Graduate Classical Mechanics

Graduate Classical Mechanics

Leon Hostetler

December 16, 2018Version 0.5

Contents

Preface iii

1 Review 11.1 Summary: Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Lagrangian Mechanics 82.1 Generalized Coordinates and Constraints . . . . . . . . . . . . . . . . . . 82.2 The Lagrangian and the Euler-Lagrange Equations . . . . . . . . . . . . . 102.3 Coordinate Independence of the Lagrangian . . . . . . . . . . . . . . . . . 152.4 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.5 The Variational Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.6 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.7 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.8 Summary: Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . . . 35

3 Scattering 383.1 Scattering Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.2 Inverse Scattering Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.3 Chaotic Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.4 Summary: Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4 Linear Oscillations 484.1 Summary: Linear Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . 58

5 Hamiltonian Mechanics 605.1 Hamilton’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605.2 Variational Principle for the Hamiltonian . . . . . . . . . . . . . . . . . . 655.3 Poisson Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 665.4 Canonical Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 685.5 Liouville’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.6 Integrability and Action-Angle Variables . . . . . . . . . . . . . . . . . . . 755.7 Adiabatic Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 785.8 Summary: Hamiltonian Mechanics . . . . . . . . . . . . . . . . . . . . . . 81

6 Nonlinear Dynamics 83

Preface

About These Notes

These are my class notes from the graduate classical mechanics class I took at MichiganState University with Professor Mohammad Maghrebi. The primary textbooks we usedwere Classical Dynamics: A Contemporary Approach by J. Jose and E. Saletan andClassical Mechanics by H. Goldstein.

My class notes can be found at www.leonhostetler.com/classnotesPlease bear in mind that these notes will contain errors. Any errors are certainly my

own. If you find one, please email me at [email protected] with the name of theclass notes, the page on which the error is found, and the nature of the error.

This work is currently licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. That means you are free to copy and distributethis document in whole for noncommercial use, but you are not allowed to distributederivatives of this document or to copy and distribute it for commercial reasons.

Updates

Last Updated: December 16, 2018Version 0.5: (Dec. 16, 2018) First upload.

www.leonhostetler.com/classnotes

mailto:[email protected]

Chapter 1

Review

In classical mechanics, we can take a vector approach and solve

~F = m~a =d2~x

dt2,

but it’s usually much easier to take scalar approach if possible. If ~F is conservative, thenwe can use

~F = −~∇U.

This implies conservation of energy

E =1

2mv2 + U(~x) = const.

Pendulum

Consider a pendulum consisting of a rigid, massless rod of length L. We want to find theequation of motion θ(t).

Using Newton’s law gets messy since we need two components. So instead, we usean energy approach. The kinetic and potential energies are

T =1

2mv2 =

1

2mL2θ2 =

1

2Iθ2

U = mgh = −mgL cos θ = −A cos θ,

where we have defined I = mL2 and A = mgL. Then conservation of energy implies that

E = T + U =1

2Iθ2 −A cos θ = const.

2 Review

It is useful to have θ as a function of θ, so we rearrange to get

θ =

√2

I

√E +A cos θ.

Since θ = dθdt , this can be written as√

I

2

1√E +A cos θ

dθ = dt.

Integrating gives us

t− t0 =

√I

2

ˆ θ

θ0

1√E +A cos θ

dθ.

This gives t(θ), which can in principle be inverted to get θ(t). Unfortunately, we cannotreally solve this integral. However, we can still obtain a physical understanding of thesolution. Let’s look at the potential U(θ).

Note that the quantity1

2mv2 = E − U,

is always positive. This implies that

E ≥ U.

That is, we cannot have a total energy E that is less than the potential energy U . Thisimplies the classical turning points ±θc.

Suppose −A ≤ E ≤ A and θc << 1. That is, suppose E is close to −A. Then wecan apply the small-angle approximation to get

E =1

2Iθ2 −A cos θ =

1

2Iθ2 −A

(1− 1

2θ2 + · · ·

)' 1

2Iθ2 +

1

2Aθ2.

This has the same form as the harmonic oscillator energy

EHO =1

2mx2 +

1

2kx2 = const.

For the harmonic oscillator, ω =√k/m. Similarly, for the pendulum, we have

ω =

√A

I.

This approach of approximating instead of getting an exact solution is common in classicalmechanics.

Consider the different possible values of E

Review 3

1. E < −A: This is not possible2. −A < E < A

• If θc is small, then we have a harmonic oscillator with frequency ω =√A/I

• If θc is not small, then we have an oscillator with a changing period

3. E = A: This is an unstable equilibrium at the top. It takes infinitely long to returnto the top.

4. E > A: In this case, the pendulum is “unbounded” and there are no turning points.This is the case in which the pendulum has so much energy that it is spinning infull circles

A useful perspective is gained by looking at the phase portrait of periodic motion.This is obtained by plotting the time derivative of the coordinate versus the coordinatefor different values of the total energy. In the case of the pendulum, we know that

E =1

2mv2 −A cos θ.

In this case, our coordinate is θ and its time derivative is proportional to v, so we wantto plot v versus θ for different values of E. Solving for v gives us

v(θ) = ±√

2

m(E +A cos θ).

If −A < E < A and E ≈ −A, then we get the equation of an ellipse.

More generally, if −A < E < A, but E is not approximately −A, then we no longer getan ellipse, but we do get a smooth, closed loop that looks vaguely elliptical.

If E = A, then we get sharp discontinuities at θ = ±π. This corresponds to the top ofthe pendulum’s motion. This shape is called the separatrix since it separates the boundmotion of −A < E < A (i.e. motion between two turning points), and the unboundmotion of E > A.

4 Review

Finally, if E > A, the velocity never changes sign, and we get:

Note that we only look at either the top curve or the bottom curve (depending on whetherthe pendulum is rotating clockwise or counterclockwise.

Finally, to get the full phase portrait, we plot a bunch of curves at different valuesof E. In this case, we are also changing the theta axis from [−π, π] to [−2π, 2π].

For any point (θ, v) in this phase portrait, there is a unique path through the point.Note, in this example, we had only one degree of freedom and one conserved quantity.

Kepler Orbit

Again, we want to use a scalar (energy) instead of vector (Newton’s laws) approach.

Review 5

In this example, we have two degrees of freedom r and θ, but we also have twoconserved quantities: energy and angular momentum.

Remember,d

dt~L =

d

dt(m~r × ~v) = ~r × ~F ,

so if ~r is pointing toward the center, then ~r × ~F = 0. That is, ~L is conserved for allcentral forces.

Recall that in polar coordinates

~r = rr

~v = rr + rθθ.

Conservation of energy gives us

E = T + U =1

2mv2 − GmM

r=

1

2m(r2 + r2θ2

)− GmM

r= const,

where m is the mass of Earth and M is the mass of the sun. Conservation of angularmomentum gives us

~L = m~r × ~v = mr2θz = Lz = const,

where L ≡ mr2θ, or θ = L/(mr2), which we can plug into the energy equation to get

E =1

2mr2 +

L2

2mr2− k

r= const,

where k = GmM . This looks like a 1D problem with kinetic energy

K =1

2mr2,

and an effective potential energy

Ueff =L2

2mr2− k

r.

To plot Ueff (r), we look at the limits

Ueff ∼L2

2mr2, if r is small

Ueff ∼ −k

r, if r is large

So the plot goes as 1/r2 for r near zero, and as −1/r for large r. So it looks somethinglike:

6 Review

Note that E < 0 corresponds to bound orbits, and E > 0 corresponds to unbound orbits.To get the phase portrait, we plot r versus r for different values of E.

1.1. Summary: Review 7

1.1 Summary: Review

Skills to Master

• Given a potential, plot it and describe qualitatively the behavior of the system• Given a potential and a total energy, identify the turning points, equilibria, their stability, and the

frequency of small oscillations about the equilibria• Plot the phase portrait of a system

Given the potential V and a total energy E for asystem, the turning points occur where

E = V.

Equilibrium points occur where

dV

dx= 0.

Equilibria occur where the force, and therefore the ac-celeration are zero, so if you have an equation of mo-tion, equilibria can be found by setting

x = 0,

and solving for x. An equilibrium point x0 is stable if

d2V

dx2> 0,

at x0.One method for obtaining the frequency of small

oscillations about equilibria is to approximate stableequilibria as harmonic potentials. To do this, expandthe potential at the equilibrium point by using a Taylorseries. For example, if the total energy is

E =1

2mv2 + V (x),

and there is an equilibrium point at x = x0, then

V (x) ' V (x0) + V ′(x0)(x0) +1

2V ′′(x0)(x− x0)2.

The first derivative term is necessarily zero at an equi-librium point, and the first term is a constant, and itcan be ignored. Comparing this with a harmonic oscil-lator potential, we note the effective spring constant

keff = V ′′(x0).

Then the frequency of small oscillations is

ω =

√keffm

.

Since you can ignore constants, the straightforwardway to approximate the frequency of oscillations fora particle of mass m about an equilibrium point x0 is

ω =

√V ′′(x0)

m.

To plot the phase portrait of a system, write downan expression for the total energy of the system interms of a coordinate and its time derivative. Solvefor the time derivative and plot this versus the coor-dinate for different values of E. A simple example iswhen

E =1

2mv2 + V (x).

Then the phase portrait is obtained by plotting

v(x) = ±√

2E

m− 2

mV (x),

for different values of E.

Chapter 2

Lagrangian Mechanics

2.1 Generalized Coordinates and Constraints

How can we use the energy approach if we have many degrees of freedom?We should find a scalar object that gives us the equations of motion. There are two

such objects—the Lagrangian and the Hamiltonian.Suppose we have a system ofN particles. Their Cartesian coordinates are ~x1,~x2, . . . ,~xN ,

where ~xi = (xi, yi, zi).Suppose we have K constraints. Any constraint condition can be written in the form

fI(~x1, . . . ,~xN , t) = 0, I = 1, . . . ,K.

Our system now has 3N −K degrees of freedom.Constraints are called holonomic if they don’t depend on the velocities of the coor-

dinates. This is the only kind of constraint we will be dealing with in this course.One way to think of a constraint condition is as a hypersurface in a 3N -dimensional

space as represented below. In this 3N -dimensional space, all motion occurs on thehypersurface. Our system of N particles represents a single point on this hypersurface atone instant in time.

If we have a set of coordinates, we can solve for the constraints. That is,

~xi = ~xi(q1, q2, . . . , q3N−K , t

)can be solved for the constraint

fI (~xiqi, . . . ,~xNqi) .

2.1. Generalized Coordinates and Constraints 9

Example 2.1.1

Suppose we have N = 1 particle, and the constraint on the particle is that itmoves on a circle x2 + y2 = R2. Then we can write the constraint in terms of thecoordinates as

f(x, y) = x2 + y2 −R2 = 0,

where x = x1 and y = x2.We can describe the motion of the particle with a single generalized coordinate

q = θ. Our Cartesian coordinates in terms of the generalized coordinate are

x = R cos θ

y = R sin θ.

To solve for the motion of the particle, we need to find x(t) and y(t).If there is no external force on the particle then the potential energy is zero,

and the total energy is

E =1

2mv2 =

1

2m(x2 + y2

),

where

x = −Rθ sin θ

y = Rθ cos θ.

Plugging these in and simplifying gives us

E =1

2mR2θ2,

which implies that

θ = ±√

2E

mR2.

We choose the positive sign to indicate the current direction of motion, thenintegrating gives us

θ(t) = θ0 + ωt, ω =

√2E

mR2.

Then we can plug this result back into the x and y equations to get x(t) and y(t)

x = R cos (θ0 + ωt)

y = R sin (θ0 + ωt) .

What if the radius of the constraint circle is expanding as

x2 + y2 = R2(t) = (vt)2?

The direction of motion of the particle is now no longer perpendicular to theconstraint force. Thus, the constraint force is now doing work. This implies thatenergy is not conserved.

In general, if constraint forces depend on time, then E is not conserved.

10 Lagrangian Mechanics

Tip

In general, if constraintforces depend on time, thentotal energy is not con-served.

2.2 The Lagrangian and the Euler-Lagrange Equations

Newton’s law tells us that

~pi = m~xi = ~F i + ~Ci,

where ~Ci is a generic constraint force. By definition, ~Ci is perpendicular to the surfacein which the particle moves.

Now, consider a virtual displacement δ~xi that takes ~xi to the new point ~xi + δ~xi. Ifthis virtual displacement were performed, what would be the work done?∑

i

~pi · δ~xi = work =∑i

(~F i · δ~xi + ~Ci · δ~xi

).

This implies that (~F i − ~pi

)· δ~xi + ~Ci · δ~xi = 0.

Note: We are implying Einstein summation here.We choose our virtual displacement to be along the constraint surface, i.e. δ~xi ‖ ~Ci.

Then ~Ci · δ~xi = 0, which implies (~F i − ~pi

)· δ~xi = 0.

Remember that ~xi = ~xi(qα), where i = 1, . . . , 3N and α = 1, . . . , 3N − K. We canwrite

δ~xi =∑α

∂~xi∂qα

∣∣∣tδqα.

Then

~F · δ~xi = ~F i ·∂~xi∂qα

δqα = Qαδqα,

where summation over α is implied, and

Qα = ~F i ·∂~xi∂qα

.

is a generalized force.Then

~pi · δ~xi =d~pidt· ∂~xi∂qα

δqα

= δqα[d

dt

(~pi ·

∂~xi∂qα

)− ~pi ·

d

dt

∂~xi∂qα

]= δqα

[d

dt

(m~xi ·

∂~xi∂qα

)−m~xi ·

d

dt

∂~xi∂qα

].

In general, if we have a function g = g(qα, t) of coordinates and time, then its totalderivative with respect to time is

d

dtg(qα, t) =

∂g

∂t+

∂g

∂qαdqα

dt.

Before proceeding further, we want to show that

d

dt

∂~xi∂qα

=∂

∂qαd

dt~xi.

2.2. The Lagrangian and the Euler-Lagrange Equations 11

Remember that ~xi = ~xi(q1, . . . , q3N , t). The left hand side expands as

d

dt

(∂~xi∂qα

)=

∂

∂t

(∂~xi∂qα

)+

∂2~xi∂qα∂qβ

qβ .

Here, we are simply using the rule given above with g(qα, t) = ∂~xi∂qα . On the right-hand

side, we haved

dt~xi =

∂

∂t~xi +

∂~xi∂qβ

qβ .

Putting the two sides together, we have

∂

∂qα

(d

dt~xi

)=

∂

∂qα∂

∂t~xi +

∂2~xi∂qα∂qβ

qβ .

So the equality we wanted to show is true provided that we define

∂qβ

∂qα= 0,

which is reasonable.One consequence of these equations is the cancellation of dots

∂~xi∂qα

=∂~xi∂qα

.

Moving on, we have that∑i

~pi · δ~xi =∑α,i

δqα

[d

dt

(mi~xi ·

∂~xi∂qα

∣∣∣qβ ,t

)−mi~xi ·

∂~xi∂qα

∣∣∣qβ ,t

].

We can write this as∑i

~pi · δ~xi =∑α

δqα[d

dt

(∂T

∂qα

∣∣∣qβ ,t

)− ∂T

∂qα

∣∣∣qβ ,t

],

where

T =∑i

1

2mi~x

2

i ,

is the kinetic energy. This implies that∑α

δqα[d

dt

(∂T

∂qα

∣∣∣qβ ,t

)− ∂T

∂qα

∣∣∣qβ ,t−Qα

]= 0.

Since this is true for arbitrary δqα, the quantity in brackets must be zero, and so

d

dt

(∂T

∂qα

∣∣∣qβ ,t

)− ∂T

∂qα

∣∣∣qβ ,t

= Qα,

where

Qα = ~F i ·∂~xi∂qα

= − ∂V∂~xi· ∂~xi∂qα

= − ∂V∂qα

,

where V is the potential energy.We then define the Lagrangian as the quantity

L = T − V.

