physics430_lecture15.ppt Dale. E. Gary Lagrange Equations

Physics 430: Lecture 15
Lagrange’s Equations
Dale E. Gary
NJIT Physics Department

October 21, 2010
 Statement of the problem:
 A surface of revolution is generated as follows: Two fixed
points (x1, y1) and (x2, y2) in the x-y plane are joined by a
curve y = y(x). [Actually, you’ll make life easier if you start
out writing this as x = x(y).] The whole curve is now rotated
about the x axis to generate a surface. Show that the curve
for which the area of the surface is stationary has the form
where xo and yo are constants. (This is often called the soap-bubble problem, since
the resulting surface is usually the shape of a soap bubble held by two coaxial
rings of radii y1 and y2.)
 Solution:
 What we have to minimize now is not a line, but rather a surface, dA = y df ds.
 We could write this
but then both Euler-Lagrange terms will be non-zero. Instead, follow the hint:
Problem 6.19
(x1,y1)
(x2,y2)
y(x)
y
x
],
/
)
cosh[( o
o
o y
x
x
y
y 

ds
,
)
(
1
2
2
2
1
2
2
1 
 



x
x
dx
x
y
y
yds
A 

,
)
(
1
2
2
2
1
2
2
1 
 



y
y
dy
y
x
y
yds
A 


October 21, 2010
 Solution, cont’d:
 We identify
so and
 Squaring, we have
 Solving for
 Integrating
 The statement of the problem wants y(x), so we have to solve for y, to get
where we identify C = yo.
Problem 6.19, cont’d
,
0



x
f
,
)
(
1 2
y
x
y
f 


C
x
x
y
x
x
y
dy
d
x
f
dy
d













2
2
1
0
1
 .
1 2
2
2
2
x
C
x
y 



.
2
2
C
y
C
x



.
arccosh
1
1
o
2
2
2
2
2
x
C
y
C
u
du
C
dy
dy
C
y
C
x
C
y













 


],
/
)
cosh[( o
o
o y
x
x
y
y 


October 21, 2010
6.4 More Than Two Variables
 Aside: Maximum and minimum vs. stationary—geodetics on surfaces.
 When we discussed the shortest path between two points, we examined the
situation in the figure below.
 However, this is not the most general path. Here is one that cannot be
expressed as a function y(x).
 To handle this case, we need to consider a path specified by an independent
variable u along the path, i.e. x = x(u), y = y(u) .
 Now the length of the path is then
 In general, the problem is
 This may seem more difficult than the previous case,
but in fact we just proceed in the same manner with
the “wrong” paths
.
)
(
)
(
2
1
2
2
 

u
u
du
u
y
u
x
L
x
y
x1 x2
y2
y1
(wrong)
)
(
)
(
)
( x
x
y
x
Y 


y = y(x) (right)
2
1
.
0
)
(
)
(
);
(
)
(
.
0
)
(
)
(
);
(
)
(
2
1
2
1








u
u
u
u
y
y
u
u
u
u
x
x






.
]
),
(
),
(
),
(
),
(
[
2
1
 

 du
u
u
y
u
x
u
y
u
x
f
S

October 21, 2010
More Than Two Variables-2
 The stationary path is the one for which
 Following the same procedure as before, we end up with two Euler-Lagrange
equations that the function f must satisfy:
 In Chapter 7, we will meet problems with several variables, with time t as
the independent variable. In these cases, there will be two or more Euler-
Lagrange equations to satisfy (for example, equations for x, y and z).
 One of the great strengths of Lagrangian mechanics is its ability to deal with
cartesian, cylindrical, spherical, and any other coordinate systems with ease.
 In order to generalize our discussion, we write (x, y, z), (r, f, z) or (r, q, f) as
generalized coordinates (q1, q2, …, qN).
.
0
and
0 







S
S
.
0
and
0 













y
f
du
d
y
f
x
f
du
d
x
f

October 21, 2010
The Lagrangian
 In the next chapter, we will use slightly different notation, and refer to our
integrand function
as the Lagrangian
 Note that because the independent variable is t, we can use rather than .
 We will then make the action integral stationary
which requires that L satisfy
 We will develop this more next time.
.
;
;
;
2
2
1
1 N
N q
dt
d
q
q
dt
d
q
q
dt
d
q 


 













