HS 2015
Prof. C. Anastasiou
Exercise 1.
The Motion of Vortices
Consider a single two-dimensional vortex of strength ω placed in the origin. It produces a 2D velocity field given by u ( r ) =
ω r e
θ
, which can be rewritten in components as u x
= −
ωy r 2
, u y
=
ωx
.
r 2
In this problem we will study the motion of many such vortices in their own velocity field.
Suppose that we have n vortices of the same strength ω at distinct positions r i
= ( x i
, y i
).
(a) Explain why the velocity of the i th vortex is given by
˙ i
= − ω
X j = i
( x i
− x j y
) i
2
− y j
+ ( y i
− y j
) 2
, y ˙ i
= ω
X j = i
( x i
− x j x i
− x j
) 2 + ( y i
− y j
) 2
(1)
Quite surprisingly, this system can be described by using the Hamiltonian formalism. For each vortex we now define a generalized coordinate momentum’.
q i
= x i and identify p i
= y i with its ‘canonical
(b) Show that the equations ( 1 ) are equivalent to the Hamilton’s equations
q ˙ k
=
∂H
∂p k
, p ˙ k
= −
∂H
∂q k with H = −
ω
4
X log ( q i i,j i = j
− q j
)
2
+ ( p i
− p j
)
2
.
Hint: After differentiating H it is convenient to write the sum as
∂H
∂q k
=
X X
A ij
δ ik
+
X X
B ij
δ jk j = i i i = j j for some objects A ij
, B ij and use the properties of the Kronecker δ ij
.
(c) Consider an infinitesimal rotation of the system given by
(2)
δq i
= − p i
, δp i
= q i
, i = 1 , 2 , ..., n (3) and show that it does not change the Hamiltonian to the order O ( ). What is the generating function ˜ ( q
1
, ..., q n
, p
1
, ..., p n
) of such transformation?
Hint: You should find the expansion log(1 + generating function is defined by δp i
2
) =
= − ∂ G/∂q i
, δq i
2
+ O (
4
) useful. Recall that the
= ∂ G/∂p i
.
(d)* Define new coordinates Q i
( ) = q i
+ δq i
, P i
( ) = p i
+ δp i and the transformed Hamiltonian
H ( ) = H ( Q i
( ) , P i
( )). Use the result from c) to deduce the value of d H/ d |
=0
. Hence show that ˜ is a conserved quantity, i.e. d ˜ d t = [ ˜ ] = 0.
1

Solution.
(a) The velocity field at ( x i
, y i
) due to the j th vortex is described by u x
= −
ω ( y i
− y j
| r i
− r j
|
)
, u y
=
ω ( x i
− x j
)
.
| r i
− r j
|
Summing contributions from all vortices j = i gives the equations of motion.
(b) First we split the sum as suggested in the hint and use the sampling property of the Kronecker δ
∂H
∂q k
= −
ω
4
X i,j i = j
2( q i
− q j
)( δ ik
− δ jk
)
( q i
− q j
) 2 + ( p i
− p j
) 2
= −
ω
2
X j = k q k
− q j
( q k
− q j
) 2 + ( p k
− p j
) 2
= −
ω
2
X X j = i i q i
− q j
( q i
− q j
) 2 + ( p i
− p j
) 2
δ ik
+
ω
X X
2 q i
− q j
( q i
− q j
) 2 + ( p i
− p j
) 2
δ jk
= i = j j
+
ω
X
2 i = k q i
− q k
( q i
− q k
) 2 + ( p i
− p k
) 2
Now we relabell i → j in the second sum and see that the result can be written as
∂H
∂q k
= − ω
X j = k
( q k q k
− q j
− q j
) 2 + ( p k
− p j
) 2 y k
.
The calculation for ∂H/∂p k is analogous.
(c) The argument of the logarithm becomes after an ICT ( 3 )
( q i
− p i
− q j
+ p j
)
2
+ ( p i
+ q i
− p j
− q j
)
2
= ( q i
− q j
)
2
+ ( p i
− p j
)
2
+ ( p i
− p j
)
2
+ ( q i
− q j
)
2 2
. (note that contributions of O ( ) cancel after expanding the brackets). Hence we see that the transformed
Hamiltonian is
H ( ) = −
ω
4
X log ( p i
− p j
)
2 i,j i = j
+ ( q i
− q j
)
2
(1 +
2
) = H (0) −
ω
4
X log 1 + i,j i = j
2
= H (0) + O (
2
) and there is no term linear in .
