Physics 312: Lecture 3 Lagrange’s Equations

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 acurve y = y(x). [Actually, you’ll make life easier if you startout writing this as x = x(y).] The whole curve is now rotatedabout the x axis to generate a surface. Show that the curvefor 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 dds. 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[( ooo yxxyy

ds

,)(1222

1

22

1 x

xdxxyyydsA

,)(1222

1

22

1 y

ydyyxyydsA

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 2yxyf

Cx

xy

x

xy

dy

d

x

f

dy

d

22 1

01

.1 2222 xCxy

.22 Cy

Cx

.arccosh 11

o2

2

222x

C

yC

u

duC

dydy

Cy

Cx

C

y

],/)cosh[( ooo yxxyy

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 withthe “wrong” paths

.)()(2

1

22 u

uduuyuxL

x

y

x1 x2

y2

y1

(wrong) )()()( xxyxY

y = y(x) (right)

21

.0)()( );()(

.0)()( );()(

21

21

uuuuyy

uuuuxx

.]),(),(),(),([2

1 duuuyuxuyuxfS

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), (, , z) or (r, , ) as generalized coordinates (q1, q2, …, qN).

. 0 and 0

SS

.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.

.;;;2211 NN qdt

d

qqdt

d

qqdt

d

q

LLLLLL

],,,,,,,,,[ 2121 tqqqqqqf NN

].,,,,,,,,[ 2121 tqqqqqq NN LL

,],,,,,,,,[ 2121 dttqqqqqqS NN L

iq iq

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).

),( 222212

212

21 zyxmmmvT r

.0

y

f

dx

d

y

f

).,,()( zyxUUU 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:

].,,,,,,[ tzyxzyx LL

. and , xx pxmx

T

xF

x

U

x

LL

,xx Fpxmxmdt

d

xdt

d

L

.xdt

d

x

LL

. and , ,zdt

d

zydt

d

yxdt

d

x

LLLLLL

Note: this last equality isonly 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

tdtS 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

tdtS 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.

].,,,,,[ 321321 qqqqqq LL

)(221 rr Um L

. and , ,332211 qdt

d

qqdt

d

qqdt

d

q

LLLLLL

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

).,()( 2221 yxUyxm L

. ,ydt

d

yxdt

d

x

LLLL

. , yx Fy

U

yF

x

U

x

LL

. , ymymdt

d

ydt

dxmxm

dt

d

xdt

d

LL

.or , aF mymFxmF yx

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.

. , yx Fy

U

yF

x

U

x

LL

.force) dgeneralize of component th (iqi

L

.momentum) dgeneralize of component th (iqi

L

,ii qdt

d

q

LL

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, ), 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

).( 22221 rrmT

. ,

LLLL

dt

d

rdt

d

r

.2 , 222

mrrmrmrdt

dUrm

r

Umr

,2 rrmFr

).,()( 22221 rUrrm L

centripetal acceleration r2.

.ˆ1

ˆ φr

U

rr

UU

angular momentum mr2.

.2

mrdt

drF

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.

. , ,

, , ,

222222

111111

zzyyxx

zzyyxx

pFpFpF

pFpFpF

. , ,

, , ,

222222

111111

zdt

d

zydt

d

yxdt

d

x

zdt

d

zydt

d

yxdt

d

x

LLLLLL

LLLLLL

. , , ,662211 qdt

d

qqdt

d

qqdt

d

q

LLLLLL

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 thenatural coordinate .

The first task is to write down the Lagrangian L = T – U in terms of . Clearly, for this problem U = mgh = mgl(1 cos ). Likewise, the kinetic energy is

The relevant Lagrange equation is and you can basically write down the solution

.22 yx

.22 xy

h

.22212

21 mmvT

,

LL

dt

d

.sin 2 mmg Equivalent to = I