MechanicsPhysics 151

Lecture 18Hamiltonian Equations of Motion

(Chapter 8)

What’s Ahead

We are starting Hamiltonian formalismHamiltonian equation – Today and 11/26Canonical transformation – 12/3, 12/5, 12/10Close link to non-relativistic QM

May not cover Hamilton-Jacobi theoryCute but not very relevant

What shall we do in the last 2 lectures?Classical chaos?Perturbation theory?Classical field theory?Send me e-mail if you have preference!

Hamiltonian Formalism

Newtonian Lagrangian HamiltonianDescribe same physics and produce same resultsDifference is in the viewpoints

Symmetries and invariance more apparentFlexibility of coordinate transformation

Hamiltonian formalism linked to the development ofHamilton-Jacobi theoryClassical perturbation theoryQuantum mechanicsStatistical mechanics

Lagrange Hamilton

Lagrange’s equations for n coordinates

n equations 2n initial conditions

Can we do with 1st-order differential equations?Yes, but you’ll need 2n equationsWe keep and replace with something similar

We take the conjugate momenta

0i i

d L Ldt q q

⎛ ⎞∂ ∂− =⎜ ⎟∂ ∂⎝ ⎠

1, ,i n= … 2nd-order differential equation of n variables

( 0)iq t = ( 0)iq t =

iq iq( , , )j j

ii

L q q tp

q∂

≡∂

Configuration Space

We considered as a point in an n-dim. spaceCalled configuration spaceMotion of the system A curve in the config space

When we take variations,we consider and asindependent variables

i.e., we have 2n independent variables in n-dim. spaceIsn’t it more natural to consider the motion in 2n-dim space?

1( , , )nq q…

( )i iq q t=

iq iq

Phase Space

Consider coordinates and momenta as independentState of the system is given byConsider it a point in the 2n-dimensional phase space

We are switching theindependent variables

A bit of mathematical trickis needed to do this

1 1( , , , , , )n nq q p p… …

( )( )

i i

i i

q q tp p t

=

=( , , ) ( , , )i i i iq q t q p t→

If and

Legendre Transformation

Start from a function of two variablesTotal derivative is

Define and consider its total derivative

i.e. g is a function of u and y

( , )f x y

g f ux≡ −

f fdf dx dy udx vdyx y

∂ ∂= + ≡ +

∂ ∂

( )dg df d ux udx vdy udx xdu vdy xdu= − = + − − = −

g vy

∂=

∂g xu

∂= −

∂f L= ( , ) ( , )x y q q=

( , ) ( , )L q q g p q L pq→ = −

This is what we need

Hamiltonian

Define Hamiltonian:Total derivative is

Lagrange’s equations say

This must be equivalent to

( , , ) ( , , )i iH q p t q p L q q t= −

i i i i i ii i

L L LdH p dq q dp dq dq dtq q t

∂ ∂ ∂= + − − −

∂ ∂ ∂

Opposite sign from Legendre transf.

ii

L pq

∂=

∂

i i i iLdH q dp p dq dtt

∂= − −

∂

i ii i

H H HdH dp dq dtp q t

∂ ∂ ∂= + +

∂ ∂ ∂

Putting them together gives…

Hamilton’s Equations

We find and

2n equations replacing the n Lagrange’s equations1st-order differential instead of 2nd-order“Symmetry” between q and p is apparent

There is nothing new – We just rearranged equationsFirst equation links momentum to velocity

This relation is “given” in Newtonian formalismSecond equation is equivalent to Newton’s/Lagrange’s equations of motion

ii

H qp

∂=

∂ ii

H pq

∂= −

∂H Lt t

∂ ∂= −

∂ ∂

Particle under Hooke’s law force F = –kx

Hamilton’s equations

Quick Example

2 2

2 2m kL x x= −

2 2

22

2 2

2 2

m kH xp L x x

p k xm

= − = +

= +

Lp mxx

∂= =

∂

Hp kxx

∂= − = −

∂H pxp m

∂= =

∂

Replace withx pm

Usual harmonic oscillator

Energy Function

Definition of Hamiltonian is identical to the energy

function

Difference is subtle: H is a function of (q, p, t)

This equals to the total energy ifLagrangian isConstraints are time-independent

This makesForces are conservative

This makes

( , , ) ( , , )ii

Lh q q t q L q q tq

∂= −

∂

0 1 2( , ) ( , ) ( , )i j kL L q t L q t q L q t q q= + +

2 ( , ) j kT L q t q q=

0 ( )V L q= −

See Lecture 4, or Goldstein Section 2.7

Hamiltonian and Total Energy

If the conditions make h to be total energy, we can skip calculating L and go directly to H

For the particle under Hooke’s law force

This works often, but not alwayswhen the coordinate system is time-dependent

e.g., rotating (non-inertial) coordinate systemwhen the potential is velocity-dependent

e.g., particle in an EM field

22

2 2p kH E T V xm

= = + = +

Let’s look at this

Particle in EM Field

For a particle in an EM field

We’d be done if we were calculating h

For H, we must rewrite it using

2

2 i i imL x q qA xφ= − + We can’t jump on H = E

because of the last term, but

i i ip mx qA= +2

( )2

ii i i

mxH mx qA x L qφ= + − = + This is in fact E

2( )( , )2

i ii i

p qAH x p qm

φ−= +

i i ip mx qA= +

Particle in EM Field

Hamilton’s equations are

Are they equivalent to the usual Lorentz force?Check this by eliminating pi

2( )( , )2

i ii i

p qAH x p qm

φ−= +

i ii

i

p qAHxp m

−∂= =

∂j j j

ii i i

p qA AHp q qx m x x

φ− ∂∂ ∂= − = −

∂ ∂ ∂

( ) ji i i

i i

Ad mx qA qx qdt x x

φ∂ ∂+ = −

∂ ∂

( ) ( )i i id mv qE qdt

= + ×v BA bit of work

Conservation of Hamiltonian

Consider time-derivative of Hamiltonian

H may or may not be total energyIf it is, this means energy conservationEven if it isn’t, H is still a constant of motion

