Theoretical Dynamics
September 24, 2010
Homework 3
Instructor: Dr. Thomas Cohen
1
1.1
Submitted by: Vivek Saxena
Goldstein 8.1
Part (a)
The Hamiltonian is given by
H(qi , pi , t) = pi q̇i − L(qi , q̇i , t)
(1)
where all the q̇i ’s on the RHS are to be expressed in terms of qi , pi and t. Now,
∂H
∂H
∂H
dqi +
dpi +
dt
∂qi
∂pi
∂t
dH =
(2)
From (1),
dH = pi dq̇i + q̇i dpi − dL
∂L
∂L
∂L
= pi dq̇i + q̇i dpi −
dqi +
dq̇i +
dt
∂qi
∂ q̇i
∂t
∂L
∂L
∂L
= −
dqi + q̇i dpi + pi −
dq̇i −
dt
∂qi
∂ q̇i
∂t
(3)
Comparing (2) and (3) we get
∂H
∂L
=−
∂qi
∂qi
= −ṗi
q̇i =
∂L
∂ q̇i
∂L
−
∂t
pi −
∂H
∂qi
= 0
=
(2nd equality from Hamilton’s equation)
(4)
(also Hamilton’s equation)
(5)
(H is not explicitly dependent on q̇i )
∂H
∂t
(6)
(7)
From (4) and (6) we have
d
dt
∂L
∂ q̇i
−
∂L
∂qi
= 0,
which are the Euler-Lagrange equations.
3-1
i = 1, 2, . . . , n
(8)
1.2
Part (b)
L0 (p, ṗ, t) = −ṗi qi − H(q, p, t)
= pi q̇i − H(q, p, t) −
(9)
d
(pi qi )
dt
d
(pi qi )
dt
= L(q, q̇, t) − ṗi qi − pi q̇i
= L(q, q̇, t) −
(10)
(11)
(12)
So,
∂L0
∂L0
∂L0
dpi +
dṗi +
dt
∂pi
∂ ṗi
∂t
∂L
dt
(from (9))
= −q̇i dpi − qi dṗi +
∂t
dL0 =
(13)
(14)
Comparing (12) and (13) we get
(15)
qi
(16)
Thus the equations of motion are
∂L0
d ∂L0
−
dt ∂ ṗi
∂pi
2
∂L0
∂pi
∂L0
= −
∂ ṗi
q̇i = −
= 0,
i = 1, 2, . . . , n
(17)
Goldstein 8.6
Hamilton’s principle is
Z
δ
L dt = 0
(18)
2L dt = 0
(19)
or equivalently
Z
δ
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
pi qi . So, the ‘modified’ Hamilton’s principle is
Z d
(20)
δ
2L − (pi qi ) dt = 0
dt
3-2
Using the Legendre transformation, this becomes
Z
δ (2pi q̇i − 2H − pi q̇i − ṗi qi ) dt = 0
Z
=⇒ δ (2H + ṗi qi − pi q̇i ) dt = 0
(21)
(22)
now,

ṗi qi − pi q̇i =
q1 . . . qn | p1 . . . pn
1×2n
0n×n 1n×n
−1n×n 0n×n
T
= η J η̇








2n×2n 




q̇1
.
.
q̇n
−−
ṗ1
.
.
ṗn







 (23)






2n×1
(24)
So (22) becomes
Z
δ
2H + η T J η̇ dt = 0
(25)
which is the required form of Hamilton’s principle.
3
Goldstein 8.9
The constraints can be incorporated into the Lagrangian L by defining a “constrained
Lagrangian” Lc , as
X
Lc (q, q̇, t) = L(q, q̇, t) −
λk ψk (q, p, t)
(26)
k
Applying Hamilton’s principle, and using the Legendre transformation for L, we get
!
Z
X
δ
pi q̇i − H(q, p, t) −
λk ψk (q, p, t) dt = 0
(27)
k
By analogy with the constrained Lagrangian, we can define a “constrained Hamiltonian”
Hc as
X
Hc (q, p, t) = H(q, p, t) +
λk ψk (q, p, t)
(28)
k
Since both the terms are functions of qi , pi and t, this is a “good” Hamiltonian. Equation
(27) can then be written as
Z
δ (pi q̇i − Hc (q, p, t)) dt = 0
(29)
3-3
This bears a resemblance to the usual variational principle in Hamiltonian mechanics, for
a Hamiltonian Hc . So the Hamilton equations are
∂Hc
∂pi
∂Hc
= −
∂qi
q̇i =
ṗi
which become
q̇i =
−ṗi =
∂H X ∂ψk
+
λk
∂pi
∂pi
k
∂H X ∂ψk
+
λk
∂qi
∂qi
(30)
(31)
k
Time as a canonical variable
If time t is treated as a canonical variable, we define qn+1 = t. By Hamilton’s equations
∂H
∂qn+1
∂H
= −
∂t
dH
= −
dt
ṗn+1 = −
(32)
(33)
(34)
and
∂H
∂pn+1
= 1
(since qn+1 = t)
q̇n+1 =
(35)
(36)
As the Hamiltonian contains terms of the form pi q̇i 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 q̇n+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(q1 , . . . , qn , qn+1 ; p1 , . . . , pn ) + pn+1 = 0
(37)
Hamilton’s principle,
Z
δ
(pi q̇i − H)dt = 0
(38)
(pi q̇i − H)t0 dθ = 0
(39)
can be written as
Z
δ
3-4
where t0 = dt/dθ and θ is some parameter.