 L
L
L
L
L
L
],
,
,
,
,
,
,
,
,
[ 2
1
2
1 t
q
q
q
q
q
q
f N
N





].
,
,
,
,
,
,
,
,
[ 2
1
2
1 t
q
q
q
q
q
q N
N





L
L 
,
]
,
,
,
,
,
,
,
,
[ 2
1
2
1

 dt
t
q
q
q
q
q
q
S N
N





L
i
q
 i
q

October 21, 2010
Advantages Over Newtonian
Mechanics
 We are now going to use the ideas of the previous lecture to develop a new
formalism for mechanics, called Lagrangian mechanics, invented by
Lagrange (1736-1813).
 There are two important advantages of the Lagrange formalism over that of
Newton.
 First, as we have seen, Lagrange’s equations take the same form for any
coordinate system, so that the method of solution proceeds in the same
way for any problem.
 Second, the Lagrangian approach eliminates the forces of constraint, which
we talked about in Chapter 4. This makes the Lagrangian formalism easier
to solve in constrained problems.
 This chapter is the heart of advanced classical mechanics, but it introduces
some new methods that will take getting used to. Once you master it, you
will find it an extraordinarily powerful way to solve mechanics problems.
 To help you master it, we will spend the next three lectures on it, and you
will be solving a number of problems. However, there are many more at
the end of the chapter, and you are advised to try a few more.

October 21, 2010
7.1 Lagrange’s Equations for
Unconstrained Motion
 Recall our general function f, that played such a large role in the Euler-
Lagrange equation
 To make the connection to mechanics, we are now going to show that a
function called the Lagrangian, L = T – U, is the function that, when used in
the Euler-Lagrange equation, leads to the equation of motion for a particle.
 Here, T is the particle’s kinetic energy
and U is the potential energy
 Note that L is NOT the total energy, E = T + U. One can ask why the
quantity T – U should give rise to the equation of motion, but there seems to
be no good answer.
 I do note, however, that the problem can be cast in terms of the total
energy, which gives rise to Hamiltonian mechanics (sec. 7.8 and chap. 13).
),
( 2
2
2
2
1
2
2
1
2
2
1
z
y
x
m
m
mv
T 


 



 r
.
0







y
f
dx
d
y
f
).
,
,
(
)
( z
y
x
U
U
U 
 r

October 21, 2010
Lagrangian
 Obviously, with the dependences of T on the x, y, z velocities, and U on the x,
y, z positions, the Lagrangian depends on both, i.e.
 Let’s look at the first two derivatives
 But note that if we differentiate the second equation with respect to time we
get
 So you can see that the Lagrange equation is manifestly true for a free
particle:
 In Cartesian coordinates (so far) in three dimensions, we have:
].
,
,
,
,
,
,
[ t
z
y
x
z
y
x 


L
L 
.
and
, x
x p
x
m
x
T
x
F
x
U
x

















L
L
,
x
x F
p
x
m
x
m
dt
d
x
dt
d











L
.
x
dt
d
x 




 L
L
.
and
,
,
z
dt
d
z
y
dt
d
y
x
dt
d
x 

 













 L
L
L
L
L
L
Note: this last equality is
only true in an inertial frame.

October 21, 2010
Connection to Euler-Lagrange
 If you compare the Lagrange equations with the Euler-Lagrange equation we
developed in the previous chapter, you see that they are identical.
 Since the Euler-Lagrange equation is the solution to the problem of
stationarity of a path integral, we see that
is stationary for the path followed by the particle.
 This integral has a special name in Physics—it is called the action integral,
and when it is a minimum it is called the principle of least action, although
that is a misnomer in the sense that this could be a maximum, or even an
inflection point.
 The principle is also called Hamilton’s Principle:


2
1
t
t
dt
S L
The actual path that a particle follows between two points 1 and
2 in a given time interval, t1 to t2, is such that the action integral
is stationary when taken along the actual path.


2
1
t
t
dt
S L

October 21, 2010
Generalized Coordinates
 We have now seen that the following three statements are completely
equivalent:
 A particle’s path is determined by Newton’s second law F = ma.
 The path is determined by the three Lagrangian equations (at least in Cartesian
coordinates).
 The path is determined by Hamilton’s Principle.
 Again, the great advantage is that we can prove that Lagrange’s equations
hold for any coordinate system such that, for any value r = (x, y, z) there is a
unique set of generalized coordinates (q1, q2, q3) and vice versa.
 Using these, we can write the Lagrangian in terms of
generalized coordinates as
 In terms of these, we then have
 These are true in any coordinate system, so long as the coordinates are
measured in an inertial frame.
].
,
,
,
,
,
[ 3
2
1
3
2
1 q
q
q
q
q
q 


L
L 
)
(
2
2
1
r
r U
m 
 
L
.
and
,
,
3
3
2
2
1
1 q
dt
d
q
q
dt
d
q
q
dt
d
q 

 













