7119251-A-Goldstein-Solved-Problems-Classical-Mechanics

7119251-A-Goldstein-Solved-Problems-Classical-Mechanics -...

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Find the one-dimensional problem equivalent to its motion. What is the condition on the particle’s initial velocity to produce circular motion? Find the period of small oscillations about this circular motion. We’ll take the paraboloid to be defined by the equation z = αr 2 . The kinetic and potential energies of the particle are m2 ˙ (r + r2 θ2 + z 2 ) ˙ ˙ 2 m ˙ ˙ ˙ = (r2 + r2 θ2 + 4α2 r2 r2 ) 2 V = mgz = mgαr 2 . T= Hence the Lagrangian is L= m ˙ (1 + 4α2 r2 )r2 + r2 θ2 − mgαr2 . ˙ 2 This is cyclic in θ, so the angular momentum is conserved: ˙ l = mr2 θ = constant. 1 Homer Reid’s Solutions to Goldstein Problems: Chapter 3 2 For r we have the derivatives ∂L ˙ = 4α2 mrr2 + mrθ2 − 2mgαr ˙ ∂r ∂L ˙ = m(1 + 4α2 r2 )r ∂r ˙ d ∂L = 8mα2 rr2 + m(1 + 4α2 r2 )¨. ˙ r dt ∂ r ˙ Hence the equation of motion for r is ˙ 8mα2 rr2 + m(1 + 4α2 r2 )¨ = 4α2 mrr2 + mrθ2 − 2mgαr ˙ r ˙ or ˙ m(1 + 4α2 r2 )¨ + 4mα2 rr2 − mrθ2 + 2mgαr = 0. r ˙ In terms of the constant angular momentum, we may rewrite this as m(1 + 4α2 r2 )¨ + 4mα2 rr2 − r ˙ l2 + 2mgαr = 0. mr3 So this is the differential equation that determines the time evolution of r. If initially r = 0, then we have ˙ m(1 + 4α2 r2 )¨ + − r l2 + 2mgαr = 0. mr3 Evidently, r will then vanish—and hence r will remain 0, giving circular motion— ¨ ˙ if l2 = 2mgαr mr3 or ˙ θ = 2gα. So if this condition is satisfied, the particle will execute circular motion (assuming its initial r velocity was zero). It’s interesting to note that the condition on ˙ θ for circular motion is independent of r. Homer Reid’s Solutions to Goldstein Problems: Chapter 3 3 Problem 3.2 A particle moves in a central force field given by the potential V = −k e−ar , r where k and a are positive constants. Using the method of the equivalent onedimensional potential discuss the nature of the motion, stating the ranges of l and E appropriate to each type of motion. When are circular orbits possible? Find the period of small radial oscillations about the circular motion. The Lagrangian is e−ar m2 ˙ r + r2 θ2 + k ˙ . 2 r As usual the angular momentum is conserved: L= ˙ l = mr2 θ = constant. We have ∂L e−ar ˙ = mrθ2 − k (1 + ar) 2 ∂r r ∂L = mr ˙ ∂r ˙ so the equation of motion for r is e−ar k ˙ r = rθ2 − (1 + ar) 2 ¨ m r l2 k e−ar = 2 3− (1 + ar) 2 . mr m r The condition for circular motion is that this vanish, which yields ˙ θ= e−ar0 /2 k (1 + ar0 ) 3/2 . m r 0 (1) (2) What this means is that that if the particle’s initial θ velocity is equal to the above function of the starting radius r0 , then the second derivative of r will remain zero for all time. (Note that, in contrast to the previous problem, in this case the condition for circular motion does depend on the starting radius.) To find the frequency of small oscillations, let’s suppose the particle is executing a circular orbit with radius r0 (in which case the θ velocity is given by (2)), and suppose we nudge it slightly so that its radius becomes r = r0 + x, where x is small. Then (1) becomes x= ¨ e−a[r0 +x] k e−ar0 k 1 + ar0 − (1 + a[r0 + x]) 2 m r0 m [r0 + x]2 (3) Homer Reid’s Solutions to Goldstein Problems: Chapter 3 4 Since x is small, we may write the second term approximately as k e−ar0 x 2 (1 + ar0 + ax)(1 − ax) 1 − 2 r m r0 0 −ar0 −ar0 k (1 + ar0 ) e ke ≈ (1 + ar0 ) 2 + a − a(1 + ar0 ) − 2 2 m r0 m r0 r0 −ar0 −ar0 k e ke 2 ≈ (1 + ar0 ) 2 − 2a + + a2 r0 x. 2 m r0 m r0 r0 ≈ x The first term here just cancels the first term in (??), so we are left with x= ¨ k e−ar0 2 m r0 2a + 2 + a 2 r0 x r0 The problem is that the RHS here has the wrong sign—this equation is satisfied by an x that grows (or decays) exponentially, rather than oscillates. Somehow I messed up the sign of the RHS, but I can’t find where–can anybody help? Problem 3.3 Two particles move about each other in circular orbits under the influence of gravitational forces, with a period τ . Their motion is suddenly stopped, and they are then released and allowed to fall into each other. Prove that they collide after a √ time τ /4 2. Since we are dealing with gravitational forces, the potential energy between the particles is k U (r) = − r and, after reduction to the equivalent one-body problem, the Lagrangian is L= µ2 k ˙ [r + r2 θ2 ] + ˙ 2 r where µ is the reduced mass. The equation of motion for r is k ˙ µr = µrθ2 − 2 . ¨ r (4) If the particles are to move in circular orbits with radius r0 , (4) must vanish at ˙ r = r0 , which yields a relation between r0 and θ: r0 = = k ˙ µθ 2 kτ 2 4π 2 µ 1/3 1/3 (5) Homer Reid’s Solutions to Goldstein Problems: Chapter 3 5 where we used the fact that the angular velocity in the circular orbit with period ˙ τ is θ = 2π/τ . When the particles are stopped, the angular velocity goes to zero, and the first term in (4) vanishes, leaving only the second term: r=− ¨ k . µr2 (6) This differential equation governs the evolution of the particles after they are stopped. We now want to use this equation to find r as a function of t, which we will then need to invert to find the time required for the particle separation r to go from r0 to 0. The first step is to multiply both sides of (6) by the integrating factor 2r: ˙ 2rr = − ˙¨ or d2 d r =+ ˙ dt dt from which we conclude r2 = ˙ 2k + C. µr (7) 2k µr 2k r ˙ µr2 The constant C is determined from the boundary condition on r. This is simply ˙ that r = 0 when r = r0 , since initially the particles are not moving at all. With ˙ the appropriate choice of C in (7), we have r= ˙ dr = dt = 2k µ 2k µ 1/2 1 1 − r r0 r0 − r . rr0 (8) 1/2 We could now proceed to solve this differential equation for r(t), but since in fact we’re interested in solving for the time difference corresponding to given boundary values of r, it’s easier to invert (8) and solve for t(r): 0 ∆t = r0 0 dt dr dr dt 1/2 dr −1 = r0 dr 0 r0 = µ 2k r r0 r0 − r 1/2 dr Homer Reid’s Solutions to Goldstein Problems: Chapter 3 6 We change variables to u = r/r0 , du = dr/r0 : = µ 2k 1/2 r0 3/2 1 0 u 1−u 1/2 du Next we change variables to u = sin2 x, du = 2 sin x cos x dx : =2 µ 2k 1/2 r0 3/2 0 π/2 sin2 x dx µ = 2k Now plugging in (5), we obtain ∆t = µ 2k τ =√ 42 1/2 3/2 π r0 . 