Sect. 3.10: Central Force Field Scattering

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Sect. 3.10: Central Force Field Scattering. Application of Central Forces outside of astronomy: Scattering of particles . Atomic scale scattering: Need QM of course! Description of scattering processes: Independent of CM or QM. 1 body formulation = Scattering of particles by a - PowerPoint PPT Presentation

Transcript of Sect. 3.10: Central Force Field Scattering

  • Sect. 3.10: Central Force Field ScatteringApplication of Central Forces outside of astronomy: Scattering of particles.Atomic scale scattering: Need QM of course! Description of scattering processes: Independent of CM or QM. 1 body formulation = Scattering of particles by a Center of Force.Original 2 body problem = Scattering of particle with the reduced mass from a center of forceHere, we go right away to the 1 body formulation, while bearing in mind that it really came from the 2 body problem (beginning of chapter!).

  • Consider a uniform beam of particles (of any kind) of equal mass and energy incident on a center of force (Central force f(r)).Assume that f(r) falls off to zero at large r. Incident beam is characterized by an intensity (flux density) I # particles crossing a unit area ( beam) per unit time (= # particles per m3 per s)As a particle approaches the center of force, it is either attracted or repelled & thus its orbit will be changed (deviate from the initial straight line path). Direction of final motion is not the same as incident motion. Particle is Scattered

  • For repulsive scattering (what we mainly look at here) the situation is as shown in the figure:

    Define: Cross Section for Scattering in a given direction (into a given solid angle d): ()d (Ns/I). With I = incident intensity Ns = # particles/time scattered into solid angle d

  • Scattering Cross Section: ()d (Ns/I) I = incident intensity Ns = # particles/time scattered into angle dIn general, the solid angle depends on the spherical angles , . However, for central forces, there must be symmetry about the axis of the incident beam () ( ()) is independent of azimuthal angle d 2 sind , ()d 2 sind, Angle between incident & scattered beams, as in the figure. cross section. It has units of area Also called the differential cross section.

  • As in all Central Force problems, for a given particle, the orbit, & thusthe amount of scattering, is determined by the energy E & the angular momentum Define: Impact parameter, s, & express the angular momentum in terms of E & s. Impact parameter s the distance between the center of force & the incident beam velocity (fig). GOAL: Given the energy E, the impact parameter s, & the force f(r), what is the cross section ()?

  • Beam, intensity I. Particles, mass m, incident speed (at r ) = v0.

    Energy conservation: E = T + V = ()mv2 + V(r) = ()m(v0)2 + V(r ) Assume V(r ) = 0 E = ()m(v0)2 v0 = (2E/m)Angular momentum: mv0s s(2mE)

  • Angular momentum mv0s s(2mE) Incident speed v0.Ns # particles scattered into solid angle d between & + d. Cross section definition Ns I()d = 2I()sind Ni # incident particles with impact parameter between s & s + ds . Ni = 2Isds Conservation of particle number Ns = Ni or: 2I()sin|d| = 2Is|ds| 2I cancels out! (Use absolute values because Ns are always > 0, but ds & d can have any sign.)

  • ()sin|d| = s|ds| (1)s = a function of energyE & scattering angle : s = s(,E) (1) () = (s/sin) (|ds|/|d|) (2)To compute () we clearly need s = s(,E)Alternatively, could use = (s,E) & rewrite (2) as: () = (s/sin)/[(|d|/|ds|)] (2)Get = (s,E) from the orbit eqtn. For general central force ( is the angle which describes the orbit r = r(); ) (r) = (/r2)(2m)-[E - V(r) - {2(2mr2)}]-dr

  • Orbit eqtn. General central force: (r) = (/r2)(2m)-[E - V(r) - {2(2mr2)}]-dr (3)Consider purely repulsive scattering. See figure:

    Closest approach distance rm. Orbit must be symmetric about rm Sufficient to look at angle (see figure): Scattering angle - 2. Also, orbit angle = - in the special case r = rm