The equations of motion are then given by the Euler-Lagrange equations

d

dt

∂L∂qα− ∂L∂qα

= 0.


Tip

Remember, in general,r, θ, φ are all functions oftime, so remember to usethe product and chain rulesas appropriate when takingderivatives.

Example 2.2.1

In 2D Cartesian coordinates, the Lagrangian is

L = T − V =1

2m(x2 + y2

)− V (x, y).

In this example, the Euler-Lagrange equations are

d

dt

∂L∂x− ∂L∂x

= 0

d

dt

∂L∂y− ∂L∂y

= 0.

Taking the relevant partial derivatives of the Lagrangian gives us

d

dtmx− ∂V

∂x= 0

d

dtmy − ∂V

∂y= 0.

Taking the time derivatives gives us

mx− ∂V

∂x= 0

my − ∂V

∂y= 0.

This is precisely Newton’s second law in 2D Cartesian coordinates

mx = Fx

my = Fy.

Example 2.2.2

Do the same in 2D polar coordinates where the Lagrangian is

L(r, θ, r, θ) =1

2m(r2 + r2θ2

)− V (r, θ).

The transformation between spherical and Cartesian coordinates is

x = r sin θ cosφ

y = r sin θ sinφ

z = r cos θ.

Note, in general, that r, θ, φ are all functions of time, and you have to keep this in mindwhen calculating the time derivatives of x, y, and z. That is, you have to use the productand chain rules as appropriate. Taking the time derivatives gives us

x = r sin θ cosφ+ rθ cos θ cosφ− rφ sin θ sinφ

y = r sin θ sinφ+ rθ cos θ sinφ+ rφ sin θ cosφ

z = r cos θ − rθ sin θ.

Squaring these, adding them together, and simplifying, allows us to write the kinetic

2.2. The Lagrangian and the Euler-Lagrange Equations 13

energy in spherical coordinates as

T =1

2mv2 =

1

2m(x2 + y2 + z2

)=

1

2m(r2 + r2θ2 + r2φ2 sin2 θ

).

Example 2.2.3

Consider a bead of mass m on a ring of radius R that is spinning with constantangular speed Ω in the presence of gravity.

We will use spherical coordinates r, θ, φ. Then

φ = Ω =⇒ φ = Ωt, r = R.

Note that energy E = T+V is not conserved in this example because the constraintforce can do work on the bead. Fortunately, the Lagrangian approach doesn’t careabout energy conservation.

We can choose the origin of our coordinate system to be at the center of thering with z measured in the downward direction. Then the potential energy is

V = mgh = −mgR cos θ.

It is negative because we are measuring z in the downward direction. Wheneveryou get confused about the sign of the gravitational potential energy, just checkthat V gets larger as the particle goes higher in the gravitational field.

The kinetic energy is

T =1

2mv2 =

1

2m(x2 + y2 + z2

)=

1

2mR2

(θ2 + Ω2 sin2 θ

).

We just used the formula derived earlier for the kinetic energy in spherical coor-dinates and used the fact that in this example, r = R and φ = Ω.

Then our Lagrangian is

L = T − V =1

2mR2

(θ2 + Ω2 sin2 θ

)+mgR cos θ.

We get two Euler-Lagrange equations—one for each of our generalized coordinates.The θ-equation gives us

d

dt

(∂L

∂θ

)− ∂L

∂θ= 0

d

dt

(mR2θ

)−mR2Ω2 sin θ cos θ +mgR sin θ = 0


Simplifying, we get

θ = Ω2 sin θ cos θ − g

Rsin θ.

What are the equilibrium positions? Remember that equilibrium means thatθ is constant. For Newton’s equation F = mx, to find the equilibrium values of x,we set mx = 0 and solve. We use the same approach here, we set θ = 0. Now ourθ-equation reduces to

sin θ(

Ω2 cos θ − g

R

)= 0,

which, for 0 ≤ θ ≤ π gives us three potential solutions. The sin θ = 0 part gives usthe solutions θ = 0 and θ = π. The (Ω2 cos θ− g/R) = 0 part gives us the solutionθ = cos−1(g/RΩ2) provided that g/RΩ2 < 1.

One way to analyze the stability of equilibrium solutions is to set θ = ε. Thatis, to add a small perturbation. Another approach is to examine the Lagrangianin a different manner. We will take the second approach here.

Notice that we can write the Lagrangian as

L = T − V ,

where

T =1

2mR2θ2

V = −1

2mR2Ω2 sin2 θ −mgR cos θ.

Note that these are not the actual kinetic and potential energies of this problem.But they can be treated as effective kinetic and potential energies. In general, ifyou see a Lagrangian of the form

L =1

2mq2 − V (q),

then you immediately know that the quantity

E =1

2mq2 + V (q) = const,

is conserved. In our case, this implies that

E = T + V = const,

is a conserved quantity. To prove this, we could show that dEdt = 0. When

energy is conserved, we can analyze the stability of equilibria by plotting thepotential energy. In our example, energy is not conserved, but effective energy Eis conserved, and we can analyze the stability of equilibria by plotting the effectivepotential V .

Plotting V gives us the picture on the left if g/Ω2R > 1. We see that theequilibrium at θ = π is unstable and the equilibrium at θ = 0 is stable. On theright is the picture we get when g/Ω2 < 1. Now θ = 0 is an unstable equilibriumand θ = cos−1(g/RΩ2) is a stable equilibrium. We have plotted from θ = −π toθ = π, but all we really care about is the region 0 ≤ θ ≤ π.

2.3. Coordinate Independence of the Lagrangian 15

Tip

Whenever you get confusedabout the sign of the grav-itational potential energy,just check that V getslarger as the particle goeshigher in the gravitationalfield.

Tip

The Lagrangian approachdoesn’t care whether or notenergy is conserved. How-ever, with the Lagrangianapproach you can typicallyfind quantities that are con-served.

Note, the Lagrangian is not unique. Different Lagrangians can yield the same equa-tions of motion. Keep this in mind when you see unfamiliar Lagrangians. For example,the following three Lagrangians

L = mbxy − kbxy

L =1

2ma(x2 + y2

)− 1

2ka(x2 + y2

)L =

1

2ma(x2 − y2

)− 1

2ka(x2 − y2

),

all yield the same two equations of motion

x = − kmx, y = − k

my.

2.3 Coordinate Independence of the Lagrangian

If you have N degrees of freedom, then the Lagrangian is generally a function of 2N + 1variables—the N coordinates, the N derivatives of the coordinates, and time.

We originally defined the Lagrangian in Cartesian coordinates (let’s denote it Lx) as

Lx(~xi, ~xi, t) = T − V =∑ 1

2mi~x

2− V (~xi).

We can write the Cartesian coordinates in terms of generalized coordinates

~xi = ~xi(qα, t)

~xi = ~xi(qα, qα, t),

where α = 1, . . . , N goes over the degrees of freedom. Then

Lx(~xi, ~xi, t) = Lx(~xi(qα, t), ~xi(q

α, qα, t), t) = L(qα, qα, t).

We can go even switch to new coordinates qγ ′ (where γ = 1, . . . , N) defined in terms ofthe old coordinates

qγ ′ = qγ ′(qα, t).

Then the Lagrangian would be written as

L′ = L′(qγ ′, qγ ′, t).


Tip

The Lagrangian is invariantunder coordinate transfor-mations.

But the Lagrangian is just a number, and it is the same number regardless of whatcoordinate system it is expressed in. The Lagrangian is a scalar, and it is invariant undercoordinate transformations including nonlinear and time-dependent transformations. Theform of the Lagrangian may look different, but it will always yield equivalent equationsof motion.

Consider the three different coordinate systems shown below. We have the regularx, y, z system, the rotated x′, y′, z′ system, and the translated x′′, y′′, z′′ system.

We can write down the Cartesian form of the kinetic energy in each of these systems as

E =1

2m(x2 + y2 + z2

)=

1

2m(x′ 2 + y′ 2 + z′ 2

)=

1

2m(x′′ 2 + y′′ 2 + z′′ 2

).

So no matter whether our coordinate system is rotated or translated, the formula for theenergy in that coordinate system is the same. That is, kinetic energy is invariant underrotation and translation in Cartesian coordinates.

However, energy is not invariant with respect to a coordinate system that is movingat a speed v. If our reference coordinate system is x, y, z and the coordinate systemX,Y, Z is moving along the y direction with speed v, then

X = x

Y = y − vtZ = z,

which implies

X = x

Y = y − vZ = z.

Then the kinetic energy is

E =1

2m(X2 + (Y + v)2 + Z2

).

Since energy is not invariant when going to a reference frame in relative motion, thequantity is not so useful in relativity. However, the Lagrangian is invariant under alltransformations, so it is a useful quantity in special and even in general relativity.

2.4. The Hamiltonian 17

Tip

The Hamiltonian is con-served if and only if the La-grangian does not explicitlydepend on time.

Differential geometry is an approach to dynamics such that the equations of motioncan be written completely independent of coordinate systems. This approach is gainingpopularity.

The general approach in theoretical physics:1. Start from a physical principle (e.g. physics should be the same in all reference

frames)2. Deduce symmetries implied by this principle3. Write down a Lagrangian that satisfies those symmetries4. Investigate the Lagrangian. What else does it imply?

2.4 The Hamiltonian

In 1D Cartesian coordinates, the Lagrangian is

L = T − V =1

2mx2 − V (x).

Compare this with the mechanical energy in Cartesian coordinates

E = T + V =1

2mx2 + V (x).

The nice thing about the energy is that it is often a conserved quantity—a constant ofthe motion. Notice that we can write the energy in terms of the Lagrangian as

E = x∂L∂x− L =

1

2mx2 + V (x).

In generalized coordinates, we write

H(qα, qα) =∑α

(qα

∂L∂qα

)− L.

We call this quantity the Hamiltonian. To see if or when this quantity is conserved,we look at its time derivative. In general, for a function f(q(t), q(t), t), the total timederivative, obtained using the chain rule, is

df

dt=∂f

∂t+∂f

∂qq +

∂f

∂qq.

Applying this to the Hamiltonian gives us

d

dtH = qα

∂L∂qα

+ qαd

dt

(∂L∂qα

)− d

dtL

= qα∂L∂qα

+ qαd

dt

(∂L∂qα

)−(qα

∂L∂qα

+ qα∂L∂qα

+∂L∂t

)= −∂L

∂t.

We cancelled two of the terms above by applying the Euler-Lagrange equation. So thetime derivative of the Hamiltonian is

dHdt

= −∂L∂t.

This implies that the Hamiltonian is conserved if and only if the Lagrangian does notexplicitly depend on time.


Tip

If the potential V does notdepend on time, and thetransformation from Carte-sian to generalized coordi-nates does not depend ontime, then the Hamiltonianis equal to the total energy.

Example 2.4.1

For the following three Lagrangians, determine if the associated Hamiltonian(i.e. “energy”) is conserved

L =1

2mx2 − V (x)

L =1

2mx2 − 1

2k (x− v0t)2

L = eγt(

1

2mx2 − 1

2kx2).

The first Lagrangian contains no explicit t, so the Hamiltonian is constant.The second Lagrangian is that of a center of potential moving to the right at aconstant speed v0. Since there is an explicit t in the Lagrangian, the Hamiltonianis not conserved. In the third Lagrangian, the time-dependent exponential factorout front is just friction. Since it contains an explicit t, the Hamiltonian is notconserved.

In all three cases you can easily confirm these by writing down the Hamiltonianusing the formula given above, and then taking its time derivative.

When is the Hamiltonian H(q, q) equal to the total energy E = T (q, q) + V (q)? Ingeneral,

H = E = T + V,

if the following two conditions are satisfied:

• The potential V does not depend explicitly on time• The transformation from Cartesian coordinates to generalized coordinates does not

depend on time

Recall our earlier example of the bead on a spinning ring. In that case, the Lagrangianwas

L =1

2mR2θ2 +

1

2mR2Ω2 sin2 θ +mgR cos θ,

where the first two terms on the right are the kinetic energy T and the last term is −V .First, see that H is conserved because L does not explicitly depend on t. Now we want toknow if the Hamiltonian is the total energy or not. The first condition is satisfied becausethere is no t anywhere in L. However, the second condition is not satisfied because thetransformation from Cartesian to spherical coordinates is not time-independent in thiscase. For example,

x = r sin θ cosφ = r sin θ cos(Ωt).

Therefore, in this case,H 6= E = T + V.

In this case, E is not conserved, but the more general quantity H is.

Example 2.4.2

Consider a free particle, i.e. V (~x) = 0, so the Lagrangian is

L =1

2m~x

2.

Now consider the particle from a frame that is rotating at constant velocity relativeto the first frame.

2.5. The Variational Principle 19

The transformations are

x = x′ cosωt− y′ sinωty = x′ sinωt+ y′ cosωt

z = z′.

In the new coordinates, the Lagrangian is

L =1

2m(x2 + y2 + z2

)=

1

2m(x′ 2 + y′ 2 + z′ 2

)+

1

2mω2

(x′ 2 + y′ 2

)+mω (y′x′ − x′y′) .

The first term is the kinetic energy in the rotating frame. The second term hasthe form of a negative Hooke’s law potential, and is the result of centrifugal force.The last term is the result of Coriolis force.

There is no explicit t in this Lagrangian, so we know that H is a conservedquantity. However, the coordinate transformations depend on t, so we know thatH 6= E, i.e. the total energy is not conserved.

The conserved quantity in our case is

H =∑

qα∂L∂qα− L

= x′∂L∂x′

+ y′∂L∂y′

+ z′∂L∂z′− L

=1

2m(x′ 2 + y′ 2 + z′ 2

)− 1

2mω2

(x′ 2 + y′ 2

).

This is the kinetic energy plus the centrifugal potential energy term. The termdue to the Coriolis force does not appear because the Coriolis force does no workon the particle.

2.5 The Variational Principle

The Euler-Lagrange equation is

d

dt

∂L

∂qα− ∂L

∂qα= 0.


When

L→ L+dF

dt,

for any F = F (qα, t), we end up with the same equation of motion. From L, we get theaction as

S =

ˆ t

t0

Ldt.

If you change L by dFdt , S only changes by a boundary term.

Consider light traveling from a point A to a point B in a material with index ofrefraction n. Fermat’s principle tells us that the path the light travels is the onethat minimizes the travel time. The speed of light in the material is v = c/n, and thendT = dl/v = (1/c)ndl, and so

T =

ˆndl.

Can we construct a similar principle for mechanics?Hamilton’s principle tells us that the trajectory that a particle takes is the one

that minimizes the action.Huygen’s principle tells us that every point on a planar wave front acts as a wave

source as shown in the image below. According to this principle, light travels all possiblepaths but with different phases.

For example, we can work out the diffraction of light by applying this principle.

So for the light ray traveling from A to B, we can think of the light as taking all possiblepaths, but that the phases are such that the waves along all but one path—the one thatminimizes travel time—cancel each other out.

Fermat’s principle treats light as a particle traveling a single path. Huygen’s principletreats it as a wave traveling all possible paths. How are the two approaches reconciled?

The Hamilton principle is similar to the Huygen’s principle, but instead of light, wehave a particle. The particle travels all possible paths, but every path cancels due tointerference from other paths except for the one that minimizes the action. All pathsare possible, but each path gets a different weight eiS . Only the path that minimizes

2.5. The Variational Principle 21

the action survives. The others cancel out. This is like the path integral approach toquantum mechanics. You can derive quantum mechanics by thinking in terms of opticsprinciples but with massive particles instead of light.

Suppose that qα(t) is the actual path that is followed by our particle, and q(t) is anarbitrary path that differs from qα(t) by δq. I.e.

δqα = qα(t)− qα(t). (2.1)

Then the variation of the action is

δS = S (qα(t))− S(qα(t)

).

The variational principle tells us that

δS = 0,

for small δqα. In other words, qα(t) corresponds to a minimum (or a maximum) of theaction. The condition δS = 0 is equivalent to the Euler-Lagrange equations.