4 1/2 kτ 2 4π 2 µ 1/2 π 4 as advertised. Problem 3.6 (a) Show that if a particle describes a circular orbit under the influence of an attractive central force directed at a point on the circle, then the force varies as the inverse fifth power of the distance. (b) Show that for the orbit described the total energy of the particle is zero. (c) Find the period of the motion. (d) Find x, y, and v as a function of angle around the circle and show that all ˙˙ three quantities are infinite as the particle goes through the center of force. Let’s suppose the center of force is at the origin, and that the particle’s orbit is a circle of radius R centered at (x = R, y = 0) (so that the leftmost point of the particle’s origin is the center of force). The equation describing such an orbit is √ r(θ) = 2R(1 + cos 2θ)1/2 so u(θ) = 1 1 =√ . r(θ) 2R(1 + cos 2θ)1/2 (9) Homer Reid’s Solutions to Goldstein Problems: Chapter 3 7 Differentiating, sin 2θ du =√ dθ 2R(1 + cos 2θ)3/2 1 du sin2 2θ 2 cos 2θ =√ +3 dθ (1 + cos 2θ)5/2 2R (1 + cos 2θ)3/2 1 1 =√ 2 cos 2θ + 2 cos2 2θ + 3 sin2 2θ . 2 2R (1 + cos 2θ)5/2 Adding (9) and (10), 1 d2 u (1 + cos 2θ)2 + 2 cos 2θ + 2 cos2 2θ + 3 sin2 2θ +u= √ 5/2 dθ2 2R(1 + cos 2θ) 1 =√ [4 + 4 cos 2θ] 2R(1 + cos 2θ)5/2 4 =√ 2R(1 + cos 2θ)3/2 = 8R2 u3 . The differential equation for the orbit is d2 u md +u=− 2 V dθ2 l du Plugging in (11), we have 8R2 u3 = − so V so f (r) = − 8l2R2 mr5 (14) 1 u =− 2l2R2 4 u m −→ V (r) = − 2l2 R2 mr4 (13) md V l2 du 1 u 1 u (12) (11) (10) which is the advertised r dependence of the force. (b) The kinetic energy of the particle is T= m2 ˙ [r + r2 θ2 ]. ˙ 2 (15) Homer Reid’s Solutions to Goldstein Problems: Chapter 3 8 We have r= √ 2R(1 + cos 2θ)1/2 r2 = 2R2 (1 + cos 2θ) √ sin 2θ ˙ θ r = 2R ˙ (1 + cos 2θ)1/2 r2 = 2R2 ˙ sin2 2θ ˙2 θ 1 + cos 2θ Plugging into (15), ˙ T = mR2 θ2 ˙ = mR2 θ2 ˙ = 2mR2 θ2 ˙ In terms of l = mr2 θ , this is just 2R2 l2 mr4 But this is just the negative of the potential energy, (13); hence the total particle energy T + V is zero. = (c) Suppose the particle starts out at the furthest point from the center of force on its orbit, i.e the point x = 2R, y = 0, and that it moves counter-clockwise from this point to the origin. The time required to undergo this motion is half the period of the orbit, and the particle’s angle changes from θ = 0 to θ = π/2. Hence we can calculate the period as π /2 sin2 2θ + 1 + cos 2θ 1 + cos 2θ sin2 2θ + 1 + 2 cos 2θ + cos2 2θ 1 + cos θ τ =2 0 π /2 dt dθ dθ dθ ˙ θ =2 0 ˙ Using θ = l/mr2 , we have =2 = m l π /2 r2 (θ) dθ 0 4R2 m π/2 (1 + 2 cos 2θ + cos2 2θ) dθ l 0 4R2 m 3π = · l 4 3πR2 m = . l Homer Reid’s Solutions to Goldstein Problems: Chapter 3 9 Problem 3.8 (a) For circular and parabolic orbits in an attractive 1/r potential having the same angular momentum, show that the perihelion distance of the parabola is one half the radius of the circle. (b) Prove that in the same central force as in part (a) the speed of a particle at √ any point in a parabolic orbit is 2 times the speed in a circular orbit passing through the same point. (a) The equations describing the orbits are 2 l mk r= 1 l2 mk 1 + cos θ (circle) (parabola.) Evidently, the perihelion of the parabola occurs when θ = 0, in which case r = l2 /2mk , or one-half the radius of the circle. (b) For the parabola, we have r= ˙ sin θ (1 + cos θ)2 ˙ sin θ = rθ 1 + cos θ l2 mk ˙ θ (16) so ˙ v 2 = r2 + r2 θ2 ˙ ˙ = r2 θ2 ˙ = r2 θ2 sin2 θ +1 (1 + cos θ)2 sin2 θ + 1 + 2 cos θ + cos2 θ (1 + cos θ)2 1 ˙ = 2r2 θ2 1 + cos θ ˙ 2mkr3 θ2 = 2 l 2k = (17) mr ˙ in terms of the angular momentum l = mr 2 θ2 . On the other hand, for the circle r = 0, so ˙ ˙ v 2 = r2 θ2 = l2 k = m2 r 2 mr (18) Homer Reid’s Solutions to Goldstein Problems: Chapter 3 10 where we used that fact that, since this is a circular orbit, the condition k/r = l2 /mr2 is satisfied. Evidently (17) is twice (18) for the same particle at the √ same point, so the unsquared speed in the parabolic orbit is 2 times that in the circular orbit at the same point. Problem 3.12 At perigee of an elliptic gravitational orbit a particle experiences an impulse S (cf. Exercise 9, Chapter 2) in the radial direction, sending the particle into another elliptic orbit. Determine the new semimajor axis, eccentricity, and orientation of major axis in terms of the old. The orbit equation for elliptical motion is r(θ) = a(1 − 2 ) . 1 + cos(θ − θ0 ) (19) For simplicity we’ll take θ0 = 0 for the initial motion of the particle. Then perigee happens when θ = 0, which is to say the major axis of the orbit is on the x axis. Then at the point at which the impulse is delivered, the particle’s momentum is entirely in the y direction: pi = piˆ. After receiving the impulse S in the radial j (x) direction, the particle’s y momentum is unchanged, but its x momentum is now px = S . Hence the final momentum of the particle is pf = Sˆ+piˆ. Since the i j particle is in the same location before and after the impulse, its potential energy is unchanged, but its kinetic energy is increased due to the added momentum: Ef = E i + S2 . 