The condition δp i simultaneously by
= − ∂
˜ i implies q i
= − ∂
∂q i
. Similarly p i
= − ∂
˜ i
. These conditions are satisfied
˜
= −
1
2
X q
2 i i
+ p
2 i
.
(d) H ( ) = H (0) + O (
2
) implies that d H ( ) / d |
=0
= lim
→ 0
1
[ H ( ) − H (0)] = 0. But at the same time d H d
=0
=
∂H
∂Q i d Q i d
=0
+
∂H
∂P i d P i d
=0
= −
∂H
∂q i
∂
∂p i
+
∂H
∂p i
∂
∂q i
= [ ˜ ] .
Thus [ ˜ ] = 0.
Exercise 2.
One more exercise about Canonical transformations...
Given the transformation for a system of one degree of freedom
Q = q cos α − p sin α
P = q sin α + p cos α (4)
(a) show that it satisfies the symplectic condition for any value of the parameter α .
(b) Find a generating function F for the transformation.
Hint: you can, for example, chose a generating function which depends on Q and q , so that p = p ( Q, q ) and P = P ( Q, q ) , and use the transformation equations: p =
∂F
∂q and P = −
∂F
∂Q
.
(5)
2

(c) What is the physical significance of the transformation for α = 0? For α = π/ 2?
Does your generating function work for both of these cases?
(d) * If not, can you find a generating function valid for the case where the one you have just found doesn’t work?
Hint: try with a generating function which depends on other variables, e.g.
q and P .
Solution.
(a) The symplectic condition is met if given the jacobian M is a symplettic matrix, i.e. it satisfies relation
M J M
T
= J . In our case
M =
∂Q
∂q
∂P
∂q
∂Q
∂p
∂P
∂p
!
= cos α − sin α
!
sin α cos α
Now we can calculate
M J M
T
= cos α − sin α
! sin α cos α
0 1
− 1 0
! cos α sin α
− sin α cos α
!
= cos α − sin α
!
− sin α cos α sin α cos α − cos α − sin α
!
=
− sin α cos α + sin α cos α
− cos 2 α − sin 2 α cos
2
α + sin
2
α sin α cos α − sin α cos α
!
=
0 1
!
= J.
− 1 0
(b) To find the generating function, we have to first make the choice on the variable F depends on. Given
Eq. ( 5 ), it is natural to look for dependence on
Q and q
. Rearranging the first of Eq. ( 4 ) to solve it for
p ( Q, q ), we have
Q p ( Q, q ) = − sin α
+ q cos α sin α
(S.1) then, integrating, we have p =
∂F
∂q
⇒ F =
Z dq p ( Q, q ) ⇒ F = −
Qq sin α
+ q
2 cos α
2 sin α
+ g ( Q ) (S.2)
After solving the second transformation of Eq. ( 4 ) for
P ( Q, q
), also using Eq. ( S.1
) to get rid of the
p dependence, we get
P ( Q, q ) = q sin α −
Q cos α
+ sin α q cos 2 α sin α
which allow us to integrate the second of Eq. ( 5 )):
P = −
∂F
∂Q
+ h ( q ) ⇒ F = −
Z dQ P ( Q, q )
= − qQ sin α +
1
2
Q
2 cot α − qQ
=
1
2
Q
2 cot α − qQ sin α
+ h ( q )
1 sin α
− sin α + h ( q )
(S.3)
(S.4)
Combining Eqs. ( S.2
) and ( S.4
) we get
F =
1
2
( q
2
+ Q
2
) cot α − qQ sin α
+ c (S.5)
(c) The transformation with α = 0 is simply the identy transformation P = p , Q = q . for α = π/ 2 is an exchange of variables P = q , Q = − p .
This particular choice of generating function, is not defined for α = 0, but it is not problematic for α = π/ 2.
3

(d) We could chose a generating function which depends on another combination of variables, for example P and q
, and the transformations equations would be different from Eq. ( 5 ), i.e.
p =
∂F
∂q and Q =
∂F
∂P
.
