Phys 7221, Fall 2006: Homework # 6


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1 Phys 7221, Fll 2006: Homework # 6 Gbriel González October 29, 2006 Problem 37 In the lbortory system, the scttering ngle of the incident prticle is ϑ, nd tht of the initilly sttionry trget prticle, which recoils, is φ; see Fig. 3.24, or Fig.1 below: wht re the differences in the figures? Figure 1: A mss M 1 scttered by n initilly sttionry trget M 2, in the lbortory system, from In the center of mss system, the scttering ngle of the incident prticle is Θ, nd tht of the trget prticle (initilly sttionry) is Φ = π Θ: see Fig 3.25, or Fig.2. Figure 2: A mss M 1 scttered by n initilly sttionry trget M 2, in the center of mss system, from 1
2 If V is the velocity of the center of mss, v 2 is the velocity of the trget prticle in the lbortory system, nd v 2 is the velocity of the trget prticle in the center of mss system, then v 2 = V + v 2, represented in Fig.3. Figure 3: Grphicl vector sum v 2 = v 2 + V. The components of the vector sum re equtions relting the mgnitudes of the vectors nd the ngles φ, Θ, similr to the equitons (3.106) for m 1 : v 2 sin φ = v 2 sin Θ v 2 cos φ = V v 2 cos Θ From these equtions, we obtin n eqution similr to (3.107): sin Θ tn φ = cos Θ (V/v 2 ) (1) The mgnitude of v 2 is relted to the mgnitude of the reltive velocity v: v 2 = µv/m 2. The mgnitude of the velocity of the center of mss V is relted to the speed of m 1 before the collision: V = µv 0 /m 2 (Eq ); the rtio is then simply V/v 2 = v 0/v. For inelstic scttering, this rtio is determined by the Q vlue of the inelstic collision (see 3.113), but for elstic collisions this rtio is unity. For elstic collisions, we see tht since V = v 2, the tringle in Fig.3 is isosceles, nd then φ = (π Θ)/2. We cn lso obtin this result from Eq.1, using some trig mgic: tn φ = sin Θ cos Θ 1 = 2 sin Θ/2 cos Θ/2 2 sin 2 Θ/2 = 1 = tn (π/2 Θ/2) tn Θ/2 Problem 330: Rutherford scttering for n ttrctive force We consider n ttrctive grvittionl force of the form F = ke r /r 2, nd follow the derivtion of the Rutherford formul in (3.98)(3.102). For prticle pproching n initilly sttionry trget with impct prmeter s, with velocity v 0 t infinity, the ngulr momentum is l = mv 0 s. The initil energy, if the trget is sttionry, is E = mv 2 0 /2, so l = s 2mE. When the energy E is greter thn zero, the orbit is hyperbol given by 1/r = (µk/l 2 )(1 e cos θ), with e 2 = 1 + 2El 2 /µk 2 = 1 + (m/µ)(2es/k) 2 > 1. 2
3 We hve chosen θ = π, like in the textbook, so θ = 0 corresponds to peripsis. Notice tht the eccentricity is the sme s in repulsive scttering, since it depends on k 2, but the orbit eqution is different, since it is proportionl to k. Since e > 1, the ngulr vlues re restricted so tht cos θ < 1/e nd the rdil coordinte is positive. The minimum rdil distnce will be when θ = 0, nd r = 1/(1+e); the symptotes with r = re t cos Ψ = 1/e (since Ψ > π/2), s shown in Fig.. For lrge energies nd thus lrge eccentricities, the symptotes re close to π/2 (the orbit is lmost stright); for smll energies, nd eccentricities close to unity, the symptotes re close to π (the orbit hs lmost 90deg ngle). The scttering ngle Θ is relted to Ψ s Θ = 2Ψ π (similr, but different thn 3.94), nd then 1/e = cos Ψ = cos((θ + π)/2) = sin(θ/2), nd cot 2 (Θ/2) = e 2 1 = (m/µ)(2es/k) 2, which results in s = k µ 2E m cot Θ 2, of similr form thn 3.101, nd therefore the scttering cross section is Problem 332 σ(θ) = 1 4 ( ) k 2 µ Θ 2E m csc4 2 We consider potentil equl to zero for r >, nd equl to negtive constnt V = V 0 for r. If the potentil energy V is constnt, the force F = V is zero, nd the motion is stright line with constnt velocity. If the potentil hs different constnt vlues in different regions, the prticle is refrcted cross the boundry ( sphere in this cse), trveling in stright line in ll regions, but chnging slope (velocity direction) nd speed s the prticle goes in nd out of the sphere. Since V/ r > 0 cross the sphere, the chnge in potentil represents n ttrctive centrl force, cting only cross the sphere boundry. 3
