Solutions to Problems in Goldstein, Classical Mechanics, Second Edition Homer Reid October 29, 2002 Chater 9 Problem 9. One of the attemts at combining the two sets of Hamilton s equations into one tries to take q and as forming a comlex quantity. Show directly from Hamilton s equations of motion that for a system of one degree of freedom the transformation Q q + i, P Q is not canonical if the Hamiltonian is left unaltered. Can you find another set of coordinates Q, P that are related to Q, P by a change of scale only, and that are canonical? Generalizing a little, we ut Q µq + i, P νq i. The reverse transformation is q 2 µ Q + ν P, 2i µ Q ν P. The direct conditions for canonicality, valid in cases like this one in which the
Homer Reid s Solutions to Goldstein Problems: Chater 9 2 transformation equations do not deend on the time exlicitly, are. 2 When alied to the case at hand, all four of these yield the same condition, namely µ 2iν. For µ ν, which is the case Goldstein gives, these conditions are clearly not satisfied, so is not canonical. But utting µ, ν 2i we see that equations are canonical.
Homer Reid s Solutions to Goldstein Problems: Chater 9 3 Problem 9.2 a For a one-dimensional system with the Hamiltonian H 2 2 2q 2, show that there is a constant of the motion D q 2 Ht. b As a generalization of art a, for motion in a lane with the Hamiltonian H n ar n, where is the vector of the momenta conjugate to the Cartesian coordinates, show that there is a constant of the motion D r n Ht. c The transformation Q λq, λp is obviously canonical. However, the same transformation with t time dilatation, Q λq, λp, t λ 2 t, is not. Show that, however, the equations of motion for q and for the Hamiltonian in art a are invariant under the transformation. The constant of the motion D is said to be associated with this invariance. a The equation of motion for the quantity D is dd dt {D, H} + D t The Poisson bracket of the second term in D clearly vanishes, so we have The first Poisson bracket is {q, H} H 2 4 { q, 2 } 4 { q, 2 } q 2 2 0 {q, q 2 } H. 3 q 2 2 2 4
i Homer Reid s Solutions to Goldstein Problems: Chater 9 4 Next, {q, q } 2 q 0 q 2 2q 3 q q q 2 Plugging 4 and 5 into 3, we obtain b We have 2 q 2 5 dd dt 2 2 2q 2 H 0. H 2 + 2 2 + 2 3 n/2 ax 2 + x2 2 + x2 3 n/2 so Then { r, H} i H x i anx i x 2 + x2 2 + x2 3 n/2 H i 2n i 2 + 2 2 + 2 3 n/2. { x + 2 x 2 + 3 x 3 H } x + 2 x 2 + 3 x 3 H x i i i x i {n 2 i 2 + 2 2 + 2 3 n/2 anx 2 i x2 + x2 2 + x2 3 n/2 } n 2 + 2 2 + 2 3 n/2 anx 2 + x 2 2 + x 2 3 n/2 6 so if we define D r/n Ht, then dd dt D {D, H} t D { r, H} n t Substituting in from 6, n ar n H 0.
Homer Reid s Solutions to Goldstein Problems: Chater 9 5 c We ut t Qt λq λ 2, P t t λ λ 2. 7 Since q and are the original canonical coordinates, they satisfy q H On the other hand, differentiating 7, we have dq dt t λ q λ 2 t λ ṗ H q 3. 8 λ 2 P t dp dt t λ 3 ṗ λ 2 λ 3 q t λ 2 Q 3 t which are the same equations of motion as 8. Problem 9.4 Show directly that the transformation Q log sin, P q cot is canonical. The Jacobian of the transformation is M q cot cot q csc 2.
Homer Reid s Solutions to Goldstein Problems: Chater 9 6 Hence MJM J q cot q cot cot 0 q csc 2 q cot 0 cot q csc 2 cot cot q csc 2 q csc 2 q cot 0 csc 2 cot 2 cot 2 csc 2 0 0 0 so the symlectic condition is satisfied. Problem 9.5 Show directly for a system of one degree of freedom that the transformation Q arctan q, P q2 + 2 2 2 q 2 is canonical, where is an arbitrary constant of suitable dimensions. so The Jacobian of the transformation is MJM J M + q 2 q 2 + q 2 0 0 + q 2 q q q 2 + q 2 q. + q 2 + q 2 + q 2 so the symlectic condition is satisfied.
Homer Reid s Solutions to Goldstein Problems: Chater 9 7 Problem 9.6 The transformation equations between two sets of coordinates are Q log + q /2 cos P 2 + q /2 cos q /2 sin a Show directly from these transformation equations that Q, P are canonical variables if q and are. b Show that the function that generates this transformation is F 3 e Q 2 tan. a The Jacobian of the transformation is M 2 q /2 cos +q /2 cos q/2 sin +q /2 cos q /2 sin + 2 cos sin 2 q /2 cos +q /2 cos q/2 sin +q /2 cos q /2 sin + sin 2 2q /2 cos + 2q cos 2 2q sin 2 2q /2 cos + 2q cos 2. Hence we have q MJM 2 /2 cos q /2 sin + sin 2 +q /2 cos q/2 sin 2q /2 cos + 2q cos 2 +q /2 cos q /2 sin + sin 2 2q /2 cos + 2q cos 2 2 J q /2 cos +q /2 cos q /2 sin +q /2 cos cos 0 2 +sin 2 +q /2 cos cos 2+q /2 sin sin 2 +q /2 cos cos2 +sin 2 +q /2 cos cos 2+q /2 sin sin 2 0 +q /2 cos 0 0 so the symlectic condition is satisfied.
