Transcript
Classical Mechanics - Homework Assignment 9 Alejandro G´ omez omez Espinosa
∗
November 29, 2012
Goldstein, Ch.9, 11 Determine whether the transformation is canonical p1 p2 +1 q 2 q 1 q 2 p2 q 1 p1 P 2 = + (q 2 + q 1 ) q 2 q 1
Q1 = q 1 q 2
− −
P 1 =
Q2 = q 1 + q 2
− −
To determine if this transformation is canonical, let us use the Poisson brackets: ∂u ∂v ∂q i ∂p i
[u, v]q,p =
∂u ∂v − ∂p ∂q i
i
Then, [Q1 , P 1 ]q,p =
∂Q 1 ∂P 1 ∂q 1 ∂p 1
= q 2
q 2
=
=
− q
2
− q
1
1 q 2
− q
1
−
−
−
− −
−
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1
1
2
∂P 1 ∂q 2
− ∂Q ∂p
1
2
∂P 2 ∂q 2
∂Q 2 ∂P 2 ∂Q 2 ∂P 2 ∂Q 2 ∂P 2 + ∂q 1 ∂p 1 ∂p 1 ∂q 1 ∂q 2 ∂p 2 q 1 q 1 + q 2 q 1 q 2 q 1
−
− ∂Q ∂p
−
= 0 ∗
∂P 1 ∂Q 1 ∂P 1 + ∂q 1 ∂q 2 ∂p 2
∂Q 1 ∂P 2 ∂Q 1 ∂P 2 ∂Q 1 ∂P 2 + ∂q 1 ∂p 1 ∂p 1 ∂q 1 ∂q 2 ∂p 2 q 1 q 2 q 2 + q 1 q 2 q 1 q 2 q 1
= 0
[Q2 , P 2 ]q,p =
1
1
1
= 0
[Q1 , P 2 ]q,p =
− ∂Q ∂p
− ∂Q ∂p
2
2
∂P 2 ∂q 2
[Q2 , P 1 ]q,p = =
∂Q 2 ∂P 1 ∂q 1 ∂p 1
− ∂Q ∂p
∂P 1 ∂Q 2 ∂P 1 + ∂q 1 ∂q 2 ∂p 2
2
1
− ∂Q ∂p
2
2
∂P 1 ∂q 2
1
1 − q − q q − q 2
1
2
1
= 0
[P 1 , P 2 ]q,p = =
∂P 1 ∂P 2 ∂P 1 ∂P 2 ∂P 1 ∂P 2 ∂P 1 ∂P 2 + ∂q 1 ∂p 1 ∂p 1 ∂q 1 ∂q 2 ∂p 2 ∂p 2 ∂q 2 1 p1 p2 q 1 p1 2 q 2 q 1 q 2 q 1 (q 2 q 1 ) (q 2 q 1 )2
−
− −
+
p1 (q 2
= 0
[Q1 , Q2 ]q,p =
−
− p − q ) 2
1
− − − − − − q 1 p − q − q − q − q (q − q ) − 1 2
2
−1
2
∂Q 1 ∂Q 2 ∂q 1 ∂p 1
2
1
− ∂Q ∂p
1
1
2
1
∂Q 2 ∂Q 1 ∂Q 2 + ∂q 1 ∂q 2 ∂p 2
2
1
− ∂Q ∂p
1
2
2
∂Q 2 ∂q 2
= 0 Since [Q1 , P 1 ] = [Q1 , P 2 ] = [Q2 , P 1 ] = [Q2 , P 2 ] = [Q1 , P 2 ] = [P 1 , P 2 ] = 0, therefore this transformation is canonical. Goldstein, Ch.9, 17 Show that the Jacobi identity is satisfied if the Poisson bracket sign stands for the commutator of two square matrices: [A, B] = AB
− BA
(1)
Show also that for the same representation of the Poisson bracket that [A, BC] BC] = [A, B]C + B[A, C] The Jacobi identity is given by: [A, [B, C ]] ]] + [B, [C, A]] + [C, [A, B ]] = 0 then, if (1 (1) is satisfied: [A, [B, C ]] = [A,BC = =
− CB ] A(BC − CB ) − (BC − CB )A ABC − ACB − BC A + CB A
[B, [C, A]] = [B , C A = =
− AC ] B (CA − AC ) − (CA − AC )B BC A − BAC − CAB + ACB 2
(2)
[C, [A, B ]] = [C,AB
− BA] C (AB − BA ) − (AB − BA )C CAB − C BA − ABC + BAC
= =
where is easy to see that all the terms will vanish. For (2 (2): [A, B ]C + B [A, C ] = ABC
− BAC + BAC − BC A ABC − BC A
=
= [A,BC ]
Goldstein, Ch.9, 22 For the point transformation in a system of two degrees of freedom, Q1 = q 12 ,
Q2 = q 1 + q 2
find the most general transformation equations for P 1 and P 2 consistent with the overall transformation being canonical. Show that with a particular choise for P 1 and P 2 the Hamiltonian H =
p1
− p
2
2
2q 1
+ p2 + (q 1 + q 2 )2
can be transformed to one in which both Q1 and Q2 are ignorable. By this means solve the problem and obtain expressions for q 1, q 2 , p1 , and p2 as functions of time and their initial values. Using the relations for a point transformation: Q1
=
Q2
=
∂F 2 = q 12 ∂P 1 ∂F 2 = q 1 + q 2 ∂P 2
Then, the generating function must be: F 2 = q 12 P 1 + (q 1 + q 2 )P 2
and the momentum coordinates are: p1
=
p2
=
∂F 2 = 2q 1P 1 + P 2 ∂q 1 ∂F 2 = P 2 ∂q 2
Solving for P 1 and P 2 , we found the most general transformations: P 2 p1 p2 P 1 = 2q 1
= p2
−
Therefore, the Hamiltonian is given by: H = P 12 + P 2 + Q22
