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Sect. 3.11: Transformation to Lab Coords

Sect. 3.11: Transformation to Lab Coords. Scattering so far : Treated as 1 body problem ! Assumed 1 particle scatters off a stationary “Center of Force”.

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Sect. 3.11: Transformation to Lab Coords

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  1. Sect. 3.11: Transformation to Lab Coords • Scattering so far: • Treated as 1 body problem! Assumed 1 particle scatters off a stationary “Center of Force”. • Central Force problem formulation  We know this means that we are doing problem in the Center of Mass coordinate system for 2 bodies & that we are looking at the behavior of the reduced mass μ. • ACTUAL SCATTERING, is (of course) a 2 body problem! 2 masses m1 & m2 scattering off each other. In “lab coordinate system” we need to account for both bodies. Everything we’ve done so far is valid also in the lab frame if m2 >> m1 = m = μ so that the recoil of m2due to m1scattering from it can be neglected. Effectively the same as assuming infinitely massive m2.Now, transform back to the lab frame.

  2. Recall:2 body problem in Center of Mass coordinates: • Center of Mass Coordinate:(M  (m1+m2)) R  (m1r1 +m2r2)(M) • Relative Coordinate: r  r1 - r2 • Define: Reduced Mass: μ (m1m2)(m1+m2) A useful relation: μ-1 (m1)-1 +(m2)-1 • Algebra  Inverse coordinate relations: r1 = R + (μ/m1)r; r2 = R - (μ/m2)r • Velocities related by: v1 = V + (μ/m1)v; v2 = V - (μ/m2)v

  3. To get from the 1 body CM frame scattering problem just discussed to the 2 body lab frame problem, just replacing m  μin what we’ve done so far is not sufficient!In particular: • The scattering angle measured in the lab  θ  angle between final & incident directions of the scattered particle in the lab coordinate system. • Scattering angle calculated in previous discussion: Θ= π -2∫dr(s/r)[r2{1- V(r)/E} -s2]-½ Angle between the initial & final directions of the relative coordinate r between m1 & m2 in the CM coordinate system.θ = Θ only if m2 is stationary (or infinitely massive) throughout the scattering. • NOTE: θ  θthe angle describing the orbit r(θ)!

  4. Kinematics of the Transformation • Assume m2isinitially at rest in the lab frame. • Clearly, after m1 scatters from it, in general it will not be at rest! It will recoil due to the scattering! • Freshman physics: Momentum IS ALWAYS conserved in a collision! Cannot get the lab scattering angle θ directly from solving the 1 body CM frame problem for Θ. • Need to take the result from the1 body CM frame scattering & transform it back to the lab frame. See figure

  5. In the lab frame, the situation looks like: m2is initially at rest • In the CM frame, the situation looks like: Looks like this to an observer moving with the Center of Mass.

  6. In the lab frame:  m2initially at rest. Connection between θ & Θ obtained by looking at detailed transform between lab & CM coordinates In the CM frame: In the CM frame, the total linear momentum of the 2 particles = 0. Before scattering, the particles move directly towards each other. Afterwards, they move off as shown. CM frame scattering angle Θ = same as scattering angle of either particle.

  7. Terminology, notation, changed slightly: • r1, v1 = position, velocity of the incident particle, m1AFTER scattering in the LAB system. • (r1)´, (v1)´,= position, velocity of m1AFTER scattering in the CM system. • R,V= position, velocity of the Center of Mass in the LAB system. From early discussion: V = constant. • By definition (any time) r1 = R + (r1)´ & v1 = V + (v1)´See figure (after scattering!):

  8. r1 = R + (r1)´ & v1 = V + (v1)´ Figure (after scattering!): v1 & (v1)´ make angles  & Θ, respectively with direction of V. Initial velocity of m1 in lab system = v0 . m2is initially at rest in the lab system  v0 = initial relative velocity (= initial v in the general formalism). Linear momentum conservation: (m1+ m2)V = m1v0  V = (μ/m2)v0 (1) From the figure: v1cos = (v1)´cosΘ + V (2) Also: v1sin = (v1)´sinΘ (3) Divide (2) by (3) & use (1) (ρ (μv0)/[m2(v1)´]):  tan = (sinΘ)/(cosΘ + ρ) (4) Note: if m2is infinite, ρ = 0 & = Θ

