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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