2. 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 Center of Force. –Original 2 body problem = Scattering of “particle” with the reduced mass μ from a center of force –Here, 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!).

3. 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 m 3 per s) –As a particle approaches the center of force, it is either attracted or repelled & thus it’s 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

4. 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  ): σ(  )d   (N s /I). With I = incident intensity N s = # particles/time scattered into solid angle d 

5. Scattering Cross Section: σ(  )d   (N s /I) I = incident intensity N s = # particles/time scattered into angle d  In general, the solid angle  depends on the spherical angles Θ, Φ. However, for central forces, there must be symmetry about the axis of the incident beam  σ(  ) (  σ(Θ)) is independent of azimuthal angle Φ  d   2π sinΘdΘ, σ(  )d   2π sinΘdΘ, Θ  Angle between incident & scattered beams, as in the figure. σ  “cross section”. It has units of area Also called the differential cross section.

6. As in all Central Force problems, for a given particle, the orbit, & thus the 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 σ(Θ)?

7. Beam, intensity I. Particles, mass m, incident speed (at r   ) = v 0. Energy conservation: E = T + V = (½)mv 2 + V(r) = (½)m(v 0 ) 2 + V(r   ) Assume V(r   ) = 0  E = (½)m(v 0 ) 2  v 0 = (2E/m) ½ Angular momentum:  mv 0 s  s(2mE) ½

8. Angular momentum  mv 0 s  s(2mE) ½ Incident speed v 0. N s  # particles scattered into solid angle d  between Θ & Θ + dΘ. Cross section definition  N s  Iσ(Θ)dΘ = 2πIσ(Θ)sinΘdΘ N i  # incident particles with impact parameter between s & s + ds. N i = 2πIsds Conservation of particle number  N s = N i or: 2πIσ(Θ)sinΘ|dΘ| = 2πIs|ds| 2πI cancels out! (Use absolute values because N’s are always > 0, but ds & dΘ can have any sign.)

9. σ(Θ)sinΘ|dΘ| = s|ds| (1) s = a function of energy E & scattering angle Θ: s = s(Θ,E) (1)  σ(Θ) = (s/sinΘ) (|ds|/|dΘ|) (2) To compute σ(Θ) we clearly need s = s(Θ,E) Alternatively, could use Θ = Θ(s,E) & rewrite (2) as: σ(Θ) = (s/sinΘ)/[(|dΘ|/|ds|)] (2´) Get Θ = Θ(s,E) from the orbit eqtn. For general central force (θ is the angle which describes the orbit r = r(θ); θ  Θ) θ(r) = ∫( /r 2 )(2m) -½ [E - V(r) - { 2  (2mr 2 )}] -½ dr

10. Orbit eqtn. General central force: θ(r) = ∫( /r 2 )(2m) -½ [E - V(r) - { 2  (2mr 2 )}] -½ dr (3) Consider purely repulsive scattering. See figure: Closest approach distance  r m. Orbit must be symmetric about r m  Sufficient to look at angle  (see figure): Scattering angle Θ  π - 2Ψ. Also, orbit angle θ = π - Ψ in the special case r = r m

11.  After some manipulation can write (3) as: Ψ = ∫(dr/r 2 )[(2mE)/( 2 ) - (2mV(r))/( 2 ) -1/(r 2 )] -½ (4) Integrate from r m to r   Angular momentum in terms of impact parameter s & energy E:  mv 0 s  s(2mE) ½. Put this into (4) & get for scattering angle Θ (after manipulation): Θ(s) = π - 2∫dr(s/r)[r 2 {1- V(r)/E} - s 2 ] -½ (4´) Changing integration variables to u = 1/r : Θ(s) = π - 2∫sdu [1- V(r)/E - s 2 u 2 ] -½ (4´´) Integrate from u = 0 to u = u m = 1/r m

12. Summary: Scattering by a general central force: Scattering angle Θ = Θ(s,E) (s = impact parameter, E = energy): Θ (s) = π - 2∫sdu [1- V(r)/E - s 2 u 2 ] -½ (4´´) Integrate from u = 0 to u = u m = 1/r m Scattering cross section: σ(Θ) = (s/sinΘ) (|ds|/|dΘ|) (2) “Recipe”: To solve a scattering problem: 1. Given force f(r), compute potential V(r). 2. Compute Θ(s) using (4´´). 3. Compute σ(Θ) using (2). –Goldstein tells you this is rarely done to get σ(Θ)!

