10
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Exercise 10.1:
The Transformation equations between two sets of coordinates are:
√
Q = log(1 + q cos p)
√
√
P = 2(1 + q cos p) q sin p
(1)
(2)
1. Show directly from this transformation equations that Q, P are canonical variables if
q and p are canonical as well.
2. Find the generating function for this transformation.
Solution:
1. To show the transformation is canonical we need to show that
[Q, Q] = [P, P ] = 0
&
[Q, P ] = 1
(3)
We show this
∂Q ∂Q ∂Q ∂Q
−
=0
∂q ∂p
∂p ∂q
∂P ∂P
∂P ∂P
[P, P ] =
−
=0
∂q ∂p
∂p ∂q
√
q sin p
cos p
sin p
√
(2 q cos p + 2q cos 2p) +
[Q, P ] = √
√
√ + sin 2p
√
1 + q cos p
q
2 q 1 + q cos p
√
1 + q cos p
=
=1
√
1 + q cos p
[Q, Q] =
2. Let us find F3 which is a function of Q and p which means we need to find q and P ,
q=
eQ − 1
cos p
2
;
P = 2eQ eQ − 1 tan p
1
(4)
so now we can look for F3
∂F3
= −q
∂p
∂F3
= −P
∂Q
⇒
Z
⇒
F3 = −
qdp =
2
= − eQ − q tan p + C(Q, t).
−2eQ eQ − q
2
tan p +
∂C
= −2eQ eQ − 1 tan p
∂Q
where we chose C = 0 since C = const and we may choose it as we wish.
So we get
2
F3 = − eQ − 1 tan p
(5)
(6)
(7)
(8)
Exercise 10.2: Phase space flow
2
2
p
+ kx2 and a rectangle
Consider a harmonic oscillator governed by the Hamiltonian H = 2m
(in phase space) with one corner at the point (x1 , p1 ) and sides of the lengths ∆x∆p/.
1. Show that the solution for x and p is
p0
sin ωt, −mωx0 sin ωt + p0 cos ωt
(x, p) = x0 cos ωt +
mω
(9)
2. Show that at any later time, the image of the rectangle is a parallelogram with area
∆x∆p. Hint: Start by finding the corners of the rectangle and then find the corners of
the parallelogram
Solution:
2
1. The Hamiltonian is H =
p2
2m
+ 21 mω 2 x2 , giving us the solution for x
x = A cos ωt + B sin ωt
(10)
Using the initial conditions x(0) = x0 , ẋ(0) = 0 we get
p0
x = x0 cos ωt +
sin ωt → mẋ = p = p0 cos ωt − mωx0 sin ωt
mω
p0
(x, p) = x0 cos ωt +
sin ωt, p0 cos ωt − mωx0 sin ωt
mω
(11)
(12)
as required.
2. The four corners of the intial rectangle have coordinates (x0 , p0 ), (x0 +∆x, p0 ), (x0 , p0 +
∆p), (x0 + ∆x, p0 ∆p).We can plug these coordinates into our solution, and if we were
to write (x, p) relative to (x0 , p0 ) we will get
(x, p)1 = (0, 0),
(x, p)2 = (∆ cos ωt, −mω∆x sin ωt) ,
∆p
(x, p)3 =
sin ωt, ∆p cos ωt ,
mω
∆p
(x, p)4 =
∆ cos ωt +
sin ωt, ∆p cos ωt − mω∆x sin ωt ,
mω
These points form the vertices of the following parallelogram
3
(13)
(14)
(15)
(16)
The area of this parallelogram equals the area of the dotted rectangle minus the area of
the four triangles, so we have
∆p
A = ∆x cos ωt +
sin ωt (mω∆ sin ωt + ∆x cos ωt)
(17)
mω
1
1 ∆p
−2
∆x cos ωtmω∆x sin ωt +
sin ωt∆p cos ωt
(18)
2
2 mω
∆p
= ∆x cos ωt∆p cos ωt +
sin ωtmω∆x sin ωt
(19)
mω
= ∆x∆p
(20)
as desired. This result for the harmonic oscillator holds for a rectangle of any size, so
in particular it holds for an infinitesimal rectangle. And since any region of arbitrary
shape can be built up from infinitesimal rectangles, we see that the area of any region
remains constant as it flows (and invariably distorts) through phase space.