Using the constrained form of Hamilton’s equations we get
∂H
,
i = 1, 2, . . . n
(40)
∂pi
∂H
ṗi = −(1 + λ)
,
i = 1, 2, . . . n
(41)
∂qi
q̇n+1 = λ
(42)
∂H
∂L
ṗn+1 = −(1 + λ)
=−
(43)
∂t
∂t
By regarding H 0 = (1 + λ)H as an equivalent Hamiltonian, these equations are the required
(2n + 2) equations of motion. Also, λ = q̇n+1 = dt/dθ.
q̇i = (1 + λ)
4
Goldstein 8.26
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 k1 is q and the compression (elongation) of spring k2 is q. The
potential energy is
1
1
1
V = k1 q 2 + k2 q 2 = (k1 + k2 )q 2
(44)
2
2
2
The kinetic energy is
1
T = mq̇ 2
(45)
2
The Lagrangian is
1
1
L = T − V = mq̇ 2 − (k1 + k2 )q 2
(46)
2
2
The momentum canonically conjugate to the coordinate q is
pq =
So the Hamiltonian is
∂L
= mq̇
∂ q̇
1
1
H = pq q̇ − L = mq̇ 2 + (k1 + k2 )q 2
2
2
(47)
(48)
that is,
p2q
1
+ (k1 + k2 )q 2
2m 2
Clearly, the Hamiltonian equals the total energy E. The energy is conserved since,
H(q, pq , t) =
(49)
dE
= mq̇ q̈ + (k1 + k2 )q q̇ = q̇(−(k1 + k2 )q) + (k1 + k2 )q q̇ = 0
(50)
dt
where we have used the equation of motion1 . In this case, the Hamiltonian is also conserved.
1 d
dt
∂L
∂ q̇
−
∂L
∂q
= 0 =⇒ mq̈ + (k1 + k2 )q = 0
3-5
4.2
Part (b)
Substituting q = Q+b sin(ωt) and q̇ = Q+bω cos(ωt) into the expression for the Lagrangian,
we get
1
1
L(Q, Q̇, t) = m(Q̇ + bω cos(ωt))2 − (k1 + k2 )(Q + b sin(ωt))2
(51)
2
2
and the momentum canonically conjugate to the coordinate Q is given by
pQ =
∂L
= m(Q̇ + bω cos(ωt))
∂ Q̇
(52)
So the Hamiltonian becomes
H(Q, pQ , t) = pQ Q̇ − L(Q, Q̇, t)
(53)
1
1
= m(Q̇ + bω cos(ωt))Q̇ − m(Q̇ + bω cos(ωt))2 + (k1 + k2 )(Q + b sin(ωt))2
2
2
2
2
2
mb ω
1
mQ̇
−
cos2 (ωt) + (k1 + k2 )(Q + b sin(ωt))2
=
2
2
2
2
pQ
1
=
− pQ bω cos(ωt) + (k1 + k2 )(Q + b sin(ωt))2
(54)
2m
2
The Hamiltonian is now explicitly dependent on time, and hence is not conserved, as is
confirmed by the fact that dH/dt 6= 0. The energy is given by
1
1
E = T + V = (Q̇ + bω cos(ωt))2 + (k1 + k2 )(Q + bω sin(ωt))2
2
2
(55)
So,
dE
dt
= m(Q̇ + bω cos(ωt))(Q̈ − bω 2 sin(ωt)) + (k1 + k2 )(Q + b sin(ωt))(Q̇ + bω cos(ωt))
= (Q̇ + bω cos(ωt))(m(Q̈ − Bω 2 sin(ωt)) + (k1 + k2 )(Q + b sin(ωt)))
= (Q̇ + bω cos(ωt))(mq̈ + (k1 + k2 )q)
(56)
= 0
(57)
(c.f. footnote on prev. page)
Therefore, the energy is conserved, as expected (the energy is still given by T + V , but the
Hamiltonian is not T +V anymore, as the relationship connecting the generalized coordinate
to the cartesian coordinate is now explicitly dependent on time).
