PHYSICS 601, EXAM 1, SOLUTIONSDisclaimer: These “solutions” were cobbled together in a somewhat helter-skelter fashion, and likely contain errors. Also, detailed steps have been leftout in some places – in favor of mere gestures at what to do. Thanks to variouspeople (most notably Tom Cohen and Ben Dreyfus) for help. -Don PerlisPROBLEM 1(a)ˆL =ˆL(p, ˙p) = ˙piqI+ H(q, p)where implicitly the vector q is a function of p and ˙p (otherwise the definitionofˆL would not make sense, as a function of p and ˙p) and where the piare theconjugate momenta of the qi; and thus we have pi= ∂L/∂ ˙qiand the Hamiltonequations:˙qi= ∂H(q, p)/∂pi−˙pj= ∂H(q, p)/∂qjWe then calculate:∂ˆL/∂ ˙pi= ∂[ ˙pjqj+ H(q, p)]/∂ ˙pi= ∂( ˙pjqj)/∂ ˙pi+ ∂H(q(p, ˙p), p)/∂ ˙pi[summingover j]= (∂ ˙pj/∂ ˙pi)qj+ ˙pj(∂qj/∂ ˙pi) + (∂H/∂qj)(∂qj/ ˙pi) + (∂H/∂pj)(∂pj/∂ ˙pi)= qi+ ˙pj(∂qj/∂ ˙pi) − ˙pj(∂qj/∂ ˙pi) + ˙qj· 0= qiPROBLEM 1(b)(d/dt)[∂ˆL/∂ ˙pi] = (d/dt)qi(from part a above)= ˙qiBut also∂ˆL/∂pi= ∂( ˙pjqj)/∂pi+ ∂H(q, p)/∂pi= (∂ ˙pj/∂pi)qj+ ˙pj∂qj/∂pi+ (∂H/∂qj)(∂qj/∂pi)= (∂pi/∂pi)qi+ ˙pj(∂qj/∂pi) − ˙pj(∂qj/∂pi)= ˙qiSo (d/dt)∂ˆL/∂ ˙pi= ˙qi= ∂ˆL/∂piPROBLEM 1(c)From part b above we see thatˆL satisfies the E-L equations, which meansthat the time-integral of L over a given time interval (i.e., the action) achievesan extreme value (max or min) with the function p.PROBLEM 1(d) We are given L(q, ˙q) = (1/2)m ˙q2− (1/4)aq4, in 1-dim.12 PHYSICS 601, EXAM 1, SOLUTIONSThen p = dL/∂ ˙q andH(q, p) = p ˙q − L = p ˙q − (1/2)m ˙q2+ (1/4)aq4SoˆL(p, ˙p) = ˙pq + H = ˙pq + p ˙q − (1/2)m ˙q2+ (1/4)aq4But p = ∂L/∂ ˙q = m ˙qso ˙q = p/mand(d/dt)∂L/∂ ˙q = ∂L/∂q(d/dt)m ˙q = −aq3m ˙q = −aq3m(p/m) = −aq3q =−[ ˙p/a]1/3Finally,ˆL = −˙p[ ˙p/a]1/3+ p2/m − (1/2)m(p/m)2+ (1/4)a[ ˙p/a]4/3= [−1/a1/3+ 1/(4a1/3)]( ˙p)4/3 + p2/2m= −3/(4a1/3)( ˙p)4/3 + p2/2mPROBLEM 2(a)We have planar motion given byH(r, θ , pr, L) = (pr)2/2m + (L + ar3)2/2mr2where L is not the Lagrangian but rather is a generalized momentum conjugateto θ. Note then that˙θ = ∂H/∂L = (L + ar3)/mr2, so L = mr2˙θ − ar3, whichis the usual angular momentum plus the extra term −ar3.Now H is independent of θ, so θ is cyclic, and thus L is a conserved quantity.More formally:dL/dt = L0= −∂H/∂θ = 0, so L is constant.PROBLEM 2(b)To find circular orbits, let L have a particular value, and suppose r = R isconstant. We will find r in terms of L, satifying the EOMs.Since r is constant,0 = ˙r = ∂H/∂pr= pr/m, so pr= 0.But then also0 = ˙pr= −∂H/∂r = −2(L + ar3)3ar2/2mr2+ 2(L + ar3)2/2mr3= [1/mr3][(L + ar3)2− (L + ar3)3ar3] i.e., (L + ar3)[(L + ar3) − 3ar3] = 0.So either L is negative and L + ar3= 0, giving the solution R = (−L/a)1/3,or L is positive and L = 2ar3, giving R = R(L) = (L/2a)1/3.Thus with r = R(L) as above, pr= 0, and θ constant, the EOM’s aresatisfied, so circular orbits exist, with R determined by L.PROBLEM 2(c)We have the initial data x(0) = x0, y(0) = 0, ~v(0) = (v0, 0).Note that although this means