  • After some manipulation can write (3) as: = (dr/r2)[(2mE)/(2) - (2mV(r))/(2) -1/(r2)]- (4)Integrate from rm to r Angular momentum in terms of impact parameter s & energy E: mv0s s(2mE). Put this into (4)& get for scattering angle (after manipulation): (s) = - 2dr(s/r)[r2{1- V(r)/E} - s2]- (4)Changing integration variables to u = 1/r: (s) = - 2sdu [1- V(r)/E - s2u2]- (4)Integrate from u = 0 to u = um = 1/rm

  • Summary: Scattering by a general central force:Scattering angle = (s,E) (s = impact parameter, E = energy): (s) = - 2sdu [1- V(r)/E - s2u2]- (4) Integrate from u = 0 to u = um = 1/rmScattering cross section: () = (s/sin) (|ds|/|d|) (2)Recipe: To solve a scattering problem: 1. Given force f(r), compute potential V(r). 2. Compute (s) using (4). 3. Compute () using (2).Goldstein tells you this is rarely done to get ()!

  • Sect. 3.10: Coulomb ScatteringSpecial case: Scattering by r-2 repulsive forces: For this case, as well as for others where the orbit eqtn r = r() is known analytically, instead of applying (4) directly to get (s) & then computing () using (2), we make use of the known expression for r = r() to get (s) & then use (2) to get (). Repulsive scattering by r-2 repulsive forces = Coulomb scattering by like chargesCharge +Ze scatters from the center of force, charge Ze f(r) (ZZe2)/(r2) - k/r2Gaussian E&M units! Not SI! For SI, multiply by (1/40)! For r(), in the previous formalism for r-2 attractive forces, make the replacement k - ZZe2

  • Like charges: f(r) (ZZe2)/(r2) k - ZZe2 in orbit eqtn r = r()Weve seen: Orbit eqtn for r-2 force is a conic section: [/r()] = 1 + cos( - ) (1) With: Eccentricity = [ 1 + {2E2(mk2)}] & 2 = [22(mk)]. Eccentricity = to avoid confusion with electric charge e. E > 0 > 1 Orbit is a hyperbola, by previous discussion.Choose the integration const = so that rmin is at = 0 Make the changes in notation noted: [1/r()] = [(mZZe2)/(2)](cos - 1) (2)

  • f(r) = (ZZe2)/(r2)[1/r()] = [(mZZe2)/(2)](cos - 1) (2)Hyperbolic orbit.Note: Typo in text Eq. (3.100)! Missing factor of e!With change of notation, eccentricity is = [ 1 + {2E2(mZ2Z2e4)}]Using the relation between angular momentum, energy, & impact parameter, 2 = 2mEs2 this is: = [ 1 + (2Es)2(ZZe2)2 ]Note: Typo in Eq. (3.99) of text! Missing factor of e!

  • [1/r(,s)] = [(mZZe2)/(2)](cos - 1) (2) = [ 1 + (2Es)2(ZZe2)2 ] (3)From (2) get (r,s). Then, use relations between orbit angle scattering angle , & auxillary angle in the scattering problem, to get = (s) & thus the scattering cross section. = - 2

    = direction of incoming asymptote. Determined by r in (2) cos = (1/). In terms of this is: sin() = (1/). (4) Focus of Hyperbola

  • = [ 1 + (2Es)2(ZZe2)2 ] (3) sin() = (1/) (4)Manipulate with (3) & (4) using trig identities, etc. (4) & trig identities 2 - 1 = cot2() (5)Put (3) into the right side of (5) & take the square root: cot() = (2Es)/(ZZe2) (6) Typo in text! Factor of e!Solve (6) for the impact parameter s = s(,E)

    s = s(,E) = (ZZe2)/(2E) cot() (7) (7), the impact parameter as function of & E for Coulomb scattering is an important result!

  • s = s(,E) = (ZZe2)/(2E) cot() (7)Now, use (7) to compute the Differential Scattering Cross Section for Coulomb Scattering. We had: () = (s/sin) (|ds|/|d|) (8)(7) & (8) (after using trig identities): () = ()[(ZZe2)/(2E)]2csc4() (9) (9) The Rutherford Scattering Cross SectionGet the same results in a (non-relativistic) QM derivation!