13. Special case: Scattering by r -2 repulsive forces: –For this case, as well as for others where the orbit eqtn r = r(θ) is known analytically, instead of applying (4´´) directly to get Θ(s) & then computing σ(Θ) using (2), we make use of the known expression for r = r(θ) to get Θ(s) & then use (2) to get σ(Θ). Repulsive scattering by r -2 repulsive forces = Coulomb scattering by like charges –Charge +Ze scatters from the center of force, charge Z´e  f(r)  (ZZ´e 2 )/(r 2 )  - k/r 2 –Gaussian E&M units! Not SI! For SI, multiply by (1/4πε 0 )!  For r(θ), in the previous formalism for r -2 attractive forces, make the replacement k  - ZZ´e 2 Sect. 3.10: Coulomb Scattering

14. Like charges: f(r)  (ZZ´e 2 )/(r 2 )  k  - ZZ´e 2 in orbit eqtn r = r(θ) We’ve seen: Orbit eqtn for r -2 force is a conic section: [α/r(θ)] = 1 + εcos(θ - θ´) (1) With: Eccentricity  ε = [ 1 + {2E 2  (mk 2 )}] ½ & 2α = [2 2  (mk)]. Eccentricity = ε to avoid confusion with electric charge e. E > 0  ε > 1  Orbit is a hyperbola, by previous discussion. Choose the integration const θ´ = π so that r min is at θ = 0 Make the changes in notation noted:  [1/r(θ)] = [(mZZ´e 2 )/( 2 )](εcosθ - 1) (2)

15. f(r) = (ZZ´e 2 )/(r 2 ) [1/r(θ)] = [(mZZ´e 2 )/( 2 )](εcosθ - 1) (2) Hyperbolic orbit. Note: Typo in text Eq. (3.100)! Missing factor of e! With change of notation, eccentricity is ε = [ 1 + {2E 2  (mZ 2 Z´ 2 e 4 )}] ½ Using the relation between angular momentum, energy, & impact parameter, 2 = 2mEs 2 this is: ε = [ 1 + (2Es) 2  (ZZ´e 2 ) 2 ] ½ Note: Typo in Eq. (3.99) of text! Missing factor of e!

16. [1/r(θ,s)] = [(mZZ´e 2 )/( 2 )](εcosθ - 1) (2) ε = [ 1 + (2Es) 2  (ZZ´e 2 ) 2 ] ½ (3) From (2) get θ(r,s). Then, use relations between orbit angle θ scattering angle Θ, & auxillary angle Ψ in the scattering problem, to get Θ = Θ(s) & thus the scattering cross section. Θ = π - 2Ψ Ψ = direction of incoming asymptote. Determined by r   in (2)  cosΨ = (1/ε). In terms of Θ this is: sin(½Θ) = (1/ε). (4) Focus of Hyperbola 

17. ε = [ 1 + (2Es) 2  (ZZ´e 2 ) 2 ] ½ (3) sin(½Θ) = (1/ε) (4) Manipulate with (3) & (4) using trig identities, etc. (4) & trig identities  ε 2 - 1 = cot 2 (½Θ) (5) Put (3) into the right side of (5) & take the square root:  cot(½Θ) = (2Es)/(ZZ´e 2 ) (6) Typo in text! Factor of e! Solve (6) for the impact parameter s = s(Θ,E)  s = s(Θ,E) = (ZZ´e 2 )/(2E) cot(½Θ) (7) (7), the impact parameter as function of Θ & E for Coulomb scattering is an important result!

18. s = s(Θ,E) = (ZZ´e 2 )/(2E) cot(½Θ) (7) Now, use (7) to compute the Differential Scattering Cross Section for Coulomb Scattering. We had: σ(Θ) = (s/sinΘ) (|ds|/|dΘ|) (8) (7) & (8) (after using trig identities):  σ (Θ) = (¼)[(ZZ´e 2 )/(2E)] 2 csc 4 (½Θ) (9) (9)  The Rutherford Scattering Cross Section Get the same results in a (non-relativistic) QM derivation !