5
5.1
Goldstein 8.23
Part (a)
The Lagrangian for the system is
1
L = m(v · v) + eA(r) · v − eV (r)
2
The canonical momentum is
p=
∂L
= mv + eA
∂v
3-6
(58)
(59)
So the Hamiltonian is
H = p·v−L
(60)
= (mv + eA) · v −
=
=
=
1
m(v · v) + eA(r) · v − eV (r)
2
m
v · v + eV (r)
2
(p − eA) · (p − eA)
+ eV (r)
2m
1
(p2 − 2ep · A + e2 A2 ) + eV (r)
2m
(61)
(62)
Now,
1
p·A = p· B×r
2
1
=
B · (r × p)
2
1
=
B·J
2
(63)
(64)
where J = r × p denotes the angular momentum. Also,
A2 =
=
1
(B × r) · (B × r)
4
1 2 2
B r
(as B is perpendicular to r)
4
(65)
So the Hamiltonian of equation (58) becomes
H =
5.2
p2
e
e2 2 2
−
B·J +
B r + eV (r)
2m 2m
8m
(66)
Part (b)
Let vlab = (ẋ, ẏ) denote the velocity of the particle in the lab frame, and v 0 = (ẋ0 , ẏ 0 )
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 = x0 cos(ωt) − y 0 sin(ωt)
0
(67)
0
y = x sin(ωt) + y cos(ωt)
(68)
Here, it has been assumed that the rotation is counterclockwise, i.e. ω > 0 for counterclockwise rotation. So the velocity components are related by
ẋ = ẋ0 cos(ωt) − ẏ 0 sin(ωt) − ω(x0 sin(ωt) + y 0 cos(ωt))
0
0
0
0
y = ẋ sin(ωt) + ẏ cos(ωt) − ω(x cos(ωt) − y sin(ωt))
3-7
(69)
(70)
Therefore
vlab 2 = ẋ2 + ẏ 2 = ẋ02 + ẏ 02 + 2ω(xẏ − ẋy) + ω 2 r2
(71)
The Lagrangian in the lab frame is
L =
=
1
mvlab 2 − eV (r)
2
1
1
m(ẋ02 + ẏ 02 ) + mω(x0 ẏ 0 − ẋ0 y 0 ) + mω 2 r2 − eV (r)
2
2
(72)
(73)
The momenta canonically conjugate to x and y in the rotating system are
p x0
=
py 0
=
∂L
= m(ẋ0 − ωy 0 )
∂ ẋ0
∂L
= m(ẏ 0 + ωx0 )
∂ ẏ 0
(74)
(75)
So the Hamiltonian in the rotating frame is
H = px0 ẋ0 + py0 ẏ 0 − L
=
=
p2x0
+
p2y0
2m
2
px0 + p2y0
2m
(76)
+ ω(y 0 px0 − x0 py0 ) + eV (r)
(77)
− Jz0 ω + eV (r)
(78)
where Jz0 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 =
p2
eB
e2 2 2
−
J+
B r + eV (r)
2m 2m
8m
as B = B ẑ and J = Jz ẑ = J ẑ. So, the Hamiltonian in the rotating frame is
p2
eB
e2 2 2
Hrotating f rame =
− ω+
Jz +
B r + eV (r)
2m
2m
8m
(80)
(81)
eB
It is interesting to note that if ω = ωc = − 2m
, 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 =
p2
eB
e2 2 2
+
Jz +
B r + eV (r)
2m 2m
8m
3-8
(83)
6
Problem 1
6.1
Part (a)
The subscript P B is suppressed for clarity.
[Li , Lj ]P B = [iαβ xα pβ , jγδ xγ pδ ]
= iαβ jγδ [xα pβ , xγ pδ ]
= iαβ jγδ (xα [pβ , xγ pδ ] + [xα , xγ pδ ]pβ )
(as [A, BC]P B = B[A, C]P B + [A, B]P B C)
= iαβ jγδ (xα [pβ , xγ ]pδ + xα xγ [pβ , pδ ] + [xα , xγ ]pδ pβ + xγ [xα , pδ ]pβ )
= iαβ jγδ (−δβγ xα pδ + δαδ xγ pβ )
= iαβ jβδ (−xα pδ ) + iαβ jγα xγ pβ
= iαβ jδβ xα pδ − iβα jγα xα pβ
= (δij δαδ − δiδ δjα )xα pδ − (δij δβγ − δiγ δjβ )xγ pβ
= (δij xα pα − xj pi ) − (δij xβ pβ − xi pj )
= xi p j − xj p i
Now, [L1 , L2 ]P B = x1 p2 − x2 p1 = L3 , [L1 , L3 ]P B = x1 p3 − x3 p1 = −L2 , [L3 , L2 ]P B =
x3 p2 − x2 p3 = 321 − L1 , etc. So, xi pj − xj pi = ijk Lk .