4 If the prticle pproches the sphere with horizontl velocity of mgnitude v 0, the incident ngle (ngle between velocity nd the norml to the sphere) is α, such tht sin α = s/ (see figure). As it gets refrcted, the prticle is in the sphere with velocity v 1, with the ngle between the velocity nd the norml to the sphere β. The ngulr momentum L = r p = l ˆk is conserved. The mgnitude of the ngulr momentum before the prticle enters the sphere is l = mv 0 sin α, nd the ngulr momentum just fter it enters the sphere is l = mv 1 sin β. Since the ngulr momentum is conserved, we obtin Snell s lw : sin α = v 1 v 0 sin β = n sin β, with n index of refrction n = v 1 /v 0. Energy E = T + V is conserved too. The energy before the prticle enters the sphere is E = mv0 2/2, nd fter it enters the sphere, is E = mv2 1 /2 V 0, so n = v 1 v 0 = E + V0 E. Since n > 1, β < α: the prticle gets refrcted down, like expected from n ttrctive force. When the prticle goes out of the sphere, it gets refrcted gin, exiting the sphere with velocity v 2 = v 0, t direction with ngle α with respect to the norml to the sphere, t the exiting point. The scttering ngle is then Θ = 2(α β) In order to get the scttering cross section, we need the function s(θ), where s is the impct prmeter. We use sin α = s/ nd sin β = sin α/n = s/n to get wht we need: s = sin α = sin(θ/2 + β) 4
5 = sin Θ/2 cos β + cos Θ/2 sin β = sin Θ/2 cos β + s cos Θ/2 n s (n cos Θ/2) = n sin Θ/2 cos β ( s ) 2 (n cos Θ/2) 2 = n 2 sin 2 Θ/2 cos 2 β ( s ) 2 (n cos Θ/2) 2 = n 2 sin 2 Θ/2(1 s2 n 2 2 ) ( s ) 2 ((n cos Θ/2) 2 + sin 2 Θ/2 ) = n 2 sin 2 Θ/2 s 2 = s = 2 n 2 sin 2 Θ/2 1 + n 2 2n cos Θ/2 n sin Θ/2 1 + n 2 2n cos Θ/2 This formul gives the impct prmeter s function of the scttering ngle Θ nd the index n, itself function of the energy E nd the potentil prmeter V 0. The function s(θ) is ntisymmetric on Θ: prticle with negtive impct prmeter s will sctter downwrds with negtive ngle Θ, equl to minus the upwrds sctter ngle for prticle pproching with positive impct prmeter s nd the sme energy. At s = 0, Θ = 0 nd the prticle is not refrcted t ll, since it enters the sphere in norml direction. The function s(θ) hs mximum s = t cos Θ mx /2 = 1/n, n sin Θ mx /2 = n 2 1. So, there is mximum ngle Θ mx = 2 cos 1 (1/n), which is the sctter ngle of prticles grzing the sphericl region (prticles wills sctter down from the top of the sphere, nd up from the bottom of the sphere). As the energy of the pproching prticle increses, the index of refrction n = V 0 /E + 1 1, nd Θ mx = 2 cos 1 (1/n) 0: the prticles get only slightly deflected. For smll incident energies, n = V 0 /E + 1 is lrge, nd Θ mx = 2 cos 1 (1/n) π: the prticle is refrcted by β = π/2 cross the first boundry, nd then goes bck in the direction it pproched. From the function s(θ), we obtin the scttering cross section (not pretty formul!): σ(θ) = ds dθ = 1 Θ/2 n)(1 n cos Θ/2) n(cos 2 (1 + n 2 2n cos Θ/2) 3/2 s ds sin Θ dθ = 2 n 2 4 cos Θ/2 (cos Θ/2 n)(1 n cos Θ/2) (1 + n 2 2n cos Θ/2) 2 The totl cross section is Θmx σ T = σ(ω)dω = 2π σ(θ) sin ΘdΘ = 4π 0 2π s ds dθ dθ = π(s2 (Θ mx ) s 2 (0)) = 2 n 2 sin 2 Θ mx /2 π 1 + n 2 2n cos Θ mx /2 = π2 5
