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Theoretical Dynamics September 24, 2010
Homework 3
Instructor: Dr. Thomas Cohen Submitted by: Vivek Saxena
1 Goldstein 8.
1.1 Part (a)
The Hamiltonian is given by
H(qi, pi, t) = pi q˙i − L(qi, q˙i, t) (1)
where all the ˙qi’s on the RHS are to be expressed in terms of qi, pi and t. Now,
dH = ∂H ∂qi
dqi + ∂H ∂pi
dpi + ∂H ∂t
dt (2)
From (1),
dH = pid q˙i + ˙qidpi − dL = pid q˙i + ˙qidpi −
∂L
∂qi
dqi + ∂L ∂ q˙i
d q˙i + ∂L ∂t
dt
∂L
∂qi^ dqi^ + ˙qidpi^ +
pi −
∂L
∂ q˙i
d q˙i −
∂L
∂t dt^ (3)
Comparing (2) and (3) we get
∂H ∂qi^ =^ −^
∂L
∂qi^ =^ −^ p˙i^ (2nd equality from Hamilton’s equation)^ (4) q˙i = ∂H ∂qi
(also Hamilton’s equation) (5)
pi −
∂L
∂ q˙i^ =^0 (H is not explicitly dependent on ˙qi)^ (6) − ∂L ∂t
= ∂H
∂t
From (4) and (6) we have
d dt
∂L
∂ q˙i
− ∂L
∂qi
= 0 , i = 1, 2 ,... , n (8)
which are the Euler-Lagrange equations.
1.2 Part (b)
L′(p, p, t˙ ) = − p˙iqi − H(q, p, t) (9) = pi q˙i − H(q, p, t) − d dt
(piqi) (10)
= L(q, q, t˙ ) −
d dt (piqi)^ (11) = L(q, q, t˙ ) − p˙iqi − pi q˙i (12)
So,
dL′^ =
∂L′
∂pi^ dpi^ +^
∂L′
∂ p˙i^ d^ p˙i^ +^
∂L′
∂t dt^ (13) = − q˙idpi − qid p˙i + ∂L ∂t
dt (from (9)) (14)
Comparing (12) and (13) we get
q˙i = − ∂L
′ ∂pi
qi = −
∂L′
∂ p˙i^ (16)
Thus the equations of motion are
d dt
∂L′
∂ p˙i
∂L′
∂pi^ =^0 ,^ i^ = 1,^2 ,... , n^ (17)
2 Goldstein 8.
Hamilton’s principle is
δ
L dt = 0 (18)
or equivalently
δ
2 L dt = 0 (19)
We can subtract the total time derivative of a function whose variation vanishes at the end points of the path, from the integrand, without invalidating the variational principle. This is because such a function will only contribute to boundary terms involving the variation of qi and pi at the end points of the path, which vanish by assumption. Such a function is piqi. So, the ‘modified’ Hamilton’s principle is
δ
2 L −
d dt (piqi)
dt = 0 (20)
This bears a resemblance to the usual variational principle in Hamiltonian mechanics, for a Hamiltonian Hc. So the Hamilton equations are
q˙i =
∂Hc ∂pi p ˙i = − ∂Hc ∂qi
which become
q˙i = ∂H ∂pi
k
λk^ ∂ψk ∂pi
− p˙i =
∂H
∂qi^ +^
k
λk
∂ψk ∂qi^ (31)
Time as a canonical variable
If time t is treated as a canonical variable, we define qn+1 = t. By Hamilton’s equations
p˙n+1 = −
∂H
∂qn+1^ (32) = −
∂H
∂t (33) = − dH dt
and
q˙n+1 = ∂H ∂pn+
= 1 (since qn+1 = t) (36)
As the Hamiltonian contains terms of the form pi q˙i for each coordinate and its canonical momentum, in order to incorporate the constraint imposed by the inclusion of time as the (n + 1)th^ canonical variable, we include a term of the form pn+1 q˙n+1 = pn+1 to the Hamiltonian to set up the constraint. Equivalently, the constraint can be obtained by integrating equation (34) above, and is given by
H(q 1 ,... , qn, qn+1; p 1 ,... , pn) + pn+1 = 0 (37)
Hamilton’s principle,
δ
(pi q˙i − H)dt = 0 (38)
can be written as
δ
(pi q˙i − H)t′dθ = 0 (39)
where t′^ = dt/dθ and θ is some parameter.
Using the constrained form of Hamilton’s equations we get
q˙i = (1 + λ) ∂H ∂pi
, i = 1, 2 ,... n (40)
p˙i = −(1 + λ) ∂H∂q i
, i = 1, 2 ,... n (41) q˙n+1 = λ (42) p˙n+1 = −(1 + λ) ∂H ∂t
= − ∂L
∂t
By regarding H′^ = (1 + λ)H as an equivalent Hamiltonian, these equations are the required (2n + 2) equations of motion. Also, λ = ˙qn+1 = dt/dθ.