2m (20) Hence the semimajor axis length shrinks accordingly: af = − k k ai =− = . 2Ef 2Ei + S 2 /m 1 + S 2 /(2mEi ) (21) Next, since the impulse is in the same direction as the particle’s distance from the origin, we have ∆L = r × ∆p = 0, i.e. the impulse does not change the particle’s angular momentum: Lf = Li ≡ L. (22) With (20) and (22), we can compute the change in the particle’s eccentricity: f = = 2Ef L2 mk 2 L2 S 2 2Ei L2 1+ + 2 2. 2 mk mk 1+ (23) Homer Reid’s Solutions to Goldstein Problems: Chapter 3 11 What remains is to compute the constant θ0 in (19) for the particle’s orbit after the collision. To do this we need merely observe that, since the location of the particle is unchanged immediately after the impulse is delivered, expression (19) must evaluate to the same radius at θ = 0 with both the “before” and “after” values of a and : af (1 − 2 ) ai (1 − 2 ) f i = 1+ i 1 + f cos θ0 or cos θ0 = 1 f af (1 − 2 ) f −1 . ai (1 − i ) Problem 3.13 A uniform distribution of dust in the solar system adds to the gravitational attraction of the sun on a planet an additional force F = −mC r where m is the mass of the planet, C is a constant proportional to the gravitational constant and the density of the dust, and r is the radius vector from the sun to the planet (both considered as points). This additional force is very small compared to the direct sun-planet gravitational force. (a) Calculate the period for a circular orbit of radius r0 of the planet in this combined field. (b) Calculate the period of radial oscillations for slight disturbances from this circular orbit. (c) Show that nearly circular orbits can be approximated by a precessing ellipse and find the precession frequency. Is the precession the same or opposite direction to the orbital angular velocity? (a) The equation of motion for r is mr = ¨ l2 + f (r) mr3 l2 k = − 2 − mCr. mr3 r k l2 3 − r 2 − mCr0 mr0 0 (24) For a circular orbit at radius r0 this must vanish: 0= (25) Homer Reid’s Solutions to Goldstein Problems: Chapter 3 12 −→ l = ˙ −→ θ = 4 mkr0 + m2 Cr0 1 l 2 = mr 2 mr0 0 = ≈ k 3 mr0 4 mkr0 + m2 Cr0 3 mCr0 k 1+ 3 mCr0 k 3 1+ mr0 2k Then the period is τ= 2π 3/2 ≈ 2πr0 ˙ θ 2 Cτ0 2 8π 3 mCr0 m 1− k 2k = τ0 1 − 3/2 where τ0 = 2πr0 m/k is the period of circular motion in the absence of the perturbing potential. (b) We return to (24) and put r = r0 + x with x mx = ¨ r0 : k l2 − − mC (r0 + x) m(r0 + x)3 (r0 + x)2 l2 k x x ≈ − 2 1−2 − mCr0 − mCx 1−3 3 mr0 r0 r0 r0 Using (25), this reduces to mx = − ¨ 3l2 2k 4 + r 3 − mC x mr0 0 or x + ω2x = 0 ¨ with ω= = 2k 3l2 − 3 −C 2 r4 m0 mr0 k 2l2 4 − mr 3 m2 r 0 0 1/2 1/2 where in going to the last line we used (25) again. Homer Reid’s Solutions to Goldstein Problems: Chapter 3 13 Problem 3.14 Show that the motion of a particle in the potential field V (r) = − h k +2 r r is the same as that of the motion under the Kepler potential alone when expressed in terms of a coordinate system rotating or precessing around the center of force. For negative total energy show that if the additional potential term is very small compared to the Kepler potential, then the angular speed of precession of the elliptical orbit is 2πmh ˙ Ω= 2 . lτ The perihelion of Mercury is observed to precess (after corrections for known planetary perturbations) at the rate of about 40 of arc per century. Show that this precession could be accounted for classically if the dimensionless quantity η= k ka (which is a measure of the perturbing inverse square potential relative to the gravitational potential) were as small as 7 × 10−8 . (The eccentricity of Mercury’s orbit is 0.206, and its period is 0.24 year). The effective one-dimensional equation of motion is mr = ¨ L2 k 2h − 2+ 3 3 mr r r k L2 + 2mh +2 = mr3 r L2 + 2mh + (mh/L)2 − (mh/L)2 k = +2 mr3 r k [L + (mh/L)]2 − (mh/L)2 +2 = mr3 r If mh write L2 , then we can neglect the term (mh/L)2 in comparison with L2 , and [L + (mh/L)]2 k +2 3 mr r mr = ¨ (26) which is just the normal equation of motion for the Kepler problem, but with the angular momentum L augmented by the additive term ∆L = mh/L. Such an augmentation of the angular momentum may be accounted for by Homer Reid’s Solutions to Goldstein Problems: Chapter 3 14 augmenting the angular velocity: ˙ L = mr2 θ −→ L 1+ mh L2 mh L2 2˙ 2˙ = mr θ + mr Ω ˙ = mr2 θ 1 + where mh 2πmh ˙ Ω= = ˙ L2 τ L2 θ is a precession frequency. If we were to go back and work the problem in the ˙ reference frame in which everything is precessing with angular velocity Ω, but 2 there is no term h/r in the potential, then the equations of motion would come ˙ out the same as in the stationary case, but with a term ∆L = mr 2 Ω added to the effective angular momentum that shows up in the equation of motion for r, just as we found in (26). To put in the numbers, we observe that ˙ Ω= = = so ˙ τΩ h = (1 − e2 ) ka 2π = (1 − e2 )τ fprec where in going to the third-to-last line we used Goldstein’s equation (3-62), and ˙ in the last line I put fprec = Ω/2π . Putting in the numbers, we find h = (1 − .2062 ) · 0.24 yr · 40 ka = 7.1 · 10−8 . 