3
but, whether we choose P 2 = p2 + (q 1 + q 2 )2 : H = P 12 + P 2
the Hamiltonian does not depend upon Q1 and Q2 . Now, solving this Hamiltonian: ˙1 = P
∂H − ∂Q =0 ⇒
P 1 = a
∂H =0 ⇒ − ∂Q
P 2 = b
1
˙2 = P
2
∂H = 2P 1 ∂P 1 ∂H Q˙ 2 = =1 ∂P 2
Q˙ 1 =
⇒
Q1 = 2 P 1 t + c
⇒
Q2 = t + d
where a,b,c,d are constant, i.e., the initial values. Replacing with the old coordinates: p1 + p2 2q 1
= a
p2 + (q 1 + q 2 )2
= b
q 12
p1
=
− p
q 1 = t+d
q 1 + q 2
2
t+c
Solving this equations, we found:
√
p1
= 2a 2at + c + b(t + d)2
p2
= b + (t + d)2
q 1
=
q 2
=
√ 2at + c √ t − 2at + c + d
Goldstein, Ch.9, 28 A charged particle moves in space with a constant magnetic field B such that the vector potential, A, is 1 A = (B r) 2
×
(a) If v j are the Cartesian components of the velocity of the particle, evaluate the Poisson brackets [vi , v j ],
i = j = 1, 2, 3
We know that the mometum of a charged particle in an electric field is given by pi = mvi + qA i
pi
vi =
⇒
− qA
i
m
then, [vi , v j ] = = = =
1 m2
1
pi [ p
− qA , p − qA ] p , p ] − [ p p , qA ] − [qA , p ] + q [A , A ]) ([ p i
i
j
j
j
i
j
m2 q pi , A j ] + [ Ai , p j ]) ([ p m2 q p j , Ai ] [ p pi , A j ]) ([ p m2
−
−
4
i
j
i
j
But, the vector potential in terms of the Levi-Civita symbols are: Ai = iab Ba xb
Calculate the first term in the previous relation: 1 [ p p j , iab Ba xb ] 2 1 iab Ba [ p p j , xb ] 2 1 iab Ba δ jb 2 1 iaj Ba 2
[ p p j , Ai ] = = = = Consequently, the second term:
1 [ p pi , A j ] = jai Ba 2
Replacing in the relation: q p j , Ai ] [ p pi , A j ]) ([ p m2 1 q 1 iaj Ba jai Ba 2 m 2 2 qB a (iaj jai ) 2m2 qB a 2iaj 2m2 qBa iaj m2
[vi , v j ] =
−
= =
−
−
= =
(b) If pi is the canonical momentum conjugate to xi , also evaluate the Poisson backets [xi , v j ],
pi , v j ], [ p
1
[xi , v j ] =
[x1 , p˙ j ],
− qA ] 1 ([x , p ] − q [x , A ]) m q 1 δ − [x , B x ] m 2 q 1 δ − B [x , x ] 2 m m
=
[xi , p j
pi , p˙ j ], [ p
i
= =
j
j
i
j
ij ij
i
jab
ij ij
jab
a
a b
i
b
δ ij ij m
=
1
[P i , v j ] =
− qA ] 1 p , p ] − q [ p p , A ]) ([ p m − mq B [ p p , x ] − mq B m
=
pi , p j [ p i
= =
5
j
j
i
jab
a
jai
a
i
b
j
For p˙i , we know that the Hamiltonian for a charge particle moving in a magnetic field is given by: 1 H = ( pi qA i )2
−
m
then then,, p˙i is: p˙i
=
− ∂H ∂x i
=
− m1 ( p − qA ) ∂A ∂x i
i
i
i
= = =
1
1 ∂x k qA i ) ijk B j ∂x i 2
− m ( p − − 21m ( p − qA ) − 21m ( p − qA ) i
i
i
ijk B j δ ki ki
i
i
iji B j
= 0
thus, [xi , p˙ j ] = 0 [ p pi , p˙ j ] = 0 Goldstein, Ch.9, 34 Obtain the motion in time of a linear harmonic oscillator by means of the formal soluti solution on for the Poisso Poisson n brack bracket et versio version n of the equation quation of motion motion as derive derived d from from Eq.(9. Eq.(9.116 116). ). Assume that at time t = 0 the initial values are x0 and p0 . The derivation of eq. (9.116) ends up in this relation: u(t) = u0 + t[u, H ]0 +
t2
2!
[[u, H ], H ]0 +
t3
3!
[[[u, H ], H ], H ]0 + ...
Knowing the Hamiltonian of the linear harmonic oscillator: H =
p2 mw2 x + 2m 2
we can use the previous relation to find the motion in time: [x, H ]0 = [[x, H ], H ]0 =
∂x ∂H ∂x ∂p
1 m
∂H p = − ∂x ∂p ∂x m
[ p, p, H ] =
1 m
∂p ∂H ∂x ∂p
−
Pluging them in the initial relation: x(t) = x0 +
6
p0 t m
2 2
− w4t
∂p ∂H ∂p ∂x
=
w2
−2