The time derivative of the variation in Eq. (2.1) is

δqα = qα(t)− qα(t) =d

dt

(qα(t)− qα(t)

)=

d

dtδqα.

That is, the variation of the time derivative is the same as the time derivative of thevariation.

In going from q to q + δq, both q and q change. How does S change?

δS =

ˆ t

t0

δL dt.

Since the variation in L can only occur as variation in qα or as variation in qα, thisbecomes

δS =

ˆ t

t0

δL dt =

ˆ t

t0

(∂L

∂qαδqα +

∂L

∂qαδqα)dt =

ˆ t

t0

(∂L

∂qαδqα +

∂L

∂qαd

dt[δqα]

)dt.

Integrating by parts gives us

δS = − ∂L

∂qαδq∣∣∣tt0−ˆ t

t0

(d

dt

∂L

∂qα− ∂L

∂qα

)δqα(t) dt.

The term in parentheses is precisely our Euler-Lagrange equation, and it is zero for thetrajectory q(t) taken by the particle. The first term is a boundary term. If we assumethat all paths have the same starting point, then the variation in initial position δq(t0)is zero. Similarly, since all paths have the same ending point, then the variation in finalposition δq(t) is zero, so the entire right side is zero, giving us

δS = 0.


For one-dimensional motion,

S =

ˆ t1

t0

L(x, x, t) dt =

ˆ t1

t0

dt

(1

2mx2 − V (x)

).

Then if x(t) is the actual path followed by the particle and x(t) is an arbitrary path, thenthe principle of least action states that

S (x(t)) ≤ S(x(t)),

for any path x(t).In quantum mechanics, S is the phase associated with a path.

Example 2.5.1

Consider a free particle. Show by minimizing the action that the path followedby the particle is a straight line.

The slope of the actual path taken by the particle—the straight line betweeninitial and final points is

v =x1 − x0t1 − t0

.

Now consider an alternate arbitrary path. We can split this arbitrary path intotwo pieces—one for the first ∆t/2 and the other segment over the remaining time.The alternate path will then have two separate slopes—v1 and v2.

The action for the actual path taken by the particle is

S(x(t)) =

ˆ t1

t0

dt

(1

2mx2

)=

1

2mv2∆t,

since V (x) = 0 for a free particle, and x = v is constant. Note that ∆t =´ t1t0dt =

t1 − t0.For the alternate path, we have two pieces with different velocities and with

∆t/2 instead of ∆t

S(x(t)) =1

2mv21

∆t

2+

1

2mv22

∆t

2.

So the difference in the actions is

S(x(t))− S(x(t)) =1

2mv21 + v22 − 2v2

2∆t.

However, v1 and v2 are not completely independent of v. We can relate them bythe average

v =v1 + v2

2.

2.6. Lagrange Multipliers 23

This implies that

S(x(t))− S(x(t)) =1

2mv21 + v22 − 2

(v1+v2

2

)22

∆t =1

8m(v1 − v2)2∆t.

But the right side is non-negative, and so

S(x(t)) ≥ S(x(t)).

Thus, by the principle of least action, the straight line x(t) is the actual path takenby the particle.

Notice that in this example, the difference in the action is quadratic in thevelocity. This is a general thing. The difference in the action is never linear.

In this example,

δ

ˆ1

2mx2 dt = 0,

is equivalent to

δ

ˆdl = 0.

I.e. the trajectory is the minimum of the path length.

2.6 Lagrange Multipliers

We know that

0 = δS =

ˆ t1

t0

δL dt =∂L

∂qαδqα∣∣∣t1t0−ˆ t1

t0

dt

(d

dt

∂L

∂qα− ∂L

∂qα

)δqα,

implies the Euler-Lagrange equations

d

dt

∂L

∂qα− ∂L

∂qα= 0,

since the above equation is true for all qα(t).Now we want to write the same physics in a different language. We can think of

q(t) as a vector—a tall stack of numbers. Then δq is also a vector. Then writing theEuler-Lagrange equation above in the shorthand

ELα = 0,

we have that0 = 〈ELα|δqα〉 , ∀δqα.

This is simply a convenient linear algebra notation. The set of δqα is a complete set, andthis implies that

|ELα〉 = 0.

Now we get a little more formal. Suppose we have a Lagrangian

L = L(qα, qα, t), α = 1, . . . , N,

with K constraintsfI(q

α, t) = 0, I = 1, . . . ,K.

If we could somehow solve for the constraints, i.e. come up with new coordinates

qα = qα(q′ γ , t), γ = 1, . . . , N −K,


then our Lagrangian would be

L = L(q′ γ , q′ γ , t).

Now imagine that we cannot solve for the constraints. If there were no constraints,

0 = δS = 〈EL|δq〉 .

With constraints, δqα is no longer completely arbitrary. We have

δfI = 0.

So by the chain rule,

0 = δfI =∑α

∂fI∂qα

δqα =

⟨∂fI∂q|δq⟩,

for all δq on the surface.In general, for a vector ~v, if we have the statement

~v · δ~q = 0,

for all δ~q, then this implies that ~v = 0. More specifically, if ~v · δ~q = 0 for all δ~q that areon the constraint surface, then this implies that ~v is perpendicular to the surface.

In our case, we have

0 =

⟨∂fI∂q|δq⟩,

for all δq that are parallel to the surface. This implies that

|EL〉 =∑I

λI

∣∣∣∣∂fI∂q⟩,

where the coefficients λI are possibly time-dependent.Note, in terms of notation,

δqα(t)→ |δq〉fα(t)→ |f〉∑

α

ˆ t1

t0

dt fα(t)δqα(t)→ 〈f |δq〉 .

Without constraints,

|EL〉 = 0.

This is just the statement that

d

dt

∂L

∂qα− ∂L

∂qα= 0,

for all α and all t.With constraints,

0 = δS = 〈EL|δq〉 ,

and

fI (qα(t), t) = 0, I = 1, . . . ,K.

Now we add a small perturbation

fI

(qα(t) + δqα(t), t

)= 0.


We can Taylor expand to get

fI

(qα(t) + δqα(t), t

)= 0 = fI

(qα(t), t

)+

∂f

∂qαδqα + · · · .

This implies that ⟨∂f

∂q|δq⟩

= 0,

and

|EL〉 =∑I

λI

∣∣∣∣∂fI∂q⟩.

In other words,d

dt

∂L

∂qα− ∂L

∂qα=∑I

λI(t)∂fI∂qα

.

Since L = T (q, q)− V (q), we can write this in terms of the kinetic and potential energiesas

d

dt

∂T

∂qα− ∂T

∂qα+∂V

∂qα=

∂

∂qα

∑I

λIfI .

Note that ∂V∂qα is a generalized external force and the sum on the right side is the gener-

alized constraint force.If we want to solve for the constraints, we can define a new Lagrangian

L = L+∑I

λIfI(qα, t),

where L = L(qα, qα, t) is our old Lagrangian. Now we solve the EL equations, whiletreating qα and λI as coordinates. In other words, we will get a set of EL equations

d

dt

∂L

∂qα− ∂L

∂qα= 0,

for each of our generalized coordinates qα, but we will also have a set of EL equations

d

dt

∂L

∂λI− ∂L

∂λI= 0,

for each of our constraint forces. However, this one simplifies. Since there’s no λI in L,the first term is zero, and so our set of EL equations for the constraint forces becomes

∂L

∂λI= 0→ gives constraints fI = 0.

If we plug L into our set of EL equations for the qα, we get

d

dt

∂L

∂qα− ∂L

∂qα−∑I

λI∂fI∂qα

= 0,

which is our previous result.

Example 2.6.1

Consider a sphere rolling down an incline of angle φ.


Let x be the distance that the ball has rolled, and θ be the angle throughwhich it has rolled. The kinetic energy of the ball is

T =1

2mv2 +

1

2Iω2 =

1

2mx2 +

1

2mR2θ2.

The potential energy, assuming the reference to be at the top, is

V = mgh = −mgx sinφ.

Then the Lagrangian is

L = T − V =1

2mx2 +

1

2mR2θ2 +mgx sinφ.

However, x and θ are not independent. They are related as

x = Rθ,

and so θ = x/R. ThenL = mx2 +mgx sinφ,

and applying the EL equations for the coordinate x gives us

x =g

2sinφ.

Alternatively, we could have plugged in x = Rθ to get

L = mR2θ2 +mgRθ sinφ,

and applied the EL equation for the coordinate θ to get

θ =g

2Rsinφ.

Now, let’s solve it using the other approach. Instead of explicitly using x =Rθ, we treat it as a constraint force (which is provided by friction). Now we writea new Lagrangian

L = L+ λf(x, θ).

Since we have only a single constraint, we will have only a single Lagrange multi-plier λ. We write the constraint force so that it is equal to zero. We have x = Rθ,so we just move both terms to the same side and write

f(x, θ) = x−Rθ.

Now our Lagrangian is

L =1

2mx2 +

1

2mR2θ2 +mgx sinφ+ λ(x−Rθ).


When doing the EL equation for the constraint, we get the trivial result

f(x,R) = x−Rθ = 0.

When we do the EL equations for the regular coordinates, we now make sure totreat x and θ as independent. I.e. we now have an EL equation for each of x andθ. We get

d

dt

∂L

∂x− ∂L

∂x= 0 =⇒ mx−mg sinφ− λ = 0

d

dt

∂L

∂θ− ∂L

∂θ= 0 =⇒ mR2θ + λR = 0.

With the constraint equation x = Rθ, this gives us three equations and threeunknowns—x, θ, λ.

To solve for the constraint force we read from the Euler-Lagrange equationsthe components of the force in terms of the Lagrange multiplier

Fx = −λ∂f∂x

= −λ

Fθ = −λ∂f∂θ

= λR.

To solve for λ, we use the three Euler-Lagrange equations, and we find

λ = −mg2

sinφ.

So the components of the constraint force are

Fx =mg

2sinφ

Fθ = −mgR2

sinφ.

The x-component Fx, for example, is provided by the friction between the balland the surface.

Example 2.6.2

Consider a mass m rolling down a hemisphere as shown below. If it startsat the top with speed zero, find the constraint force and the angle θ at which itleaves the surface of the hemisphere.

The constraint condition is

f(r, θ) = r −R = 0.


Using polar coordinates, the modified Lagrangian is

L =1

2m(r2 + r2θ2

)−mgr cos θ + λ(t)(r −R).

The coordinates are now r, θ, λ, and we have to assume they’re all functions oftime. The three Euler-Lagrange equations are

mr −mr2θ2 +mg cos θ − λ = 0

d

dt

(mr2θ

)−mgr sin θ = 0

r −R = 0.

Since r = R, we know that r = r = 0, and we get the simplified pair of equations

−mR2θ2 +mg cos θ − λ = 0

mR2θ −mgR sin θ = 0

Thenλ = −mR2θ2 +mg cos θ,

plays the role of the normal force on the sphere. The angle at which the massleaves the surface of the sphere is the angle at which this equation is zero. Buthow do we find that angle?

We can write

θ =dθ

dt=dθ

dt

dθ

dθ= θ

dθ

dθ=

1

2

d

dθ

(θ2).

0 = mR2θ −mgR sin θ =1

2mR2 d

dθ

(θ2)−mgR sin θ.

Then we integrate from 0 to θ

1

2mR2

ˆ θ

0

d

dθ

(θ2)dθ = mgR

ˆ θ

0

sin θ dθ.

Since the initial condition is that θ = 0 at θ = 0, the integral on the left gives usθ2 and the one on the right gives us 1− cos θ. Then

1

2mR2θ2 = mgR (1− cos θ) .

Plugging this result into the equation for λ gives us

λ = mg (3 cos θ − 2) .

This is the reaction (constraint) force.The angle at which the particle leaves the sphere occurs when the force λ

vanishes, i.e. when

θ = cos−1(

2

3

).

2.7. Symmetry 29

2.7 Symmetry

Consider the Lagrangian of a central force

L =1

2m(r2 + r2θ2

)− V (r).

Then the Euler-Lagrange equation for θ is

d

dt

(mr2θ

)= 0.

This implies that the quantity in parentheses, a generalized momentum, is constant

mr2θ = const.

This is a general feature. Whenever a Lagrangian contains the derivative q of a coordinateq, but it does not explicitly include q itself, we say that q is a cyclic coordinate, andthe generalized momentum

∂L

∂q= pq = const,

is conserved.If the Lagrangian does not explicitly contain t, i.e. if

∂L

∂t= 0,

then the Hamiltonian

H =∑i

∂L

∂qiqi − L = const,

is conserved. Remember that the Hamiltonian is not always equal to the total energyE = T + V . We can also write the Hamiltonian in terms of the generalized momenta as

H =∑i

piqi − L.

For a central force, there is no θ-dependence, meaning the system is unchanged if itis rotated within the orbital plane. This means θ is a cyclic coordinate in the Lagrangian,and this rotational symmetry in the plane implies angular momentum conservation. Sim-ilarly, energy conservation comes from time translational symmetry. In general, symme-tries imply conservation laws. This is Noether’s theorem: Any continuous symmetryyields a conserved quantity.

Suppose we have the Lagrangian

L =1

2m(r2 + r2θ2

)− V (r),

and we transform to new coordinates

r = r

θ = θ − θ0.

Then ˙r = r and˙θ = θ. Our new Lagrangian is

L =1

2m(

˙r2 + r2˙θ2)− V (r).

Notice that our new Lagrangian L has exactly the same form as our first Lagrangian L.This tells us that our transformation from the original coordinates to the tilde coordinatesis a symmetry transformation. Now we make this idea more general.


The Lagrangian is just some number. It is invariant under transformations. Supposewe have a Lagrangian

L(q, q, t),

and we make some transformation to new tilde coordinates. Then if we solve the trans-formation equations for the old coordinates q and q and plug them into the Lagrangian,we get a form of the Lagrangian which is now expressed in the new coordinates

L(q, ˙q, t).

Since the Lagrangian is just some invariant number,

L(q, q, t) = L(q, ˙q, t),

even if they may look different.Case 1: We have a simple Lagrangian in Cartesian coordinates

L =1

2m(x2 + y2

).

Now we make a transformation to new coordinates x = x+x0 and y = y. Our Lagrangianin the new coordinates is

L =1

2m(

˙x2 + ˙y2).

Since the Lagrangian is always invariant under transformations, we know that

L (x, y, x, y) = L(x, y, ˙x, ˙y

).

However, there’s a second kind of invariance here—the two Lagrangians look the same.I.e. they have the same form. If we simply removed all tildes from the second, we wouldget precisely the first. We say that

L (x, y, x, y) = L(x, y, ˙x, ˙y

).

I.e. given the Lagrangian, we can directly replace the original coordinates with thetransformed coordinates.

Case 2: We have the same original Lagrangian

L =1

2m(x2 + y2

).

Now we transform to polar coordinates, and then the new form of our Lagrangian is

L =1

2m(r2 + r2θ2

).

Since the Lagrangian is always invariant under coordinate transformations, we again havethe invariance

L(x, y, x, y) = L(r, r, θ, θ

).

However, these two Lagrangians look different—they have a different form. They do notsatisfy the second kind of invariance. I.e.

L(x, y, x, y) 6= L(r, r, θ, θ

).

They are different in the sense that we cannot directly replace the x’s and y’s in L withr’s and θ’s.

In Case 1 (but not Case 2), we have that

L(q, q, t) = L(q, ˙q, t).

2.7. Symmetry 31

When this is true, we say the transformation q → q is a symmetry transformation.In summary, since the Lagrangian is just an invariant number, we know that under

any transformation q → q,

L(q, q, t) = L(q, ˙q, t).

If in addition to being the same, they also look the same, i.e. if,

L(q, q, t) = L(q, ˙q, t),

then we know that there is a symmetry. Conversely, if we perform a symmetry transfor-mation q → q, then we will get a Lagrangian that looks the same.