Homer Reid s Solutions to Goldstein Problems: Chater 9 8 b For an F 3 function the relevant relations are q F/, P F/. We have F 3, Q e Q 2 tan so P F 3 2eQ e Q tan q F 3 eq 2 sec 2. The second of these may be solved to yield Q in terms of q and : Q log + q /2 cos and then we may lug this back into the equation for P to obtain as advertised. P 2q /2 sin + q sin 2 Problem 9.7 a If each of the four tyes of generating functions exist for a given canonical transformation, use the Legendre transformation to derive relations between them. b Find a generating function of the F 4 tye for the identity transformation and of the F 3 tye for the exchange transformation. c For an orthogonal oint transformation of q in a system of n degrees of freedom, show that the new momenta are likewise given by the orthogonal transformation of an n dimensional vector whose comonents are the old momenta lus a gradient in configuration sace. Problem 9.8 Prove directly that the transformation Q q, P 2 2 Q 2 2, P 2 2q q 2 is canonical and find a generating function. After a little hacking I came u with the generating function F 3, Q, q 2, Q 2 2Q 2 Q + q 2 Q 2
Homer Reid s Solutions to Goldstein Problems: Chater 9 9 which is of mixed F 3, F tye. This is Legendre-transformed into a function of the F tye according to The least action rincile then says F q, Q, q 2, Q 2 F 3 + q. q + 2 q 2 Hq i, i P Q + P 2 Q 2 KQ i, P i + F 3 ṗ + F 3 Q whence clearly + F 3 2 q 2 + F 3 2 Q 2 + q + q ṗ q F 3 Q P F 3 2Q 2 2 2 2 F 3 2 Q 2 P 2 F 3 2 2Q q 2 2q q 2. Problem 9.4 By any method you choose show that the following transformation is canonical: x 2P sin Q + P 2, x 2 2P cos Q Q 2 y 2P cos Q + Q 2, y 2 2P sin Q P 2 where is some fixed arameter. Aly this transformation to the roblem of a article of charge q moving in a lane that is erendicular to a constant magnetic field B. Exress the Hamiltonian for this roblem in the Q i, P i coordinates, letting the arameter take the form 2 qb c. From this Hamiltonian obtain the motion of the article as a function of time. We will rove that the transformation is canonical by finding a generating function. Our first ste to this end will be to exress everything as a function
Homer Reid s Solutions to Goldstein Problems: Chater 9 0 of some set of four variables of which two are old variables and two are new. After some hacking, I arrived at the set {x, Q, y, Q 2 }. In terms of this set, the remaining quantities are y 2 x 2 y cot Q + Q 2 9 2 x cot Q 2 Q 2 0 4 x 2 y 2 x 2 P 8 2 x y + 2 2 2 y csc 2 Q P 2 2 x + y 2 We now seek a generating function of the form F x, Q, y, Q 2. This is of mixed tye, but can be related to a generating function of ure F character according to F x, Q, y, Q 2 F x, Q, y, Q 2 y y. Then the rincile of least action leads to the condition x ẋ + y ẏ P Q + P 2 Q 2 + F x ẋ + F y ṗ y + F Q + F 2 Q 2 + yṗ y + y ẏ from which we obtain x F x 3 y F y 4 P F 5 P 2 F 2. 6 Doing the easiest first, comaring 2 and 6 we see that F must have the form F x, Q, y, Q 2 2 xq 2 yq 2 + gx, Q, y. 7 Plugging this in to 4 and comaring with 4 we find gx, Q, y 2 x y + 2 2 2 y cot Q + ψx, Q. 8 Plugging 7 and 8 into 3 and comaring with 0, we see that ψ x 2 4 x cot Q
Homer Reid s Solutions to Goldstein Problems: Chater 9 or ψx, Q 2 x 2 cot Q. 9 8 Finally, combining 9, 8, 7, and 5 and comaring with we see that we may simly take φq 0. The final form of the generating function is then F x, Q, y, Q 2 2 x + 2 x 2 y Q 2 + 8 2 x y + 2 2 2 y cot Q and its existence roves the canonicality of the transformation. Turning now to the solution of the roblem, we take the B field in the z direction, i.e. B B 0ˆk, and ut Then the Hamiltonian is Hx, y, x, y 2m A B 0 2 2m 2m y î + x ĵ. q c A 2 [ x + qb 2 0 2c y + y qb 0 2c x [ 2 2 ] x + 2 2 y + y 2 2 x where we ut 2 qb/c. In terms of the new variables, this is 2 ] HQ, Q 2, P, P 2 [ 2 2P cos Q + ] 2 2P sin Q 2m 2 m P ω c P where ω c qb/mc is the cyclotron frequency. From the Hamiltonian equations of motion alied to this Hamiltonian we see that Q 2, P, and P 2 are all constant, while the equation of motion for Q is Q H ω c Q ω c t + φ for some hase φ. Putting r 2P /, x 0 P 2 /, y 0 Q 2 / we then have x rsin ω c t + φ + x 0, x mω c 2 [r cosω ct + φ y 0 ] y rcos ω c t + φ + y 0, y mω c 2 [r sinω ct + φ + x 0 ] in agreement with the standard solution to the roblem.