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Mechanics
Physics 151
Lecture 18
Hamiltonian Equations of Motion
(Chapter 8)
What’s Ahead
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We are starting Hamiltonian formalism
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May not cover Hamilton-Jacobi theory
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Hamiltonian equation – Today and 11/26
Canonical transformation – 12/3, 12/5, 12/10
Close link to non-relativistic QM
Cute but not very relevant
What shall we do in the last 2 lectures?
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Classical chaos?
Perturbation theory?
Classical field theory?
Send me e-mail if you have preference!
Hamiltonian Formalism
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Newtonian Æ Lagrangian Æ Hamiltonian
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Describe same physics and produce same results
Difference is in the viewpoints
„ Symmetries and invariance more apparent
„ Flexibility of coordinate transformation
Hamiltonian formalism linked to the development of
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Hamilton-Jacobi theory
Classical perturbation theory
Quantum mechanics
Statistical mechanics
Lagrange Æ Hamilton
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Lagrange’s equations for n coordinates
d ⎛ ∂L ⎞ ∂L
= 0 i = 1,… , n
⎜
⎟−
dt ⎝ ∂qi ⎠ ∂qi
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2nd-order differential
equation of n variables
n equations Æ 2n initial conditions qi (t = 0) qi (t = 0)
Can we do with 1st-order differential equations?
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Yes, but you’ll need 2n equations
We keep qi and replace qi with something similar
∂L(q j , q j , t )
We take the conjugate momenta pi ≡
∂qi
Configuration Space
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We considered (q1 ,… , qn ) as a point in an n-dim. space
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Called configuration space
Motion of the system Æ
A curve in the config space
qi = qi (t )
When we take variations,
we consider qi and qi as
independent variables
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i.e., we have 2n independent variables in n-dim. space
Isn’t it more natural to consider the motion in 2n-dim space?
Phase Space
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Consider coordinates and momenta as independent
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State of the system is given by (q1 ,… , qn , p1 ,… , pn )
Consider it a point in the 2n-dimensional phase space
We are switching the
independent variables
(qi , qi , t ) → (qi , pi , t )
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A bit of mathematical trick
is needed to do this
qi = qi (t )
pi = pi (t )
Legendre Transformation
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Start from a function of two variables f ( x, y )
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Total derivative is
∂f
∂f
df = dx + dy ≡ udx + vdy
∂x
∂y
Define g ≡ f − ux and consider its total derivative
dg = df − d (ux) = udx + vdy − udx − xdu = vdy − xdu
„ i.e. g is a function of u and y
∂g
If f = L and ( x, y ) = ( q, q )
∂g
= −x
=v
∂u
∂y
L(q, q ) → g ( p, q ) = L − pq
This is what
we need
Hamiltonian
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Opposite sign from
Legendre transf.
