave
23.2 Properties of the dispersion1relation
Z
+1
(553)
So we start with an arbitrary disturbance, and this perturbation just moves without changing shape (although in three dimensions there would be a decay in amplitude due to power
conservation). This is not so for water waves, which have a di↵erent dispersion relation, and
we can highlight the di↵erence between these two cases by considering the wakes behind an
airplane and a boat.
23.2.1
The wake of an airplane
From our previous work with sound waves we know that the equation governing the propagation of a 2D disturbance in air is the wave equation
✓ 2
◆
2
2
@
@
2 2
2 @
=
c
r
=
c
+
,
(555)
@t2
@x2
@y 2
where is some scalar quantity representing the disturbance (e.g., the velocity potential,
the density or the pressure). For an airplane moving through the air we anticipate a solution
that is constant in the frame of reference of the plane. Thus
(x, y, t) = ˜(x
and we have
U2
@2 ˜
@x2
= c2
@2 ˜
@x2
U t, y),
+
@2 ˜
@y 2
!
(556)
.
(557)
Defining the Mach number M = U 2 /c2 , the above equation becomes
(M
2
@2 ˜ @2 ˜
1) 2 + 2 = 0.
@x
@y
(558)
p
If M < 1 we can make a simple change of variables X = x/ 1 M 2 and regain Laplace’s
equation. Thus everything can be solved using our conformal mapping techniques. However,
if M > 1 then the original equation now looks like a wave equation, with y replacing t,
yielding solutions of the form
p
˜(x, y) = (x y M 2 1)
(559)
Thus disturbances are confined to a wake whose half angle is given by
✓ 2
◆
2
@2
@
@
˜
2 2y, t) =
2
(x,
(x + U t, y),
=
c
r
=
c
,
@t2
@x2
@y 2
(555)
and wewhere
have is some scalar quantity representing the disturbance (e.g.,
! the velocity potential,
2˜
2˜
2˜
@
@
@
the density or the pressure). For an
airplane
moving
through
the air we anticipate a solution
2
2
U
=
c
+
.
2
2
2
that is constant in the frame of reference
of
the
plane.
Thus
@x
@x
@y
˜
(x,
2 y,2t) = (x
U t, y),
Defining the Mach number M = U /c , the above equation becomes
and we have
2
@2 ˜
U (M2
@x
!
˜
@
1) @x22 +
+ @y2 2 .= 0.
@x
@y
2 = c2
@@22 ˜
@ 2 2˜
(556)
(557)
(556)
(557)
(558)
p
Defining the Mach number M = U 2 /c2 , the above equation becomes
If M < 1 we can make a simple change of variables X = x/ 1 M 2 and regain Laplace’s
2˜
2˜
@
@
2
equation. Thus everything can be(M
solved
mapping techniques.
1) using
+ our
=conformal
0.
(558) However,
2
2
@x
@y
if M > 1 then the original equation now looks like apwave equation, with y replacing t,
If M
< 1 we can
a simple change of variables X = x/ 1 M 2 and regain Laplace’s
yielding
solutions
of make
the form
equation. Thus everything can be solved using our conformal
mapping techniques. However,
p
if M > 1 then the original equation
with y replacing t,
˜(x, y)now
= looks
(x likey a wave
M 2 equation,
1)
(559)
yielding solutions of the form
p
Thus disturbances are confined ˜to
a
wake
whose
2
(x, y) = (x y Mhalf
1)angle is given by
Thus disturbances are confined to a wake whose half1angle is given by
tan ✓ = p
.
2
1
1M
tan ✓ = p
M2
.
(559)
(560)
(560)
1 exists, and the air ahead doesn’t know
Only a narrow region behind the plane knows it
a narrow region behind the plane knows it exists, and the air ahead doesn’t know
what’sOnly
coming!
what’s coming!
107107
Solitons
credit: Christophe Finot