# Sect. 3.10: Central Force Field Scattering

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05-Feb-2016Category

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### 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!

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