Define Hamiltonian: H (q, p, t ) = qi pi − L(q, q, t )
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Total derivative is
∂L
∂L
∂L
dH = pi dqi + qi dpi −
dqi −
dqi −
dt
∂qi
∂qi
∂t
„
∂L
Lagrange’s equations say
= pi
∂qi
„
∂L
dH = qi dpi − pi dqi −
dt
∂t
This must be equivalent to
∂H
∂H
∂H
dH =
dpi +
dqi +
dt
∂pi
∂qi
∂t
Putting them
together gives…
Hamilton’s Equations
∂H
∂H
∂H
∂L
=−
= qi
= − pi and
„ We find
∂t
∂t
∂pi
∂qi
„ 2n equations replacing the n Lagrange’s equations
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1st-order differential instead of 2nd-order
“Symmetry” between q and p is apparent
There is nothing new – We just rearranged equations
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First equation links momentum to velocity
„ This relation is “given” in Newtonian formalism
Second equation is equivalent to Newton’s/Lagrange’s
equations of motion
Quick Example
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Particle under Hooke’s law force F = –kx
L=
m 2 k 2
x − x
2
2
p=
∂L
= mx
∂x
m 2 k 2
H = xp − L = x + x
2
2
p2 k 2
=
+ x
2m 2
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Hamilton’s equations
∂H
∂H p
= − kx
p=−
x=
=
∂x
∂p m
Replace x with
p
m
Usual harmonic
oscillator
Energy Function
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Definition of Hamiltonian is identical to the energy
function h(q, q, t ) = qi ∂L − L(q, q, t )
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∂qi
Difference is subtle: H is a function of (q, p, t)
This equals to the total energy if
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Lagrangian is L = L0 (q, t ) + L1 (q, t )qi + L2 (q, t )q j qk
Constraints are time-independent
„ This makes T = L2 ( q, t ) q j qk
See Lecture 4, or
Forces are conservative
Goldstein Section 2.7
„ This makes V = − L0 ( q )
Hamiltonian and Total Energy
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If the conditions make h to be total energy, we can
skip calculating L and go directly to H
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For the particle under Hooke’s law force
p2 k 2
H = E = T +V =
+ x
2m 2
This works often, but not always
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when the coordinate system is time-dependent
„ e.g., rotating (non-inertial) coordinate system
when the potential is velocity-dependent
„ e.g., particle in an EM field
Let’s look at this
Particle in EM Field
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For a particle in an EM field
m 2
L = xi − qφ + qAi xi
2
pi = mxi + qAi
We can’t jump on H = E
because of the last term, but
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mxi2
H = (mxi + qAi ) xi − L =
+ qφ
This is in fact E
2
We’d be done if we were calculating h
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For H, we must rewrite it using pi = mxi + qAi
( pi − qAi ) 2
H ( xi , pi ) =
+ qφ
2m
Particle in EM Field
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Hamilton’s equations are
∂H pi − qAi
xi =
=
∂pi
m
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( pi − qAi ) 2
H ( xi , pi ) =
+ qφ
2m
p j − qAj ∂Aj
∂H
∂φ
pi = −
=q
−q
∂xi
∂xi
∂xi
m
Are they equivalent to the usual Lorentz force?
Check this by eliminating pi
∂Aj
d
∂φ
(mxi + qAi ) = qxi
−q
∂xi
∂xi
dt
A bit of work
d
(mvi ) = qEi + q ( v × B)i
dt
Conservation of Hamiltonian
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Consider time-derivative of Hamiltonian
dH (q, p, t ) ∂H
∂H
∂H
q+
p+
=
∂q
∂p
∂t
dt
∂H
= − pq + qp +
∂t
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Hamiltonian is
conserved if it does not
depend explicitly on t
H may or may not be total energy
„ If it is, this means energy conservation
„ Even if it isn’t, H is still a constant of motion
Cyclic Coordinates
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A cyclic coordinate does not appear in L
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By construction, it does not appear in H either
H ( q , p, t ) = qi pi − L( q , q, t )
Hamilton’s equation says
∂H
Conjugate momentum of a
=0
p=−
cyclic coordinate is conserved
∂q
Exactly the same as in the Lagrangian formalism
Cyclic Example
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Central force problem in 2 dimensions
m 2 2 2
L = (r + r θ ) − V (r )
2
pr = mr pθ = mr 2θ
1 ⎛ 2 pθ2 ⎞
H=
⎜ pr + 2 ⎟ + V (r )
r ⎠
2m ⎝
1 ⎛ 2 l2 ⎞
=
⎜ pr + 2 ⎟ + V (r )
r ⎠
2m ⎝
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θ is cyclic
pθ = const = l
Hamilton’s equations
pr
l2
∂V (r )
r=
pr = 3 −
m
mr
∂r
Cyclic variable drops off by itself
Going Relativistic
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Practical approach
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Purist approach
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Find a Hamiltonian that “works”
„ Does it represent the total energy?
