lecture notes 205

Lecture Notes

Advanced Dynamics

Alejandro L. Garcia

San Jose State University

December 5, 2012

PHYSICS 205 / Fall 2012

c2012 by Alejandro L. GarciaDepartment of Physics, San Jose State University, San Jose CA 95192-0106

Creative Commons Attribution- Noncommercial-Share Alike 3.0 United States License

Preface

These lecture notes should supplement your notes, especially for those of you (like myself) who are poor

note takers. But I don't recommend reading these notes while I lecture; when you go to see a play you

don't take the script to read along. Continue taking notes in class but relax and don't worry about

catching every little detail. WARNING: THESE NOTES ARE IN DRAFT FORM AND

PROBABLY HAVE NUMEROUS TYPOS; USE AT YOUR OWN RISK.

Alejandro L. Garcia

1

Chapter 1

Equations of Motion

Generalized Coordinates (x1)Lecture 1

We begin with mechanical systems composed of one or more particles; rigid body mechanics will be

considered in Chapter VI.

A particle has position r, velocity v = dr=dt = _r, and acceleration, a = _v = v. Sometimes we'll use

a Cartesian coordinate system so

r = x^i+ y^j+ zk^

but we'll often also use other systems (polar, spherical, etc.) or not specify any specic coordinates.

The interactions on the particles are:

Explicit forces (e.g., gravity) Implicit constraints (e.g., rigid supports)

The simple pendulum is a good example with both types of interactions. We will only consider con-

straints that are holonomic, that is, they may be expressed as

f(r1; : : : ; rN ; t) = 0

For example, for a simple pendulum of length ` the constraint is jrj ` = 0. An example of a non-holonomic constraint would be for a rigid ball bouncing o a rigid surface since the constraint is an

inequality (distance between the center of the ball and the surface is greater than or equal to the

radius of the ball). Constraints may be independent of time (scleronomous) or dependent on time

(rheonomous).

A system of N particles has 3N Cartesian coordinates but due to implicit constraints these may

not all be independent. Suppose that, given the constraints, the positions of the particles are uniquely

specied by the generalized coordinates, q1; q2; : : : ; qs, where s 3N . The system then has s degreesof freedom.

Example For a system of two particles (N = 2) the cartesian coordinates are (x1; y1; z1; x2; y2; z2).

Suppose that the particles are constrained to move in the xy-plane (see Figure 1.1); then the number

of degrees of freedom is reduced from 6 to 4.

I encourage you to introduce these terms into your daily conversation (e.g., \My love for you is scleronomous.")

2

CHAPTER 1. EQUATIONS OF MOTION 3

Figure 1.1: Two masses in the xy-plane with a constraint of xed separation `.

Let's introduce a further constraint that the separation between the particles is xed to be a distance

`, that is, p(x2 x1)2 + (y2 y1)2 = `

We can now take the s = 3 generalized coordinates to be,

q1 = x1; q2 = y1; q3 = = arctanx2 x1y2 y1

where is simply the angle between the +y-axis and the line connecting the particles.

We can also express the original coordinates in terms of the generalized coordinates,

x1 = q1; y1 = q2; x2 = q1 + ` sin q3; y2 = q2 + ` cos q3

This allows us to relate the Cartesian velocities to the generalized velocities, for example,

_x2 = _q1 + ` _q3 cos q3

Expressions like this are convenient for transforming quantities like kinetic and potential energy, which

are easier to express in Cartesian coordinates, into functions of the generalized coordinates.

Note that the q's can be a mixture of position, angle, or even more complicated forms. They do not

have to have units of length; that's what we mean by \generalized."

One of the tasks (though not the only one) of classical mechanics is to determine the motion of a

system from a given initial condition. That is, we'd like to be able to nd q1(t); : : : ; qs(t) given the

initial positions, q1(0); : : : ; qs(0), and velocities, _q1(0); : : : ; _q(0). The general approach for this task will

be,

Step A Specify the implicit interactions (i.e., constraints) and establish the convenient generalized

coordinates q1; : : : ; qs.

Step B Specify the explicit interactions (i.e., forces) and establish the mechanical properties by for-

mulating the Lagrangian for the system, L(q1; : : : ; qs; _q1; : : : ; _qs; t).


Step C Using the Lagrangian, obtain the equations of motion, which will be a set of ordinary dier-

ential equations (ODEs) of the form, fj(qi; _qi; qi; t) = 0 for i; j = 1; : : : ; s.

Step D Solve this set of second order ODEs, xing the 2s constants in the solution using the initial

conditions. For complicated systems the solution is obtained numerically.

Example Let's take our previous example and simplify it further to formulate the simple pendulum.

Our mechanical system is a particle of mass m constrained to move in the xy-plane at a xed distance

` from the origin; there is a constant force on the particle equal to mgj^. Following the recipe outlinedabove,

Step A: To satisfy the constraint we'll use as our generalized coordinate q = where the angle is

from the +y-axis. Notice that q = 0 is highest position and q = is the lowest position. Comparing

with our previous example we're taking q1 = 0; q2 = 0, q3 = q.

Step B: The Lagrangian for this system happens to be,

L(q; _q; t) =1

2m`2 _q2 mg` cos q

Later you'll see how this is formulated; note that the rst term is given by the rotational kinetic energy

and the second is from the gravitational potential energy. For this system the Lagrangian does not

depend explicitly on time.

Step C: In the next section we'll show that the equations of motion are obtained from Lagrange's

equation,d

dt

@L

@ _q

@L

@q= 0

For the Lagrangian in our example the equation of motion turns out to be,

m`2q mg` sin q = 0

Later in this chapter you'll see how we obtain this from the Lagrangian.

Step D: This ODE is easy to solve numerically but not so simple to solve analytically. However,

suppose we're interested in the motion of small oscillations when the pendulum swings near the lowest

position, q0 = . We can introduce a new coordinate, x = q q0 and make use of the \small angle"approximation,

sin q = sin(x+ ) = sinx xsince we assume that x is a small displacement.y The equation of motion simplies to,

x = g`x

so the solution is,

x(t) = C1 sin(!t) + C2 cos(!t)

where ! =pg=`. The constants C1 and C2 are xed by the initial conditions. For example, if x(0) = 0,

_x(0) = 0 then C1 = 0, C2 = 0. We'll use the small oscillation approximation extensively in Chapter V

for a variety of systems.

yThis x is not the Cartesian coordinate.


We won't always do all of these steps when studying a particular system. In Chapter I we focus on

steps A and B but will often skip the explicit calculation of step C since it's just \turning the crank",

evaluating derivatives of the Lagrangian. Step D is the hardest and in Chapters II through V we'll

explore a variety of tools for either solving the equations of motion approximately (as in the example

above) or for predicting certain properties, such as using conservation of energy.

Finally, we will discover some unexpected properties in mechanical systems. Let's return to the

example we've been looking at and modify it so that the pendulum's pivot point oscillates up and

down. That is, our original generalized coordinates are rewritten as, q1 = 0; q2 = f(t); q3 = q, where

f(t) is a given periodic function (e.g., f(t) = a sin!at). The Lagrangian is easy to formulate and

we'll study this system in sections x27 and x30. The results will surprise you; for a preview, see:http://www.youtube.com/watch?v=rwGAzy0noU0 .

Principle of Least Action (x2)Lecture 2

Consider a mechanical system with generalized coordinates q1; : : : ; qs and velocities _q1; : : : ; _qs (or briey,

q and _q). The Principle of Least Actionz states that for every mechanical system there exists a function,

L(q; _q; t) = L(q1; : : : ; qs; _q1; : : : ; _qs; t)

called the Lagrangian from which we dene the action, S, as

S =

Z t2t1

L(q; _q; t)dt

This action is a minimum when the integral is evaluated taking the path from q(1) = q(t1) to q(2) = q(t2)

that the mechanical system follows by its physical motion.

For example, consider free-fall motion in one-dimensional motion (e.g., a particle falling straight

downward). Let's look at two dierent paths, qA and qB , which have the same starting and ending

positions. Path A happens to be the correct physical motion while Path B is the kind of motion you

might see in a cartoon (the particle remains at rest for a time and then falls with constant speed).

These two paths are illustrated in Fig. 1.2.

The Lagrangian for this system (as we'll later derive) happens to be,

L(q; _q; t) = 12m _q2 mgq

Notice that the rst term is the kinetic energy and the second term is the gravitational potential energy.

By the Principle of Least Action,Z t2t1

L(qA(t); _qA(t); t)dt Z t2t1

L(qB(t); _qB(t); t)dt

While this integral form of Hamilton's principle is beautiful and fundamental, it's not very practical

if we want to nd qA(t), the physical trajectory of the motion. Landau and Lifshitz show that by the

calculus of variations the Principle of Least Action leads to the result that qA(t) is the solution to

Lagrange's equations,d

dt

@L

@ _qi @L@qi

= 0 i = 1; : : : ; s

zAlso known as Hamilton's Principle.


Figure 1.2: Illustration of two possible paths, A and B, for falling motion.

I'll sketch out the proof at the end of this lecture but rst let's see some examples of how Lagrange's

equations give us the ordinary dierential equations (ODEs) that are the equations of motion for the

dynamics.

Example Let's return to the example of free-fall motion. We'll evaluate each term in Lagrange's

equations step-by-step,

@L

@ _q=

@

@ _q

12m _q

2 mgq = m _qd

dt

@L

@ _q

= mq

@L

@q= mg

sod

dt

@L

@ _q @L

@q= 0

gives the ODE,

mq +mg = 0

or simply, q = g. The general solution of this ODE is

q(t) = 12gt2 + C1t+ C2

where the constants C1 and C2 may be xed by the initial conditions, q(0) and _q(0).