(S.6) and found a different generating function. In this case we need to rewrite the second transformation of
Eq. ( 4 ) as
p =
P cos α
− q sin α cos α
,
which can be used to integrate the first of Eq. ( S.6
):
F =
Z dq p ( q, P ) + h ( P ) = q P cos α
− q
2
2 tan α + h ( P )
We also have that
Q = q cos α − p sin α = q cos α −
P cos α
− q sin α cos α
= q cos
2
α + sin
2
α
− P tan α = cos α q cos α
− P tan α sin α which leads to
F =
Z dP Q ( q, P ) + g ( q ) = q P cos α
−
P
2
2 tan α + g ( q ) .
Combining Eq. ( S.8
) with Eq. ( S.10
) we get
F =
Z dP Q ( q, P ) + g ( q ) = q P cos α
−
1
2
( P
2
+ q
2
) tan α + c , which is singular for α = π/ 2 but is fine for α = 0.
(S.7)
(S.8)
(S.9)
(S.10)
(S.11)
Exercise 3.
Conserved tensor for harmonic oscillator
We have seen that the Kepler problem features, in addition to the total energy and angular momentum, a conserved quantity known as Laplace-Runge-Lenz vector. It turns out that the existence of conserved Laplace-Runge-Lenz-like vectors are not specific to the Kepler potential.
Instead, it is rather a feature of radial force problems.
As such it is not surprising that also the three-dimensional isotropic harmonic oscillator allows for such a conserved quantity, which in this case is the (six component) tensor
A ij
=
1
λ
( p i p j
+ λ
2 r i r j
) , (6) where i, j ∈ { x, y, z } .
(a) Consider a particle of mass m in the harmonic potential
V ( ~ ) =
1
2
2
, ~ ∈
R
3
.
(7)
Write down the Lagrangian and find the Hamiltonian from the Legendre transformation.
4

Solution.
The kinetic energy is given by
T =
1
2 m r
˙ 2
the potential energy according to Eq. ( 7 ). This results in the Lagrangian
L =
1
2 m r
˙ 2
−
1
2
2
To find the Hamiltonian, we introduce the canonical momenta
.
∂L p i
=
∂ q ˙ i
= m r ˙ i and obtain
H =
X p i q ˙ i
− L =
1
2 m
( p x
2
+ p y
2 i
+ p z
2
) +
1
2 u ( r x
2
+ r y
2
+ r z
2
) .
(b) Using Poisson brackets [ · , · ] and a suitable choice of λ , show that the tensor A ij
Eq. ( 6 ) is conserved, i.e.,
[ A ij
, H ] = 0 , ∀ i, j .
defined in
(8)
Hint.
Although the problem has radial symmetry, note that it might be simpler to carry out the calculation in cartesian coordinates.
Solution.
Using the definition of the Poisson bracket we find
[ A ij
, H ] =
X k
∂A
∂q ij k
We consider therefore
∂A ij
∂q k
=
∂A ij
∂r k
= λ ( δ ik r j
+ δ jk r i
) ,
∂H
∂p k
−
∂A ij
∂p k
∂H
∂q k
∂A
∂p ij k
=
1
λ
.
( δ ik p j
+ δ jk p i
) , and
∂H
∂q k
Putting everything together, we obtain
=
∂H
∂r k
= ur k
,
∂H
∂p k
=
1 m p k
.
[ A ij
, H ] =
X
λ m
( δ ik r j
+ δ jk r i
) p k
− u
λ
( δ ik p j
+ δ jk p i
) r k k
λ
= ( r j p i
+ r i p j
) m
1 − um
λ 2
= 0 ⇔ λ
2
=
λ m
( r j p i
= um .
+ r i p j
) − u
λ
( p j r i
+ p i r j
) ,
(c) The tensor A ij has six components, corresponding to six constants of motion. The energy and the angular momentum are additional constants of motion, because the harmonic oscillator is a central force problem.
Hence there are more constants of motion than degrees of freedom. For this reason some of them must be interlinked.
In this context, relate tr A ij
=
X
A ii i
(9) to another conserved quantity.
Solution.
Using the definition of A ij we find tr A ij
=
X
A ii
=
X i i
1
λ
( p i p i
+ λ
2 r i r i
) = r m
X u i
1 m p i p i
+ ur where in the last step we used that the energy is conserved and equal to E .
i r i
= 2 r m u
E ,
5