6 which is of course the cross section re of the incoming bem of prticles incident on the sphere, the only ones tht re scttered. Problem 335 Consider potentil of the form V (r) = k/r k/ if r, nd V (r) = 0 if r, truncted Coulomb potentil. This is centrl potentil so energy nd momentum re conserved. Since the potentil is continuous cross the sphericl boundry, there is no refrction s there ws in Problem Outside the sphere, the potentil is zero, so the prticle moves in stright line. Inside the sphere, the orbit is n hyperbol, s shown in the figure: r(θ) = l2 1 mk e cos θ 1 = 1 u(θ) (2) where we hve chosen the origin for ngle thet t the closest point to the center, where r = r min = (l 2 /mk)1/(e 1). 6
7 When the prticle enters the sphere, r =, θ = φ, with the ngle φ shown in the figure. When the prticle exits the sphere, r = nd θ = φ. The scttering ngle is Θ = π 2(α + φ), or sin Θ/2 = cos(α + φ). We need then to find expressions for φ nd α relting those ngles to the scttering prmeter s nd the energy E. From the drwing, we see tht sin α = s/. The ngulr momentum is l 2 = m 2 v 2 0 s2 = 2mEs 2. From the orbit eqution, evluted t θ = ±φ where r =, we hve 1 = mk (e cos φ 1) = k l2 (e cos φ 1) 2Es2 The x nd y coordintes of the trjectory inside the sphere re given by x = r cos( θ + φ + α + Θ) = r sin(θ Θ/2), y = r sin( θ + φ + α + Θ) = r cos(θ Θ/2). When the prticle enters the sphere, r =, θ = φ, nd the velocity is horizontl, so ẏ = (dy/dθ) θ = 0, providing nother reltionship between φ nd α: dy = dr cos(θ Θ/2) r sin(θ Θ/2) dθ dθ ( ) e sin θ = r cos(θ Θ/2) sin(θ Θ/2) e cos θ 1 dy e sin φ dθ = 0 = cos( φ Θ/2) sin( φ Θ/2) θ= φ e cos φ 1 tn(φ + Θ/2) = e sin φ e cos φ 1 tn(π/2 + α) = e sin φ e cos φ 1 tn α = e cos φ 1 e sin φ e sin φ sin α = e cos φ cos α cos α cos α = e cos(φ + α) = e cos(π/2 Θ/2) cos α = e sin Θ/2 This provides the desired reltionship between s nd Θ, since cos α = 1 (s/) 2, nd the eccentricity e is relted to the energy E = T + V nd ngulr momentum l = mr 2 θ: E = 1 2 mṙ mr2 θ2 + V (r) = 1 ( ) dr 2 2 m θ2 + 1 l 2 dθ 2 mr 2 + k r k = 1 l 2 ( ) dr 2 2 mr l 2 dθ 2 mr 2 + k r k = 1 l 2 ( ) du 2 + l2 2 m dθ 2m u2 + ku k 7
8 = 1 l 2 ( ) mk 2 ( ) 2 m l 2 e 2 sin 2 θ + l2 mk 2 2m l 2 (e cos θ 1) 2 + mk2 l 2 = 1 mk 2 2 l 2 (e2 1) k e 2 2(E + k/)l2 = 1 + mk 2 = 1 + λ(s/) 2 with λ = 4E/k(1 + E/k). We finlly hve then formul for s(θ): cos 2 α = e 2 sin 2 Θ/2 (1 + λ s2 2 1 s2 2 = ) sin 2 Θ/2 s 2 = 2 1 sin2 Θ/2 1 + λ sin 2 Θ/2 = 2 cos 2 Θ/2 1 + λ sin 2 Θ/2 (e cos θ 1) k The function s(θ) is monotoniclly decresing, from s = t Θ = 0, to s = 0 t Θ = π. A prticle pproching with impct prmeter s = (1 ɛ), grzing the sphere, will be only slightly scttered up, with scttering ngle sin Θ/2 ɛ/λ. A prticle pproching with no ngulr momentum (s = 0), will hve scttering ngle Θ = π: the prticle is bounced bck ( bck scttered ) from the sphere. Prticles pproching with very smll ngulr momentum will lso be bckscttered. The minimum impct prmeter for which there will be forwrd scttering is given by cos π/4 s 0 = s(θ = π/2) = 1 + λ sin 2 π/4 = / λ/2 If the prticle s initil kinetic energy E is very lrge compred with the mximum potentil energy k/: E k/, then λ 1 nd most prticles will hve forwrd scttering. If the energy is very low, λ 1, nd the prticles with n impct prmeter s s 0 / 2 will be forwrds scttered: hlf the cross section re of the incident bem will be bck scttered, nd hlf will be forwrd scttered. The scttering cross section is σ(θ) = = 2 4 s sin Θ ds dθ 1 + λ (1 + λ sin 2 Θ/2) 2 8
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