4 Goldstein 8.
4.1 Part (a)
In the given configuration, both springs elongate or compress by the same magnitude. Suppose q denotes the position of the mass m from the left end. At t = 0, q(0) = a/2, but the unstretched lengths of both springs are given to be zero. Therefore, the elongation (compression) of spring k 1 is q and the compression (elongation) of spring k 2 is q. The potential energy is
V =
2 k^1 q
2 +^1
2 k^2 q
2 =^1
2 (k^1 +^ k^2 )q
The kinetic energy is
T =
2 m^ q˙
The Lagrangian is
L = T − V =^1 2
m q˙^2 − 1 2
(k 1 + k 2 )q^2 (46)
The momentum canonically conjugate to the coordinate q is
pq =
∂L
∂ q˙ =^ m^ q˙^ (47)
So the Hamiltonian is H = pq q˙ − L =
2 m^ q˙
2 +^1
2 (k^1 +^ k^2 )q
that is,
H(q, pq, t) =
p^2 q 2 m +
2 (k^1 +^ k^2 )q
Clearly, the Hamiltonian equals the total energy E. The energy is conserved since,
dE dt
= m q˙ ¨q + (k 1 + k 2 )q q˙ = ˙q(−(k 1 + k 2 )q) + (k 1 + k 2 )q q˙ = 0 (50)
where we have used the equation of motion^1. In this case, the Hamiltonian is also conserved.
(^1) dtd^ (^ ∂L∂ q˙^ ) − ∂L∂q = 0 =⇒ mq¨ + (k 1 + k 2 )q = 0
So the Hamiltonian is
H = p · v − L (60) = (mv + eA) · v −
m(v · v) + eA(r) · v − eV (r)
= m 2
v · v + eV (r)
= (p^ −^ eA)^ ·^ (p^ −^ eA) 2 m
= 1 2 m
(p^2 − 2 ep · A + e^2 A^2 ) + eV (r) (62)
Now,
p · A = p ·
2 B^ ×^ r
2 B^ ·^ (r^ ×^ p)^ (63) = 1 2
B · J (64)
where J = r × p denotes the angular momentum. Also,
A^2 = 14 (B × r) · (B × r)
=
4 B
(^2) r (^2) (as B is perpendicular to r) (65)
So the Hamiltonian of equation (58) becomes
H = p
2 2 m
− e 2 m
B · J + e
2 8 m
B^2 r^2 + eV (r) (66)
5.2 Part (b)
Let vlab = ( ˙x, y˙) denote the velocity of the particle in the lab frame, and v′^ = ( ˙x′, y˙′) denote the velocity in the rotating frame. Without loss of generality, we may assume that motion is confined to the xy-plane. We first derive a relationship between the Hamiltonian in a rotating frame with that in a non-rotating frame (in this case, the lab frame). The coordinates are related by
x = x′^ cos(ωt) − y′^ sin(ωt) (67) y = x′^ sin(ωt) + y′^ cos(ωt) (68)
Here, it has been assumed that the rotation is counterclockwise, i.e. ω > 0 for counter- clockwise rotation. So the velocity components are related by
x˙ = x˙′^ cos(ωt) − y˙′^ sin(ωt) − ω(x′^ sin(ωt) + y′^ cos(ωt)) (69) y = x˙′^ sin(ωt) + ˙y′^ cos(ωt) − ω(x′^ cos(ωt) − y′^ sin(ωt)) (70)
Therefore
vlab^2 = ˙x^2 + ˙y^2 = x˙′^2 + ˙y′^2 + 2ω(x y˙ − xy˙ ) + ω^2 r^2 (71)
The Lagrangian in the lab frame is
L =
2 mvlab
(^2) − eV (r) (72)
= 1 2
m( ˙x′^2 + ˙y′^2 ) + mω(x′^ y˙′^ − x˙′y′) +^1 2
mω^2 r^2 − eV (r) (73)
The momenta canonically conjugate to x and y in the rotating system are
px′^ =
∂L
∂ x˙′^ =^ m( ˙x
′ (^) − ωy′) (74)
py′ = ∂L ∂ y˙′^
= m( ˙y′^ + ωx′) (75)
So the Hamiltonian in the rotating frame is
H = px′ x˙′^ + py′ y˙′^ − L (76)
=
p^2 x′ + p^2 y′ 2 m +^ ω(y
′px′ (^) − x′py′ (^) ) + eV (r) (77)
p^2 x′ + p^2 y′ 2 m −^ J z′ ω^ +^ eV^ (r)^ (78)
where J z′ denotes the angular momentum in the z-direction (direction of ω) as measured in the rotating frame. This means that for counterclockwise rotation along the z-axis,
Hrotating f rame = Hlab f rame − ωJz (79)
This is the general relationship between Hamiltonians in the lab frame and rotating frame.
