PHY 5246: Theoretical Dynamics, Fall Assignment # 7, Solutions. Θ = π 2ψ, (1)

Transcription

1 PHY 546: Theoretical Dynamics, Fall 05 Assignment # 7, Solutions Graded Problems Problem ψ ψ ψ Θ b (.a) The scattering angle satisfies the relation Θ π ψ, () where ψ is the angle between the direction of the incoming asymptote and the periapsis (the direction of closest approach), and can be obtained from the equation of the orbit θ r dr r 0 (r ) +θ Em mv 0, () l l (r ) setting θ 0 π for r 0 (the incoming direction), such that θ π θ for r r min, i.e. ψ r min The equation of the hyperbolic orbit is then r Em l dr mv l r. (3) r m m (+ǫcos(θ π)) ( ǫcosθ), (4) l l

2 where ǫ +El /(m ) > E > 0 (eccentricity of the hyperbolic orbit) and ψ is defined by the limit In terms of the scattering angle the previous condition becomes which implies and finally for GMm and γ v 0/(GM). (.b) r cosψ ǫ. (5) ( π cos Θ ) sin Θ ǫ, (6) tan Θ ǫ cot Θ ǫ El m m v 4 0 b (7) cot Θ mv 0 b v 0 b GM b γ cot Θ, (8) Using the notation of your boo, Using the result in (.a), we can write it as dω b sinθ db dθ. (9) dω b sinθ γ sin Θ γ cot Θ sin Θ cos Θ γ sin Θ 4γ sin 4 Θ. (0) (.c) For Θ 80, i.e. for the case of bacward scattering, the differential scattering cross section is dω 4γ v 4 0, () and dω for v 0 0, () dω 0 for v 0. Indeed, for v 0 0 (i.e. if particles approach with very low speed) we have that E 0, i.e. the orbit degenerates into a parabolic orbit and all particles, after having approached the center of force, move bac from where they approached from. Onthe other hand, if v 0 (i.e. if particles approach with very high speed), their inetic energy is very large, the total energy of the orbit is very mildly affected by the center of force potential energy, and the particles orbits are very mildly deflected, such that almost no particles are deflected bacward.

3 (.d) For Θ 0, i.e. for the case of forward scattering, we see that dω, (3) i.e. the cross section for forward scattering seems to be infinite. This is due to the fact that all impact parameters can contribute to the cross section, up to infinity. Of course, the larger the impact parameter of a given trajectory, the milder the deviation of of trajectory from the initial direction. All particles coming in with very large impact parameter are scattered in the forward direction, and, if all impact parameters contribute, the cross section for forward scattering is infinite. The only way to prevent such an unphysical situation is to cut off the impact parameter. Is this a tric? Not quite. Indeed in nature all scattering problems have this property. Objects scattering with large impact parameter do not feel the center of force because this is screened by other interactions (thin for instance to the case of Rutherford scattering and the effect of electrons in screening the nuclei of atoms if the incoming particles are at a distance larger than the atomic distance), and that prevents any forward scattering physical cross section from being infinite. (.e) For Coulomb interaction the only difference is that in all previous formulas can be replaced by Kq q (for K a given constant, in this case proportional to the electron charge square) and we get γ e mv 0 /(Kq q ) E /(Kq q ), such that dω K q q 6E sin 4 Θ. (4) Problem (Goldstein 3.3) Calculate the potential due to a force f(r) r 3. r V(r) dr ( r ) 3 r r r. Using equation (3.96) in Goldstein we can calculate the deflection angle with impact parameter s as sdr Θ(s) π r m Notice that from conservation of energy: π r m r r ( V(r) E ) s sdr r r ( ). E +s E E( ) mv 0 E(r m) + rm mr m θ

4 r m r m + Therefore this integral is l + mr m m rm 4 s me mrm 4 E rm r m E +s. ( ) E +s ( ) rm E +s dr/r Θ(s) π s r m r rm dr/r π s r m ( ) r mr π s [ ( )] rm arccos π s π r m r r m r m ( π s ) s π. (5) r m E +s Now in terms of x Θ/π the above becomes s x ( x) +s E +s s ( x) ( E +s E ) s s [ ( x) ] E ( x). Therefore we can solve for the impact parameter x s E ( x). The differential cross section can be expressed in terms of the scattering angle by eq (3.93) in Goldstein, σ(θ)dθ s ds dθ. (6) sinθ dθ From (5) we can calculate: and inverting this we find dθ ds E +s s s E π +s E +s E π +s s π ( ) 3/ ( ) E 3/, E +s E +s ( ) 3/ ds dθ E π E +s.