 L
L
L
L
L
L

October 21, 2010
Example 7.1
 For a single particle in two dimensions, in Cartesian coordinates, under some
arbitrary potential energy U(x, y), the Lagrangian is
 In this case, there are two Lagrange equations
 The left side of each equation is just
 The right side of each equation, in turn, is just
 Equating these, we have Newton’s second law
).
,
(
)
( 2
2
2
1
y
x
U
y
x
m 

 

L
.
,
y
dt
d
y
x
dt
d
x 
 








 L
L
L
L
.
, y
x F
y
U
y
F
x
U
x













 L
L
.
, y
m
y
m
dt
d
y
dt
d
x
m
x
m
dt
d
x
dt
d















 L
L
.
or
, a
F m
y
m
F
x
m
F y
x 

 




October 21, 2010
Generalized Force and Momentum
 Notice that the left side term in Cartesian coordinates gives the force
 When this is expressed in terms of generalized coordinates (which, for
example, could be angular coordinates) this term is not necessarily a force
component, but it plays a role very like a force, and indeed is called the
generalized force.
 Likewise, the right side in terms of generalized coordinates plays the role of
a momentum, and is called the generalized momentum.
 Thus, the Lagrange equation can be read as
generalized force = rate of change of generalized momentum.
.
, y
x F
y
U
y
F
x
U
x













 L
L
.
force)
d
generalize
of
component
th
(i
qi


L
.
momentum)
d
generalize
of
component
th
(i
qi




L
,
i
i q
dt
d
q 




 L
L

October 21, 2010
Example 7.2
 Let’s repeat example 7.1, but in polar coordinates. In this case, the potential
energy is U(r, f), and you should be able to write down the kinetic energy
directly without much thought as
so the Lagrangian is just
 In this case, there are two Lagrange equations
 Inserting the above Lagrangian into these equations gives
 The first of these equations says
 The second equation is best interpreted by recalling the gradient in polar
coordinates
).
( 2
2
2
2
1
f

 r
r
m
T 

.
,
f
f 
 








 L
L
L
L
dt
d
r
dt
d
r
  .
2
, 2
2
2
f
f
f
f
f 






 mr
r
mr
mr
dt
d
U
r
m
r
U
mr 









 ,
2
f


 r
r
m
Fr 

).
,
(
)
( 2
2
2
2
1
f
f r
U
r
r
m 

 

L
centripetal acceleration rw2.
.
ˆ
1
ˆ φ
r
f







U
r
r
U
U
angular momentum mr2w.
 .
2
f
f
f

mr
dt
d
rF
U





torque.

October 21, 2010
N Free Particles
 It should be obvious how to extend these ideas to a larger number of
particles. In the case of two particles, at positions r1 and r2, for example,
Newton’s second law says each component of each particle obeys the
equations
 In the Lagrangian formalism, we have the equivalent Lagrange equations
 We could likewise write these in generalized coordinates as
 An important example we will use repeatedly in Chapter 8 is to replace r1
and r2 with RCM = (m1r1 + m2r2)/(m1 + m2), and r = r1  r2.
 For N particles, then, there are 3N Lagrange equations.
.
,
,
,
,
,
2
2
2
2
2
2
1
1
1
1
1
1
z
z
y
y
x
x
z
z
y
y
x
x
p
F
p
F
p
F
p
F
p
F
p
F












.
,
,
,
,
,
2
2
2
2
2
2
1
1
1
1
1
1
z
dt
d
z
y
dt
d
y
x
dt
d
x
z
dt
d
z
y
dt
d
y
x
dt
d
x




































L
L
L
L
L
L
L
L
L
L
L
L
.
,
,
,
6
6
2
2
1
1 q
dt
d
q
q
dt
d
q
q
dt
d
q 


 













 L
L
L
L
L
L

October 21, 2010
7.2 Constrained Systems Example
 We have said that the great power of the Lagrange formalism is that
constraint forces disappear from the problem.
 Before getting into the general case, let’s do a familiar problem—instead of a
free particle, let’s consider one that is tied to the ceiling, i.e. the simple
pendulum.
 The pendulum bob moves in both x and y, but it moves under the constraint
 A perfectly valid way to proceed might be to eliminate the y variable,
and express everything in terms of x, e.g. replace y with
 However, it is much simpler to express the problem in terms of the
natural coordinate f.
 The first task is to write down the Lagrangian L = T – U in terms of f.
Clearly, for this problem U = mgh = mgl(1  cos f). Likewise, the kinetic
energy is
 The relevant Lagrange equation is and you can basically write
down the solution
.
2
2
y
x 


.
2
2
x
y 
  f
h
.
2
2
2
1
2
2
1
f


m
mv
T 

,
f
f 




 L
L
dt
d
.
sin 2
f
f 


 m
mg 
 Equivalent to G = I

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