1◦ 3600 1 revolution 360◦ 1 century−1 100 yr−1 yr−1 2π τ 2π τ 2π τ m (h) L2 mka h L2 ka 1 h 1 − e2 ka Homer Reid’s Solutions to Goldstein Problems: Chapter 3 15 Problem 3.22 In hyperbolic motion in a 1/r potential the analogue of the eccentric anomaly is F defined by r = a(e cosh F − 1), where a(1 − e) is the distance of closest approach. Find the analogue to Kepler’s equation giving t from the time of closest approach as a function of F . We start with Goldstein’s equation (3.65): t= m 2 m 2 r r0 r r0 k r dr − l2 2mr 2 +E l2 2m = r dr E r2 + kr − . (27) With the suggested substitution, the thing under the radical in the denominator of the integrand is Er2 + kr − l2 l2 = Ea2 (e2 cosh2 F − 2e cosh F + 1) + ka(e cosh F − 1) − 2m 2m l2 2 22 2 = Ea e cosh F + ae(k − 2Ea) cosh F + E a − ka − 2m It follows from the orbit equation that, if a(e − 1) is the distance of closest approach, then a = k/2E . Thus = k 2 e2 k 2 e2 l2 cosh2 F − − 4E 4E 2m k2 2El2 2 2 = e cosh F − 1 + 4E mk 2 k 2 e2 k 2 e2 = cosh2 F − 1 = sinh2 F = a2 e2 E sinh2 F. 4E 4E Plugging into (27) and observing that dr = ae sinh F dF , we have t= m 2E F F0 a(e cosh F − 1) dF = ma2 [e(sinh F − sinh F0 ) − (F − F0 )] 2E and I suppose this equation could be a jumping-off point for numerical or other investigations of the time of travel in hyperbolic orbit problems. Homer Reid’s Solutions to Goldstein Problems: Chapter 3 16 Problem 3.26 Examine the scattering produced by a repulsive central force f = kr −3 . Show that the differential cross section is given by σ (Θ)dΘ = k (1 − x)dx 2E x2 (2 − x)2 sin πx where x is the ratio Θ/π and E is the energy. The potential energy is U = k/2r 2 = ku2 /2, and the differential equation for the orbit reads d2 u m dU mk +u=− 2 =− 2 u dθ2 l du l or d2 u mk + 1+ 2 u=0 2 dθ l with solution u = A cos γθ + B sin γθ where γ= 1+ mk . l2 (29) (28) We’ll set up our coordinates in the way traditional for scattering experiments: initially the particle is at angle θ = π and a great distance from the force center, and ultimately the particle proceeds off to r = ∞ at some new angle θs . The first of these observations gives us a relation between A and B in the orbit equation (28): u(θ = π ) = 0 −→ −→ A cos γπ + B sin γπ = 0 (30) A = −B tan γπ. The condition that the particle head off to r = ∞ at angle θ = θs yields the condition A cos γθs + B sin γθs = 0. Using (30), this becomes − cos γθs tan γπ + sin γθs = 0 Homer Reid’s Solutions to Goldstein Problems: Chapter 3 17 or − cos γθs sin γπ + sin γθs cos γπ = 0 −→ sin γ (θs − π ) = 0 −→ or, in terms of Goldstein’s variable x = θ/π , γ= Plugging in (29) and squaring both sides, we have 1+ 1 mk = . l2 (x − 1)2 1 . x−1 (31) γ (θs − π ) = π Now l = mv0 s = (2mE )1/2 s with s the impact parameter and E the particle energy. Thus the previous equation is 1+ or s2 = − 1 k = 2Es2 (x − 1)2 k (x − 1)2 . 2E x(x − 2) Taking the differential of both sides, 2s ds = − (x − 1)2 (x − 1)2 k 2(x − 1) −2 − dx 2E x(x − 2) x (x − 2) x(x − 2)2 k 2x(x − 1)(x − 2) − (x − 1)2 (x − 2) − x(x − 1)2 =− 2E x2 (x − 2)2 k 2(1 − x) =− . 2E x2 (x − 2)2 | s ds | . sin θ (32) The differential cross section is given by σ (θ)dΩ = Plugging in (32), we have σ (θ)dΩ = as advertised. k (1 − x) dx 2 (x − 2)2 sin θ 2E x Solutions to Problems in Goldstein, Classical Mechanics, Second Edition Homer Reid April 21, 2002 Chapter 7 Problem 7.2 Obtain the Lorentz transformation in which the velocity is at an infinitesimal angle dθ counterclockwise from the z axis, by means of a similarity transformation applied to Eq. (7-18). Show directly that the resulting matrix is orthogonal and that the inverse matrix is obtained by substituting −v for v . We can obtain this transformation by first applying a pure rotation to rotate the z axis into the boost axis, then applying a pure boost along the (new) z axis, and then applying the inverse of the original rotation to bring the z axis back in line with where it was originally. Symbolically we have L = R−1 KR where R is the rotation to achieve the new z axis, and K is the boost along the z axis. Goldstein tells us that the new z axis is to be rotated dθ counterclockise from the original z axis, but he doesn’t tell us in which plane, i.e. we know θ but not φ for the new z axis in the unrotated coordinates. We’ll assume the z axis is rotated around the x axis, in a sense such that if you’re standing on the positive x axis, looking toward the negative x axis, the rotation appears to be counterclockwise, so that the positive z axis is rotated toward the negative y 1 Homer Reid’s Solutions to Goldstein Problems: Chapter 7 2 axis. Then, using the real metric, 1 10 0 0 1 0 0 0 0 cos dθ 0 0 0 sin dθ 0 0 1 L= 0 − sin dθ cos dθ 0 0 0 γ −βγ 0 0 0 0 −βγ γ 0 0 0 1 1 0 0 0 1 0 0 0 cos dθ sin dθ 0 0 cos dθ − sin dθ = 0 − sin dθ cos dθ 0 0 γ sin dθ γ cos dθ 0 0 0 1 0 −βγ sin dθ −βγ cos dθ 1 0 0 0 0 cos2 dθ + γ sin2 dθ (γ − 1) sin dθ cos dθ −βγ sin dθ = 0 (γ − 1) sin dθ cos dθ sin2 dθ + γ cos2 dθ −βγ cos dθ 0 −βγ sin dθ −βγ cos dθ γ 0 cos dθ sin dθ 0 0 0 −βγ γ . 