PHY 711 Classical Mechanics and
Mathematical Methods
10-10:50 AM MWF Olin 103
Plan for Lecture 33:
Chapter 10 in F & W:
Surface waves
1. Water waves in a channel
2. Wave-like solutions; wave speed
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
1
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
2
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
3
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
4
Physics of incompressible fluids and their surfaces
Reference: Chapter 10 of Fetter and Walecka
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
5
Consider a container of water with average height h and
surface h+z(x,y,t); (h  z0 on some of the slides)
p0
z
z
h
y
x
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
6
Euler' s equation for incompressible fluid :
dv
p
p
 f applied 
  gzˆ 
dt


1 p
Assume that v z  v x , v y
 g 
0
ρ z
 p ( x, y, z , t )  p0  g z ( x, y, t )  h  z 
Horizontal fluid motions :
dv x v x
1 p
z


 g
dt
t
 x
x
dv y v y
1 p
z


 g
dt
t
 y
y
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
7
dz(x,y,t)/dt
v(x,y,t)
z0
v(x+dx,y+dy,t)
Continuity condition for flow of incompressible fluid :
z
 z0  v  0
t
v
From horizontal flow relations :
  gz
t
 2z
2
Equation for surface function :

gz

z 0
0
2
t
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
8
Digression:
The form of the continuity equation given on previous (and
subsequent) sides assumes that the transverse cross section of
the channel is either very large or at least uniform. Your text
also considers the case where there is a channel of varying
width:
zx,t
v(x,t)
v(x+dx,t)
h(x)
dx
11/12/2014
Continuity condition:
z

b( x )
   h( x)b( x)v( x, t ) 
t
x
b(x)
For constant b and h:
z

  h  v ( x, t ) 
t
x
PHY 711 Fall 2014 -- Lecture 33
9
For uniform channel:
Surface wave equation:
 2z
2 2
c  z 0
2
t
c  gh
2
More complete analysis finds:
g
2
2
c  tanh( kh)
where k 
k

11/12/2014
PHY 711 Fall 2014 -- Lecture 33
10
More details: -- recall setup -Consider a container of water with average height h
and surface h+z(x,y,t)
p0
z
z
h
y
x
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
11
Equations describing fluid itself (without boundaries)
Euler' s equation for incompressible fluid :
dv v
v
p
1 2

 v  v 
  2 v  v    v   U 
dt t
t

Assume that   v  0 (irrotational flow)  v  
 
  1 2
p
  
 2 v  U    0

 t
 1 2
p

 2 v  U   constant (within the fluid)
t

For the same system, the continuity condition becomes
  v   2   0
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
12
p0
z
z
h
y
x
Within fluid:
0  z  h z
 1 2

 2 v  g  z  h   constant
t
 2  0
At surface:
z  h z
dz z
z
z

 vx
 vy
dt
t
x
y
11/12/2014
(We have absorbed p0
in “constant”)
with z  z  x, y, t 
where vx , y  vx , y  x, y, h  z ,t 
PHY 711 Fall 2014 -- Lecture 33
13
Full equations:
Within fluid:
0  z  h z
 1 2

 2 v  g  z  h   constant
t
 2  0
At surface:
z  h z
dz z
z
z

 vx
 vy
dt
t
x
y
(We have absorbed p0
in “constant”)
with z  z  x, y, t 
where vx , y  vx , y  x, y, h  z , t 
Linearized equations:

For 0  z  h  z : 
 g z  h   0
  2  0
t
dz z
At surface : z  h  z

 v z  x, y , h  z , t 
dt
t
  x, y, h  z , t 

 gz  0
t
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
14
For simplicity, keep only linear terms and assume that
horizontal variation is only along x:
2
2




2
For 0  z  h  z :
    2  2   ( x, z , t )  0
x 
 z
Consider and periodic waveform:  ( x, z, t )  Z ( z ) cos  k ( x  ct ) 
 d2
2
  2  k  Z ( z)  0
 dz

Boundary condition at bottom of tank:
dZ

0  0
dz
11/12/2014
vz ( x,0, t )  0
Z ( z )  A cosh(kz )
PHY 711 Fall 2014 -- Lecture 33
15
For simplicity, keep only linear terms and assume that
horizontal variation is only along x – continued:
At surface :
z  h z
z
  x, h  z , t 
 v z  x, h  z , t   
t
z
  x, h  z , t 

 gz  0
t
 2   x, h  z , t 
z
 2   x, h  z , t 
  x, h  z , t 

g

g
0
2
2
t
t
t
z
For
 ( x, h  z , t )  A cosh(k h  z ) cosk ( x  ct ) 
 2 2
sinh(k h  z ) 
  0
A cosh(k h  z ) cosk ( x  ct )  k c  gk
cosh(k h  z ) 

11/12/2014
PHY 711 Fall 2014 -- Lecture 33
16
For simplicity, keep only linear terms and assume that
horizontal variation is only along x – continued:
g sinh( k h  z ) g
c 
 tanh k h  z 
k cosh(k h  z ) k
g
2
Assuming z  h :
c  tanh kh 
k
2
h=20 m
c
h=10 m

11/12/2014
PHY 711 Fall 2014 -- Lecture 33
17
For simplicity, keep only linear terms and assume that
horizontal variation is only along x – continued:
g
c  tanh kh 
k
2
For   h, c 2  gh
 ( x, z , t )  A cosh kz  cosk ( x  ct ) 
1   x, h  z , t  kc
z ( x, t ) 

A cosh(kh) sin k  x  ct 
g
t
g
Note that for   h, c 2  gh
(solutions are consistent with previous analysis)
11/12/2014
PHY 711 Fall 2014 -- Lecture 33
18
General problem
including
non-linearities
p0
z
z
h
y
x
Within fluid :
0  z  h z
 1 2

 2 v  g  z  h   constant (We have absorbed
t
  2  0
p0 in our constant.)
At surface : z  h  z
dz z
z
z

 vx
 vy
dt
t
x
y
11/12/2014
with z  z  x, y, t 
where v x , y  v x , y  x, y, h  z , t 
PHY 711 Fall 2014 -- Lecture 33
19