Example Let's consider a more complicated example: the co-planar double pendulum (see pg. 11 in

Landau and Figure 1.3). The generalized coordinates are the angles from the vertical, that is, q1 = 1


Figure 1.3: Double pendulum illustration.

and q2 = 2. Landau writes the Lagrangian for this system as,

L = 12 (m1 +m2)`21_21 +

12m2`

22_22

+m1`1`2 _1 _2 cos(1 2)+(m1 +m2)g`1 cos1 +m2g`2 cos2

Later we'll see how to construct this Lagrangian ourselves but for now we'll take it as given.

There are two equations of motion; let's work out the rst one by evaluating

d

dt

@L

@ _1 @L@1

= 0

The expressions are a bit messy so let's do this one step at a time. First,

@L

@ _1= (m1 +m2)`

21_21 +m2`1`2

_2 cos(1 2)

sod

dt

@L

@ _1= (m1 +m2)`

2121 +m2`1`2

2 cos(1 2)m2`1`2 _2 sin(1 2)( _1 _2)

Finally, we need to evaluate,

@L

@1= m2`1`2 _1 _2 sin(1 2) (m1 +m2)g`1 sin1

Combining all the above into Lagrange's equation gives,

(m1 +m2)`2121 +m2`1`2

2 cos(1 2)m2`1`2 _2 sin(1 2)( _1 _2)+m2`1`2 _1 _2 sin(1 2) + (m1 +m2)g`1 sin1 = 0


which is a rather complicated ODE containing 1, 2, _1, _2, 1, and 2. But that's not all because we

need to evaluate the second Lagrange equation, which yields a similarly complicated ODE. Together this

pair of ODEs describes the motion; although they're too complicated to yield an analytic solution (in

fact, the motion can be chaotic) they can easily be solved numerically to compute 1(t) and 2(t). See

the Wikipedia entry for \double pendulum" for numerical animations and photographs from laboratory

experiments.

Although the general solution is rather complicated it's interesting to check certain limiting cases.

For example, in the limit that `2 ! 0 you can verify that the equation of motion becomes,

(m1 +m2)`211 + (m1 +m2)g`1 sin1 = 0

or1 = g

`1sin1

which, as expected, is the equation for a simple pendulum. What do you get if m2 ! 0?

General Properties of Lagrangians

The next few sections in Landau discuss how to formulate the Lagrangian for various mechanical

systems. Before doing so, let's outline some of the general properties of Lagrangians,

* If systems A and B are independent then the two systems, taken together as a single system, have

a Lagrangian, LA+B = LA + LB . This can be useful for checking limiting cases, such as neglecting the

coupling between systems.

* The Lagrangian for a system is dened up to a multiplicative constant, c. This is easily seen by

inspection from Lagrange's equation,

d

dt

@L

@ _qi @L@qi

= 0 i = 1; : : : ; s

which gives the same equations of motion if we replace L) cL. This constant may be viewed as xingthe units of mass.

* Similarly, the Lagrangian is dened up to an additive constant; the equations of motion are

unchanged if we replace L ) L + c. This is the same arbitrary additive constant that appears inpotential energy.

* Landau establishes the more general result that the Lagrangian is dened up to an additive function

of the form,

L ) L+ ddtf(q1; : : : ; qs; t)

= L+dq1dt

@f

@q1+ : : :+

dqsdt

@f

@qs+@f

@t

= L+ _q1@f

@q1+ : : :+ _qs

@f

@qs+@f

@t

This result is useful for simplifying Lagrangians by allowing us to drop any terms of this form. An

important special case is that the Lagrangian is dened up to an additive function of time, so the

equations of motion are unchanged if L) L+ g(t).


Optional Proof Consider two Lagrangians that dier by the total time derivative of an arbitrary

function of the form f(q; t). That is, we may write them as,

L0(q; _q; t) = L(q; _q; t) +d

dtf(q; t)

(see eqn. (2.8)). Using L0 we obtain the equation of motion from Lagrange's equation,

d

dt

@L0

@ _q @L

0

@q= 0

But since L0 = L+ df=dt,

d

dt

@L

@ _q @L

@q+

d

dt

@

@ _q

df

dt

@

@q

df

dt

= 0

But@

@ _q

df

dt=

@

@ _q

@f

@q_q +

@f

@t

=

@f

@q

sod

dt

@L

@ _q @L

@q+

d

dt

@f

@q

@

@q

df

dt

= 0

The last two terms cancel since

d

dt

@f

@q=

d

dtg(q; t) =

@g

@q_q +

@g

@t

=@2f

@q2_q +

@2f

@q@t

and

@

@q

df

dt

=

@

@q

@f

@q_q +

@f

@t

=

@2f

@q2_q +

@2f

@q@t

so the equation of motion obtained from L0 is the same as that obtained from L.

Derivation of Lagrange's Equations

Now that we're a bit familiar with Lagrangians and Lagrange's equations, let's go back a few steps

and see how Lagrange's equations are a consequence of the Principle of Least Action. This is only a

sketch of the derivation since details are found in every mechanics book; for variety we'll formulate the

derivation a bit dierently than Landau.

Take a system with a single generalized coordinate, q. Call qA(t) the physical trajectory and qB(t)

some other (unphysical) trajectory for the motion. The dierence between these two paths is

q(t) = qB(t) qA(t)

which is called the variation. Both paths start and end at the same points, that is, qA(t1) = qB(t1) and

qA(t2) = qB(t2), which means q(t1) = q(t2) = 0.


The variation in the action is

S = SB SA =Z t2t1

L(qB(t); _qB(t); t)dtZ t2t1

L(qA(t); _qA(t); t)dt

=

Z t2t1

[L(qB(t); _qB(t); t) L(qA(t); _qA(t); t)]dt

=

Z t2t1

[L(qA(t) + q(t); _qA(t) + _q(t); t) L(qA(t); _qA(t); t)]dt

This expression is just begging to be Taylor expanded so let's do so,

L(qA(t) + q(t); _qA(t) + _q(t); t) = L(qA(t); _qA(t); t) + q@L

@q+ _q

@L

@ _q+ : : :

The Principle of Least Action tells us that the action, S, is a minimum when qA is truly the physical

path so we can consider small variations and neglect higher order terms.

Combining the above gives,

S =

Z t2t1

q@L

@q+ _q

@L

@ _q

dt

=

Z t2t1

q@L

@q+

d

dtq

@L

@ _q

dt

To pull the q out of the square brace we perform integration by parts on the second term and get,

S =

q@L

@ _q

t2t1

+

Z t2t1

q

@L

@q+

d

dt

@L

@ _q

dt

The rst term on the right hand side is zero since q(t1) = q(t2) = 0. Furthermore, by the Principle

of Least Action S = 0; this is the same as the slope being zero at an extremum since,

f = f(x0 + x) f(x0) = dfdx

x=x0

x

so f = 0 if x0 is an extremum. Since S = 0 for arbitrary q(t) this implies that the expression in

the square brace is zero. By now that term in the square brace should be familiar since it's Lagrange's

equation.

Inertial Frames and Free Particle Lagrangian (x3,4)Lecture 3

In these two sections Landau establishes the Lagrangian for a free particle, that is, a particle with no

forces acting on it. Along the way he also obtains some fundamental results in mechanics such as the

law of inertia and galilean relativity.

The general form for the Lagrangian for a single particle is L(r;v; t). We will consider the motion

of a free particle in an inertial frame of reference and use the following three properties:

Space is homogenous; the equations of motion are unchanged if we shift the origin. Space is isotropic; the equations of motion are unchanged if we rotate the coordinate axes. Time is homogeneous; the equations of motion are unchanged if we shift the time origin (t = 0).


Figure 1.4: Free particle motion as measured by two dierent coordinate systems in single frame of

reference.

Consider two coordinate systems in our inertial frame (see Figure 1.4).x We'll write the velocity ofthe particle in spherical coordinates but the position in Cartesian coordinates. Since the equations of

motion are the same in the two coordinate systems then we have,

L(r;v; t) = L(r;v; t)

or

L(x; y; z; v; ; ; t) = L(x; y; z; v; ; ; t)

Due to the spatial shift x 6= x, y 6= y, and z 6= z; due to the rotation of the coordinates 6= and 6= ; nally, due to the time shift t 6= t. Yet the magnitude of the velocity is the same for thetwo coordinate systems so v = v, which means that for a free particle the Lagrangian can only be afunction of v.

Law of Inertia

Recall Lagrange's equations,d

dt

@L

@ _qi

@L@qi

= 0

Taking q1 = x, q2 = y, and q3 = z, we have,

d

dt

@L

@ _x

@L

@x= 0

xThe two coordinate systems are in the same inertial frame; later we'll look at transforming from one inertial frameto another.


with similar expressions for y and z. We may write these component equations in a compact form as,

d

dt

@L

@v

@L

@r= 0

where the vector derivative notation means,

@f

@r=

@f

@x

i+

@f

@y

j+

@f

@z

k

which is the same as the gradient operator, rf . We just established that for a free particle L(v) so@L=@r = 0; from Lagrange's equation this implies that,

d

dt

@L

@v

= 0 so

@L

@v= constant

This presents three possibilities for the free particle Lagrangian:

L = constant independent of v. But this is equivalent to L = 0 (why?), which is the null result. L = (constant)v. But this contradicts our result that L depends only on v = jvj. L(v) is an arbitrary function of v but v is constant in time.

The last option is the one we're forced to take. Furthermore, since the direction of the velocity is absent

from the equation of motion then it must be constant so not only is v constant but so it v. This is

Newton's First Law of Motion, that is, the law of inertia.

Galilean Relativity

Landau reminds us that there isn't a single \holy" inertial frame of reference. Consider our original

frame of reference F and another frame of reference F 0 such that F 0 moves with constant velocityV relative to F . Both are inertial frames and the coordinates in the two are related by the galileantransformations,

r = r0 +Vt v = v0 +V a = a0

We'll make use of this result to nish the derivation of the Lagrangian for a free particle.