0 − sin dθ cos dθ 0 0 0 0 1 Problem 7.4 A rocket of length l0 in its rest system is moving with constant speed along the z axis of an inertial system. An observer at the origin observes the apparent length of the rocket at any time by noting the z coordinates that can be seen for the head and tail of the rocket. How does this apparent length vary as the rocket moves from the extreme left of the observer to the extreme right? Let’s imagine a coordinate system in which the rocket is at rest and centered at the origin. Then the world lines of the rocket’s top and bottom are xt = {0, 0, +L0/2, τ } µ xb = {0, 0, −L0 /2, τ } . µ where we are parameterizing the world lines by the proper time τ . Now, the rest frame of the observer is moving in the negative z direction with speed v = βc relative to the rest frame of the rocket. Transforming the world lines of the rocket’s top and bottom to the rest frame of the observer, we have xt = {0, 0, γ (L0 /2 + vτ ), γ (τ + βL0 /2c)} µ (1) (2) xb µ = {0, 0, γ (−L0/2 + vτ ), γ (τ − βL0 /2c)} . Now consider the observer. At any time t in his own reference frame, he is receiving light from two events, namely, the top and bottom of the rocket moving past imaginary distance signposts that we pretend to exist up and down the z axis. He sees the top of the rocket lined up with one distance signpost and the bottom of the rocket lined up with another, and from the difference between the two signposts he computes the length of the rocket. Of course, the light that he sees was emitted by the rocket some time in the past, and, moreover, the Homer Reid’s Solutions to Goldstein Problems: Chapter 7 3 light signals from the top and bottom of the rocket that the observer receives simultaneously at time t were in fact emitted at different proper times τ in the rocket’s rest frame. First consider the light received by the observer at time t0 coming from the bottom of the rocket. Suppose in the observer’s rest frame this light were emitted at time t0 − ∆t, i.e. ∆t seconds before it reaches the observer at the origin; then the rocket bottom was passing through z = −c∆t when it emitted this light. But then the event identified by (z, t) = (−c∆t, t0 − ∆t ) must lie on the world line of the rocket’s bottom, which from (2) determines both ∆t and the proper time τ at which the light was emitted: −c∆t t0 − ∆ t = γ (−L0 /2 + vτ ) = γ (τ + βL0 /2c) 1+β 1−β 1/2 =⇒ τ= t0 − L0 2c ≡ τb (t0 ). We use the notation τb (t0 ) to indicate that this is the proper time at which the bottom of the rocket emits the light that arrives at the observer’s origin at the observer’s time t0 . At this proper time, from (2), the position of the bottom of the rocket in the observer’s reference frame was zb (τb (t0 )) = −γL0 /2 + vγτb (t0 ) = −γL0 /2 + vγ 1+β 1−β 1/2 t0 − L0 2c (3) Similarly, for the top of the rocket we have τt (t0 ) = and zt (τt (t0 )) = γL0 /2 + vγ 1+β 1−β 1/2 1+β 1−β 1/2 t0 + L0 2c t0 + L0 2c (4) Subtracting (3) from (4), we have the length for the rocket computed by the observer from his observations at time t0 in his reference frame: L(t0 ) = γ (1 + β )L0 = 1+β 1−β 1/2 L0 . Homer Reid’s Solutions to Goldstein Problems: Chapter 7 4 Problem 7.17 Two particles with rest masses m1 and m2 are observed to move along the observer’s z axis toward each other with speeds v1 and v2 , respectively. Upon collision they are observed to coalesce into one particle of rest mass m3 moving with speed v3 relative to the observer. Find m3 and v3 in terms of m1 , m2 , v1 , and v2 . Would it be possible for the resultant particle to be a photon, that is m3 = 0, if neither m1 nor m2 are zero? Equating the 3rd and 4th components of the initial and final 4-momentum of the system yields γ 1 m1 v 1 − γ 2 m2 v 2 = γ 3 m3 v 3 γ 1 m1 c + γ 2 m2 c = γ 3 m3 c Solving the second for m3 yields m3 = γ1 γ2 m1 + m2 γ3 γ3 (5) and plugging this into the first yields v3 in terms of the properties of particles 1 and 2: γ 1 m1 v 1 − γ 2 m2 v 2 v3 = γ 1 m1 + γ 2 m2 Then γ 1 m 1 β1 − γ 2 m 2 β2 v3 = c γ 1 m1 + γ 2 m2 2 2 2 2 2 2 γ1 m2 + 2γ1 γ2 m1 m2 + γ2 m2 − [γ1 m2 β1 + γ2 m2 β2 − 2γ1 γ2 m1 m2 β1 β2 ] 1 2 1 2 2 1 − β3 = (γ1 m1 + γ2 m2 )2 22 2 22 2 γ m (1 − β1 ) + γ2 m2 (1 − β2 ) + 2γ1 γ2 m1 m2 (1 − β1 β2 ) =11 (γ1 m1 + γ2 m2 )2 2 2 m + m2 + 2γ1 γ2 m1 m2 (1 − β1 β2 ) =1 (γ1 m1 + γ2 m2 )2 β3 = and hence 2 γ3 = (γ1 m1 + γ2 m2 )2 1 2 = m2 + m2 + 2γ γ m m (1 − β β ) . 1 − β3 12 1 2 12 1 2 (6) Now, (5) shows that, for m3 to be zero when either m1 or m2 is zero, we must have γ3 = ∞. That this condition cannot be met for nonzero m1 , m2 is evident from the denominator of (6), in which all terms are positive (since β1 β2 < 1 if m1 or m2 is nonzero). Homer Reid’s Solutions to Goldstein Problems: Chapter 7 5 Problem 7.19 A meson of mass π comes to rest and disintegrates into a meson of mass µ and a neutrino of zero mass. Show that the kinetic energy of motion of the µ meson (i.e. without the rest mass energy) is (π − µ)2 2 c. 2π Working in the rest frame of the pion, the conservation relations are πc2 = (µ2 c4 + p2 c2 )1/2 + pν c µ 0 = p µ + pν (energy conservation) (momentum conservation). From the second of these it follows that the muon and neutrino must have the same momentum, whose magnitude we’ll call p. Then the energy conservation relation becomes πc2 = (µ2 c4 + p2 c2 )1/2 + pc −→ (πc − p)2 = µ2 c2 + p2 −→ p= π 2 − µ2 c. 2π Then the total energy of the muon is Eµ = (µ2 c4 + p2 c2 )1/2 = c 2 µ2 + = (π 2 − µ2 )2 4π 2 1/2 c2 4π 2 µ2 + (π 2 − µ2 )2 2π c2 2 = (π + µ2 ) 2π 1/2 Then subtracting out the rest energy to get the kinetic energy, we obtain K = Eµ − µc2 = c2 2 (π + µ2 ) − µc2 2π c2 2 = (π + µ2 − 2πµ) 2π c2 = (π − µ)2 2π as advertised. Homer Reid’s Solutions to Goldstein Problems: Chapter 7 6 Problem 7.20 A π + meson of rest mass 139.6 MeV collides with a neutron (rest mass 939.6 MeV) stationary in the laboratory system to produce a K + meson (rest mass 494 MeV) and a Λ hyperon (rest mass 1115 MeV). What is the threshold energy for this reaction in the laboratory system? We’ll put c = 1 for this problem. The four-momenta of the pion and neutron before the collision are pµ,π = (pπ , γπ mπ ), pµ,n = (0, mn ) and the squared magnitude of the initial four-momentum is thus pµ,T pµ = −|pπ |2 + (γπ mπ + mn )2 T = m2 + m2 + 2γπ mπ mn n π 2 = −|pπ |2 + γπ m2 + m2 + 2γπ mπ mn π n = (mπ + mn )2 + 2(γπ − 1)mπ mn (7) The threshold energy is the energy needed to produce the K and Λ particles at rest in the COM system. In this case the squared magnitude of the fourmomentum of the final system is just (mK + mΛ )2 , and, by conservation of momentum, this must be equal to the magnitude of the four-momentum of the initial system (7): (mK + mΛ )2 = (mπ + mn )2 + 2(γπ − 1)mπ mn =⇒ γπ = 1 + (mK + mΛ )2 − (mπ + mn )2 = 6.43 2mπ mn Then the total energy of the pion is T = γπ mπ = (6.43 · 139.6 MeV) = 898 MeV, while its kinetic energy is K = T − m = 758 MeV. The above appears to be the correct solution to this problem. On the other hand, I first tried to do it a different way, as below. This way yields a different and hence presumably incorrect answer, but I can’t figure out why. Can anyone find the mistake? The K and Λ particles must have, between them, the same total momentum in the direction of the original pion’s momentum as the original pion had. Of course, the K and Λ may also have momentum in directions transverse to the original pion momentum (if so, their transverse momenta must be equal and opposite). But any transverse momentum just increases the energy of the final system, which increases the energy the initial system must have had to produce the final system. Hence the minimum energy situation is that in which the K and Λ both travel in the direction of the original pion’s motion. (This is equivalent to Goldstein’s conclusion that, just at threshold, the produced particles are at Homer Reid’s Solutions to Goldstein Problems: Chapter 7 7 rest in the COM system). Then the momentum conservation relation becomes simply pπ = p K + p λ and the energy conservation relation is (with c = 1) (m2 + p2 )1/2 + mn = (m2 + p2 )1/2 + (m2 + p2 )1/2 . π π K K Λ Λ (9) (8) The problem is to find the minimum value of pπ that satisfies (9) subject to the constraint (8). To solve this we must first resolve a subquestion: for a given pπ , what is the relative allocation of momentum to pK and pΛ that minimizes (9) ? Minimizing Ef = (m2 + p2 )1/2 + (m2 + p2 )1/2 . K K Λ Λ subject to pK + pΛ = pπ , we obtain the condition pΛ pK = (m2 + p2 )1/2 (m2 + p2 )1/2 K K Λ Λ Combining this with (8) yields pΛ = mΛ pπ mK + m Λ pK = mK pπ . mK + m Λ (11) =⇒ pK = mK pΛ mΛ (10) For a given total momentum pπ , the minimum possible energy the final system can have is realized when pπ is partitioned between pK and pΛ according to (11). Plugging into (8), the relation defining the threshold momentum is (m2 π + p2 )1/2 π + mn = m2 K + mK mK + m Λ 2 1/2 p2 π + m2 Λ + mΛ mK + m Λ 2 1/2 p2 π Solving numerically yields pπ ≈ 655 MeV/c, for a total pion energy of about 670 MeV. Homer Reid’s Solutions to Goldstein Problems: Chapter 7 8 Problem 7.21 A photon may be described classically as a particle of zero mass possessing nevertheless a momentum h/λ = hν/c, and therefore a kinetic energy hν . If the photon collides with an electron of mass m at rest it will be scattered at some angle θ with a new energy hν . Show that the change in energy is related to the scattering angle by the formula θ λ − λ = 2λc sin2 , 2 where λc = h/mc, known as the Compton wavelength. Show also that the kinetic energy of the recoil motion of the electron is T = hν 2 1+2 λc λ λc λ sin2 sin θ/2 θ 2 2 . Let’s assume the photon is initially travelling along the z axis. Then the sum of the initial photon and electron four-momenta is 0 0 0 0 0 0 . (12) pµ,i = pµ,γ + pµ,e = h/λ + 0 = h/λ mc + h/ mc h/λ Without loss of generality we may assume that the photon and electron move in the xz plane after the scatter. If the photon’s velocity makes an angle θ with the z axis, while the electron’s velocity makes an angle φ, the four-momentum after the collision is pe sin φ (h/λ ) sin θ + pe sin φ (h/λ ) sin θ 0 0 0 = pµ,f = pµ,γ + pµ,e = (h/λ ) cos θ + (h/λ ) cos θ + pe cos φ pe cos φ h/λ m 2 c2 + p 2 (h/λ ) + m2 c2 + p2 e e (13) . Equating (12) and (13) yields three separate equations: (h/λ ) sin θ + pe sin φ = 0 (h/λ ) cos θ + pe cos φ = h/λ h/λ + From the first of these we find h sin θ =⇒ cos φ = 1 + sin φ = − λ pe h λ pe 2 1/2 (14) (15) (16) m 2 c2 + p2 e = mc + h/λ sin θ 2 Homer Reid’s Solutions to Goldstein Problems: Chapter 7 9 and plugging this into (15) we find p2 = e h2 h2 h2 + 2 −2 cos θ. λ2 λ λλ (17) On the other hand, we can solve (16) to obtain p2 = h2 e 1 1 − λλ 2 + 2mch 1 1 − λλ . Comparing these two determinations of pe yields cos θ = 1 − or sin2 mc (λ − λ) h mc 1 θ = (λ − λ) = (λ − λ) 2 2h 2λc so this is advertised result number 1. Next, to find the kinetic energy of the electron after the collision, we can write the conservation of energy equation in a slightly different form: h h = γmc + λ λ 1 1 − =⇒ (γ − 1)mc = K = h λλ λ −λ =h λλ mc + =h = where we put χ = λc /λ. h λ 2λc sin2 (θ/2) λ[λ + 2λc sin2 (θ/2)] 2χ sin2 (θ/2) 1 + 2χ sin2 (θ/2) Problem 7.22 A photon of energy E collides at angle θ with another photon of energy E . Prove that the minimum value of E permitting formation of a pair of particles of mass m is 2m2 c4 . Eth = E (1 − cos θ) We’ll suppose the photon of energy E is traveling along the positive z axis, while that with energy E is traveling in the xz plane (i.e., its velocity has Homer Reid’s Solutions to Goldstein Problems: Chapter 7 10 spherical polar angles θ and φ = 0). Then the 4-momenta are p1 = p2 = 0, 0, EE , cc E E E sin θ, 0, cos θ, c c c E + E cos θ E + E E pt = p 1 + p 2 = sin θ, 0, , c c c It’s convenient to rotate our reference frame to one in which the space portion of the composite four-momentum of the two photons is all along the z direction. In this frame the total four-momentum is pt = 0, 0, 1 c E 2 + E 2 + 2E E cos θ, E+E c . (18) At threshold energy, the two produced particles have the same four-momenta: p3 = p4 = 0, 0, p, (m2c2 + p2 )1/2 (19) and 4-momentum conservation requires that twice (19) add up to (18), which yields two conditions: 2p = 2 m 2 c2 + p 2 = 1 c √ E +E c E 2 + E 2 + 2E E cos θ −→ p 2 c2 = = −→ m2 c4 + p2 c2 1 2 4 (E 1 2 4 (E + E 2 + 2E E cos θ) + E 2 + 2E E ) Subtracting the first of these from the second, we obtain m2 c 4 = or E= as advertised. 