( , , )dH q p t H H Hq pdt q p t

Hpq qpt

∂ ∂ ∂= + +

∂ ∂ ∂∂

= − + +∂

Hamiltonian is conserved if it does not depend explicitly on t

Cyclic Coordinates

A cyclic coordinate does not appear in LBy construction, it does not appear in H either

Hamilton’s equation says

Exactly the same as in the Lagrangian formalism

(H q , , ) (i ip t q p qL= − , , )q t

0Hpq

∂= − =

∂Conjugate momentum of a

cyclic coordinate is conserved

Cyclic Example

Central force problem in 2 dimensions

Cyclic variable drops off by itself

2 2 2( ) ( )2mL r r V rθ= + −

rp mr= 2p mrθ θ=

22

2

22

2

1 ( )2

1 ( )2

r

r

pH p V rm r

lp V rm r

θ⎛ ⎞= + +⎜ ⎟

⎝ ⎠⎛ ⎞

= + +⎜ ⎟⎝ ⎠

θ is cyclic constp lθ = =

Hamilton’s equations

rprm

=2

3

( )r

l V rpmr r

∂= −

∂

Going Relativistic

Practical approachFind a Hamiltonian that “works”

Does it represent the total energy?

Purist approachConstruct covariant Hamiltonian formalism

For one particle in an EM field

Don’t expect miraclesFundamental difficulties remain the same

Practical Approach

Start from the relativistic Lagrangian that “works”

It does equal to the total energyHamilton’s equations

2 21 ( )L mc Vβ= − − − x

21i

ii

mvLpv β

∂= =

∂ −Did this last time

2 2 2 4 ( )H h p c m c V= = + + x

2

2 2 2 4i i

ii

p c pHxp mp c m c γ

∂= = =

∂ +i i

i i

H Vp Fx x

∂ ∂= − = − =

∂ ∂

Practical Approach w/ EM Field

Consider a particle in an EM field

Hamiltonian is still total energy

Difference is in the momentum

2

2 2 2 2 2 4

H m c q

m v c m c q

γ φ

γ φ

= +

= + +

Can be easily checked

2 21 ( ) ( )L mc q qβ φ= − − − + ⋅x v A

i i ip m v qAγ= +

2 2 2 4( )H q c m c qφ= − + +p A

Not the usual linear momentum!

Practical Approach w/ EM Field

Consider H – qφ

It means that is a 4-vector,and so is

This particular Hamiltonian + canonical momentum transforms as a 4-vector

True only for well-defined 4-potential such as EM field

2 2 2 4( )H q c m c qφ= − + +p A

2 2 2 2 4( ) ( )H q q c m cφ− − − =p A constant

( , )H q c q cφ− −p A( , )H cp Similar to 4-momentum (E/c, p) of

a relativistic particle

Remember p here is not the linear momentum!

Purist Approach

Covariant Lagrangian for a free particle

We know that p0 is E/cWe also know that x0 is ct…

Generally true for any covariant LagrangianYou know the corresponding relationship in QM

12 mu uν

νΛ =

p muu

µ µ

µ

∂Λ= =

∂ 2p p

Hm

µµ=

Energy is the conjugate “momentum” of time

Purist Approach

Value of Hamiltonian is

What is important is H’s dependence on pµ

Hamilton’s equations

Time components are

dx H pd p m

µ µ

µτ∂

= =∂

0dp Hd x

µ

µτ∂

= − =∂

2

2 2p p mcH

m

µµ= =

4-momentum conservation

This is constant!

( )d ct E cd mc

γτ

= =( ) 0d E cdτ

= Energy definitionand conservation

Purist Approach w/ EM Field

With EM field, Lagrangian becomes

Hamilton’s equations are

A bit of work can turn them into

12( , )x u mu u qu Aµ µ µ µ

µ µΛ = + p mu qAµ µ µ= +

( )( )2 2

mu u p qA p qAH

m

µ µ µµ µ µ− −

= =

dx H p qAd p m

µ µ µ

µτ∂ −

= =∂

( )p qAdp H Ad x m x

µ νν ν

µ µτ−∂ ∂

= − = −∂ ∂

du A Am q u Kd x x ν

µ ν µµ

µ ντ⎛ ⎞∂ ∂

= − =⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠4-force

EM Field and Hamiltonian

In Hamiltonian formalism, EM field always modify the canonical momentum as

A handy rule:

Often used in relativistic QM to introduce EM interaction

0Ap p qAµ µ µ= +With EM field Without EM field

Hamiltonian with EM field is given by replacing pµ

in the field-free Hamiltonian with pµ – qAµ

Summary

Constructed Hamiltonian formalismEquivalent to Lagrangian formalism

Simpler, but twice as many, equationsHamiltonian is conserved (unless explicitly t-dependent)

Equals to total energy (unless it isn’t) (duh)Cyclic coordinates drops out quite easily

A few new insights from relativistic HamiltoniansConjugate of time = energypµ – qAµ rule for introducing EM interaction