






Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Not my copyright. I found it on the internet and I see that it has not been uploaded here.
Typology: Exercises
1 / 11
This page cannot be seen from the preview
Don't miss anything!
On special offer
Theoretical Dynamics September 24, 2010
Instructor: Dr. Thomas Cohen Submitted by: Vivek Saxena
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 −
∂qi
dqi + ∂L ∂ q˙i
d q˙i + ∂L ∂t
dt
∂qi^ dqi^ + ˙qidpi^ +
pi −
∂ q˙i
d q˙i −
∂t dt^ (3)
Comparing (2) and (3) we get
∂H ∂qi^ =^ −^
∂qi^ =^ −^ p˙i^ (2nd equality from Hamilton’s equation)^ (4) q˙i = ∂H ∂qi
(also Hamilton’s equation) (5)
pi −
∂ q˙i^ =^0 (H is not explicitly dependent on ˙qi)^ (6) − ∂L ∂t
∂t
From (4) and (6) we have
d dt
∂ q˙i
∂qi
= 0 , i = 1, 2 ,... , n (8)
which are the Euler-Lagrange equations.
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′^ =
∂pi^ dpi^ +^
∂ p˙i^ d^ p˙i^ +^
∂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 = −
∂ p˙i^ (16)
Thus the equations of motion are
d dt
∂ p˙i
∂pi^ =^0 ,^ i^ = 1,^2 ,... , n^ (17)
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
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 =
∂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 = −
∂qn+1^ (32) = −
∂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
∂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.
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 k^2 q
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 =
∂ q˙ =^ m^ q˙^ (47)
So the Hamiltonian is H = pq q˙ − L =
2 m^ q˙
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^ ×^ p)^ (63) = 1 2
where J = r × p denotes the angular momentum. Also,
A^2 = 14 (B × r) · (B × r)
=
(^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)
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′^ =
∂ 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)
[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 ).
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
∂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.