2m2 c4 E (1 − cos θ) EE (1 − cos θ) 2 Solutions to Problems in Goldstein, Classical Mechanics, Second Edition Homer Reid October 29, 2002 Chapter 9 Problem 9.1 One of the attempts at combining the two sets of Hamilton’s equations into one tries to take q and p as forming a complex quantity. Show directly from Hamilton’s equations of motion that for a system of one degree of freedom the transformation Q = q + ip, 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 put Q = µ(q + ip), The reverse transformation is q= 1 2 1 1 Q+ P µ ν , p= 1 2i 1 1 Q− P µ ν . P = ν (q − ip). (1) The direct conditions for canonicality, valid in cases (like this one) in which the 1 Homer Reid’s Solutions to Goldstein Problems: Chapter 9 2 transformation equations do not depend on the time explicitly, are ∂Q ∂q ∂Q ∂p ∂P ∂q ∂P ∂p ∂p ∂P ∂q =− ∂P ∂p =− ∂Q ∂q = . ∂Q = (2) When applied to the case at hand, all four of these yield the same condition, namely 1 µ=− . 2iν For µ = ν = 1, which is the case Goldstein gives, these conditions are clearly 1 not satisfied, so (1) is not canonical. But putting µ = 1, ν = − 2i we see that equations (1) are canonical. Homer Reid’s Solutions to Goldstein Problems: Chapter 9 3 Problem 9.2 (a) For a one-dimensional system with the Hamiltonian H= p2 1 − 2, 2 2q pq − Ht. 2 show that there is a constant of the motion D= (b) As a generalization of part (a), for motion in a plane with the Hamiltonian H = |p|n − ar−n , where p is the vector of the momenta conjugate to the Cartesian coordinates, show that there is a constant of the motion D= p·r − Ht. n (c) The transformation Q = λq, p = λP is obviously canonical. However, the same transformation with t time dilatation, Q = λq, p = λP, t = λ2 t, is not. Show that, however, the equations of motion for q and p for the Hamiltonian in part (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 ∂D dD = {D, H } + dT ∂t The Poisson bracket of the second term in D clearly vanishes, so we have 1 {pq, H } − H 2 1 1 = pq, p2 − 4 4 = The first Poisson bracket is pq, p2 = ∂ (pq ) ∂ (p2 ) ∂ (pq ) ∂ (p2 ) − ∂q ∂p ∂p ∂q = (p)(2p) − 0 = 2p2 pq, 1 q2 − H. (3) (4) Homer Reid’s Solutions to Goldstein Problems: Chapter 9 4 Next, pq, 1 q2 ∂ (pq ) ∂ q2 ∂ (pq ) ∂ q2 − ∂q ∂p ∂p ∂q 2 =0− − 3 q q 2 =2 q = 1 1 (5) Plugging (4) and (5) into (3), we obtain dD p2 1 = − 2 −H dt 2 2q = 0. (b) We have H = (p2 + p2 + p2 )n/2 − a(x2 + x2 + x2 )−n/2 1 2 3 1 2 3 so ∂H = anxi (x2 + x2 + x2 )−n/2−1 1 2 3 ∂xi ∂H = 2npi (p2 + p2 + p2 )n/2−1 . 1 2 3 ∂pi Then {p · r, H } = = i i ∂ (p1 x1 + p2 x2 + p3 x3 ) ∂H ∂ (p1 x1 + p2 x2 + p3 x3 ) ∂H − ∂xi ∂pi ∂pi ∂xi np2 (p2 + p2 + p2 )n/2−1 − anx2 (x2 + x2 + x2 )−n/2−1 i 1 2 3 i 1 2 3 (6) = n(p2 + p2 + p2 )n/2 − an(x2 + x2 + x2 )−n/2 1 2 3 1 2 3 so if we define D = p · r/n − Ht, then ∂D dD = {D, H } − dT ∂t ∂D 1 = {p · r, H } − n ∂t Substituting in from (6), = |p|n − ar−n − H = 0. Homer Reid’s Solutions to Goldstein Problems: Chapter 9 5 (c) We put Q(t ) = λq t λ2 , P (t ) = 1 p λ t λ2 . (7) Since q and p are the original canonical coordinates, they satisfy ∂H =p ∂p ∂H 1 p=− ˙ = 3. ∂q q q= ˙ On the other hand, differentiating (7), we have dQ 1 =q ˙ dt λ 1 =p λ = P (t ) dP 1 t = 3p ˙ dt λ λ2 1 1 =3 t λ q λ2 1 =3 Q (t ) which are the same equations of motion as (8). t λ2 t λ2 (8) Problem 9.4 Show directly that the transformation Q = log is canonical. The Jacobian of the transformation is M= = ∂Q ∂q ∂P ∂q ∂Q ∂p ∂P ∂p 1 sin p , p P = q cot p 1 −q cot p cot p −q csc2 p . Homer Reid’s Solutions to Goldstein Problems: Chapter 9 6 Hence ˜ MJM = = = = =J 1 −q cot p cot p −q csc2 p cot p −1 q cot p −q csc2 p 2 01 −1 0 0 csc2 p − cot2 p 2 cot p − csc p 0 cot p −q csc2 p 1 − cot p q 01 −1 0 cot p −1 q cot p −q csc2 p so the symplectic condition is satisfied. Problem 9.5 Show directly for a system of one degree of freedom that the transformation Q = arctan αq , p P= αq 2 2 1+ p2 α2 q 2 is canonical, where α is an arbitrary constant of suitable dimensions. The Jacobian of the transformation is M= = α p 1 1+( αq p ∂Q ∂q ∂P ∂q ∂Q ∂p ∂P ∂p α p 1 1+( αq ) p 2 − αq p2 1 1+( αq ) p p α 2 αq so ˜ MJM = = =J . p α ) 2 αq 2 − 0 αq p2 1 1+( αq p ) p α αq − α p 1 1+( αq p ) 2 + αq p2 1 1+( αq p ) 2 1 −1 0 so the symplectic condition is satisfied. Homer Reid’s Solutions to Goldstein Problems: Chapter 9 7 Problem 9.6 The transformation equations between two sets of coordinates are Q = log(1 + q 1/2 cos p) P = 2(1 + q 1/2 cos p)q 1/2 sin p (a) Show directly from these transformation equations that Q, P are canonical variables if q and p are. (b) Show that the function that generates this transformation is F3 = −(eQ − 1)2 tan p. (a) The Jacobian of the transformation is M= ∂Q ∂q ∂P ∂q ∂Q ∂p ∂P ∂p 1 2 ˜ MJM = Hence we have q −1/2 sin p + 2 cos p sin p 2q 1/2 cos p + 2q cos2 p − 2q sin2 p q −1/2 cos p q 1/2 sin p 1 − 1+q1/2 cos p 2 1+q 1/2 cos p . = q −1/2 sin p + sin 2p 2q 1/2 cos p + 2q cos 2p q −1/2 cos p 1+q 1/2 cos p q 1/2 sin p − 1+q1/2 cos p 1 2 = q −1/2 cos p 1+q 1/2 cos p q sin p − 1+q1/2 cos p 1/2 q −1/2 sin p + sin 2p 2q 1/2 cos p + 2q cos 2p = = =J × q −1/2 sin p + sin 2p 2q 1/2 cos p + 2q cos 2p − 2 1 2 q −1/2 cos p 1+q 1/2 cos p q 1/2 sin p 1+q 1/2 cos p 0 p+sin2 p+q 1/2 cos p cos 2p+q 1/2 sin p sin 2p 1+q 1/2 cos p cos2 p+sin2 p+q 1/2 cos p cos 2p+q 1/2 sin p sin 2p 1+q 1/2 cos p − cos 0 0 1 −1 0 so the symplectic condition is satisfied. Homer Reid’s Solutions to Goldstein Problems: Chapter 9 8 (b) For an F3 function the relevant relations are q = −∂F/∂p, P = −∂F/∂Q. We have F3 (p, Q) = −(eQ − 1)2 tan p so ∂F3 = 2eQ (eQ − 1) tan p ∂Q ∂F3 q=− = (eQ − 1)2 sec2 p. ∂p The second of these may be solved to yield Q in terms of q and p: P =− Q = log(1 + q 1/2 cos p) and then we may plug this back into the equation for P to obtain P = 2q 1/2 sin p + q sin 2p as advertised. Problem 9.7 (a) If each of the four types 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 F4 type for the identity transformation and of the F3 type for the exchange transformation. (c) For an orthogonal point 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 components are the old momenta plus a gradient in configuration space. Problem 9.8 Prove directly that the transformation Q1 = q 1 , Q2 = p 2 , P1 P2 = p1 − 2p2 = −2q1 − q2 is canonical and find a generating function. After a little hacking I came up with the generating function F13 (p1 , Q1 , q2 , Q2 ) = −(p1 − 2Q2 )Q1 + q2 Q2 Homer Reid’s Solutions to Goldstein Problems: Chapter 9 9 which is of mixed F3 , F1 type. This is Legendre-transformed into a function of the F1 type according to F1 (q1 , Q1 , q2 , Q2 ) = F13 + p1 q1 . The least action principle then says ˙ ˙ p1 q1 + p2 q2 − H (qi , pi ) = P1 Q1 + P2 Q2 − K (Qi , Pi ) + ˙ ˙ + whence clearly ∂F13 = Q1 ∂p1 ∂F13 P1 = − = −p1 − 2Q2 ∂Q1 = −p1 − 2p2 ∂F13 = Q2 p2 = ∂q2 ∂F13 P2 = − = −2Q1 − q2 ∂Q2 q1 = − ∂F13 ∂F13 ˙ p1 + ˙ Q1 ∂p1 ∂Q1 ∂F13 ∂F13 ˙ q2 + ˙ Q 2 + p 1 q1 + q 1 p 1 ˙ ˙ ∂q2 ∂Q2 = −2q1 − q2 . Problem 9.14 By any method you choose show that the following transformation is canonical: x= 1 ( 2P1 sin Q1 + P2 ), α 1 y = ( 2P1 cos Q1 + Q2 ), α px py α ( 2P1 cos Q1 − Q2 ) 2 α = − ( 2P1 sin Q1 − P2 ) 2 = where α is some fixed parameter. Apply this transformation to the problem of a particle of charge q moving in a plane that is perpendicular to a constant magnetic field B. Express the Hamiltonian for this problem in the (Qi , Pi ) coordinates, letting the parameter α take the form α2 = qB . c From this Hamiltonian obtain the motion of the particle as a function of time. We will prove that the transformation is canonical by finding a generating function. Our first step to this end will be to express everything as a function Homer Reid’s Solutions to Goldstein Problems: Chapter 9 10 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, Q1 , py , Q2 }. In terms of this set, the remaining quantities are 1 1 1 x − 2 py cot Q1 + Q2 2 α α α α2 1 px = x − py cot Q1 − Q2 4 2 2 22 1 1 αx − xpy + 2 p2 csc2 Q1 P1 = 8 2 2α y α 1 P2 = x + p y 2 α y= (9) (10) (11) (12) We now seek a generating function of the form F (x, Q1 , py , Q2 ). This is of mixed type, but can be related to a generating function of pure F1 character according to F1 (x, Q1 , y, Q2 ) = F (x, Q1 , py , Q2 ) − ypy . Then the principle of least action leads to the condition ∂F ∂F ∂F ˙ ∂F ˙ ˙ ˙ px x + p y y = P 1 Q1 + P 2 Q2 + ˙ ˙ x+ ˙ py + ˙ Q1 + Q2 + y p y + p y y ˙ ˙ ∂x ∂py ∂Q1 ∂Q2 from which we obtain ∂F ∂x ∂F y=− ∂py ∂F P1 = − ∂Q1 ∂F . P2 = − ∂Q2 px = (13) (14) (15) (16) Doing the easiest first, comparing (12) and (16) we see that F must have the form α 1 F (x, Q1 , py , Q2 ) = − xQ2 − py Q2 + g (x, Q1 , py ). (17) 2 α Plugging this in to (14) and comparing with (14) we find g (x, Q1 , py ) = 1 1 − xpy + 2 p2 cot Q1 + ψ (x, Q1 ). 2 2α y (18) Plugging (17) and (18) into (13) and comparing with (10), we see that α2 ∂ψ = x cot Q1 ∂x 4 Homer Reid’s Solutions to Goldstein Problems: Chapter 9 11 α 2 x2 cot Q1 . (19) 8 Finally, combining (19), (18), (17), and (15) and comparing with (11) we see that we may simply take φ(Q1 ) ≡ 0. The final form of the generating function is then ψ (x, Q1 ) = F (x, Q1 , py , Q2 ) = − α 1 x + py Q2 + 2 α α 2 x2 1 1 − xpy + 2 p2 cot Q1 8 2 2α y or and its existence proves the canonicality of the transformation. Turning now to the solution of the problem, we take the B field in the z ˆ direction, i.e. B = B0 k, and put A= Then the Hamiltonian is H (x, y, px , py ) = 2 1 q p− A 2m c qB0 1 px + y = 2m 2c B0 2 − yˆ + xˆ . i j 2 + py − 2 qB0 x 2c 2 2 = 1 2m px + α2 y 2 + py − α2 x 2 where we put α2 = qB/c. In terms of the new variables, this is H (Q1 , Q2 , P1 , P2 ) = 1 α 2m α2 = P1 m = ω c P1 2P1 cos Q1 2 +α 2P1 sin Q1 2 where ωc = qB/mc is the cyclotron frequency. From the Hamiltonian equations of motion applied to this Hamiltonian we see that Q2 , P1 , and P2 are all constant, while the equation of motion for Q1 is ∂H ˙ Q1 = = ωc −→ Q 1 = ωc t + φ ∂P1 √ for some phase φ. Putting r = 2P1 /α, x0 = P2 /α, y0 = Q2 /α we then have x = r(sin ωc t + φ) + x0 , y = r(cos ωc t + φ) + y0 , px py = mωc [r cos(ωc t + φ) − y0 ] 2 mωc = [r sin(ωc t + φ) + x0 ] 2 in agreement with the standard solution to the problem. 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