Consider a small transformation

qα = qα(t) + δqα(t),

where δqα = δqα(q, q, t). Then qα = ddtq

α(q, q, t). By invariance of the first kind, we knowthat both of the following are true:

d

dt

∂

∂qαL(qα, qα)− ∂

∂qαL(qα, qα) = 0

d

dt

∂

∂ ˙qαL(qα, ˙qα)− ∂

∂qαL(qα, ˙qα) = 0,

because L and L are the same but expressed in different coordinate systems. Moreimportantly, if the coordinate transformation q → q is a symmetry transformation, then

d

dt

∂

∂qαL(qα, qα)− ∂

∂qαL(qα, qα) = 0

d

dt

∂

∂ ˙qαL(qα, ˙qα)− ∂

∂qαL(qα, ˙qα) = 0.

Note the L instead of L in the second equation. I.e., in the case of a symmetry transfor-mation, both qα(t) and qα(t) are solutions to the same Euler-Lagrange equation.

The action is the integral of the Lagrangian, and it is invariant under symmetrytransformation. Assuming δ designates a symmetry transformation, we have that

0 = δS = δ

ˆ t1

t0

Ldt =

ˆ t1

t0

[∂L

∂qαδqα +

∂L

∂ ˙qαδqα]

=

ˆ t1

t0

[∂L

∂qα− d

dt

∂L

∂qα

]δqα dt+

∂L

∂qαδqα∣∣∣t1t0.

To get to the second line, we used integration by parts. This is the same process wefollowed when we derived the Euler-Lagrange equation from the variational principle.When we did that, we argued that there was no variation in the initial point δqα(t0) orthe final point δqα(t0), and hence that entire boundary term is zero. This implied thatthe integral must also be zero, which implied that the quantity in brackets—the Euler-Lagrange equation must be zero. Now we are doing something different. Now δqα is somesymmetry transformation, so we cannot argue that the boundary term is zero. However,instead of wanting to derive the Euler-Lagrange equation, we can now assume it is true(since we’ve already proved so), and so we know the integral must be zero, and so

0 =∂L

∂qαδqα∣∣∣t1t0.

This implies that∂L

∂qαδqα(t1) =

∂L

∂qαδqα(t0).


We haven’t made any assumptions about t0 or t1, so we can choose any values for thetime. This implies that

∂L

∂qαδqα(t) = const.

What we’ve shown is that for a symmetry transformation qα → qα = qα + δqα,Hamilton’s principle 0 = δS implies that the quantity

∂L

∂qαδqα(t),

is a constant of the action.For example, suppose we have space variation for the quantity q1, such that

q1 → q1 = q1 + ε.

The other qα are not varied. Then the conserved quantity is

N∑β=1

∂L

∂qβδqβ = const.

The δqβ are zero except for β = 1, so the conserved quantity is

∂L

∂q1δq1 =

∂L

∂q1ε = const.

This is just the momentum in the 1-direction.Consider the example in which we have an N -particle system with the Lagrangian

L =

N∑i=1

1

2mi~x

2

i −∑i,j

V (|~xi − ~xj |).

The potential just depends on the distance between the particles. What is the symmetryin this case? Notice that L remains unchanged if we make the transformation

~xi → ~xi = ~xi + ~ε.

Here, ~ε plays the role of δ~xi. Since it is constant, it drops out of both ~x2

and V . I.e. Lhas the exact same form as L. Then the conserved quantity is

N∑α=1

∂L

∂qαδqα = mi~xi~ε = const.

This is just conservation of total momentum.Consider a central force problem with the Lagrangian

L =1

2m(x2 + y2

)− V

(√x2 + y2

).

Now we rotate the whole system (while keep it in the orbital plane) by an angle θ. Thistransformation is

x→ x = x cos θ + y sin θ

y → y = −x sin θ + y cos θ.

Then the Lagrangian in these coordinates simplifies to

L =1

2m(

˙x2 + ˙y2)− V

(√x2 + y2

),

which has the exact same form as L. So our transformation is a symmetry transformation.

2.7. Symmetry 33

If we perform only a small rotation, the we can use the small angle approximation, andwe get

x→ x = x+ yθ

y → y = −xθ + y,

which implies that

δx = yθ, δy = −xθ.

Notice that the variation now depends on the coordinate. The conserved quantity is∑i

∂L

∂qiδqi =

∂L

∂xδx+

∂L

∂yδy = mxyθ −myxθ = (pxy − pyx) θ,

which is angular momentum conservation.Suppose the action is invariant up to a boundary term. I.e.

S → S + Φ(q, q, t)∣∣∣t1t0.

Now instead of writing δS = 0, we write

δS =

ˆ(EL) dt+

∂L

∂qαδqα∣∣∣t1t0

= Φ(q, q, t)∣∣∣t1t0.

The Euler-Lagrange equation is zero, i.e. the integral is zero, so we get

∂L

∂qαδqα − Φ(q, q, t) = const. (2.2)

That is, we moved everything that depends on t0 to one side and everything that dependson t1 to the other side, which implies that both sides must be constant. And since t0 andt1 are arbitrary, we can simply replace either with t.

Now let us consider a small translation in time

qα(t) = qα(t+ ε) = qα(t).

We can expand qα(t) = qα(t− ε) in a Taylor series as

qα(t) ' qα(t)− εqα(t).

Now −εqα(t) is playing the part of δq,

δqα = −εqα(t).

Then the variation in the action is

δS =

ˆ t1

t0

L(q, ˙q) dt−ˆ t1

t0

L(q, q) dt,


which becomes

δS =

ˆ t1−ε

t0−εL(q, q) dt−

ˆ t1

t0

L(q, q) dt = −εL(q, q)∣∣∣t1t0.

I.e. all but the boundary terms cancel each other out. From Eq. (2.2), we have that

∂L

∂qαδqα + εL(q, q)

∣∣∣t1t0

= const

− ∂L

∂qαεqα + εL(q, q) = const

−ε(∂L

∂qαqα − L

)= const.

This is conservation of the Hamiltonian (i.e. the generalized energy), which is what wefound previously.

2.8. Summary: Lagrangian Mechanics 35

2.8 Summary: Lagrangian Mechanics

Skills to Master

• Be comfortable working with partial and total derivatives• Know the kinetic energy in all of the common coordinate systems• For a given problem, identify convenient generalized coordinates and write down the Lagrangian• Obtain the equilibria and analyze their stability• Given a Lagrangian, identify any conserved quantities• Calculate the Hamiltonian• Understand the variational principle and be able to prove that δS = 0• Find the constraint force(s) by applying the method of Lagrange multipliers

The Lagrangian

The Lagrangian of a system is the kinetic energy minusthe potential energy

L = T − V.

The equations of motion are then given by the Euler-Lagrange equations

d

dt

∂L∂qα− ∂L∂qα

= 0.

We get one equation for each of the generalized coor-dinates.

You should have the kinetic energy

T =1

2mv2,

memorized for the common coordinate systems. In 2DCartesian and polar coordinates, the kinetic energy is

T =1

2m(x2 + y2

)T =

1

2m(r2 + r2θ2

).

In 3D Cartesian, spherical, and cylindrical coordinates,it is

T =1

2m(x2 + y2 + z2

)T =

1

2m(r2 + r2θ2 + r2φ2 sin2 θ

)T =

1

2m(ρ2 + ρ2φ2 + z2

).

Whenever you get confused about the sign of thegravitational potential energy, just check that V getslarger as the particle goes higher in the gravitationalfield.

If you get stuck writing the Lagrangian, follow thisguaranteed procedure:

1. Write down the Lagrangian in Cartesian coordi-nates

L =1

2m(x2 + y2 + z2

)− V (x, y, z).

If the only external force is gravity, thenV (x, y, z) = mgh.

2. Identify the generalized coordinates you want touse

3. Write x, y, and z in terms of your generalizedcoordinates

4. Differentiate them with respect to time andplug them into the Lagrangian. Remember thatyou’re taking the total time derivative here, soyou need to use the product and chain rules sinceyour generalized coordinates are implicitly func-tions of time

The Lagrangian is not unique. Different La-grangians can yield the same equations of motion.Keep this in mind when you see unfamiliar La-grangians. When

L→ L+dF

dt,

for any F = F (qα, t), we end up with the same equa-tion of motion.

The Lagrangian is a scalar, and it is invariant un-der coordinate transformations including nonlinear andtime-dependent transformations. The form of the La-grangian may look different, but it will always yieldequivalent equations of motion. Kinetic energy is in-variant under rotation and translation in Cartesian co-ordinates, but it is not invariant with respect to a co-ordinate system that is moving at a speed v. The La-grangian, however, is invariant under all transforma-tions.

Conserved Quantities

The Lagrangian approach doesn’t care whether or notenergy is conserved. However, with the Lagrangian ap-


proach you can typically find quantities that are con-served.

In general, a quantity A is conserved if

dA

dt= 0.

If a generalized coordinate q does not explicitly ap-pear in the Lagrangian (whether or not its time deriva-tive q appears), then the generalized momentum asso-ciated with this coordinate

pq =∂L∂q

= const,

is conserved.The Hamiltonian is defined as

H(qα, qα) =∑α

(qα

∂L∂qα

)− L.

The Hamiltonian is the generalization of total mechan-ical energy. Its time derivative is

dHdt

= −∂L∂t.

This implies that the Hamiltonian is conserved if andonly if the Lagrangian does not explicitly depend ontime.

The Hamiltonian is equal to the total energy E =T + V , if and only if both of the following are true:

• The potential V does not depend explicitly ontime

• The transformation from Cartesian coordinatesto generalized coordinates does not depend ontime

To identify conserved quantities from the La-grangian:

• Are there any cyclic coordinates, i.e. coordinatesq that do not appear explicitly in the Lagrangian?If so, then for each of these, the associated mo-mentum

pq =∂L∂q,

is conserved.• Does time appear explicitly in the Lagrangian?

If no, then

H =∑α

(qα

∂L∂qα

)− L,

is conserved. If the transformation(s) fromCartesian to the generalized coordinates do notdepend on time, then H = E = T + V .

Equilibria

Given a Lagrangian, you will often be asked to findequilibria and analyze their stability. There are sev-eral approaches you can take:

• Given a generic Lagrangian L, if you can arrangeit in the form

L = Teff (q)− Veff (q),

where Teff (q) = 12Cq

2 (and does not depend onq), and Veff does not depend on q, then you canthink of Teff and Veff as effective kinetic and po-tential energies and C as an effective mass. ThenEeff = Teff +Veff is conserved. To find equilib-ria and their stability, just plot Veff and identifyextrema.

• If you’ve simplified the Euler-Lagrange equationsand have the equations of motion, then you canfind equilibria in the coordinates q by settingq = 0.

The frequency of small oscillations about q0 canbe obtained by expanding the Euler-Lagrange equationabout the equilibrium point by defining q(t) = q0+ε(t),where ε is a small perturbation. Then q = q0+ε, q = ε,and q = ε. Since ε and its derivatives are small, anypowers of them should be discarded as they are verysmall. In the end, you should get a result

ε = −ω2 ε,

from which you can read off the frequency ω.To find turning points, set q = 0.

Variational Principle

Hamilton’s principle tells us that the trajectory thata particle takes is the one that minimizes the action.This is an example of a variational principle.

The action is defined as

S =

ˆ t

t0

Ldt.

The variational principle tells us that

δS = 0,

for small changes in the path δqα. A particle’s actualtrajectory is the trajectory with the smallest action.This condition is equivalent to the Euler-Lagrangeequations.

2.8. Summary: Lagrangian Mechanics 37

Lagrange Multipliers

If there are constraints, the usual approach is to usethose constraints to reduce the number of generalizedcoordinates needed to solve the problem. However, ifwe want to know the constraint forces themselves, wecannot use this approach. Instead, we use the methodof Lagrange multipliers.

1. If you have K constraints, write them all in theform

fI (qα(t), t) = 0, I = 1, . . . ,K.

2. Define a new Lagrangian

L = L+∑I

λIfI(qα, t),

where L = L(qα, qα, t) is our old Lagrangian.3. Solve the Euler-Lagrange equations as before but

now with the modified Lagrangian

d

dt

∂L

∂qα− ∂L

∂qα= 0.

4. Now we have K equations from our constraintconditions and N equations from our Euler-Lagrange equations. Solve this system of equa-tions for the Lagrange multipliers λI .

5. To get the constraint forces, we plug the La-grange multiplier and the partial derivative of the

constraint condition into

Nq = −λI∂fI∂q

.

Alternatively, we can simply read these termsfrom the Euler-Lagrange equations.

If you’re using the method of Lagrange multipliers,make sure you are including all constraints.

Derivatives

In general, if we have a function g = g(qα, t) of coor-dinates and time, then its total derivative with respectto time is

d

dtg(qα, t) =

∂g

∂t+

∂g

∂qαdqα

dt.

A useful derivative rule is the cancellation of dots

∂~xi∂qα

=∂~xi∂qα

.

Don’t forget that the Euler-Lagrange equation in-cludes partial derivatives and a total derivative. Re-member, in general, r, θ, φ are all functions of time, soremember to use the product and chain rules as appro-priate when taking derivatives.

Chapter 3

Scattering

3.1 Scattering Theory

In scattering problems, we have some particle or beam of particle impinging on sometarget.

One quantity of interest is the intensity or flux density of the incoming particles,which is the number of particles per unit area per unit time

I = Intensity =number of particles

area× time.

Another quantity of interest is the cross-section

σ = Cross-section.

For example, if the target is a hard sphere of radius R, then the cross-section is πR2,which is the cross-section of the sphere that is perpendicular to the incoming particles.

Think of the total cross-section as the extent of the potential. For example, for thepotential

V (r) =1

r,

the cross-section is infinite, since the potential extends to infinity. For a potential like

V (r) =

V0 if r < a

0 if r > a

the cross-section is πa2 since the piece-wise defined potential only extends that far. Like-wise,

V (r) =

0 if r < a

−V0 if r > a

and

V (r) =

1r −

1a if r < a

0 if r > a

have total cross-section πa2.

3.1. Scattering Theory 39

Suppose we have N targets which are spaced sparsely enough so that they don’tobscure each other. Then the total cross-section is Nσ.

The scattering rate is the number of particles scattered by the target per unit time,and can be calculated as

S = INσ.

The total number of scattering events in time t is then St.We will now assume that we have N = 1 arbitrary target, and we have a detector

set up, such that if the target is at the origin of our spherical coordinate system and theparticles are moving in the +z direction, then the detector is at angles Θ and φ.

We can writedS = NI σ(Θ, φ) dΩ,

where dΩ = sin Θ dΘ dφ is the solid angle of interest (e.g. the solid angle subtendedby our detector), and σ(Θ, φ) is the differential cross-section. The differential cross-section quantifies how many particles are scattered in a certain direction. Integrating thedifferential cross section gives us the total cross-section

σtot =

ˆσ(Θ, φ) dΩ =

ˆσ(Θ, φ) sin Θ dΘ dφ.

Now suppose our single target has spherical symmetry. For example, it could be a particlegenerating a Coulomb potential. If there’s spherical symmetry, then the differential cross-section has no φ-dependence. Then

σtot =

ˆσ(Θ) dΩ.

In general, the differential cross-section σ(Θ) is what we’re interested in calculating or inmeasuring experimentally.

ThendS

I= 2π σ(Θ) sin Θ dΘ.

If you know what the potential is, then you can use these relations to find σ(Θ). Thisis generally what is called “scattering theory”, and it is the focus of this first section. Onthe other hand, if you know σ(Θ) and are trying to deduce what potential produces thatcross-section, we call this inverse scattering theory. That is the focus of the nextsection.

The impact parameter is one of the parameters we need for our analysis. It iswhat the distance of closest approach would be if no scattering occurred. As shown

40 Scattering

in the drawing above, the scattering angle Θ is asymptotically, the angle at which thescattered particle leaves the scattering region.

In principle, we can always find Θ as a function of b. We can also think of b as afunction of Θ.