Hence, [Li , Lj ]P B = ijk Lk .
6.2
Part (b)
For each i = 1, 2, 3,
[Li , L2 ]P B = [Li , Lj Lj ]
(sum over j)
= Lj [Li , Lj ] + [Li , Lj ]Lj
= Lj (ijk Lk ) + (ijk Lk )Lj
= 2ijk Lj Lk
= 0
(as ijk is antisymmetric under j ↔ k, while Lj Lk is symmetric.)
So,
[L, L2 ]P B = êi [Li , L2 ]P B = 0
6.3
(84)
Part (c)
For each i = 1, 2, 3,
[Li , f (r)]P B =
Now, r =
√
∂Li ∂f (r) ∂Li ∂f (r)
−
∂xα ∂pα
∂pα ∂xα
(85)
∂r ∂f
xα ∂f
=
∂xα ∂r
r ∂r
(86)
xi xi and Li = ijk xj pk , so
∂f
∂xα
∂f
∂pα
=
= 0
3-9
(87)
So,
∂Li ∂f (r)
∂pα ∂xα
xα ∂(ijk xj pk ) ∂f (r)
−
r
∂pα
∂r
ijk xα xj δα,k ∂f
−
r
∂r
ijk xj xk ∂f
−
r
∂r
(r × r)i ∂f
−
r
∂r
0
[Li , f (r)]P B = −
=
=
=
=
=
7
7.1
(88)
Problem 2
Part (a)
[ξi , ξj ] =
∂ξj
∂ξi
Jαβ
∂ηα
∂ηβ
(89)
So,
d
[ξi , ξj ] =
d
=
=
=
=
=
∂ξj
d ∂ξi
Jαβ
d ∂ηα
∂ηβ
∂ξj
d ∂ξi
∂ξi
d ∂ξj
+
Jαβ
Jαβ
d ∂ηα
∂ηβ
∂ηα
d ∂ηβ
∂ξj
dξj
∂
dξi
∂ξi
∂
Jαβ
+
Jαβ
∂ηα d
∂ηβ
∂ηα
∂ηβ d
∂ξj
∂
∂ξi
∂
([ξi , g]) Jαβ
+
Jαβ
([ξj , g])
∂ηα
∂ηβ
∂ηα
∂ηβ
∂ξj
∂ξj
∂
∂ξi
∂g
∂ξi
∂
∂g
Jγδ
Jαβ
+
Jαβ
Jωθ
∂ηα ∂ηγ
∂ηδ
∂ηβ
∂ηα
∂ηβ ∂ηω
∂ηθ
2
2
∂ξj
∂g
∂ξi
∂ g
∂ ξi
Jγδ
+
Jγδ
Jαβ
∂ηα ∂ηγ
∂ηδ
∂ηγ
∂ηα ηδ
∂ηβ
2
∂ ξj
∂ξj
∂ξi
∂g
∂2g
+
Jαβ
Jωθ
+
Jωθ
∂ηα
∂ηβ ∂ηω
∂ηθ
∂ηω
∂ηβ ∂ηθ
Now, for = 0, ξ = η, so the second order terms
∂ 2 ξj ∂ 2 ξi ,
∂ηα ∂ηγ =0
∂ηβ ∂ηω =0
3 - 10
(90)
equal zero. So,
d
[ξi , ξj ]
=
d
=0
∂ξj
∂ξj
∂2g
∂2g
∂ξi
∂ξi
Jγδ
Jαβ
Jωθ
Jαβ
+
∂ηγ
∂ηα ∂ηδ
∂ηβ
∂ηα
∂ηω
∂ηβ ∂ηθ
(91)
=
∂ξj
∂ξj
∂2g
∂2g
∂ξi
∂ξi
Jαβ
Jαβ
Jγδ
+
Jδγ
∂ηα
∂ηγ ∂ηβ
∂ηδ
∂ηα
∂ηγ ∂ηβ
∂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,
∂ξj
∂ξj
∂2g
∂2g
d
∂ξi
∂ξi
Jαβ
Jαβ
[ξi , ξj ]
Jγδ
−
Jγδ
(93)
=
d
∂ηα
∂ηγ ∂ηβ
∂ηδ
∂ηα
∂ηγ ∂ηβ
∂ηδ
=0
= 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
= [ξi , g]P B
dt
(96)
This condition is seen to be satisfied if g is taken as the Hamiltonian H, for in this case,
∂ξi
∂H
Jkj
∂ξk
∂ξj
∂H
= δik Jkj
∂ξj
∂H
= Jij
∂ξj
= ξ˙i
[ξi , H]P B =
(97)
(98)
(99)
(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.
3 - 11