Free-particle Lagrangian

Consider a galilean transformation V = where is innitesimally small. The equations of motionof a free particle are the same in the two frames of reference so L(v) and L0(v0) must be equivalent. Aswe know, that doesn't mean that they are identical since the Lagrangian is specied up to an additive

function of the form,

L0(v0) = L(v) +d

dtf(r; t) = L(v) +

@f

@r drdt

+@f

@t

= L(v) +@f

@r v + @f

@t()

where f(r; t) is arbitrary.


The next step is to express L0(v0) as L0(v + g(v) ) in order to Taylor expand this and relate andL0(v0) to L(v). To determine the form of the function g from our galilean transformation we note that

(v0)2 = v0 v0 = (v + ) (v + )= v v + 2v + = v2 + 2v

since jj is innitesimally small. Thus,

v0 =pv2 + 2v = v

p1 + 2v =v2 = v + v =v

where we used the fact thatp1 + x = 1+ 12x in the limit that jxj ! 0. We now see that for this galilean

transformation g(v) = v=v so

L0(v0) = L0(v + v =v) = L(v) + v v

dL

dv()

where we performed a Taylor expansion and discarded the terms that are quadratic in . Note that

L0(v) = L(v) thought L0(v0) 6= L(v) (why?).Comparing equations (*) and (**) we see that,

@

@rf(r; t) =

v

d

dvL(v)

The left hand side is only a function of r and t while the right hand side is only a function of v. The

only (non-null) solution for L(v) occurs if the left hand side is a constant so

1

v

d

dvL(v) = constant

or

L(v) = Cv2

where C is a constant. In order to preserve our traditional denition of mass (given by F = ma) it

turns out that C = 12m.

L(v) =1

2mv2 (Free particle Lagrangian)

Finally, for N non-interacting particles the Lagrangian is

L =1

2m1v

21 + : : :+

1

2mNv

2N

by the additivity of Lagrangians of independent systems.

The Lagrangian for a free particle in cartesian coordinates is,

L = 12mv2 = 12m( _x

2 + _y2 + _z2)

Writing Lagrange's equations in these coordinates,

d

dt

@L

@ _x

@L

@x= 0

we get x = y = z = 0, as expected from the law of inertia.


In cylindrical coordinates we replace x and y with r and using the transformation,

x = r cos y = r sin

To nd v2 in these coordinates we evaluate

_x = _r cos r _ sin_y = _r sin+ r _ cos

so

v2 = _x2 + _y2 = _r2 + r2 _2

Finally, this gives the Lagrangian of a free particle in cylindrical coordinates as

L =1

2m( _r2 + r2 _2 + _z2)

A similar calculation in spherical coordinates gives,

L =1

2m( _r2 + r2 _2 + r2 _2 sin2 )

There are more complicated coordinate systems but the three above are the most common.

In some cases we impose a constraint on the coordinates, for example, if a free particle is constrained

to move along the x-axis then the Lagrangian is simply,

L =1

2m _x2

since the constraint implies _y = _z = 0. Another example would be a free particle that's constrained to

move in a circle of radius R; from the result above for cylindrical coordinates,

L =1

2mR2 _2

since the constraint implies r = R and _r = 0. From Lagrange's equation we nd that = 0 so

the particle moves with constant angular speed ( _ = constant). Let's nish the lecture with a more

complicated example of motion with a constraint.

Example A particle (mass m) is constrained such that it is a constant distance `, from a pivot

point. That pivot point rotates in a circle (radius a) with constant angular velocity . The motion is

further constrained to be in the x y plane. Find the Lagragian (note that this is similar to Problem3a, pg. 11, in Landau but without gravity).

Solution The geometry is illustrated in Figure 1.5 The Lagrangian for a particle in the x y planeis

L =1

2m( _x2 + _y2)

By some basic trig we can convert to the generalized coordinate by the transformation

x = x1 + x2 = a cos t+ ` sin

y = y1 + y2 = a sin t ` cos


Figure 1.5: Particle constrained to move a xed distance from a pivot that uniformly rotates in a circle.


so

_x = a sin t+ ` _ cos_y = a cos t+ ` _ sin

Thus

L(; _; t) =1

2ma22 sin2 t 2a` _ sin t+ `2 _2 cos2

+a22 cos2 t+ a` _ cos t+ `2 _2 sin2

=1

2ma22 + 2a` _ sin( t) + `2 _2

This expression may be simplied further; rst, the term a22 may be dropped since it's a constant.

Second, we may rewrite the middle term using,

_ sin( t) = sin( t) ddt

cos( t)

then we may drop the cosine term since it is of the form d=dtf(q; t). Finally, we nd that the Lagrangian

simplies to the form,

L(; _; t) = am`2 sin( t) + 12m`2 _2

which matches Landau's expression.

Lagrangian for a System of ParticlesLecture 4

We already derived the Lagrangian for a single free particle, L = 12mv2 and by the additivity of

independent Lagrangians, for a system of N independent (i.e., non-interacting) free particles,

L =1

2m1v

21 + : : :

1

2mNv

2N

We now consider closed systems in which particles interact among themselves but with no external

interactions. We take the form of this interaction to be such that the Lagrangian for the closed system

of particles may be expressed as,

L =1

2m1v

21 + : : :

1

2mNv

2N U(r1; : : : ; rN )

= T (v1; : : : ; vN ) U(r1; : : : ; rN )

We call T the kinetic energy and U the potential energy. In classical, non-relativistic mechanics all

physical interactions of interest for a closed system may be represented by a potential energy of this

form.

Example The Lagrangian for two similar particles interacting by an elastic (spring) attraction is,

L =1

2mv21 +

1

2mv22

1

2k(jr1 r2j `)2


where k is the spring constant and ` is the rest length (recall that the spring potential energy is U = 12kx2

where x is the extension from the rest position).

From Lagrange's equations,

d

dt

@L

@va

@L@ra

= 0 a = 1; : : : ; N

and the form of the Lagrangian as L = T (v) U(r),d

dt

@T

@va= @U

@ra

Since@

@va

1

2mv2a

=

@

@va

1

2mva va

= mva

we have

m _va = @U@ra

so we identify the net force on particle a as

Fa = @U@ra

in agreement with Newton's Second Law of Motion.

Example From the earlier example of two particles with an elastic interaction, the force on particle

1 is,

F1 = @U@r1

= @@r1

1

2k(jr1 r2j `)2

= 1

2k

@

@x1

p(x1 x2)2 + (y1 y2)2 + (z1 z2)2 `

2i+ (Similar terms for y and z)

= k (jr1 r2j `) r1 r2jr1 r2j= k(jr12j `)r^12

where r12 = r1 r2 is the separation vector. Note that evaluating the force on the other particle givesF2 = F1, as expected from Newton's Third Law of Motion.

Generalized Coordinates

The simple form of the Lagrangian in vector form does not necessarily carry over to generalized coor-

dinates. Given that the constraints may be written in the form,

xa = fa(q1; : : : ; qs)

with similar expressions for ya and za we have that

_xa =sX

k=1

@fa@qk

dqkdt

=sX

k=1

@fa@qk

_qk


To nd the expression for the kinetic energy in terms of the generalized coordinates we use,

_x2a =sX

i=1

sXk=1

@fa@qi

@fa@qk

_qi _qk =

sXi=1

sXk=1

ga;i;k(q1; : : : ; qs) _qi _qk

and similar expressions for the y and z components in

L =1

2

NXa=1

ma( _x2a + _y

2a + _z

2a) U(x1; : : : ; zN )

From the above we have,

L =1

2

sXi=1

sXk=1

aik(q1; : : : ; qs) _qi _qk U(q1; : : : ; qs)

A simple example is the Lagrangian for a free particle in cylindrical coordinates,

L =1

2m_r2 + r2 _2 + _z2

The generalized coordinates are q1 = r, q2 = , and q3 = z; by inspection

a11 = m; a22 = mq21 ; a33 = m

and aik = 0 if i 6= k. For more complicated examples, see the worked problems on pages 11 and 12 ofLandau.

Open Systems

We can extend the formulation above to open systems by the decomposition:

System A The system of interest

System B A large external system coupled to system A

System A+B A closed system that is the combination of systems A and B

Systems A and B interact but the motion of system B is assumed to be unaected (in any signicant

way) by system A (e.g., A is an apple and B is the Earth).

The Lagrangian for system A+B is

LA+B = TA(qA; _qA) + TB(qB ; _qB) U(qA; qB)

Since system B is unaected by system A we may take qB(t) and _qB(t) as given functions of time and

write,

LA+B = TA(qA; _qA) + TB(qB(t); _qB(t)) U(qA; qB(t))= TA(qA; _qA) + TB(t) U(qA; t)= TA(qA; _qA) U(qA; t)

where we may drop the term TB(t) (why?). So the Lagrangian for an open system is like that of a

closed system except that U may be an explicit function of time (Problem 3 on page 11 of Landau is a

good example).


A single particle moving in an external eld has a Lagrangian,

L =1

2mv2 U(r; t)

where U is the potential for the external eld. The equation of motion is

mr = @U@r

= F(r; t)

where F is the force on the particle due to the external eld. If the force is constant then

U(r) = F r (Constant force)

For example, near Earth's surface the force of gravity is F = mgk^ so U = mgz.

Worked Examples

Landau closes the chapter with some worked examples; he leaves out many of the steps so let's do one

of them in detail.