To begin our analysis, we consider a ring of incoming particles as shown in thedrawing below. These are particles with

b < impact parameter < b+ db.

The number of incoming particles going through this ring per unit time is I · 2πbdb. Allof these particles will exit the region between some angle Θ and Θ + dΘ. That is

Θ < scattering range < Θ + dΘ.

All particles coming in through the ring on the left will exit through the curved ringon the right. The number of particles exiting between Θ and Θ + dΘ per unit time isI · 2πσ(Θ) sin Θ dΘ.

Since the number of incoming particles equals the number of outgoing particles, we havethat

I · 2πbdb = −I · 2πσ(Θ) sin Θ dΘ.

The negative sign is there by convention since the smaller the b, the greater the scatteringangle. This implies that

σ(Θ) = − b(Θ)

sin Θ

db

dΘ.

Consider now an attractive target potential V (r), e.g. a Coulomb potential.

What is σ(Θ) as a function of the potential V (r). In spherical coordinates, the energy ofa particle is

E =1

2mr2 +

`2

2mr2+ V (r).

What is the trajectory ~r(t) of the particle? By the chain rule, and conservation of angularmomentum (` = mr2θ),

r =dr

dt=dr

dθθ =

`2

mr2dr

dθ.

3.1. Scattering Theory 41

So we can write

E =1

2

`2

mr4

(dr

dθ

)2

+`2

2mr2+ V (r),

which give us

dθ = ± 1√E − `2

2mr2 − V (r)

√`2

2mr4dr.

Integrating gives us

θ(r) = ±√

`2

2m

ˆ r

∞

1

r2√E − V (r)− `2

2mr2

dr.

Now we want to find the scattering angle Θ.We know that θ, the angle made by r(t) starts at 0, since the particle comes in from

infinity, and ends at π + Θ since the particle leaves along the scattering angle. Then wecan write

Θ = π − θ(∞).

When r → ∞, what is θ? Note that θ(∞) = 2θ(r0), where r0 is the distance of closestapproach of the particle.

By conservation of energy, we know the total energy is equal to the kinetic energy whenthe particle is at r =∞. That is,

E =1

2mv2∞.

By conservation of angular momentum,

` = mv∞b.

So we can write

Θ(b) = π − 2b

ˆ ∞r0

1

r

√r2(

1− V (r)E

)− b2

dr.

For example, if we have the Coulomb potential,

V (r) = −αr,

we can plug this into the integral to get

b(Θ) =α

mv2cot

(Θ

2

).

Then

σ(Θ) =α2

4m2v21

sin4 (Θ/2).

Note that the total cross-section diverges

σtot = 2π

ˆdΘ sin Θσ(Θ) =∞.

42 Scattering

3.2 Inverse Scattering Theory

In inverse scattering theory, we want to solve the problem of finding V (r) when we knowΘ(b).

Starting from

Θ(b) = π − 2b

ˆ ∞r0

1

r

√r2(

1− V (r)E

)− b2

dr,

we let

y(r) ≡ r√

1− V (r)

E,

then

Θ(b) = π − 2b

ˆ ∞r0

1

r√y2(r)− b2

dr.

Note that y(r0) = b. We can write

dr =dr

dydy = r′ dy,

then

Θ(b) = π − 2b

ˆr′

r√y2 − b2

dy,

and then

π −Θ = 2b

ˆdy√y2 − b2

d

dyln r.

If V (r) were zero, then y = r and there would be no scattering angle Θ. Then

π = 2b

ˆdy√y2 − b2

d

dyln y.

So we can convince ourselves that

Θ = 2b

ˆ ∞b

dy√y2 − b2

d

dyln

y

r(y).

Next we define

T (y) ≡ 1

π

ˆ ∞y

db√b2 − y2

Θ(b)

=1

π

ˆ ∞y

2b√b2 − y2

[ˆ ∞b

du√u2 − b2

d

duln

u

r(u)

]du

=1

π

ˆ ∞y

[d

duln

u

r(u)

]du

ˆ u

y

2b db√(b2 − y2)(u2 − b2)

= − lny

r(y).

Note, when we switched the order of the integrals, we had to keep in mind that y < b < u.Now we have that

r(y) = yeT (y).

So if we have access to experimental data, then T (y) is known. From that we can obtainr(y), which we can then plug that back into the definition of y to get V (r).

3.3. Chaotic Scattering 43

3.3 Chaotic Scattering

Now we consider scattering off of multiple hard spheres. We will start by consideringscattering from a pair of disks of radius a and separated by a distance D, as shown below.

There is one possible periodic orbit for a scattered particle, and that is if the particleis trapped so that it bounces back and forth between the disks as illustrated below.However, this would require the particle to start between the two disks.

There are also non-periodic orbits. Consider the two cases shown below. In case, 1,the particle leaves at the top. In case 2, the particle leaves at the bottom.

The only change for these two cases is a change in the scattering angle. Somewherebetween these two cases, there must exist an incident angle which leaves the particletrapped—i.e. never leaving from the top or the bottom.

In general, we can specify a scattering incident by two angles as shown in the imagebelow. The angle φn is the incident angle (measured from the normal) of the nth scatteringevent by the particle. Notice that we label the two disks as L and R.

44 Scattering

There must exist some mapping M , such that

(θn+1, φn+1) = M(θn, φn).

Given (θ0, φ0), you can successively list subsequent hitting angles by repeatedly applyingthis mapping M

(θ0, φ0)→ (θ1, φ1)→ (θ2, φ2)→ · · ·We can make a phase space diagram of this system. The periodic orbit (where it

bounces infinitely between the two spheres) is at the point (0, 0) since it always hits sphereL at θ = 0 and φ = 0. The line S is stable in the sense that if the system starts anywhereon this line, it will be asymptotically attracted to (0, 0). The line U is the opposite. Wecan think of U as the time-reverse of S.

The stable center point and the lines of attraction and repulsion are reminiscent ofa similar point and lines on the phase diagram of a pendulum shown below.

3.3. Chaotic Scattering 45

Now we add a third disk to the scattering problem. We label them K, L, and M .Now there are infinitely many periodic orbits. Two of them are shown below.

Now we turn to the Cantor set. Recall that the Cantor set is constructed byrepeatedly removing middle thirds from the real interval [0, 1]. This is illustrated below.

Notice how we label the different levels as I0, I1, and so on. Then the Cantor set is givenby

C = ∩∞n=0In = I∞.

The Cantor set, I∞ is not empty. For example, any endpoint of any Ii is contained in C.In fact, C contains infinitely many points. The length of the nth interval In is defined as

ˆIn

1 dx =

(2

3

)n.

Then the length of I∞ is zero since (2

3

)n→ 0,

as n → ∞. At the same time, C has the power of the continuum. It can be mapped toevery real number in [0, 1].

We can label each individual subinterval using a string of r’s and `’s as shown in theimage above. So every point in I∞ can be represented by a string fo r’s and `’s. Start at 0and replace every ` with 0 and every r with 1, and then you have a binary representation

`r``r . . .→ 0.01001 . . .

Any number in [0, 1] can be written in binary form as a point followed by 0’s and 1’s,which means it can be represented by a string of r’s and `’s.

What is the fractal dimension of the Cantor set C? If you have intervals of lengthε, it takes 1/ε of them to fill the interval [0, 1]. To tile a 2D region, you would need 1/ε2

of them. In 3D, you would need 1/ε3 to fill the space. In d-dimensions, you need

N ∼ 1/εd.

46 Scattering

Can we use this to deduce the “dimension” of a fractal? Solving for d gives us

d = − logN

log ε.

Interval In contains 2n subintervals. Then we need εn = 1/3n. Then

d ' n′ log 2

n′ log 3=

log 2

log 3≈ 0.631.

This is the fractal dimension of the Cantor set.Can we find something similar to the Cantor set in the scattering from 3 disks? We

set our initial interval I0 to be the initial range of impact parameters. We set I1 to bethe range of impact parameters that yields at least 1 scattering event, I2 to be the rangeof impact parameters that yields at least 2 scattering events, and so on.

ThenIn = IKn ∪ ILn ∪ IMn .

We can again use the labeling convention like LRLL . . ..Chaos occurs in systems that are very sensitive to initial conditions and have low

predictability. The real world is deterministic, but not predictable.Why is self-similarity (scale invariance) a result of chaotic systems?The divergence between “identically” prepared chaotic systems is exponential with

time. For example, if two identically prepared systems follow some trajectory x(t), thentheir divergence over time is

|δx(t)| ∼ |δx(0)|eλt,

where λ is the Lypunov exponent.

3.4. Summary: Scattering 47

3.4 Summary: Scattering

Skills to Master

•

Chapter 4

Linear Oscillations

For a generic potential with equilibrium at q0, a system perturbed from equilibrium willundergo oscillation.

We can Taylor expand the potential about the equilibrium point as

V (q) = V (q0) + V ′(q0)(q − q0) +1

2V ′′(q0)(q − q0)2 + · · ·

= V (q0) +1

2V ′′(q0)(q − q0)2 + · · ·

= V (q0) +1

2k(q − q0)2 + · · · ,

wherek = V ′′(q0),

is the effective spring constant. Then the Lagrangian is

L = T − V =1

2mq2 − V (q0)− 1

2k(q − q0)2 −O((q − q0)3).

Dropping the constant term and the higher order terms gives us

L ≈ 1

2mq2 − 1

2k(q − q0)2.

Substituting q ≡ q0+x, i.e. x ≡ q−q0, then we see the Lagrangian for a regular harmonicoscillator

L =1

2mx2 − 1

2kx2,

which leads us to the equation of motion

mx+ kx = 0, =⇒ ω =

√k

m.

Linear Oscillations 49

In general, we can have more than one degree of freedom. Suppose we have nDOF, q1, . . . , qn, with a potential V (q) = V (q1, . . . , qn), and an equilibrium positionq0 = (q10 , . . . , q

n0 ).

Taylor expanding V (q) about the equilibrium position q0,

V (q) = V (q0) +∂V

∂qα(qα − qα0 ) +

1

2

∂2V

∂qα∂qβ(qα − qα0 )

(qβ − qβ0

)+ · · · .

Keep in mind that repeated indices are summed over. The first derivative term is neces-sarily zero since the slope is zero at equilibrium. In the second derivative term, we canmake the definitions

Kαβ ≡∂2V

∂qα∂qβ

xα ≡ qα − qα0xβ ≡ qβ − qβ0 .

Now we can write the potential as

V (x) ' const +1

2Kαβx

αxβ .

We can write a more general form of the Lagrangian as

L =1

2mαβ(q)qαqβ − V (q).

Since the motion is happening around equilibrium, it is safe to expand L about q0. Noticethat the mass term looks like a matrix. Expanding the mass term gives us

mαβ(q) = mαβ(q0) + · · · (q − q0) + · · · (q − q0)2.

The first term determines a constant matrix—a matrix whose elements are constant. I.e.the elements are numbers which do not depend on the coordinates q. We can neglectall powers of q − q0 in this expansion because in the Lagrangian every term is alreadymultiplied by ∼ (q − q0)2, and since q − q0 is small, we don’t need to take the expansionto even higher powers. In other words, we keep only the first term in the expansion ofmαβ(q). So we define the mass matrix as

Mαβ = mαβ(q0).

Now our Lagrangian is now written as

L =1

2Mαβ x

αxβ − 1

2Kαβx

αxβ .

50 Linear Oscillations

Tip

The matrices M and K aresymmetric by construction.

For example, consider the mass term of the Lagrangian in polar coordinates: 12m(r2+

r2θ2). In the new notation, we can write it as

1

2mαβ q

αqβ =1

2mrr r

2 +1

2mθθ θ

2.

In this example, the mass matrix is

M =

[m 0

0 mr2

].

If we have two DOF and

M =

[m1 0

0 m2

],

then the mass term in our Lagrangian would look like

1

2Mαβ x

αxβ =1

2m1x

21 +

1

2m2x

22.

Keep in mind that when you have velocity-dependent forces such as in magneticfields or Coriolis force, then you can have cross-terms like xαxβ , and the mass matrix willnot be diagonal.

Note: The matrices M and K are symmetric by construction.Applying the Euler-Lagrange equations

d

dt

∂L

∂xα− ∂L

∂xα= 0,

gives us the equation of motion

Mαβ xβ +Kαβx

β = 0.

Remember, we are summing over the repeated index β, so this designates n equationssince α = 1, . . . , n. We can think of this as a matrix equation

M ~x+K~x = 0,

where M and K are n× n matrices, and ~x and ~x are vectors of length n. Assuming Mis invertible, then

~x+M−1K~x = 0,

which gives us the eigenvalue equation

~x+ Λ~x = 0, Λ = M−1K.

So all of the information is contained in the single matrix—Λ. This matrix equation islinear, which means we can solve it. In general,

~x = −Λ~x,

has the solution

~x(t) = eiωt ~Xω + e−iωt ~Xω,

where the ~Xω are eigenvectors of Λ, and ω2 are the associated eigenvalues. That is,

Λ ~Xω = ω2 ~Xω.


Tip

The dimension of the Mand K matrices is thesame as the number ofmasses (not the number ofsprings).

We will write the eigenvalues as ω2 and the eigenvectors as ~a, then

Λ~a = ω2~a,

and our general solution is the linear combination

~x(t) =∑ω

~aω(αωe

iωt + α∗ωe−iωt) .

The coefficients αω are obtained from the initial conditions. Since we have n degrees offreedom, we expect 2n initial conditions—the initial coordinates and the initial velocities.

Example 4.0.1

Consider a triatomic molecule modeled as three masses connected by springsof spring constant k as shown below.

In this example, we have three degrees of freedom. Let x1, x2, x3 be thedisplacements of the masses from equilibrium, then the Lagrangian is

L = T − V =1

2mx21 +

1

2Mx22 +

1

2mx23 −

1

2k(x2 − x1)2 − 1

2k(x3 − x2)2.

This gives us three Euler-Lagrange equations

mx1 = −k(x1 − x2)

Mx2 = −k(x2 − x1)− k(x2 − x3)

mx3 = −k(x3 − x2).

The next step is to identify the matrices M and K by comparing the Euler-Lagrange equations with the matrix equation

M ~x = −K~x.

Note, I am using “M” both for the middle mass and the overall mass matrix.Context should tell you which I mean in any specific usage. We see that

m 0 0

0 M 0

0 0 m

x1

x2

x3

= −

k −k 0

−k 2k −k0 −k k

x1

x2

x3

.Now we can easily see what the matrices M and K are.

Next we would calculate Λ = M−1K, and then find the eigenvalues andeigenvectors of Λ.


Example 4.0.2

Now we will repeat the same triatomic molecule problem, but with a differentsetup. Suppose in equilibrium, the distance between the neighboring atoms is `.

We will define coordinates in terms of distance from the equilibrium position ofthe middle mass as shown below.

Notice that in equilibrium, the coordinates have the values

q1 = −`q2 = 0

q3 = `.

Our Lagrangian is

L =1

2mq21 +

1

2Mq22 +

1

2mq23 −

1

2k(q2 − q1 − `)2 −

1

2k(q3 − q2 − `)2.

Now we define new coordinates in terms of displacement from equilibrium

q1 = −`+ x1

q2 = 0 + x2

q3 = `+ x3.

Notice that the first term on the right for each line is the equilibrium position forthat coordinate. Now, q1 = x1 and so on, so

L =1

2mx21 +

1

2Mx22 +

1

2mx23 −

1

2k(x2 − x1)2 − 1

2k(x3 − x2)2.

Notice that we end up with the same Lagrangian as before.Since we have three oscillators, we have three degrees of freedom, so we know

our matrices M and K will be 3 × 3. Since all the x’s are squared in the kineticpart of the Lagrangian, we can easily see that the matrix M is diagonal, and wecan easily read off those diagonal elements—m,M,m. By writing the potential inthe form

−V = −1

2

∑i,j

Kijxixj ,

as it is in the general form of the Lagrangian, we can deduce the elements of K.We find that

M =

m 0 0

0 M 0

0 0 m

, K = k

1 −1 0

−1 2 −1

0 −1 1

.