For this problem, from equation (62) above, we have
Hlab f rame = p
2 2 m −^
eB 2 m J^ +^
e^2 8 m B
(^2) r (^2) + eV (r) (80)
as B = B zˆ and J = Jz zˆ = J ˆz. So, the Hamiltonian in the rotating frame is
Hrotating f rame =
p^2 2 m −
ω +
eB 2 m
Jz +
e^2 8 m B
(^2) r (^2) + eV (r) (81)
It is interesting to note that if ω = ωc = − eB 2 m , then the term linear in the magnetic field vanishes. In this problem, it is given that
ω = −
eB m (82)
which is twice the frequency ωc. So, in this case, the Hamiltonian becomes
Hrotating f rame =
p^2 2 m +^
eB 2 m Jz^ +^
e^2 8 m B
(^2) r (^2) + eV (r) (83)
So,
[Li, f (r)]P B = −
∂Li ∂pα
∂f (r) ∂xα = − x rα
∂(ijkxj pk) ∂pα
∂f (r) ∂r = −
ijkxαxj δα,k r
∂f ∂r = − ijkxj^ xk r
∂f ∂r = − (r^ ×^ r)i r
∂f ∂r = 0 (88)
7 Problem 2
7.1 Part (a)
[ξi, ξj ] = ∂ξi ∂ηα
Jαβ^ ∂ξj ∂ηβ
So,
d d [ξi, ξj^ ]^ =^
d d
∂ξi ∂ηα^ Jαβ
∂ξj ∂ηβ
= d d
∂ξi ∂ηα
Jαβ^ ∂ξj ∂ηβ
Jαβ^ d d
∂ξj ∂ηβ
= (^) ∂η∂ α
dξi d
Jαβ
∂ξj ∂ηβ^ +^
∂ξi ∂ηα^ Jαβ
∂ηβ
( (^) dξ j d
∂ηα^ ([ξi, g])^ Jαβ
∂ξj ∂ηβ^ +^
∂ξi ∂ηα^ Jαβ
∂ηβ^ ([ξj^ , g])
= ∂ ∂ηα
∂ξi ∂ηγ
Jγδ^ ∂g ∂ηδ
Jαβ^ ∂ξj ∂ηβ
Jαβ^ ∂ ∂ηβ
∂ξj ∂ηω
Jωθ^ ∂g ∂ηθ
∂^2 ξi ∂ηα∂ηγ^ Jγδ
∂g ∂ηδ^ +^
∂ξi ∂ηγ^ Jγδ
∂^2 g ∂ηαηδ
Jαβ
∂ξj ∂ηβ
Jαβ
( (^) ∂ (^2) ξ j ∂ηβ ∂ηω^ Jωθ
∂g ∂ηθ^ +^
∂ξj ∂ηω^ Jωθ
∂^2 g ∂ηβ ∂ηθ
Now, for = 0, ξ = η, so the second order terms
∂^2 ξi ∂ηα∂ηγ
=
∂^2 ξj ∂ηβ ∂ηω
=
equal zero. So,
d d [ξi, ξj^ ]
=
∂ξi ∂ηγ^ Jγδ
∂^2 g ∂ηα∂ηδ^ Jαβ
∂ξj ∂ηβ^ +^
∂ξi ∂ηα^ Jαβ
∂ξj ∂ηω^ Jωθ
∂^2 g ∂ηβ ∂ηθ^ (91)
= (^) ∂η∂ξi α
Jαβ^ ∂
(^2) g ∂ηγ ∂ηβ^ Jγδ
∂ξj ∂ηδ^ +^
∂ξi ∂ηα^ Jαβ
∂^2 g ∂ηγ ∂ηβ^ Jδγ
∂ξj ∂ηδ^ (92)
which is obtained by switching γ ↔ α, δ ↔ β in the first term and θ ↔ γ, ω ↔ δ in the second term. As J is antisymmetric, Jδγ = −Jγδ. So,
d d [ξi, ξj^ ]
=
∂ξi ∂ηα^ Jαβ
∂^2 g ∂ηγ ∂ηβ^ Jγδ
∂ξj ∂ηδ^ −^
∂ξi ∂ηα^ Jαβ
∂^2 g ∂ηγ ∂ηβ^ Jγδ
∂ξj ∂ηδ^ (93) = 0 (94)
So upto O(^2 ), we have d[ξi, ξj ]/d = 0 and so [ξi, ξj ] is constant up to O(^2 ). As [ξi, ξj ]|=0 = [ηi, ηj ] = Jij , we have [ξi, ξj ] = Jij upto O(^2 ).
So, ξ is a canonical transformation upto O(^2 ).
7.2 Part (b)
First of all, η(ξ, 0) = ξ (95)
because at t = 0, the canonical coordinates overlap in phase space. So, using the result of part (a), η(ξ, t) can be treated as a one-parameter family of canonical transformations (parametrized by t), provided there exists some function g satisfying
dξi dt = [ξi, g]P B^ (96)
This condition is seen to be satisfied if g is taken as the Hamiltonian H, for in this case,
[ξi, H]P B = ∂ξ ∂ξi k
Jkj^ ∂H∂ξ j
= δikJkj
∂H
∂ξj^ (98) = Jij^ ∂H ∂ξj
= ξ˙i (100)
where the last equality is obtained via Hamilton’s equations. So, taking g = H satisfies the Poisson bracket relation with H acting as the generator of time translations.