Exercise 10.3:
Find the generating function for the canonical transformation Q(t) = q (t + τ ) and P (t) =
p (t + τ ) where τ is a constant for the following case:
1. A free particle.
2. Motion in a uniform force field U = −F q
3. A Harmonic oscillator.
Guidance: Expand the two transformations in a Taylor series
Solution:
we can use the expansion:
τ2
+ ...
2
τ2
P = p(t + τ ) = p(t) + ṗ(t)τ + p̈ + ...
2
1. For a free particle there is no force, this means that q̈ = ṗ = 0.
In addition, from Hamilton’s equations it follows that: q̇ = mp and we have:
p
Q = q(t + τ ) = q(t) + τ
m
P = p(t + τ ) = p(t)
Q = q(t + τ ) = q(t) + q̇(t)τ + q̈
4
(21)
(22)
(23)
(24)
We look for a generating function of type 1:
m
m
q2
∂F1 (q, Q)
=
(Q − q) → F1 (q, Q) =
Qq −
+ G(Q) (25)
p =
∂q
τ
τ
2
∂F1 (q, Q)
m
m
m
−P =
= q + G0 (Q) =
(q − Q) → G(Q) = − Q2
(26)
∂Q
τ
τ
2τ
such that
F1 (q, Q) = −
m
(Q − q)2
2τ
2. From Hamilton’s equations it follows that: q̇ =
a uniform field ṗ = F we have:
p
,
m
(27)
and using the equation of motion in
p
Fτ2
τ+
m
2m
P = p(t + τ ) = p(t) + F τ
Q = q(t + τ ) = q(t) +
(28)
(29)
We look for a generating function of type 1:
m
Fτ
m
∂F1 (q, Q)
q2 F τ q
= (Q − q)
→ F1 (q, Q) =
+ g(Q)
p =
Qq
∂q
τ
2
τ
2
2
∂F1 (q, Q)
m
m
Fτ
m
Fτ2
−P =
= q + G0 (Q) =
(q − Q) −
→ G(Q) = − Q2 −
q
∂Q
τ
τ
2
2τ
2
such that
F1 (q, Q) = −
m
Fτ
(Q − q)2 −
(q + Q)
2τ
2
(30)
3. We now use the Lagrangian of a Harmonic oscillator
1
1
L = mq̇ 2 − kq 2
2
2
(31)
q̈ = −ω 2 q
(32)
for which the EOM are
We use this to find that
p
ω2τ 2
ω3τ 3 p
p
Q = q(t + τ ) = q(t) +
ωτ +
q−
= q cos ωτ +
sin ωτ (33)
mω
2
3! mω
mω
ω2τ 2
P = p(t + τ ) = p(t) − mωqωτ −
p + ... = p cos ωτ − mωq sin ωτ
(34)
2
5
We look for a generating function of type 1:
mω
∂F1 (q, Q)
=
(Q − q cos ωτ )
p=
∂q
sin ωτ
→
mω
F1 (q, Q) =
sin ωτ
q2
Qq − cos ωτ
2
+ G(Q)(35)
and similarly we have:
cos2 ωτ
−P = −p cos ωτ + mωq sin ωτ = −mωQ cot ωτ + mωq
+ mωq sin ωτ =
sin ωτ
mω
=
q + G0 (Q)
sin ωτ
we find
G(Q) = −mω
Q2
cot ωτ
2
(36)
such that
mω
F1 (q, Q) =
sin ωτ
q 2 + Q2
Qq −
cos ωτ
2
6
(37)