Then

Λ = M−1K = k

1m − 1

m 0

− 1M

2M − 1

M

0 − 1m

1m

.Remember, the inverse of a diagonal matrix is just the matrix with its elementsreplaced by their reciprocals.

Next, we calculate the eigenvalues and eigenvectors of Λ. The first one wefind is the eigenvalue and eigenvector pair

ω1 =

√k

m, ~a1 =

1

0

−1

.Notice that the second component in the eigenvector is zero. So in this mode, themiddle ball does not move at all. The other components have equal magnitudeand opposite sign, implying that the two outside balls move with opposite phase.

For the second mode, we find

ω2 =

√k

(1

m+

2

M

), ~a2 =

1

− 2mM

1

.In this mode the outer balls are moving in one direction while the middle one ismoving in the other direction.

For the third mode, we find

ω3 = 0, ~a3 =

1

1

1

.In this mode all three balls are moving in the same direction. The frequency ofoscillation is ω = 0 implying that there is no oscillation, just translational motion.Zero modes like this are often associated with symmetry, and in this case, that istranslational symmetry.

The three modes are illustrated by the arrows in the drawing below:

The general solution is a linear combination of the specific solutions

~X(t) =

x1(t)

x2(t)

x3(t)

=

DOF∑n=1

~an(αne

iωnt + α∗ne−iωnt

).


Tip

The inverse of a diagonalmatrix is just the same ma-trix with its elements re-placed by their reciprocals.

We could also write the solution as a linear combination of the formAn cos(ωnt) + Bn sin(ωnt). In both cases, we have two arbitrary coefficients foreach n.

For our example,

~X(t) =

1

0

−1

(α1eiω1t + α∗1e

−iω1t)+

1

− 2mM

1

(α2eiω2t + α∗2e

−iω2t)+

1

1

1

(a+ vt) .

We have six arbitrary coefficients since we have 3 DOF. These are determined bythe initial conditions. For our problem, assume the initial conditions are

x1(0) = −A

x2(0) =Am

Mx3(0) = 0

x1(0) = 0

x2(0) = 0

x3(0) = 0.

Then our final solution is

~X(t) =1

2A

cos(ω1t) + cos(ω2t)

− 2mM cos(ω2t)

− cos(ω1t) + cos(ω2t)

.

Example 4.0.3

Consider a pair of oscillating masses of mass m separated from each otherand two walls with three springs of spring constant k as shown below.

At equilibrium, the distances between the masses and the masses and thewalls are `. When not in equilibrium, their positions q1 and q2 are measured fromthe left wall.

The Lagrangian is

L =1

2m(q21 + q22

)− 1

2k(q1 − `)2 −

1

2k(q2 − q1 − `)2 −

1

2k(2`− q2)2.

When writing down the potential energy in the Lagrangian, you will always haveone term of the form 1

2k∆x2 for each spring. Then ∆x is just the current lengthof the spring minus its equilibrium length.


From the Lagrangian, it is easy to see that

M = m

[1 0

0 1

].

To find K, it is easier to switch coordinates to

q1 = `+ x1

q2 = 2`+ x2.

Notice that the first term on the right for each equation is just the equilibriumposition of the mass in the old coordinates. The second term on the right for eachequation is the new coordinate. Since q1 = x1 and q2 = x2, the form of the kineticpart of the Lagrangian is unchanged, and so the mass matrix m is unchanged bythis coordinate change. Now the Lagrangian reads

L =1

2m(x21 + x22

)− 1

2kx21 −

1

2k(x2 − x1)2 − 1

2kx22.

Expanding the quadratic in the potential part of the Lagrangian and then gath-ering terms, we can read off the coefficients to get

K = k

[2 −1

−1 2

].

Then

Λ = M−1K =k

m

[2 −1

−1 2

].

Then we find the eigenvalues and eigenvectors of Λ.The first is

ω1 =

√k

m, ~a1 =

[1

1

].

This is the motion in which both masses move in phase with each other. Thesecond is

ω2 =

√3k

m, ~a2 =

[1

−1

].

In this mode, the two masses move opposite each other.

The general solution is

~X(t) =

[1

1

] (α1e

iω1t + α∗1e−iω1t

)+

[1

−1

] (α2e

iω2t + α∗2e−iω2t

).


Example 4.0.4

Analyze the system given below containing two identical masses m and threeidentical springs. What are the normal modes and frequencies. Assume thatgravity is zero (g = 0).

Our Lagrangian in the coordinates shown is

L =1

2m(z21 + z22

)− 1

2k (z1 − a)

2 − 1

2k (z2 − z1 − a)

2 − 1

2k (2a− z2)

2.

We transform to equilibrium coordinates

z1 = a+ x1 =⇒ x1 = z1 − az2 = 2a+ x2 =⇒ x2 = z2 − 2a.

Now our Lagrangian is

L =1

2m(x21 + x22

)− 1

2kx21 −

1

2k (x2 − x1)

2 − 1

2kx22.

Applying the Euler-Lagrange equations, we get the equations of motion

mx1 + kx1 + k(x1 − x2) = 0

mx2 + k(x2 − x1) + kx2 = 0.

At this point, make a habit of stopping and checking the signs of terms containingk—it should be a restoring force.

Rearranging,

x1 = − km

(2x1 − x2)

x2 = − km

(2x2 − x1) .

We can write this in the form ~x = −Λ~x as

d2

dt2

[x1

x2

]= − k

m

[2 −1

−1 2

][x1

x2

],

where our matrix of interest is

Λ =k

m

[2 −1

−1 2

].


Tip

If you get an imaginary fre-quency, you’ve made a mis-take somewhere.

Tip

Adding a constant externalforce to a system of oscil-lators changes their equi-librium positions, but doesnot affect the normal modesor frequencies.

Calculating the eigenvalues ω2 and the eigenvectors gives us the normal frequenciesand normal modes for the system

ω1 =

√k

m, ~a1 =

[1

1

]

ω2 =

√3k

m, ~a2 =

[1

−1

].

Notice that the higher frequency is associated with the less symmetric mode. Thisis a general feature.

Example 4.0.5

Repeat the previous problem, but now include gravity g.This adds two terms: −mgz1 − mgz2 to the Lagrangian. Then we get the

equations of motion

mz1 = −k(z2 − z1)−mgmz2 = −k(2z2 − z1 − 3a)−mg.

Next, identify the equilibrium points by setting the forces equal to zero:mz1 = mz2 = 0. This gives the equilibria

zeq1 = a− mg

k, zeq2 = 2a− mg

k.

Then we transform to new equilibrium coordinates via

z1 = zeq1 + x1

z2 = zeq2 + x2.

Then the equations of motion we get are exactly the same as the equations ofmotion we found when we didn’t include gravity. In other words, including gravitydoesn’t change the behavior of the system—it only shifts the equilibrium positionsof the masses. In general, a constant force changes the equilibrium points but notthe normal modes or frequencies.

A way to see this without performing any calculations is to recall that

Kij =∂2V

∂zi ∂zj

∣∣∣eq.

I.e. the K matrix doesn’t care about linear terms in the Lagrangian, so linearterms in the Lagrangian will not contribute to the normal modes and frequencies.


4.1 Summary: Linear Oscillations

Skills to Master

• Given an arbitrary potential with an equilibrium point, perform a harmonic approximation, andcalculate the effective spring constant and the frequency of small oscillations about the equilibrium

• Know how to calculate the inverse of a 2× 2 matrix• Be able to calculate the eigenvalues, and eigenvectors of 2× 2 and 3× 3 matrices• Identify frequencies and normal modes for coupled oscillators

Given a generic potential V (x) with an equilib-rium at x0, we can expand the potential in a Taylorseries about the equilibrium point. Then

L ≈ 1

2mq2 − 1

2k(q − q0)2,

where the effective spring constant is

k = V ′′(q0).

This is the Lagrangian of a simple harmonic oscillatorwith frequency ω =

√k/m.

Coupled Oscillators

To analyze a coupled oscillator problem, follow thisprocedure:

1. Write down the Lagrangian for the system. Itshould contain one kinetic term of the form 1

2mx2

for each mass, and one potential term of the form12k∆x2 for each spring. There may also be anexternal force present (e.g. gravity), which youhave to include in the potential

2. If you’re not already in such coordinates, trans-form to coordinates in terms of displacement fromequilibrium

old coord. = equil. + new coord.

This should simplify your Lagrangian3. Obtain the Λ-matrix. There are two ways to do

this:

• Compare your Lagrangian with

L =1

2Mαβ x

αxβ − 1

2Kαβx

αxβ ,

and identify the M and K matrices. Then,

Λ = M−1K.

The matrices M and K must be symmetric.Their dimension is the same as the num-ber of masses (not the number of springs).

Note that this approach requires calculatinga matrix inverse, so don’t use this approachif you have three or more oscillating massesand you have to do calculations by hand.

• Obtain the equations of motion via theEuler-Lagrange equations and write themdown as a matrix equation

~x = −Λ~x.

Note the negative sign!

4. Calculate the eigenvalues and eigenvectors of Λ5. Write down the general solution, which is the sum

of the specific solutions to the eigenvalue problemΛ~a = ω2~a,

~x(t) =∑ω

~aω(αωe

iωt + α∗ωe−iωt) ,

where ~aω is the eigenvector of Λ associated withthe eigenvalue ω2, and the α are arbitrary con-stants set by the initial conditions. You can alsowrite this as sines and cosines instead of as com-plex exponentials. Note, you will have as manyeigenvalues and terms in this sum as there aremasses in the physical problem. If an eigenvalueis ω2 = 0, then use a + bt instead of the twoexponentials for that specific solution.

6. Set the arbitrary constants via the initial condi-tions

Miscellaneous

Remember the following time-saving facts about ma-trices:

• The inverse of a diagonal matrix is just the samematrix with its elements replaced by their recip-rocals

• The inverse of a 2× 2 matrix is[a b

c d

]−1=

1

ad− bc

[d −b−c a

]• Know how to quickly calculate the eigenvalues

and eigenvectors of block matrices

4.1. Summary: Linear Oscillations 59

• The determinant of a constant times a matrix isthe same as the constant raised to the power ofthe dimension of the matrix times the determi-nant of the matrix

Det(aK) = an Det(K).

But if you’re setting the determinant equal tozero in the end anyway such as when calculatingeigenvalues, you can discard the constant fromthe beginning.

Chapter 5

Hamiltonian Mechanics

5.1 Hamilton’s Equations

The Hamiltonian formalism is useful in quantum mechanics and in general when dealingwith conserved quantities.

Consider a free particle in 2D with Lagrangian

L =1

2m~x

2=

1

2mr2 +

1

2mr2θ2.

We know there’s a conserved quantity—` = mr2θ. If we plug this into L, we get

L =1

2mr2 +

`2

2mr2,

Applying the Euler-Lagrange equation

d

dt

∂L

∂r− ∂L

∂r= 0,

gives us the equation of motion

mr +`2

mr3= 0.

However, because of the negative sign in

mr = − `2

mr3,

this implies a central force. But we started with a free particle—there should be no centralforce, so somewhere we made an error. There’s something illegal about our substitution ofthe conserved quantity ` into the Lagrangian. The problem occurs because of the partialderivatives in the Euler-Lagrange equation—they’re not actually holding ` constant. Theproper way to solve this problem in Lagrangian mechanics is to find the equation ofmotion

mr −mrθ2 = 0,

before substituting in the conserved quantity ` = mr2θ. Then we get the equation ofmotion with the right sign

mr =`2

mr3.

Note that∂L

∂qα

∣∣∣qβ6= ∂L

∂qα

∣∣∣p.

5.1. Hamilton’s Equations 61

I.e., the partial of L with respect to a coordinate while holding a velocity constant isnot the same as the partial of L with respect to a coordinate while holding a momentumconstant. How different are these two partial derivatives? Since L = L(q, q), we can write

dL =∂L

∂qα

∣∣∣qdqα +

∂L

∂qα

∣∣∣qdqα

=∂L

∂qα

∣∣∣pdqα +

∂L

∂pα

∣∣∣qdpα.

Remember that the vertial bar with a subscript identifies the quantity being held constantwhen the partial derivative is taken. This implies that

∂L

∂qα

∣∣∣p

=∂L

∂qα

∣∣∣q

+∂L

∂qβ∂qβ

∂qα

∣∣∣p

∂L

∂qα

∣∣∣q

=∂L

∂qα

∣∣∣p− pβ ∂q

β

∂qα

∣∣∣p

=∂

∂qα[L− pβ qβ

] ∣∣∣p

= − ∂H∂qα

∣∣∣p,

where

H(q, p) = pαqα − L,

is the Hamiltonian. To write the Hamiltonian H as a function of q and p, rewrite anyq in terms of p.

Recall the Euler-Lagrange equation

d

dt

∂L

∂qα− ∂L

∂qα

∣∣∣q

= 0.

In the first term, we can write ∂L∂qα = pα, and using the relation we found above, we can

write the second term in terms of the Hamiltonian

d

dtpα +

∂

∂qαH(q, p)

∣∣∣p

= 0.

This gives us the first of two Hamilton equations

pα = − ∂H∂qα

∣∣∣p.

By the chain rule,

∂L

∂pα=

∂L

∂qβ

∣∣∣q

∂qβ

∂pα+∂L

∂qβ

∣∣∣q

∂qβ

∂pα= pβ

∂qβ

∂pα.

With this, we can rewrite

∂H

∂pα=

∂

∂pα[pβ q

β − L] ∣∣∣q

= pβ∂qβ

∂pα+ qα − pβ ∂q

β

∂pα

= qα.

62 Hamiltonian Mechanics

This gives us the second of Hamilton’s equations

qα =∂H

∂pα

∣∣∣q.

The whole point of the derivation of Hamilton’s equations was to see if we can dealproperly with conserved quantities. So let’s return to our original problem where

L =1

2mr2 +

1

2mr2θ2.

The Hamiltonian is

H(q, p) = pαqα − L(q, q)

= pr r + pθ θ − L(r, r, θ, θ)

= pr r + pθ θ −1

2mr2 − 1

2mr2θ2.

Since H = H(q, p), there should be no velocities in the Hamiltonian. So wherever we seea velocity, we need to rewrite it in terms of the associated momentum. In general,

pα ≡ ∂L

∂qα.

In our case,

pr =∂L

∂r= mr =⇒ r =

prm

pθ =∂L

∂θ= mr2θ =⇒ θ =

pθmr2

.

Substituting these in gives us

H =p2r2m

+p2θ

2mr2.

As we did with the Lagrangian approach earlier, we want to treat pθ as constant, so wewrite

H =p2r2m

+`2

2mr2.

The Hamilton equations are then

r =∂H

∂pr=prm

pr = −∂H∂r

=`2

mr3.

These imply that

mr =`2

mr3.

Now we have the right sign for this force.If we didn’t know from the beginning that ` was conserved, how would this Hamil-

tonian approach give us that conservation law? From the Hamiltonian

H =p2r2m

+p2θ

2mr2,

the Hamilton equation for θ gives us

d

dtpθ = −∂H

∂θ= 0,

5.1. Hamilton’s Equations 63

which tells us that pθ is conserved.What about energy conservation? The total time derivative of H is

dH

dt=∂H

∂qαqα +

∂H

∂pαpα +

∂H

∂t

= −pαqα + qαpα +∂H

∂t

=∂H

∂t.

So if H has no explicit time-dependence, then

dH

dt= 0,

and so energy conservation is a natural part of the Hamiltonian formulation.

Example 5.1.1: 3D Spherically Symmetric Potential

For a particle in a 3D spherically-symmetric potential, the Lagrangian is

L =1

2m~x

2− V (r) =

1

2m(r2 + r2θ2 + r2 sin2 θ φ2

)− V (r).

For each coordinate qα = r, θ, φ, we could write an Euler-Lagrange equation toget the equations of motion. How do we find the equations of motion in theHamiltonian formalism?