Example (Problem 4, pg. 12) This example is a mechanical system similar to the centrifugal

governor used in steam engines (see http://en.wikipedia.org/wiki/Centrifugal_governor). Two

similar particles (mass m1) on the sides are connected by four rigid rods of equal length a (see Fig. 1.6)

The upper rods are linked to a xed hinge; the lower rods are linked by a hinge to a particle (mass

m2) that can slide up and down. The central axis spins with constant angular speed _ = ; by the

constraint the masses on the sides rotate at the same angular speed. Our task is to nd the Lagrangian,

L(; _; t) where the generalized coordinate is the angle between the side masses and the (vertical)

z-axis.

The kinetic energy for a free particle in spherical coordinates is

T =1

2m( _r2 + r2 _2 + r2 _2 sin2 )

For the particles on the sides the constraints make r = a, _r = 0, and _ = so the kinetic energy for

each of them is,

T1 =1

2m(a2 _2 + a22 sin2 )

For the particle constrained to move along the vertical axis the kinetic energy is T2 =12m2 _z

22 . To

transform this into our generalized coordinate we use

z2 = 2a cos so _z2 = 2a _ sin

which gives T2 = 2m2a2 _2 sin2 . Note that we're taking the top hinge as the origin.

The potential energy for each of the particles on the sides is

U1 = m1gz1 = m1ga cos

For the particle moving along the vertical axis,

U2 = m2gz2 = 2m2ga cos


Figure 1.6: Simple centrifugal governor (Problem 4, pg. 12)


Putting it all together (and remembering that there are two particles on the sides) we have,

L = 2T1 + T2 (2U1 + U2)= m1a

2( _2 +2 sin2 ) + 2m2a2 _2 sin2 + 2(m1 +m2)ga cos

which matches Landau's result (but his approach is much more elegant and compact). Finally, notice

that we may rearrange the terms and write this as,

L = a2(m1 + 2m2 sin2 ) _2 + 2(m1 +m2)ga cos +m1a

2 sin2 2

The last term may be viewed as an eective potential energy for the centrifugal force

Fc = @Uc@R

= 2m1

2R

where R is the radius vector in cylindrical coordinates; recall that the masses on the sides move in a

circle of radius R = a sin .

Chapter 2

Conservation Laws

Energy (x6)Lecture 5

In this chapter we consider various integrals of motion, which are various functions of co-ordinates and

velocities that, under certain conditions, are constants of the motion. We focus on the seven additive

integrals of motion:

Energy (scalar) Linear momentum (3 components) Angular momentum (3 components)

These quantities are called conserved variables and their constancy arises from homogeneity and isotropy

of space and time. Other integrals of motion are specic to particular systems (e.g., three body gravi-

tational system) so they're not as important.

The rst conserved variable we'll study is the energy. Consider a system whose Lagrangian does not

depend explicitly on time, that is, L(q; _q). Important examples are closed systems but in some cases

open systems (e.g., Problem 4, pg. 12) have Lagrangians of this form. By the chain rule,

d

dtL(q1; : : : ; qs; _q1; : : : ; _qs) =

@L

@q1

dq1dt

+ : : :@L

@qs

dqsdt

+@L

@ _q1

d _q1dt

+ : : :@L

@ _qs

d _qsdt

=

sXi=1

@L

@qi_qi +

sXi=1

@L

@ _qiqi

From Lagrange's equation,@L

@qi=

d

dt

@L

@ _qi

so using this in the rst summation gives,

dL

dt=

sXi=1

d

dt

@L

@ _qi

_qi +

sXi=1

@L

@ _qiqi

=

sXi=1

d

dt

@L

@ _qi

_qi

22

CHAPTER 2. CONSERVATION LAWS 23

Finally, collect everything to the left hand side and we may write,

d

dt

"(sX

i=1

@L

@ _qi

_qi

) L

#

This means that dE=dt = 0 where

E =

sXi=1

@L

@ _qi

_qi L

is the denition of the energy.

Example For projectile motion in 2D,

L =1

2m( _x2 + _y2)mgy

so

E =

@L

@ _x

_x+

@L

@ _y

_y L

= m _x2 +m _y2 1

2m( _x2 + _y2)mgy

=

1

2m( _x2 + _y2) +mgy

This example illustrates the general result that for a closed system (or one in a constant eld) the

energy is simply E = T +U . However, note that the open system in Problem 4, pg. 12, is an example of

a system in which E 6= T + U however E is conserved since the Lagrangian does not depend explicitlyon time.

Conservation of energy is a consequence of the homogeneity of time, that is, if the Lagrangian does

not depend explicitly on time then the equations of motion are unchanged if t! t+ t0.

Momentum (x7)Consider now the case where the Lagrangian of a system is unchanged by a spatial shift of positions,

r! r+ where is an innitesimal vector displacement. For a Lagrangian with this property we have,

L(r1 + ; : : : ; rN + ;v1; : : : ;vN ; t) = L(r1; : : : ; rN ;v1; : : : ;vN ; t)

By Taylor expansion of the left hand side we have,

L(r1 + ; : : : ; rN + ;v1; : : : ;vN ; t) = L(r1; : : : ; rN ;v1; : : : ;vN ; t) + @L@r1

+ @L@rN

+ (h:o:t:)

Since is innitesimal we drop the higher order terms, which gives,

NXa=1

@L

@ra= 0


If the Lagrangian is unchanged by spatial shift in any direction (i.e., if the direction of is arbitrary)

thenNXa=1

@L

@ra= 0 ()

Two important results follow from this:

* For a closed system of particles the Lagrangian has the general form,

L =

NXa=1

1

2mav

2a U(r1; : : : ; rN )

so the net force on particle a is

Fa = @U@ra

=@L

@ra

From our result (*) above,NXa=1

Fa = 0

For two particles this is simply F1 + F2 = 0 so F1 = F2, which is Newton's Third Law of Motion.* Lagrange's equations may be written as,

d

dt

@L

@va

@L@ra

= 0

so from (*)

d

dt

NXa=1

@L

@va= 0

Dene the momentum of a particle and the total momentum as,

pa =@L

@vaand P =

NXa=1

pa

so we have that the total momentum is constant (i.e., dP=dt = 0).

We may also dene a generalized momentum and generalized force as,

pi =@L

@ _qiand Fi =

@L

@qi

so by Lagrange's equations for generalized coordinates,

d

dt

@L

@ _qi

@L@qi

= 0

which gives us the generalized form of Newton's Second Law of Motion,

_pi = Fi

Note that this result is just a way to re-write Lagrange's equations in terms of pi and Fi and it applies

to arbitrary Lagrangian. If the Lagrangian does not contain a given coordinate qi then @L=@qi = 0,

implying that the corresponding momentum pi is constant (i.e., _pi = 0). In that case we say that qi is

a cyclic coordinate.


Figure 2.1: Pendulum (mass m2) with a pivot (mass m1) that slides horizontally.

Example The Lagrangian for a particle in a constant gravity eld is

L =1

2m( _x2 + _y2 + _z2)mgz

Since this Lagrangian does not explicitly depend on x or y these are cyclic coordinates which means

that

px =@L

@ _x= m _x and py =

@L

@ _y= m _y

are constant. Note that pz = @L=@ _z = m _z is not constant.

Example Consider the system in Problem 2, pg. 11, which is a pendulum whose pivot slides freely

on a horizontal wire (see Figure 2.1). The Lagrangian is

L =1

2(m1 +m2) _x

2 +1

2m2(`

2 _2 + 2` _x _ cos) +m2g` cos

By inspection x is a cyclic coordinate so its corresponding momentum,

px =@L

@ _x= (m1 +m2) _x+m2` _ cos

is constant. To understand this result, consider that the horizontal position and velocity of the pendulum

mass are,

x2 = x+ ` sin and _x2 = _x+ ` _ cos

so the generalized momentum px may be written as,

px = (m1 +m2) _x+m2(` _ cos)

= (m1 +m2) _x+m2( _x2 _x)= m1 _x1 +m2 _x2

which is the total horizontal momentum of the particles.


Note that we may write this momentum as

px = _X

where = m1 +m2 is the total mass and

X =m1x1 +m2x2m1 +m2

is the horizontal coordinate of the center of mass. Since px is constant the velocity of the center of

mass, _X, is also constant. In the next section we'll see more general results for the motion of the center

of mass.

Example Landau solves the following problem on page 16: A particle moving freely with velocity

v1 on the xy-plane encounters a steep ramp parallel to the x-axis. The potential energy of the particle

is U1 and U2 before and after the encountering the ramp, respectively. The direction of motion changes,

as shown in Figure 2.2; nd the ratio sin 1= sin 2.

Solution The particle's Lagrangian is

L =1

2m( _x2 + _y2) + U(y) whereU(y) =

U1 y 0U2 y < 0

By inspection the x component is a cyclic coordinate so the component of momentum in the x direction

is conserved. Equating this component of momentum before and after the particle passes the ramp

gives,

v1 sin 1 = v2 sin 2

By inspection the Lagrangian does not depend explicitly on time so energy is conserved so

1

2mv21 + U1 =

1

2mv22 + U2

With a little algebra this gives,

v2 = v1

s1 +

2

mv21(U1 U2) = v1

r1 U

T1

where T1 is the initial kinetic energy. Finally, collecting the above gives,

sin 1sin 2

=

r1 U

T1

Notice that this result holds even if the ramp is not a step function as long as the gradient is perpendic-

ular to the x-axis (i.e., as long as x is a cyclic coordinate). Finally, this result is the Newtonian model

for the refraction of light.


Figure 2.2: Particle deected by crossing a ramp.

Center of Mass (x8)Lecture 6

The total momentum for a system of N particles is,

P =

NXa=1

mava

If the Lagrangian is unchanged by a shift in the coordinates in a specied direction n^ then _P n^ = 0;if it's unchanged for all coordinate shifts then _P = 0.