that the angular momentum is initially zero,L 6= 0 since L is not the angular momentum; it includes the term −ar3inaddition to the angular momentum. Initially, then, L = mr2˙θ − ar3= −ax30,and since L is conserved, then L = −ax30at all times.Now pr= m ˙r, and so E = H = (m ˙r)2/2m + (L + ar3)2/2mr2PHYSICS 601, EXAM 1, SOLUTIONS 3and(1/2)m ˙r2= E − (L + ar3)2/2mr2so˙r = [2E/m − (L + ar3)2/m2r2]1/2ordr/[2E/m − (L + ar3)2/m2r2]1/2= dtand thenT =RT0dt =RRx0dr/[2E/m − (L + ar3)2/m2r2]1/2where R = R(L) as in part b, and T is the time to reach r = R from r = x0.We now can eliminate E and L in the integral, first usingE = E(0) = (1/2)m(v0)2+ (L + a(x0)3)2/2m(x0)2to getT =RRx0dr/[2(1/2)m(v0)2+ (L + a(x0)3)2/2m(x0)2/m −(L + ar3)2/m2r2]1/2=RRx0dr/[(v0)2+ (L + a(x0)3)2/m2(x0)2− (L + ar3)2/m2r2]1/2and then using L = −ax30to getT =Rx0Rdr/[(v0)2]1/2PROBLEM 2(d) Given: a magnetic field~B = ˆzB(r) = ∇×~A, and charge qon the particle.The associated Hamiltonian isH = (~p − q~A)2/2m = p2/2m − q(~p ·~A)/m + q2A2/2m= (pr)2/2m + (pt)2/2m − qprAr/m − qptAt/m + q2(Ar)2/2m + q2(At)2/2mBut L = rmvt, sopt= L/r= (pr)2/2m + L2/2mr2− qprAr/m −qLAt/mr + q2(Ar)2/2m + q2(At)2/2m= (pr)2/2m + (L − qAtr)2/2mr2− qprAr/m + q2(Ar)2/2m[radial and tangential components; note pt= mvt,But we knowH = (pr)2/2m + (L + ar3)2/2mr2so the two expressions will match if Ar= 0 and −qAtr = ar3. We then setAt= −ar2/q and Ar= 0. This means that the vector potential~A is in thetangential direction, ˆz × ˆr =ˆφ. In fact, we can set~A =~A(x, y) with Az= 0;then~A =~A(r) = −(ar2/q)ˆφ = −(ar2/q)(−y, x, 0)/r = −(ar/q)(−y, x, 0),where of course r = |~r| = |(x, y, 0)| =p(x2+ y2).Now we can calculate~B = ∇ ×~A = ∇ × −ar/q(−y, x, 0)= −a/q∇ × r(−y, x, 0) = −a/q(3r) = −3ar/qˆz.PROBLEM 2(e)Separability gives ∂S/∂t = −H(q1, .., qn, ∂S/∂q1, ..., qn).In this case we haveS(r, θ, E, L) = W1(r, E, L) + W2(θ, E, L) − Etwith q1= r, q2= θ, p1= pr= ∂S/∂r, and p2= L = ∂S/∂θ. So4 PHYSICS 601, EXAM 1, SOLUTIONS∂S/∂r = ∂W1/∂r = pr=p2m(E − V ) =p2mE − (L + ar3)2/r2PROBLEM 3(a)Given: ∂ρ/∂t = −∂ρ/∂ξi˙ξi− ρ∂˙ξi/∂ξiTo show: ∂ρ/∂t = −{ρ, H}, i.e., show{ρ, H} = ∂ρ/∂ξi˙ξi+ ρ∂˙ξi/∂ξi, where i ranges from 1 to 2n.We use Hamilton’s equations in symplectic form,˙ξi= ωij∂H/∂ξj. Then{ρ, H} = (∂ρ/∂ξi)ωij(∂H/∂ξj) = ∂ρ/∂ξi˙ξiBut then ∂ρ/∂t = −(∂ρ/∂ξi)˙ξi− ρ∂˙ξi/∂ξi= {ρ, H} − ρ∂˙ξi/∂ξiThus we must show ρ∂˙ξi/∂ξi= 0. But again using Hamilton’s equations,ρ∂˙ξi/∂ξi= ρ∂(ωij∂H/∂ξj)∂ξi= ρ(∂2H/∂ξi∂ξj)ω(ij)But for each pair i, j with i/neqj, along with the expression (∂2H/∂ξi∂ξj)ωijin the RHS above also appears the expression(∂2H/∂ξj∂ξi)ωjiand since ωij= −ωjithen all these terms cancel. Moreover,when i = j ωij= 0. So the RHS is simply zero, which is what we needed toshow.PROBLEM 3(b)We check that ρ0(q, p) satisfies the equation −∂ρ/∂t = {ρ, H}, where ξ1= qand ξ2= p.{ρ0, H} = (∂ρ0/∂q)(∂H/∂p) − (∂ρ0∂p)(∂H/∂q)= −λ2m2ω2qρ0p/m + λ2pρ0mω2q = 0 = −∂ρ/∂tBut since ∂ρ0/∂t = 0 then∂ρ0/∂t = 0 =