Construct covariant Hamiltonian formalism
„ For one particle in an EM field
Don’t expect miracles
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Fundamental difficulties remain the same
Practical Approach
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Start from the relativistic Lagrangian that “works”
L = −mc 2 1 − β 2 − V (x)
mvi
∂L
=
pi =
∂vi
1− β 2
H =h=
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Did this last time
p 2 c 2 + m 2 c 4 + V ( x)
It does equal to the total energy
Hamilton’s equations
∂H
∂V
pi c 2
pi
∂H
pi = −
=−
= Fi
=
=
xi =
∂xi
∂xi
∂pi
p 2 c 2 + m2 c 4 mγ
Practical Approach w/ EM Field
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Consider a particle in an EM field
L = −mc 2 1 − β 2 − qφ (x) + q ( v ⋅ A)
„ Hamiltonian is still total energy
H = mγ c 2 + qφ
Can be easily checked
= m 2γ 2 v 2 c 2 + m2 c 4 + qφ
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Difference is in the momentum pi = mγ vi + qAi
H = (p − qA) 2 c 2 + m 2 c 4 + qφ
Not the usual linear momentum!
Practical Approach w/ EM Field
H = (p − qA) 2 c 2 + m 2 c 4 + qφ
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Consider H – qφ
( H − qφ ) 2 − (p − qA)2 c 2 = m2 c 4
constant
It means that ( H − qφ , pc − qAc) is a 4-vector,
and so is ( H , pc)
Similar to 4-momentum (E/c, p) of
a relativistic particle
Remember p here is not the linear momentum!
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This particular Hamiltonian + canonical momentum
transforms as a 4-vector
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True only for well-defined 4-potential such as EM field
Purist Approach
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Covariant Lagrangian for a free particle Λ = 12 muν uν
∂Λ
µ
= mu µ
p =
∂uµ
H=
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We know that p0 is E/c
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We also know that x0 is ct…
pµ p µ
2m
Energy is the conjugate “momentum” of time
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Generally true for any covariant Lagrangian
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You know the corresponding relationship in QM
Purist Approach
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Value of Hamiltonian is
pµ p µ mc 2
=
H=
This is constant!
2m
2
What is important is H’s dependence on pµ
Hamilton’s equations
µ
µ
dx
∂H p
=
=
dτ ∂pµ
m
„
µ
dp
∂H
=−
=0
dτ
∂xµ
Time components are
d (ct ) E
d ( E c)
=
=γc
=0
dτ
mc
dτ
4-momentum
conservation
Energy definition
and conservation
Purist Approach w/ EM Field
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With EM field, Lagrangian becomes
Λ( x µ , u µ ) = 12 muµ u µ + qu µ Aµ
H=
„
„
muµ u µ
2
=
p µ = mu µ + qAµ
( pµ − qAµ )( p µ − qAµ )
2m
Hamilton’s equations are
( pν − qAν ) ∂Aν
∂H
dx µ ∂H p µ − qAµ dp µ
=−
=−
=
=
dτ
∂xµ
m
∂xµ
dτ ∂pµ
m
A bit of work can turn them into
⎛ ∂Aν ∂Aµ ⎞
du µ
= q⎜
−
m
uν = K µ
⎟
4-force
⎜ ∂x
⎟
∂
dτ
x
ν ⎠
⎝ µ
EM Field and Hamiltonian
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In Hamiltonian formalism, EM field always modify
the canonical momentum as p Aµ = p0µ + qAµ
With EM field
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Without EM field
A handy rule:
Hamiltonian with EM field is given by replacing pµ
in the field-free Hamiltonian with pµ – qAµ
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Often used in relativistic QM to introduce EM interaction
Summary
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Constructed Hamiltonian formalism
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Equivalent to Lagrangian formalism
„ Simpler, but twice as many, equations
Hamiltonian 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 Hamiltonians
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Conjugate of time = energy
pµ – qAµ rule for introducing EM interaction