Consider a Galilean transformation from the current frame of reference (call it K) to another inertial

frame of reference (called K 0) moving with velocity V relative to K (see Figure 2.3).By galilean relativity the particle velocities in the two frames are related by

va = v0a +V

The total momentum in frame K is

P =NXa=1

mava =NXa=1

ma(v0a +V)

=NXa=1

mav0a +V

NXa=1

ma

= P0+ V

where is the total mass. Note that P0= 0 if V = P=; in that case we call K

0the \center of

mass frame of reference." In a sense, the global motion of the system is zero in this frame, though the

What happens if U > T1?


Figure 2.3: Galilean transformation of particle velocities in two frames of reference K and K 0.

particles move relative to each other. Call V the velocity of the center of mass in frame K; we may

write it as,

V =

PamavaPama

=d

dt

PamaraPama

= _R

where

R =

PamaraPama

is the position of the center of mass.

Example Consider two particles moving on the x-axis with an elastic attraction; we'll take the

Lagrangian to be,

L =1

2m1 _x

21 +

1

2m2 _x

22

1

2k(x1 x2)2

We expect that the total momentum is conserved since the Lagrangian is unchanged if x1 ! x1 + ,x2 ! x2 + .

The equations of motion turn out to be,

m1x1 + k(x1 x2) = 0m2x2 k(x1 x2) = 0

Adding them gives,

m1x21 +m2x

22 = 0

sod

dt

m1 _x1 +m2 _x2m1 +m2

=d

dtV = 0

which means that the center of mass velocity is constant, as we expect since the total momentum is

conserved for this system.


Figure 2.4: Innitesimal rotation of the position of a particle (not to scale).

Angular Momentum (x9)The derivation for conservation of angular momentum is similar to that of linear momentum except that

we consider Lagrangians that are invariant under an innitesimal rotation instead of a displacement

(See Figure 2.4). After a few manipulations, Landau arrives at the result,

Xa

ra pa = 0

Dene the angular momentum as

M =Xa

ra pa

and our result may be expressed as,d

dt( M) = 0

If the Lagrangian is invariant under any rotation thenM is constant. If the Lagrangian is invariant only

for a rotation around a specic direction, say the z-axis, then only that component of M is constant.

In frame K 0 is the center of mass frame then

M =M0+ RV =M0 +RP

The termM0is called the intrinsic angular momentum and the termRP gives the angular momentum

of the motion as a whole.


Figure 2.5: Two particles moving together with equal velocities.

Example Consider two particles moving freely in the xy-plane. The z-component of angular mo-

mentum is

Mz = k^ M = k^ Xa

ra pa

= m1(x1 _y1 y1 _x1) +m2(x2 _y2 y2 _x2)

Now suppose that we take the special case that the particles move in a straight line with the same

velocity, specically, let's take _x1 = _x2 and y1 = y2 (see Figure 2.5). In that case,

Mz = y1(m1 _x1 +m2 _x2) = Y _X = (RV) k^

where R = Xi^+ Y j^ is the center of mass position and V = _R. This simple example demonstrates our

result that

Mz =M0z + (RV) k^

since obviously M0z = 0 for this system (particles are at rest in the center of mass frame). By the way,

it's easy to show that Mx =My = 0 since the motion is in the xy-plane so _z1 = _z2 = 0.

For a given coordinate axis, if is the angle about that axis (e.g., the z-axis in cylindrical and

spherical coordinates) then

Mz =Xa

@L

@ _a

For example, for a free particle in cylindrical coordinates Mz = mr2 _ since the Lagrangian may be

written as L = 12m( _r2 + r2 _2 + _z2).

Example Consider a particle that moves freely in the interior of a 45 degree cone (so z = r); there

is a constant downward gravitational force acting on the particle. See Figure 2.6 Using conservation of

energy and angular momentum, express _z and _ as a functions of z.


Figure 2.6: Particle moving freely inside a 45 degree cone.

Solution We have the general result that for a closed system (or one in a constant eld) the energy

is simply E = T + U . In cylindrical coordinates the kinetic energy of a free particle,

T =1

2m( _r2 + r2 _2 + _z2)

The potential energy in a constant gravitational eld with acceleration g is U = mgz. If you have any

doubts you can use the general denition of energy and verify that this result is correct. Using the

constraint (r = z, _r = _z) the energy is,

E =1

2mz2 _2 +m _z2 +mgz

Since the Lagrangian does not depend explicitly on time the energy is constant.

For the angular momentum only the z component is conserved since the Lagrangian is invariant

under rotation about the z axis but not about any other axis. In cylindrical coordinates Mz = mr2 _

and with our constraint this is Mz = mz2 _.

By inspection we may write

E =M2z2mz2

+m _z2 +mgz

or

_z =

rE

m M

2z

2m2z2 gz

From our expression for Mz,

_ =Mzmz2


Figure 2.7: Graph indicating maximum and minimum heights for particle rolling inside a cone for a

given value of angular momentum, Mz, and energy, E.

This pair of rst-order ODEs for z and may (in theory) be solved to nd z(t) and (t). The initial

conditions z(0), _z(0), (0), and _(0) x the two constants of integration of these ODEs as well as the

values of E and Mz.

Before leaving this example, let's examine some of the qualitative features about the motion that

we can extract from the result above. From the expression for _z we see that as the particle's height

decreases this vertical velocity may increase (due to the gravity term) or decrease (due to the angular

momentum term). Physically, the particle spins in the cone rising to a maximum height, constrained

by the gravity term, and diving to a minimum height, constrained by the angular momentum term.

The turning points occur when _z = 0, which means

E =M2z2mz2

+mgz = f(z)

This cubic equation has two real roots, zmin and zmax; we won't bother with their explicit form but

instead just consider the graphical solution shown in Figure 2.7. For a given value of Mz the turning

points depend on the energy E, as shown in the gure. Notice that the curve of f(z) has a minimum

value, which is the case when the particle orbits in a circle of constant height.

Finally, in the absence of gravity the particle will move out towards z !1 (with _z !pE=m and_ ! 0) unless the initial condition is _z(0) = 0, in which case the orbit is circular; Landau considersthis system in Problem 2 on page 34. The next chapter extends the ideas in this example to obtain

properties of the motion using the constants of the motion.


Mechanical Similarity (x10)Landau points out the a number of interesting results may be inferred about the motion of a system

when the potential energy is a homogeneous function of the coordinates, that is, if,

U(r1; : : : ; rn) = kU(r1; : : : ; rn)

For example, for Newtonian gravitational attraction among n particles of mass m,

U(r1; : : : ; rn) =1

2

nXi=1

nXj=1

Gm2

jri rj j

In this case,

U(r1; : : : ; rn) =1

2

nXi=1

nXj=1

Gm2

jri rj j =1

U(r1; : : : ; rn)

so = 1. Similarly, for a potential of the form U / jri rj jc we have k = c (e.g., for harmonicoscillator potential k = 2).

Starting from the Lagrangian for a system of particles,

L(r;v; t) =nX

a=1

1

2majvaj2 U(r1; : : : ; rn)

Now consider the transformation of the system such that coordinates and the time are changed as,

ra ) ra ; t) t

This means that all of the velocities va = dra=dt are changed by a factor of = so the kinetic energy

is changed by a factor of 2=2. The potential energy is multiplied by a factor of k. In other words,

L0(r; (=)v; t) =2

2

nXa=1

1

2majvaj2 kU(r1; : : : ; rn)

Now if we choose =p2k then L0 and L dier by a multiplicative constant and thus they have the

same equation of motion so the two have mechanical similarity. As Landau says, \if the potential energy

of the system is a homogeneous function of degree k in the (Cartesian) coordinates, the equations of

motion permit a series of geometrically similar paths, and the times of the motion between corresponding

points are in the ratio,

t0=t = (`0=`)112k

where `0=` is the ratio of linear dimensions of the two paths."

Example For the motion of free particles in a constant gravitational eld U(r) = gzz^ so k = 1and

t0=t = (`0=`)12

For example, a particle's parabolic trajectory is mechanically similar for a length scale (height and

range) magnied by 16 if the time scale is magnied by a factor of 4.

Mechanical similarity is used by Hollywood to create special eects using high-speed cameras. Nor-

mal lm speed is 24 frames per second so when a scene is lmed at 96 frames per second then played


Figure 2.8: Scene from Jason and the Argonauts (1963).

back at normal speed what appears to be a boulders falling o a 64 foot cli is really small rocks falling

from a 4 foot height (see Fig. 2.8). In modern lms these special eects are usually done using computer

graphics some are still produced using high-speed cameras. Furthermore, the visual eects artists who

use computer animation need to duplicate this relationship between timing and scale.

Chapter 3

Integration of the Equations of

Motion

Lecture 7

In Chapter 1 we formulated Lagrangian dynamics; in Chapter 2 we derived various important conserved

quantities, such as energy and momentum. We now consider some important cases where it's possible

to solve, at least in part, the equations of motion.

Motion in One-Dimension (x11)Consider a Lagrangian of the form,

L =1

2m _x2 U(x)

where x is a single coordinate. Many problems reduce to this form (e.g., motion in a central eld, x14).The physical picture for the motion for such a Lagrangian is that of a particle sliding on a roller coaster

whose shape is the graph of U(x) versus x (see Figure 3.1).

The total energy is

E =1

2m _x2 + U(x)

Figure 3.1: Analogy between the shape of a potential and the motion of a particle sliding in a similarly

shaped landscape.

35

CHAPTER 3. INTEGRATION OF THE EQUATIONS OF MOTION 36

Figure 3.2: Motion in a potential between endpoints x1 and x2.

Since the total energy is constant we may obtain an integral expression for the motion without mak-

ing use of the equations of motion (i.e., without applying Lagrange's equations to the Lagrangian).

Specically, the approach is to write1

2m _x2 = E U(x)

sodx

dt=

r2

m(E U(x)) Note : E U(x)

or

dt =dxq

2m (E U(x))

Integrating both sides,

t =

rm

2

Zdxp

E U(x)The initial conditions (initial position and velocity) specify the constant of integration and the value of

E.