5 Plugging this into (6) we get the desired result: Problem 3 σ(θ)dθ s sin(πx) E E ( ) 3/ E π E +s dθ x E ( x) x ( ( x) ) ( ) 3/ [ ] 3/ + ( x) dx E ( x) sin(πx) dx sin(πx) E x x ( x) dx sin(πx). Starting from the relation between the velocity of the incident particle after scattering in the laboratory frame (v ) and in the center-of-mass frame (v ), where V is the velocity of the center of mass, we can derive that v V+v, (7) v cosθ v cosθ+v cosθ v cosθ+v v, (8) as one obtains by projecting Eq. (7) along the direction of the approaching incident particle, and v (v ) +V +v V cosθ cosθ v (v ) V v V, (9) as obtained by squaring Eq. (7). Substituting Eq. (9) into Eq. (8) we can write that cosθ v (v ) +V v V. (0) Furthermore, conservation of momentum (and assuming that the second particle is initially at rest) tells us that m (m +m )V m v 0 V v 0, () m +m while, using the inematic of a two-body system in the center-of-mass frame, we can write that r m m +m r v ṙ m m +m v, () where v ṙ is the relative velocity after the collision. Finally, we can trade velocities for energies by using that: the energy of the incident particle before scattering in the laboratory frame (E 0 ) is E 0 m v 0 v 0 E0 m, (3)

6 the energy of the incoming particle after scattering in the laboratory frame (E ) is from conservation of energy we have that E m v v E m, (4) E i Q E 0 Q E f E CM + µv, (5) where E CM (m +m )V, µ m m (m +m is the reduced mass of the system, and v ṙ is ) the relative velocity after the collision. Using simple manipulations Eq. (5) gives Substituting Eqs. ()-(6) into Eq. (0) we get, cosθ m +m m Non-graded Problems Problem 5 (Goldstein 3.3) The potential has the form µv 0 Q µv v v 0 µ Q. (6) E + m m E0 + E 0 m E { V 0 r > a V 0 r a m Q m E0 E. (7) First consider what happens on the interface with the spherical surface that separates the region with V 0 from the region with V V 0. Given the geometry of the problem, it seems obvious to use spherical coordinates. However, since the problem has azimuthal symmetry, we can use polar coordinates. Notice that the force acting on the incoming particle is all in the ê r direction (given V(r)). Thus the linear momentum in the ê θ direction is conserved. We can then write v v cosθ ê r +v sinθ ê θ v v cosθ ê r +v sinθ ê θ v sinθ v sinθ or v sinθ sinθ v. We can also use conservation of energy for the incoming particle, E mv mv V 0 mv msin θ sin θ v V 0 sin θ sin + V 0 + V 0 θ mv E.

7 ê θ ê R v θ θ θ θ s v θ θ θ a θ θ Thus this has exactly the form of Snell s law! The refractive index is n sinθ sinθ Now refer to figure. The scattering angle is and the impact parameter s satisfies Θ (θ θ ) s asinθ. E +V0 E. (8) We need to derive s s(θ) and insert it into the standard formula for the differential cross-section σ(θ) s ds sinθ dθ. (9) Using (8) we have n sinθ sinθ Inverting this and using some trig. identities we find Now since sinθ s/a we can write sinθ sin(θ Θ ). n sinθ cos Θ cosθ sin Θ sinθ cos Θ cotθ sin Θ. (30) cotθ cosθ sinθ ( s a s/a ) (a ). s

8 Now inserting this relationship into (30), we can solve for s: now we calculate the quantity n cos Θ ( s a (a s ) sin Θ, (a ) cos Θ n s sin Θ, ( ( a cos Θ + s) ) n sin Θ, ) sin Θ s sin Θ + ( cos Θ n a sin Θ + n n cos Θ s ds dθ a n { ( ) sin Θ n + ncos Θ cos Θ (n +) nsin Θ a n sin Θ ( ( ) ncos Θ )( n + ncos Θ n cos Θ ). ). a n sin Θ n + ncos Θ. ( cos Θ Θ ) } +sin Now plugging this last expression into (9) we find the required result, s a n sin Θ ( σ(θ) sin Θ cos Θ ( ) s ncos Θ )( n + ncos Θ n cos Θ ) ( ) a n ncos Θ )( n cos Θ 4cos Θ (. n + ncos ) Θ