We wantH = pαq

α − L = pr r + pθ θ + pφφ− L(q, q).

Since the Hamiltonian must be written in terms of the coordinates and momenta(no velocities), we need to express all velocities in terms of momenta.

pr =∂L

∂r= mr =⇒ r =

prm

pθ =∂L

∂θ= mr2θ =⇒ θ =

pθmr2

pφ =∂L

∂φ= mr2 sin2 θ φ =⇒ φ =

pφ

mr2 sin2 θ.

Plugging L into H and then replacing all velocities with what we found here interms of momenta gives us

H =p2r2m

+p2θ

2mr2+

p2φ

2mr2 sin2 θ+ V (r).


Then we calculate Hamilton’s equations

r =∂H

∂pr=prm

pr = −∂H∂r

=1

mr3

(p2θ +

p2φ

sin2 θ

)− V ′(r)

θ =∂H

∂pθ=

pθmr2

pθ = −∂H∂θ

=p2φ cos θ

mr2 sin3 θ

φ =∂H

∂pφ=

pφ

mr2 sin2 θ

pφ = −∂H∂φ

= 0.

These are the equations of motion. In particular, notice that pr = mr and that pφis conserved. If we’re given the inital conditions r(0), θ(0), φ(0), pr(0), pθ(0), pφ(0),then we can solve these in principle.

Whereas the Lagrangian approach would have given us three second-orderODEs, the Hamiltonian approach gives us six first-order ODEs.

Example 5.1.2: Relativistic Free Particle

Relativistic energy is

E = γmc2 =mc2√

1− v2/c2=

mc2√1− x2/c2

,

and relativistic momentum is

p = γmv =mx√

1− x2/c2.

We can see thatE2 = p2c2 +m2c4,

which implies that the Hamiltonian for a relativistic particle is

H =√p2c2 +m2c4.

Can we reverse engineer this to find the corresponding Lagrangian?The Hamiltonian is defined in terms of the Lagrangian as H = px− L, so

L = px−H.

Now we want to write L = L(x, x) as opposed to having p as an independentvariable.

L = px−H

=

(mx√

1− x2/c2

)x−

√√√√( mx√1− x2/c2

)2

c2 +m2c4

= −mc2√

1− x2/c2.

5.2. Variational Principle for the Hamiltonian 65

This is the Lagrangian for a relativistic free particle.The action is

S =

ˆ t

t0

Ldt = −mc2ˆ t

t0

√1− x2/c2 dt = −mc2

ˆ τ

τ0

dτ,

wheredτ =

√1− x2/c2 dt,

is the proper time. Then

S = −mc2ˆ τ

τ0

dτ = −mc2∆τ.

5.2 Variational Principle for the Hamiltonian

Recall the action

S =

ˆ t1

t0

L dt.

Minimization of the action, δS = 0 implies the Euler-Lagrange equations which giveus the equations of motion. In this variation, q(t) → q(t) + δq(t) with the constraintδq(t0) = δq(t1) = 0. I.e, the initial and final points of the trajectory are fixed. ForHamiltonians H = H(q, p), we have to deal with momenta in addition to the coordinates.The momentum p depends on q, but for now, we will pretend that p is independent of q.We can write the action as

S =

ˆ t1

t0

L dt =

ˆ t1

t0

[pαqα −H(q, p)] dt,

where the integrand is now considered to be a function of q and p. Now we want to applyδS = 0 along with some constraints for the coordinates like δq(t0) = δq(t1) = 0, andperhaps also similar constraints for the momenta. Using an analogue of the chain rule,we can write

δS =

ˆ t1

t0

[(δpα(qα) + pα

d

dtδqα)− ∂H

∂qαδqα − ∂H

∂pαδpα]dt.

Doing an integration by parts gives us

δS =

ˆ t1

t0

δpα

(qα − ∂H

∂pα

)dt+ pαδq

α∣∣∣t1t0−ˆ t1

t0

(pα +

∂H

∂qα

)δqα dt.

Notice that to have δS = 0, we require

qα − ∂H

∂pα= 0, pα +

∂H

∂qα= 0.

These are just Hamilton’s equations. We also need the boundary term to be zero. To getthat, we can choose δq to be zero at the endpoints. I.e., δq(t0) = δq(t1) = 0.

With the Lagrangian, our trajectory was q(t), and the actual trajectory taken by theparticle is the one that minimizes the action. With the Hamiltonian, our trajectory is thepair p(t) and q(t). The real p(t) and q(t) is the pair that minimizes the action.


5.3 Poisson Brackets

Recall that∂L

∂x= 0 =⇒ p =

d

dt

∂L

∂x=∂L

∂x= 0,

so momentum is conserved. For energy conservation,

d

dtH = 0 if

∂H

∂t= 0 ≡ ∂L

∂t= 0,

sinced

dtH =

∂H

∂qαqα +

∂H

∂pαpα +

∂H

∂t= −pαqα + qαpα +

∂H

∂t=∂H

∂t.

For any quantity A(q, p, t), the total time derivative is

dA

dt=

∂A

∂qαqα +

∂A

∂pαpα +

∂A

∂t=

∂A

∂qα∂H

∂pα− ∂A

∂pα

∂H

∂qα,

where we are assuming that ∂H∂t = 0, i.e. A has no explicit dependence on t. We write

define the Poisson bracket as

A,H ≡ ∂A

∂qα∂H

∂pα− ∂A

∂pα

∂H

∂qα.

Note that there is an implied sum over the α. Then the total time derivative can bewritten as

dA

dt= A,H+

∂A

∂t.

So if A does not depend explicitly on t, then

A,H = 0 ⇐⇒ A is conserved.

Following are some useful relations to remember:qα, qβ

= 0

pα, pβ

= 0qα, pβ

= δαβ .

These are essentially the same as the commutation relations of q and p in quantummechanics. The Poisson bracket also satisfies the antisymmetry property

A,B = −B,A .

If k is a constant over phase space, meaning it does not depend on any of the coor-dinates or momenta, then

A, k = 0.

The Poisson bracket of something with itself is zero

A,A = 0.

If k is a real number, then

kA,B = A, kB = kA,B.

5.3. Poisson Brackets 67

More generally, we can say the Poisson bracket is bilinear. If a, b, c ∈ R, then

aA+ bB, cC = aA, cC+ bB, cC = acA,C+ bcB,C.

The Leibniz rule

A,BC = B A,C+ A,BC,

and the Jacobi identity

A, B,C+ C, A,B+ B, C,A = 0.

These properties imply that Poisson brackets form a Lie algebra.Consider A(q, p) and B(q, p) which are both conserved so that A,H = B,H = 0.

Then by the Jacobi identity, this implies

A,B , H = 0.

That is, there is a new quantity

C = A,B ,

that is conserved since C,H = 0. That is, given two conserved quantities (other thanH itself), we can generate a third conserved quantity. We say that conserved quantitiesform an algebra. If we have the set of conserved quantities A1, A2, . . ., then Ai, Ajgives us another conserved quantity that must be in the set.

The time evolution of the Poisson bracket is given by

d

dtA,B = A,B, H+

∂

∂tA,B .

By the Jacobi identity, this becomes

d

dtA,B = A,H, B+ H,B, A+

∂A

∂t,B

+

A,

∂B

∂t

.

Using the linearity of the Poisson bracket, and grouping terms, we can write this as

d

dtA,B =

A,H+

∂A

∂t,B

+

A, B,H+

∂B

∂t

=

dA

dt,B

+

A,

dB

dt

.

Sod

dtA,B =

A, B

+A, B

.

This tells us that if A and B are both conserved, then A = B = 0, which implies thewhole Poisson bracket is conserved, d

dtA,B = 0. I.e.,

dA

dt=dB

dt= 0 =⇒ d

dtA,B = 0.

Consider angular momentum ~L = ~r × ~p, which has components

Lx = ypz − zpyLy = zpx − xpzLz = xpy − ypx.


Then

Lx, z = ypz − zpy, z= ypz, z − zpy, z= ypz, z+ y, zpz − zpy, z − z, zpy= −y.

Similarly, it can be shown that

Lx, Ly = Lz

Ly, Lz = Lx

Lz, Lx = Ly.

We can summarize these in the rules

Li, xj = εijkxk

Li, Lj = εijkLk

Consider a particle moving under a central force with Hamiltonian

H =p2

2m+ V (r), r = |~r|.

Then H,Lx = 0 implies that ddtLx = 0. In other words, rotational symmetry implies

conservation of angular momentum. Suppose you don’t know H, but you are given thatH,Lx = 0 and H,Ly = 0. Then you can infer that Lz is also conserved, sinceLx, Ly = Lz implies that H,Lz = 0.

5.4 Canonical Transformations

Suppose we have the Hamiltonian H(qα, pα) where α = 1, . . . , n. Suppose that q1 is acyclic variable, then

∂L

∂q1= 0 =⇒ ∂H

∂q1= 0 =⇒ p1 = −∂H

∂q1= 0 =⇒ p1 = const.

SoH (qα, pα)→ H

(p1, q

β , pβ), β = 2, . . . , n.

That is, we can ignore the cyclic variables—our problem is simplified since every cyclicvariable reduces the degrees of freedom by one.

Consider a canonical transformation

Qα = Qα(q, p), Pα = Pα(q, p).

Previously, we hadqα = qα, Hq,p , pα = pα, Hq,p ,

where

A,Bq,p ≡ ∂A

∂qα∂B

∂pα− ∂A

∂pα

∂B

∂qα.

Note that the superscript q, p on the Poisson bracket indicates that the derivatives arewith respect to the old variables q and p rather than the new variables Q and P . Now,we define

Qβ =Qβ , H

Q,P, Pβ = Pβ , HQ,P ,

5.4. Canonical Transformations 69

What is the condition for the new variables Q and P to satisfy these relations?Assume that Q and P are not explicitly dependent on time, then

Qβ =Qβ , H

q,p, Pβ = Pβ , Hq,p .

Notice that the derivatives in the Poisson bracket are now with respect to the old variablesq and p. These equations are true because they denote the time dependence of any twovariables. I.e. we could replace the Q and P above with any other variables which arenot explicitly dependent on time, and the equations would hold true. This suggests thatthe condition that we are looking for is

A,Bq,p = A,BQ,P .

For example, since we know that

Qα, Pβq,p = δαβ

Qα, Qβq,p = 0

Pα, Pβq,p = 0,

we would require the same relations to hold if we replaced the q and p in the superscriptwith Q and P .

For example, suppose we have the transformation

Q = Q(q, p) = p, P = P (q, p) = −q,

then

Q,Pq,p =∂Q

∂q

∂P

∂p− ∂Q

∂p

∂P

∂q= (0)(0)− (1)(−1) = 1.

and

Q,PQ,P =∂Q

∂Q

∂P

∂P− ∂Q

∂P

∂P

∂Q= (1)(1)− (0)(0) = 1.

Similarly, Qα, Qβq,p = Qα, QβQ,P = 0 and Pα, Pβq,p = Pα, PβQ,P = 0. Ourtransformation satisfies the necessary conditions, so it is a canonical transformation.

For a Hamiltonian like

H(p, q) =p2

2m,

q is a cyclic variable, which means ∂H∂q = 0, which implies p is conserved. This makes the

problem much easier. Now suppose we have some complicated Hamiltonian. However, ifwe can make a canonical transformation to turn it into a Hamiltonian of the form

H(Q,P ) =P 2

2m,

then we’re back to the easy problem. The transformation

Q = Q(q, p), P = P (q, p)

has to preserve the form of the Hamilton equations

Q =∂H

∂P= Q,HQ,P

P = −∂H∂Q

= P,HQ,P .

We require

Q,Pq,p =∂Q

∂q

∂P

∂p− ∂Q

∂p

∂P

∂q= 1. (5.1)


If this is satisfied, then the form of Hamilton’s equations is preserved. This is called acanonical transformation from (q, p) to (Q(q, p), P (q, p)).

Consider the transformation

Q = p, P = −q.

This satisfies Eq. (5.1), so it is a canonical transformation.A more complicated example is the transformation

Q =1

2p2, P = −q

p.

This is also a canonical transformation. For example, it is the transformation for which

H(q, p) =

(q

p

)5

→ H(Q,P ) = P 5.

Is there a systematic way of generating canonical transformations? We can use thevariational principle. Finding something that doesn’t change the Hamilton’s equations isequivalent to finding something that doesn’t change the variational principle, since theHamilton equations were derived from the variational principle. We want

δ

ˆ (PαQα −H(Q,P )

)dt = 0.

We want to get the same Hamilton equations from this as with the old form of thevariational principle

δ

ˆ(pαqα −H(q, p)) dt = 0.

What are the conditions for this to occur? One possibility is that the two integrands arethe same:

PαQα −H(Q,P ) = pαqα −H(q, p).

But this is not the most general condition. We know that we can add a total timederivative without changing the result

PαQα −H(Q,P ) = pαqα −H(q, p)− dF

dt,

where F = F (q, p). The Hamiltonians are the same—just written in new coordinates.That is, H(Q,P ) = H(q, p). So, our condition is that

pαqα − PαQα =dF

dt.

We can write this aspα dq

α − Pα dQα = dF. (5.2)

For example, for the transformation Q = p and P = −q, we have that

p dq − (q) d(−p) = p dq + q dp = d(pq).

Thus, in this case, F (q, p) = pq.For the transformation Q = p2/2, and P = −q/p, we actually get the same—

F (q, p) = pq.What if instead of F (q, p), we look at F (q,Q)? The differential is

dF (q,Q) =∂F

∂qαdqα +

∂F

∂QαdQα.

5.4. Canonical Transformations 71

Comparing with Eq. (5.2), we see that

pα =∂F

∂qα, −Pα =

∂F

∂Qα. (5.3)

In a sense, F now becomes a generating function. Choose any F , and put it into Eq. (5.3),and it gives the canonical transformation.

For example, if F (q,Q) = qQ, then Eq. (5.3) gives us

p =∂F

∂q= Q, P = −q.

This implies the canonical transformation

p→ Q, q → −P.

Example 5.4.1

Consider a harmonic oscillator with Hamiltonian

H(q, p) =1

2m

(p2 +m2ω2q2

)→ H(Q,P ),

whereH(Q,P ) really only depends on P . This makes the transformed Hamiltoniansimpler since Q is then a cyclic variable.

Consider a transformation of the form

p = f(P ) cosQ, mωq = f(P ) sinQ.

Squaring both of these and adding them together implies that the Hamiltonianwill transform as

H(q, p)→ H(Q,P ) =1

2mf2(P ).

If we divide one by the other, we see that

p

mωq= cotQ =⇒ p = mωq cotQ.

But we also know that p = ∂F∂q , and so

F =

ˆdq mωq cotQ =

1

2mωq2 cotQ.

We also have that

P = −∂F∂Q

=mωq2

2 sin2Q=⇒ q =

√2P

mωsinQ.

Comparing p = f(P ) cosQ with p = mωq cotQ, we see that

f(P ) =mωq

sinQ=

mω

sinQ

√2P

mωsinQ =

√2mωP.

Then the Hamiltonian becomesH = ωP.

Note that

Q =∂H

∂P= ω.


5.5 Liouville’s Theorem

The volume in phase space is invariant under dynamics.Consider a volume dV = dq dp in phase space. When we transform from the coordi-

nates (q, p) to the coordinates (Q,P ), our new volume element is dV ′ = dQdP .

If it is a canonical transformation, then the volumes are the same. The volume elementstransform by way of the Jacobian determinant as

dV ′ = dQdP = J dq dp =

∣∣∣∣∣ ∂Q∂q ∂Q∂p

∂P∂q

∂P∂p

∣∣∣∣∣ dq dp = Q,Pq,p dq dp = dq dp,

since Q,Pq,p = 1 for a canonical transformation.Given a gas with N ∼ 1023 particles, our phase space consists of 3N p’s and 3N q’s.