This result has various useful corollaries, for example we may use it to nd the period of oscillation

for periodic motion. Consider the graph of a potential U(x) such as in Figure 3.2. For a given value of

energy, E and an initial position between points x1 and x2 we know that the motion is periodic between

these turning points (why?). By denition, the time it takes the particle to go from x1 to x2 is half the

period so the period of oscillation, T , is,

T =p2m

Z x2x1

dxpE U(x)

Example Consider an innite square-well potential,

U(x) =

0 if 0 x L1 otherwise

Landau writes the period as T but I'll write T to distinguish it from the kinetic energy.


The turning points are x1 = 0 and x2 = L for all values of E; since the potential is zero in this interval,

T =p2m

Z L0

dxpE

=

r2m

EL

Since U = 0 in the interval E = 12mv2 so T = 2L=v, that is, the period is the time it takes a particle

with speed v to travel a distance of 2L.

Landau works through some more complicated examples on pages 26 and 27, including the simple

pendulum without the small angle approximation. In general the resulting integrals can be dicult to

solve in closed form but are easy to evaluate numerically. One important result that has a complicated

but closed form solution is for the potential U(x) = Ajxjn (see Problem 2(a) on page 27); the period ofoscillation is

T = 2n

r2m

E

E

A

1=n(1=n)

( 12 + 1=n)/ E 1n 12

where is the gamma function. An important special case is n = 2, which is the potential for simple

harmonic motion; by inspection it is the only exponent n that gives a period that's independent of E,

which means it's independent of the initial amplitude.

Finding U(x) from T (E) (x12)We just saw how to get the period, T for a given potential; is it possible to do the reverse? Namely,from knowing T (E) is it possible to obtain U(x)? The answer is yes but the potential is not uniqueunless we also impose the condition that it is symmetric, say about x = 0.

After a few manipulations (see page 28) Landau arrives at the result,

x(U) =1

2p2m

Z U0

T (E)dEpU E

from which we may get U(x) with U(x) = U(x). Let's see how this works in an example,

Example Consider the case for which T (E) = T0 = constant. That is, the period is independentof the energy. In this case,

x(U) =T0

2p2m

Z U0

dEpU E

=T0

2p2m

h2pU E

iU0=

T0p2m

pU

so

U(x) =22m

T 20x2

You should recognize this as the potential for simple harmonic motion (SHM), which is usually written

as U(x) = 12kx2 so in our case k = 42m=T 20 . Recall that the period for SHM is

TSHM = 2rm

k= 2

rm

42m=T 20= T0


Figure 3.3: Symmetric potential, U(x), and asymmetric potential, U(x), with the same period, T (E).

which conrms the expected result.

Landau shows that an asymmetric potential, U(x) will have the same period as a symmetricpotential, U(x) as long as the distance between points at the same energy is equal for the two potentials.

This is illustrated graphically in Figure 3.3; mathematically the condition is that

x2(U) x1(U) = 2x(U) when U = U:

Example A simple way to build an asymmetric potential is to take x1(U) = 0 for all values of U

and x2(U) = 2x(U). The condition on x1 implies that the asymmetric potential U(x) =1 for x < 0.

The condition on x2 simply means that U(x) is twice as wide as U(x) for all x > 0 (see Figure 3.4).

As an example, consider a simple pendulum of length `; with the pivot at the origin the (symmetric)

potential may be written as,

U = mg(`+ y) = mg`mgp`2 x2

mg`mg`1 1

2

x2

`2

=

1

2

mg

`x2

using the small angle approximation and the Taylor expansionp1 + x = 1+ 12x+: : :. The corresponding

asymmetric potential would be U(x) =1 for x < 0 and

U(x) =1

2

mg

`(x=2)2 =

1

2

mg

4`x2

Let's interpret this result: The asymmetric potential corresponds to a pendulum that is 4 times longer

and that swings against a rigid, elastic wall located at x = 0. We know that in the small angle

approximation the period of a pendulum is T = 2pg=` so a pendulum that's 4 times longer wouldhave twice the period but due to the rigid wall it only travels half the distance so U(x) and U(x) have


Figure 3.4: Simple pendulum example illustrating a symmetric potential, U(x), and an asymmetric

potential, U(x), with the same period, T (E).

the same period for all values of E (i.e., for all initial conditions). Finally, note that we may construct a

similar expression for U without using the small angle approximation but then the resulting potentialis not equivalent to an interrupted pendulum.

Reduced Mass (x13)Lecture 8

In this section and the next two we analyze in detail the two-body problem, that is, a closed system

with two interacting particles. The general form of the Lagrangian is,

L =1

2m1 _r

21 +

1

2m2 _r

22 U(jr1 r2j)

Notice that the potential can only depend on the magnitude of the separation due to the homogeneity

and isotropy of space (both total linear momentum and angular momentum are constant).

Dene

r = r1 r2 (relative position)m =

m1m2m1 +m2

(reduced mass)

In the center of mass frame of reference m1 _r1 +m2 _r2 = 0 so

m21 _r21 +m

22 _r

22 + 2m1m2(_r1 _r2) = 0

which means that

1

2m _r2 =

1

2m(_r1 _r2)2

=1

2

m1m2m1 +m2

(_r21 + _r22 2(_r1 _r2))

=1

2

m1m2m1 +m2

(_r21 + _r22 +

m21m1m2

_r21 +m22

m1m2_r22)

=1

2m1 _r

21 +

1

2m2 _r

22


This simplies the Lagrangian to be

L =1

2m _r U(jrj)

so the two body problem is equivalent to the problem of a single particle with reduced mass m moving

in a central potential.

One last thing: Be careful to remember that

_r2 = _r _r 6= _r2

For example, in cylindrical coordinates,

_r2 = _r2 + r2 _2 + _z2

Motion in a Central Field (x14)Now we turn to the general problem of a single particle moving in a eld with potential U(jrj) = U(r).The force on the particle is,

F = @U@r

= rU = dU

dr

r^

where r^ = r=jrj is the radial unit vector. Notice that if dU=dr > 0 then the direction of F is r^ so thepotential is attractive; if dU=dr > 0 then the potential is repulsive.

The angular momentum M is constant (why?) and M is perpendicular to r since M = r p. Thismeans that the motion of the particle, r(t) is in a plane; we'll take it to be the xy-plane by having the

z-axis parallel to M. Using polar coordinates (i.e., cylindrical coordinates with z = _z = 0) the energy

is,

E =1

2m( _r2 + r2 _2) + U(r)

Using M =Mz = mr2 _ = constant we havey

E =1

2m _r2 +

1

2

M2

mr2+ U(r)

=1

2m _r2 + Ue(r)

where

Ue(r) =1

2

M2

mr2+ U(r)

is the eective potential.

Note that we have now reduced the problem to one-dimensional motion in a potential Ue(r), which

allows us to use the results from Section 11. For example, the motion r(t) may be found by evaluating

t =

Zdrq

2m (E Ue(r))

()

with the initial conditions setting the constant of integration as well as xing the values of E and M.

The eective potential leads to an eective force,

Fe = @Ue@r

= @U@r

+M2

mr2r^ = F+ Fc

yRecall that Mx = My = 0 since the motion is in the xy-plane.


Figure 3.5: Sketch of eective potential Ue(r) versus r. Notice that for small r the centrifugal repulsion,M2

2mr2 , dominates but for large r we have Ue(r) U(r).

where Fc is the centrifugal force, which is an eective force of repulsion.

If U(r) is an attractive potential (i.e., dU=dr > 0) then graphically the eective potential is illustrated

in Figure 3.5. If the potential U(r) has an upper bound (e.g., if U ! 0 as r ! 1) then the motionis bounded if the energy is less than this upper bound. If the potential is unbounded (e.g., the space

oscillator potential, U(r) = 12kr2) then the motion is bounded for all values of E. Bounded motion is

conned between r = rmin and rmax where U(rmin) = U(rmax) = E.

Using M = mr2 _ we may write d = (M=mr(t)2)dt and then use (*) to get

=

ZM dr

r2p2m(E Ue(r))

The change in angle during one period of the motion (r = rmin then r = rmax then back to r = rmin) is

=

Z rmaxrmin

M dr

r2p2m(E Ue(r))

In general 6= 2 so orbits are typically not closed (see Figure 9 on page 33 in Landau). Theonly power-law potentials with closed orbits are U(r) / r1 (Kepler problem) and U(r) / r2 (spaceoscillator). We consider the former in detail in the next section (and next lecture) while Landau discusses

the latter in Problem 3 on page 70.

Example Consider the attractive central potential,

U(r) = r2

Note that this is not the Kepler problem. For this example we'll take the initial condition to be such

that M2 = 2m; for this unique case

Ue(r) =1

2

M2

mr2+ U(r) = 0

which greatly simplies all the calculations. Notice that since E = 12m _r2+Ue(r) we know that E 0.


Figure 3.6: Trajectory for particle in potential U(r) = r2 for the initial conditions r(0) = 0 andM =

p2m. Plots are for short (left) and long (right) times; parameter values are E = 103 and

= 1.

First we'll consider the case where E > 0 and calculate the motion using (*), which simplies to,

t =

Zdrq2m E

=1q2m E

(r(t) r(0))

so

r(t) =

r2

mE t+ r(0)

We obtain the same result from the observation that since in this example Ue(r) = 0 then E =12m _r

2.

But energy is conserved so _r =p2E=m is constant. Physically, the particle moves away from the origin

with constant radial velocity.

The relation between the angle and radius is found using,

=

ZM dr

r2p2mE

= Mp2mE

1

r+ C

where C is the constant of integration. We'll take r(0) = 0 to see how the motion looks as the particle

ies away from the origin. For this initial condition we may arbitrarily set C = 0 so

= Mp2mE

1

r= Mp

2mE

1q2m E t

= mM4E2

1

t

The resulting motion is a spiral, as shown in Figure 3.6.

To nish the example, what about the case E = 0? From the above we know that since Ue(r) = 0

that _r =p2E=m so r is constant. Recall that M = mr2 _ so

_ =M

mr2=

r2

m

1

r2


Physically, the orbits are circular with a constant angular velocity _ that goes as r2 so their period,T = 2= _, goes as r2. Compare this with the period of circular orbits in the Kepler problem for whichT / r3=2 by Kepler's Third Law.

One nal aside: The result for the period of orbits may be obtained from the principle of mechanical

similarity (see Section x10). Specically, if we take r0 = ar, where a is a constant, thenU(r0) =

(r0)2=

(ar)2= a2U(r)

Landau shows that when a potential scales as

U(ar) = akU(r)

then geometrically similar paths (e.g., circular orbits) scale as

t0

t=

`0

`

For the potential in this example k = 2 so the period of orbits goes as r2; for the Kepler problemk = 1 so the period goes as r3=2.

Kepler Problem (x15)Lecture 9

We now turn to an important mechanics problem, motion in a central potential where U(r) / 1=r;the gravitational and Coulomb potentials for point objects are well-known cases. We will rst consider

attractive potentials so

U(r) = r

( > 0)

so the eective potential for the radial motion (as formulated in the previous section) is

Ue(r) = r+

M2

2mr2

Recall that M is constant with direction perpendicular to the plane of the motion. A sketch of the

eective potential is shown in Figure 3.7.

If E < 0 then the motion is bounded between the turning points rmin and rmax. If E 0 then themotion is unbounded so r ! 1 as t ! 1. If E = minfUe(x)g then the radius is constant, that is,the orbit is circular.

From conservation of angular momentum, M , and energy, E, we derived in the previous section

that,

=

ZM dr

r2p2m[E Ue(r)

For our case the integration yields,

= arccos

"Mr1 m=Mp2mE +m22=M2

#+ 0

Taking 0 = 0 and dening the quantities,

p =M2

m(Semi latus rectum)

e =

r1 +

2EM2

m2(Eccentricity)


Figure 3.7: Eective potential Ue(r) for the attractive potential, U(r) = r , of the Kepler problem.

Figure 3.8: Geometry of an: ellipse (left); parabola or hyperbola (right)


we may rewrite the expression above as,

r =p

1 + e cos

which is the equation for a conic section. By inspection, for E < 0 the eccentricity e < 1; the orbit is

an ellipse (Kepler's First Law) with semi-major and semi-minor axes,

a =p

1 e2 b =pp

1 e2The perihelion and aphelion are

rmin =p

1 + ermax =

p

1 eNote that e = 0 for a circular orbit.

Given that angular momentum is conserved in a central eld we may use M = mr2 _ to write

dA =1

2r2d =

M

2mdt

where dA is the area swept out by an orbital increment d. This result (A line joining a planet and

the Sun sweeps out equal areas during equal intervals of time) is Kepler's Second Law (see Figure 3.9).

Integrating over the period of an orbit we have

T = 2mM

A

Since the area of an ellipse is A = ab from the previous results we nd that T / a3=2, which is Kepler'sThird Law. Why did Kepler express this as a relation between T and a instead of T and A? Becausethe angular momentum M diers for each planet.

For E > 0 the eccentricity e > 1 and the trajectory is a hyperbola with perihelion rmin = p=(1 + e);

if E = 0 then e = 1 and the trajectory is a parabola with rmin = p=2.

Landau shows that we may formulate expressions for the position versus time but only in parametric

form. For example, for an ellipse,

r = a(1 e cos )!t = e sin

where ! =p=ma3. Note that (the eccentric anomaly) and !t (the mean anomaly) go from 0 to

2 (see Figure 3.10). Note that the period is T = 2=! = 2pm=a3=2; for gravity / m1m2 andthe sun is massive so m= / (m1 + m2)1 1=msun. The calculation of for a given time t is atranscendental problem that was an early application for numerical root nding algorithms.

For a repulsive potential, U(r) = +=r, the results are similar as for an attractive potential except

that there is no bounded motion, that is, the trajectory is either a parabola or a hyperbola. The main

application is the study of scattering due to Coulombic repulsion.

Laplace-Runge-Lenz Vector

We'll nish this chapter by deriving a special constant of the motion that's specic to the 1=r potential.

Consider the general case of a central potential; the Lagrangian is,

L =1

2m _r2 U(r)


Figure 3.9: Illustration of Kepler's three laws with two planetary orbits. (1) The orbits are ellipses,

with focal points 1 and 2 for the rst planet and 1 and 3 for the second planet. The Sun is placed in

focal point 1. (2) The two shaded sectors A1 and A2 have the same surface area and the time for planet

1 to cover segment A1 is equal to the time to cover segment A2. (3) The total orbit times for planet 1

and planet 2 have a ratio a13=2 : a23=2. (From Wikipedia)

Figure 3.10: Parametric solution for an elliptical orbit. Note that tan(=2) =p(1 + e)(1 e) tan(=2).


so using Lagrange's equations (in vector form),

d

dt

@L

@ _r

@L

@r= 0

gives,

mr+

dU

dr

r^ = 0

or

_p+

1

r

dU

dr

r = 0

Note that we used the vector calculus identity,

@f

@r= rf(r; ; ) = @f

@rr^+

1

r

@f

@^ +

1

r sin

@f

@^

in spherical coordinates.

Applying M to the previous expression gives,

_pM+1

r

dU

dr

rM = 0 ()

yet

rM = mr (r _r) = m[r(r _r _r(r r)]using the identity A (BC) = B(A C)C(A B). This expression simplies to

rM = m[r _rr r2 _r]

Since r _r = 12 ddt (r r) = r _r. Finally, we rewrite this as

rM = mr3[r2 _rr r1 _r] = mr3 ddt

hrr

iPutting this result back into (*) and recalling that M is a constant gives,

d

dt[pM]

r2dU

dr

d

dt

hmr

r

i= 0

Up to now this result holds for any central potential U(r). But notice that if the term in the curly

brackets is a constant then we can combine the terms into ddt [: : :] = 0, yielding a new constant of the

motion. This is the case for the U(r) = =r potential since r2dU=dr = so,d

dt

hpMmr

r

i= 0

our constant of the motion is

A = v M rr

which is called the Laplace-Runge-Lenz vector. Since A M = 0 the vector A lies in the plane of motion.At perihelion the vector v M is parallel to r so A points along the semi-major axis; with a bit morealgebra you can show that it has magnitude A = e.

Finally, it seems that we have now established seven constants of integration for the Kepler problem:

E, the three components of M and of A. However we really only have ve constants since A M = 0


and jAj may be expressed in terms of E and M (via e). Can take these ve constants of the motionas E, the three components of M, and A, the direction angle of A (which gives the direction of the

semi-major axis). The direction ofM xes the plane of the motion while E andM determine the shape

and size of the trajectory. One more value is needed to uniquely specify the motion and we can take

that as the initial angle (0). These six parameters are uniquely specied by the initial conditions r(0)

and v(0) (and vice versa).

Chapter 4

Collisions

In the rst two sections of this chapter we ask the question: What can be said in general about the Lecture 10

motion of interacting particles in certain \before" and \after" situations without explicitly evaluating

the complete trajectories (see Figure 4.1) Particles are assumed to have no interaction in the \before"

and \after" situations but have an arbitrary interaction in between (e.g., repulsion during a collision).

Disintegration (x16)First consider the case where in the \before" situation we have a single particle of mass m1 +m2 and

in the \after" situation there are two particles, m1 and m2, that are non-interacting. In the center of

mass frame of reference (C system) the initial particle is at rest. By conservation of momentum the

particles have \equal but opposite" momenta after disintegration so p1 = p2 or m1v1 = m2v2.By conservation of energy,

Ei = E1i +p202m1

+ E2i +p202m2

where Ei is the internal energy; note that p0 = jp1j = jp2j. Call = Ei (E1i+E2i) the disintegrationenergy (i.e., the internal energy released in the disintegration) then

=1

2p20

1

m1+

1

m2

=

1

2p20

m1 +m2m1m2

=

p202m

where m is the reduced mass.

Going to a general frame of reference in which the initial particle moves with velocity V (call this

the lab frame or L system) then for either particle we have v = V+v0, where v is in the L system and

v0 is in the C system. Since v V = v0,

(v V) (v V) = v0 v0

or

v2 + V 2 2vV cos = v20where is the angle between v and V.

This result has a simple geometric picture: First, take the case where the two particles have the

same mass so their \after" velocities are v0. For V < v0 the direction of v1 in the L system can go

49

CHAPTER 4. COLLISIONS 50

Figure 4.1: Illustration of \before" and \after" situations for: (Top) Disintegration (x16); (Bottom)Elastic collisions (x17)

Figure 4.2: Diagrams of the post-disintegration velocities in the laboratory frame of reference for (left)

V > v0 and (right) V < v0; note that if V = v0 then V equals the radius of the circle.

from = 0 to depending on 0, the direction of v0 (see Figure 4.2 and Fig. 14 in Landau). For V > v0the maximum value of is given by,

sin max =v0V

(again, see Figure 4.2 and Fig. 14 in Landau). Notice that for v0 = V that max = =2; furthermore

the relation between 0 and is not unique if V > v0 (see points B and C in the Figures). Finally, in

the case that the particles have dierent masses then the geometric construction is the same but each

particle has its own circle of radius v0;1 = p0=m1 and v0;2 = p0=m2.

Example From observations of numerous similar disintegrations (i.e., particle masses are always

m1 and m2, center of mass velocity is always V) we nd that that the angle for particle 1 never exceeds

1;max however the angle for particle 2 can be arbitrary. What does this tell us about the ratio m2=m1?