In statistical mechanics, instead of tracking the system as a single point in this phasespace, we average over an ensemble of systems. I.e., we look at a volume in phase space.

Over time, the points move through phase space as illustrated below.

Liouville’s theorem tells us that the volume of the region is preserved under Hamil-tonian dynamics.

5.5. Liouville’s Theorem 73

There can be no trajectory collisions, because that would contradict determinism.If there were ever a collision of trajectories, then a system exactly at the collision pointwould have an undeterminable future. I.e. there are two separate trajectories into thefuture from that point.

The fact that there are no collisions of trajectories implies that points can never leavea region. If they did, that would require a collision with the trajectory of a point on theboundary of the region.

Dynamics itself is a canonical transformation

(q, p)→ (q(t), p(t)).

We can approximate

q(t+ ∆t) ≈ q(t) + q∆t = q(t) + q(t), H∆tp(t+ ∆t) ≈ p(t) + p∆t = p(t) + p(t), H∆t.

Then

q(t+ ∆t), p(t+ ∆t) = q(t), p(t)+ ∆t [q,H, p+ q, p,H] +O((∆t)2

).

Since q, p = 1, this becomes

q(t+ ∆t), p(t+ ∆t) = 1 + ∆t [q, p, H] +O((∆t)2

)= 1 +O

((∆t)2

).

Thus, dynamics is a canonical transformation, and we can think of the Hamiltonian as agenerator of canonical transformation from time 0 to time t.

In phase space, the density of points can vary. The density is given by

D(q, p, t) =∆n

∆V=

# states

volume.


The time evolution of the density is given by

dD

dt= D,H+

∂D

∂t.

By Liouville’s theorem, the volume is constant. This implies that at the microscopic level,the density is constant. Thus,

dD

dt= 0.

Eventually, the system will reach equilibrium, and nothing changes in time anymore.Then,

∂D

∂t= 0.

So in equilibrium, we must haveD,H = 0.

Note, in practice, phase space volumes are not conserved due to friction which is notincluded in the Hamiltonian.

Liouville’s theorem tells us that the area enclosed in phase space does not change.Consider the three cases shown in the graphic below.

• For a time-independent Hamiltonian, both the area and the shape are conserved inphase space, although the shape may be rotated.• For a time-dependent Hamiltonian, the shape is not preserved• For a chaotic system, the shape becomes complex

However, in all three cases, by Liouville’s theorem, the area is preserved.

5.6. Integrability and Action-Angle Variables 75

5.6 Integrability and Action-Angle Variables

A system with n degrees of freedom is integrable if you can treat it as n systems eachwith one degree of freedom.

An integrable system is the opposite of a chaotic system.For example, given a 2-body system with a central force, we can go to center-of-mass

coordinates—taking the 2D problem to an effective 1D problem. An example of this isthe Kepler problem.

If we have a system with n degrees of freedom, then if it is integrable, we can breakit into n 1D problems.

For general systems, we are interested in making the canonical transformation

(q, p)→ (φ, J),

with φ, Jq,p = 1. In general,

J =1

2π

˛p dq =

1

2π· (area enclosed).

For a more complicated problem with 2n degrees of freedom, want to make thetransformation to n different action-angle pairs (φα, Jα) where α = 1, . . . , n. Then theaction variables are calculated as

Jα =1

2π

ˆpα dq

α.

The transformed Hamiltonian is now only a function of the action variables

H = H(Jα).

Then

φα =∂H

∂Jα= ωα = const.

A system is completely integrable if you have n independent constants of motionIα that commute with each other in the sense that Iα, Iβ = 0. This is the Liouvilledefinition of integrability.

What does independence mean in this context? Suppose we have a 2n-dimensionalphase space (q, p). Constants are Iα(q, p) = Cα. These give us an n-dimensionalconstraint surface. They are independent if the individual constraint surfaces are inde-pendent of each other. For example, x2 + y2 = 1 and (x2 + y2)2 = 1 are not independentbecause they describe the same surface in 3D.

Harmonic Oscillator

Consider the harmonic oscillator with Hamiltonian

H(q, p) =p2

2m+

1

2mω2q2.


We can rearrange this as

1 =p2

2mE+mω2

Eq2,

illustrating that the orbit in phase space is an ellipse.

Since the harmonic oscillator has a closed orbit in phase space, can we describe its motionwith an angle?

Let

p =√

2mE sinφ, mωq =√

2mE cosφ,

which implies φ = tan−1(p/(mωq)). If we square both of these and add them together,we find that we get the right energy equation.

We know that φ = ω, which implies φ(t) = φ0 + ωt. Thus, we can write

q(t) = C1 cos(φ0 + ωt).

More formally, we want to make the transformation

(q, p)→ (φ, J),

where φ is the angle, and J is the action. The transformation needs to be canonical,which means we require

φ, Jq,p = 1.

It turns out that J =¸p dq is just the area under the orbit. By convention, we

include a factor of 2π so that

J =1

2π

˛p dq.

For the hamornic oscillator, the orbit is an ellipse. The area of an ellipse is πab wherea and b are the two radii of the ellipse. For the harmonic oscillator, we find that

area =

˛p dq =

2Eπ

ω,

which implies that

J =1

2π

˛p dq =

E

ω=H

ω.

The Poisson bracket gives us

φ, J =1

ωφ,H =

φ

ω=ω

ω= 1,

as required for a canonical transformation.We can now write

H = ωJ,

5.6. Integrability and Action-Angle Variables 77

for the harmonic oscillator, then

φ =∂H

∂J= ω

J = −∂H∂φ

= 0 =⇒ J = const.

After making our transformation, the angle variable became cyclic, which implies thatthe action variable is conserved. This is a general feature and appeal of the action-angleapproach.

In summary, for the harmonic oscillator, we made the transformation

(q, p)→ (φ, J) = (φ,H/ω),

with φ, J = 1, and φ = ω and φ = ωt + φ0. The point is that instead of describingharmonic oscillator motion with the variables q and p, we can also describe it in terms ofthe action J and angle φ variables.

Central Force

For a more complicated problem with 2n degrees of freedom, want to make the trans-formation to n different action-angle pairs (φα, Jα) where α = 1, . . . , n. Then the actionvariables are calculated as

Jα =1

2π

ˆpα dq

α.

For a central force problem,

H =p2r2m

+p2θ2m

+ V (r).

θ is cyclic, so pθ is conserved.This motion takes place in a 4D space (r, θ, pr, pθ). We can write the motion in terms

of r and pr as

E =p2r2m

+p2θ2m

+ V (r),

where E and pθ are constant. This gives us a closed orbit in the phase space of pr versusr, which should be surprising since for a generic problem in the variables (r, θ, pr, pθ), thepicture in the phase space of pr versus r would be a complicated mess.

Since we have a closed orbit in pr versus r, we can in principle define an angularvariable and make the transformation

(r, pr)→ (φr, Jr),

with φr = ωr.More generally, the action variables are

Jθ =1

2π

˛pθ dθ

Jr =1

2π

˛pr dr.

Since pθ is constant,

Jθ =1

2π

ˆpθ dθ =

pθ2π

ˆdθ = pθ = const.


Tip

In general, once you trans-form using action-anglevariables, the Hamiltonianwill depend only on theaction variables and not onthe angle variables.

Since H = E is constant and pθ is conserved, we can in principle solve H = E for pr toget pr(r), then we can integrate

Jr =1

2π

˛pr dr =

√2m

E

˛ √E − V (r)−

J2θ

2mr2dr.

If our potential is V (r) = −k/r, then we can show (using the commutation require-ment for canonical transformations) that

φr = ωr

√m

2

ˆ (E +

k

r− J2

θ

2mr2

)−1/2dr

φθ = θ +

√m

2

ˆ (E +

k

r− J2

θ

2mr2

)−1/2(ωθ −

Jθmr2

)dθ.

Then it can be shown that

H(Jr, Jθ) = − k2m

2(Jr + Jθ)2.

Now the Hamiltonian contains only the two action variables—it is free of the angle vari-ables. So now, φθ and φr are cyclic, and

φr =∂H

∂Jr=

k2m

(Jr + Jθ)3= ωr = const

φθ =∂H

∂Jθ=

k2m

(Jr + Jθ)3= ωθ = const.

If ωr 6= ωθ, then the orbit is not necessarily closed. If ωr/ωθ ∈ Z, then the orbit isclosed.

5.7 Adiabatic Theorem

Consider the harmonic oscillator, but now with a time-dependent frequency ω(t). TheHamiltonian is

H =p2

2m+

1

2mω2(t)q2.

Now we no longer have energy conservation since ∂H∂t 6= 0.

Now we assume that the change happens slowly, i.e. adiabatically. This means that

∆ω

∆t,

is small. Then we can use the adiabatic theorem, which states that for a harmonicoscillator,

E(t)

ω(t)= const.

5.7. Adiabatic Theorem 79

We want to show this.More generally, consider a generic Hamiltonian with a time-dependent parameter λ,

i.e. H = H(q, p, λ(t)). Then

dH

dt=∂H

∂qq +

∂H

∂pp+

∂H

∂λλ =

∂H

∂λλ,

since the first two terms in the middle are zero by Hamilton’s equations. Then the totalchange of the energy is

∆E = ∆H =

ˆ t1

t0

∂H

∂λλ dt.

The adiabatic theorem assumes that λ is small.Fact 1: With the adiabatic assumption (i.e. that λ is small), particles on a shared

orbit with energy E0 at time t = t0 will be on a shared orbit with energy E1 at time t1.This is only true for adiabatic motion. If it’s not adiabatic, the particles could end

up on orbits with different energies.

In adiabatic motion, the periodic cycle repeats many times before the overall orbitmoves significantly in phase space. This is illustrated in the graphic below.

This implies that we can take the average over one cycle, and

∆E =

ˆ t1

t0

∂H

∂λλ dt ≈

ˆ t1

t0

∂H

∂λλ dt.


Suppose λ is not small. I.e. suppose the adiabatic theorem does not apply. Then howmight the energies change for particles? Suppose we initially have a number of particleson a shared orbit (i.e. they all have the same energy). The adiabatic theorem tells us thatif the change is adiabatic, then at the later time, the particles will still be on a sharedorbit even though that orbit may be different from the initial one. Compare that with anabrupt (i.e. not adiabatic) change. After an abrupt change, the particle distribution inphase space still has the initial shape, but now they lie on different orbits. I.e. differentparticles now have different energy.

Fact 2: Liouville’s theorem says that the area enclosed by the system of particles inphase space does not change. This is always true (classically).

The adiabatic invariant of motion is

J =1

2π

ˆR(t)

p dq = const =E

ω=E(t)

ω(t),

where R(t) is the particle’s orbit in phase space. So the integral above is just the area ofthe particle’s orbit in phase space.

I.e. under adiabatic change, the area of any single orbit is conserved. What Liou-ville’s theorem says is that the area of an ensemble of systems is conserved—not that thearea of a single orbit is conserved.

5.8. Summary: Hamiltonian Mechanics 81

5.8 Summary: Hamiltonian Mechanics

Skills to Master

• Given a Lagrangian, calculate the Hamiltonian• Given a physical system, write down the Hamiltonian• Given a Hamiltonian, calculate Hamilton’s equations of motion• Given a Lagrangian, show that the equations of motions derived from the Euler-Lagrange equations

are consistent with the equations of motion derived from Hamilton’s equations• Given a Hamiltonian H(p, q), derive the Lagrangian L(q, q)• Identify conserved quantities• Be comfortable simplifying Poisson brackets• Identify conserved quantities using Poisson brackets• Determine whether or not a proposed transformation is canonical• Perform canonical transformations of the Hamiltonian

Hamiltonian Formalism

To write down the Hamiltonian for a given physical sys-tem, write down the Lagrangian, and then write downthe Hamiltonian using the definition

H(q, p) = pαqα − L.

Note that there is an implied sum over the α. TheHamiltonian is a function of p and q—there should beno velocities appearing in the Hamiltonian. If thereare, rewrite them in terms of the momenta and coor-dinates. A useful tool for doing this is the definition ofthe momentum

pα ≡ ∂L

∂qα.

To get the equations of motion from the Hamil-tonian, start by writing down the Hamilton equationsfor each momentum and coordinate pair. Be sure thatthe Hamiltonian is written only in terms of coordinatesand momenta before calculating Hamilton’s equations.


∣∣∣p

qα =∂H

∂pα

∣∣∣q.

This gives us a system of first-order ODEs, which canbe solved to get the coordinates as functions of time.To compare with the equations of motion from theEuler-Lagrange equations, differentiate the q equationsagain to get q. Then plug in the p equations as neces-sary to get q as functions of coordinates only.

Conserved quantities:

• If L contains a velocity q but not the associatedcoordinate q, then q is said to be “cyclic,” andthe associated momentum

pq =∂L

∂q,

is conserved.• If


∣∣∣p

= 0,

then pα is conserved.• If H (or L) does not explicitly contain t, then H

is conserved

Poisson Brackets

For any quantity A(q, p, t), the total time derivative is

dA

dt=

∂A

∂qαqα +

∂A

∂pαpα +

∂A

∂t= A,H+

∂A

∂t,

where the Poisson bracket is defined as

A,H ≡ ∂A

∂qα∂H

∂pα− ∂A

∂pα

∂H

∂qα.

So if A does not depend explicitly on t, then

A,H = 0 ⇐⇒ A is conserved.

Following are some useful relations to remember:qα, qβ

= 0

pα, pβ

= 0qα, pβ

= δαβ

A,B = −B,AA,BC = B A,C+ A,BC

0 = A, B,C+ C, A,B+ B, C,A .

Given any two conserved quantities A and B (notincluding H itself), then we know that A,H =B,H = 0. Then there is a third conserved quan-tity

C = A,B.


A corollary is that

dA

dt=dB

dt= 0 =⇒ d

dtA,B = 0.

If k is a constant over phase space, meaning it doesnot depend on any of the coordinates or momenta, then

A, k = 0.

If k is a real number, then

kA,B = A, kB = kA,B.

More generally, we can say the Poisson bracket is bi-linear. If a, b, c ∈ R, then

aA+ bB, cC = aA, cC+ bB, cC= acA,C+ bcB,C.

For angular momentum defined as ~L = ~r× ~p, it isoften useful to consider the components

Li = εijkrjpk.

Then

Li, xj = εijkxk

Li, Lj = εijkLk

Canonical Transformations

A transformation from variables q, p to new variablesQ(q, p), P (q, p) is canonical if

Q,Pq,p =∂Q

∂q

∂P

∂p− ∂Q

∂p

∂P

∂q= 1.

If this is satisfied, then the form of Hamilton’s equa-tions is preserved, meaning that

Q =∂H

∂P, P = −∂H

∂Q.

Chapter 6

Nonlinear Dynamics

Index

Action-angle variables,75

Adiabatic theorem, 78

Canonical transforma-tions, 68

Cantor set, 45Chaos, 43Conservation of momen-

tum, 29Constraint forces, 8Cross-section, 38Cyclic coordinate, 29

Differential cross-section, 39

Euler-Lagrange equa-tion, 11

Fermat’s principle, 20Flux density, 38

Generalized coordi-nates, 8

Generalized momentum,29

Hamilton’s equations,60–62

Hamilton’s principle, 20Hamiltonian, 17, 61Hamiltonian mechanics,

60Harmonic oscillator, 75Huygen’s principle, 20

Integrability, 75Intensity, 38Inverse scattering the-

ory, 39, 42

Jacobi identity, 67

Lagrange multipliers, 23Lagrangian, 11Leibniz rule, 67Lie algebra, 67Linear oscillations, 48

Liouville’s theorem, 72Lypunov exponent, 46

Noether’s theorem, 29Nonlinear Dynamics, 83

Oscillation, 48

Path integral, 21Poisson bracket, 66Principle of least action,

20Proper time, 65

Relativity, 64

Scattering, 38, 43Scattering rate, 39Special relativity, 64Symmetry, 29Symmetry transforma-

tion, 29

Variational principle,19, 21, 65

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