Solution Since m1v01 = m2v02 then v01 = (m2=m1)v02. Furthermore,

sin 1;max =v01V

=m2m1

v02V


orm1m2

sin 1;max =v02V

But since particle 2 doesn't have a maximum angle then v02 < V so

m1m2

sin 1;max < 1

so m2m1 < sin 1;max. Physically, when particle 1 is massive compared with particle 2 the deection of

its motion due to disintegration is limited by the ratio of masses; for example if 1;max = 45 then

m2 < m1=p2.

Elastic Collisions (x17)Elastic collisions of two particles is a problem that's similar to disintegration but with a few more

elements. Again, we start the analysis in the C system where before the collision,

m1v10 +m2v20 = 0

so again p10 = p20 and p10 = p20 = p0. By conservation of momentum the total momentum after thecollision is still zero so p010 = p020 and p010 = p020 = p00. By denition the energy in an elastic collisionis conserved so E = p20=2m = E

0 = p200 =2m so p0 = p00. This means that in the C system (center of

mass system) the elastic collision changes the direction of the momentum but not the magnitude.

Using relative velocity, v = v10 v20 we may write,

v10 =m2

m1 +m2v

v20 = m1m1 +m2

v

From the above v = v0 but v 6= v0. Landau denes a unit vector, n^0 in the direction of the post-collisionrelative velocity so v0 = vn^0 and

v010 =m2

m1 +m2v0 =

m2m1 +m2

vn^0

v020 = m1

m1 +m2vn^0

We now shift to the laboratory frame of reference (L system) by the galilean transformation,

v1 = v10 +V v2 = v20 +V

where the center of mass velocity is

V =m1v1 +m2v2m1 +m2

Some minor algebra yields,

p01 = mvn^0 +m1

m1 +m2(p1 + p2) = mv

0 +Q1

p02 = mvn^0 +m2

m1 +m2(p1 + p2) = mv0 +Q2


Figure 4.3: Diagram of the pre- and post-collision momenta in the laboratory frame of reference for an

elastic collision. Note that the circle has radius mv = mv0.

This result has a geometric interpretation, as shown in Figure 4.3 and Fig. 16 in Landau. In the

gure the vectors Q1 and Q2 are oriented on the x-axis; notice that they are parallel and the ratio of

their magnitudes is m1=m2. Recall that an elastic collision only changes the direction of the relative

velocity, not its magnitude; in the Figure we see how changing this direction changes the direction and

magnitude of the momenta after the collision.

A special case of the result above is the case where (in the L system) particle 2 is initially at rest

(v2 = p2 = 0, v = v1), which gives,

Q2 =m2

m1 +m2p1 =

m1m2m1 +m2

v1 = mv

In this case point B in the diagram lies on the circle. If m1 < m2 then point A is inside the circle and

1 can be any angle. If m1 > m2 then point A is outside the circle since Q1 > Q2. The angles 1 and

2 of the post-collision velocities may be expressed in terms of the collision angle of the post-collision

relative velocity (see Figure 4.4).

Since OC = OB then the angle formed by OCB equals 2 and from the diagram it's clear that

+ 22 = 180. With a little trig we arrive at,

tan 1 =m2 sin

m1 +m2 cos

(Note that these expressions for 1 and 2 remain valid if m1 m2). From the diagram we see thatthe maximum deection of particle 1 occurs when p01 is perpendicular to the circle (point C

) and this


Figure 4.4: Diagram of the collision when particle 2 is initially at rest and m1 > m2.

Figure 4.5: Diagram of the collision when particle 2 is initially at rest and m1 = m2.

maximum angle is,

sin 1;max =jOCjjOAj =

mv

Q1=

m1m2m1+m2

jv1jm21

m1+m2jv1j

=m2m1

Physically, when the mass of particle 1 is very large then the radius of the circle (mv) is small compared

with Q1 so the maximum deection angle is small.

Finally, if m1 = m2 then both points A and B lie on the circle so 21 +22 = 180 (see Figure 4.5).

This means that p01 is always perpendicular to p02 (and v

01?v02). This result is well-known to those of

you who play billiards.

Example A very light particle (mass m1) moving in the +x direction collides with a very massive

stationary particle (mass m2, m2 m1). The light particle is deected by the collision such that itmoves in the +y direction. Find v02, that is, the magnitude and direction of the velocity of the massiveparticle after collision, in terms of m1, m2, and v1, the magnitude of the velocity of the light particle

before collision.

Solution The post-collision velocity of the second particle has magnitude

v02 =p02m2

=

p2p1m2

=

p2m1m2

v1

with a direction of 45 below the x-axis. These results may be seen directly from the collision diagramin Figure 4.6.


Figure 4.6: Collision of a light particle with a stationary massive particle in which the light particle is

deected by 90 degrees.

Scattering (x18)Lecture 11

In the last two sections we worked out all that can be predicted for the interaction of two particles without

specifying the interaction potential. We now consider what can be determined regarding collisions for a

given potential and, conversely, what can we say about an unknown potential based on measurements

of collisions.

Although we're interested in the interaction of two particles undergoing a collision we formulate the

problem as that of a single particle scattering o of a xed cental potential U(r). We can later map

the results into the collision of two particles in the center of mass frame with the center of mass at the

origin.

From Section 14 we have the result illustrated in Figure 4.7 (also see Fig. 18 in Landau). The

trajectory is symmetric about the apse line, which is the line from the origin to the nearest point on

the trajectory. The apse line makes an angle 0 with respect to the incoming velocity (which we may

take as being in the -x direction). The scattering angle, , is the angle of deection of the outgoing

velocity (i.e., = 0 for no deection and = if the direction is reversed by the scattering). From

Figure 4.7 it's clear that,

= j 20jThe angle of the apse line is useful since from the results in Section 14 we have that

0 =

Z 1rmin

(M=r2) drp2m[E U(r)]M2=r2 =

Z 1rmin

(M=r2) drp2m[E Ue(r)]

()

since = 0 when the particle is initially at r = 1. Note that E = Ue(rmin) since rmin is a turningpoint of the motion.


Figure 4.7: Geometry of scattering illustrating the impact parameter, , and the scattering angle .

(See also Fig. 18 in Landau)

As Landau points out, for comparison with scattering experiments it's convenient and traditional

to reformulate this result, replacing E and M with the initial particle speed, v1, and the impactparameter, , which is the distance from the tangent line of v and the origin (see Figure 4.7 again).

The energy and angular momentum are constant so we may use the initial values of position and

velocity to write,

E =1

2mv21 M = mv1

The result for M is evident if we consider the case of U(r) = 0, for which v1 is constant and = rmin.

Using these results we may write (*) as,

0 =

Z 1rmin

=r2 drp1 (=r)2 (2U(r)=mv21)

An important example is the case U(r) = +=r (e.g., Coulomb scattering) for which the result is,

0 = cos1"

A=p1 + (A=rho)2

#A =

mv21

so

= A tan0 = A cot(=2)This example is treated at length in Section 19 (Rutherford's formula).

Note that the value of M is independent of U(r).


Figure 4.8: Geometry of scattering from a sphere of radius a (See also Fig. 19 in Landau)

Example (Problem 1, pg. 50) Consider scattering of particles from a perfectly rigid sphere of radius

a, that is, for the potential,

U(r) =

0 r > a

1 r aSee Figure 4.8.

The trajectories of particles bouncing o the sphere are straight lines so using simple trigonometry

we have, sin0 = =a or

= a sin0 = a sin

1

2( )

= a cos(=2)

or

() = 2 cos1a

for a and = 0 for > a (particle misses the sphere).

Finally, we can use the formal, integral expression and evaluate,

0 =

Z 1a

=r2 drp1 (=r)2) =

Z 1a

dr

rpr2 2) = sin

1(=a)

which leads to the same results. Notice that due to the unique form of this potential is only a function

of and does not depend on v1.


Figure 4.9: Geometry of scattering illustrating the relation particles with impact parameter between

and + d scattering into angles between and + d.

In scattering experiments one often cannot easily measure the impact parameter ; what's typically

done is to re a beam of particles (with known velocity v1) at a target (the scattering potential atthe origin) and the distribution is measured with a detector placed at a given scattering angle . Call

dN the number of particles per unit time that are scattered between and + d, where d may be

thought of as the angular aperture of the detector. We normalize this by n, the number of particles per

unit time per unit area and write,

d = dN()=n

To quote Landau, \This ratio has the dimensions of area and is called the eective scattering cross-

section. It is entirely determined by the form of the scattering eld and is the most important charac-

teristic of the scattering process."

All the particles with impact parameter between and + d scattering into angles between and

+ d (see Figure 4.9). We will assume (as is typically the case) that this is a one-to-one relation so

not only is () but also (). The number of particles scattering into (; + d) equals the product

of n and the area between the two circles of radii and + d so,

dN = (2d)n and d() = 2()d

Because it's easier to experimentally control (which depends on the position of the detector) than

(since we shoot the target with a beam of particles) we instead use

d() = 2()

dd d

As Landau mentions, sometimes this expression is written in terms of solid angle o instead of polar

angle ; since the scattering from a central potential is symmetric in azimuthal angle we have do =

2 sind.


Example (Problem 1, pg. 50, cont.) Determine the eective cross-section for scattering of particles

from a perfectly rigid sphere of radius a.

Solution Since we already found that = a cos(=2) so

d = 2

dd d

= 2[a cos(=2)]

12a sin(=2) d

=

2a2 sind =

1

4a2do

Notice that the scattering is isotropic (since d = (constant) do). Furthermore, d is independent of

v1 for this rigid potential. The integral over solid angle immediately gives,

=

Z1

4a2do =

1

4a2Z 20

d

Z 0

sind =1

4a2(2)(2) = a2

which i

lecture notes 205

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