  9. r1 = R + (r1)´ & v1 = V + (v1)´Figure (after scattering!): Alternative relation from the Law of Cosines. From the figure: (v1)2 = [(v1)´]2 + V2 + 2(v1)´VcosΘ Also: v1sin = (v1)´sinΘ & V = (μ/m2)v0 Combine & get (ρ (μv0)/[m2(v1)´]):  cos = (cosΘ + ρ)/[1+2ρcosΘ + ρ2]½ (4´)

  10. Relations between scattering angles in the lab & CM frames: tan = (sinΘ)/(cosΘ + ρ) (4) cos = (cosΘ + ρ)/[1+2ρcosΘ + ρ2]½ (4´) • Considerρ (μv0)/[m2(v1)´]: From the CM definition, (v1)´ = (μ/m1)v, v = |r|= relative speed after collision:ρ = (m1/m2)(v0/v) • Elastic (KE conserving) scattering: v0 = v, ρ = (m1/m2) • Inelastic (KE non-conserving) scattering: (E = (½)μ(v0)2) (½)μv2 - (½)μ(v0)2 Q  “Q value” of collision. Clearly, since KE is lost, Q < 0 Algebra gives (M =m1+m2): (v/v0) = [1 +(M/m2)(Q/E)]½ ρ = (m1/m2)[1 +(M/m2)(Q/E)]-½ (5)  Analyze scattering kinematics: Combine (5) & (4) or (4´)

  11. Transforming  • To analyze scattering cross sections in the lab frame, its not sufficient to do simple kinematics!Also need to transform the cross sectionσitself from a function of Θto a function of . σ(Θ) σ´() • Connection:Obtained by conservation of particle number: # particles scattered into a given differential solid angle d must be the same, whether measured in the lab or CM frame. So: 2πIσ(Θ)sinΘ|dΘ| = 2πIσ´()sin|d|  σ´() = σ(Θ)(sinΘ/sin)(|dΘ|/|d|) Rewrite as:σ´() = σ(Θ)(|dcosΘ|/|dcos|) Use kinematic result: cos = (cosΘ + ρ)/[1+2ρcosΘ + ρ2]½ Take derivative & get: (ρ = (m1/m2)[1 +(M/m2)(Q/E)] ]-½) σ´()= σ(Θ)[1+2ρcosΘ + ρ2]½(cos Θ + ρ)-1 (6)

  12. σ´()= σ(Θ)[1+2ρcosΘ + ρ2]½(cos Θ + ρ)-1 (6) • Note: σ´()&σ(Θ)are both measuredin the lab frame! They’re expressed in terms of different coordinates. • Special Case #1:Elastic scattering with m1 = m2: ρ = 1 cos = [(½)(1+ cosΘ)]½ = cos(½Θ)  = (½Θ) • Since Θ π, in this case, cannot have  > ½π  In the lab system, all scattering is in forward hemisphere. • In this case, (6) becomes: σ´() = 4cosΘσ(Θ)  Even in the very special case where σ(Θ) = constant, σ´()still depends on angle! • Special Case #2:Elastic scattering with m1 << m2 (effectively, m2is infinite) ρ 0 σ´()  σ(Θ)

  13. More Details • Obviously, scattering slows down the incident particle! • More kinematics: We had (v1)2 = [(v1)´]2 + V2 + 2(v1)´VcosΘ Also, ρ = (μv0)/[m2(v1)´] and V = (μ/m2)v0 Combine these to get (algebra): [(v1)2/(v0)2] = [μ2/(m2ρ)2][1+ 2ρcosΘ + ρ2](a) • Special case: Elastic scatteringρ = (m1/m2) • Let E0 (½)m1(v0)2 = initial KE of m1 before scattering • Let E1 (½)m1(v1)2 = final KE of m1 after scattering (a)  (E1/E0) = [1+2 ρcosΘ + ρ2]/(1+ ρ)2 If m1 = m2 , (E1/E0) = (½)(1+ cosΘ) = cos2 (Typo in text, forgot the square!). For max Θ = π,  = (½)π  (E1/E0) = 0. The incident particle stops in the lab system!! Principle behind “moderator” in neutron scattering.

  14. Classical Mech vs. QM • Some final thoughts on classical scattering discussion. • All we’ve used is simple conservation of momentum & energy. The cross section results are classical. • However, as long as we know the Q value & momentum is conserved, it doesn’t really matter if it is QM or classical scattering! • Why? Because we’ve analyzed the outgoing particle beam (mostly, except for Coulomb scattering) without caring what the details of the scattering were! Details of the scattering, of course, usually require QM analysis!  The results ofMOSTof Sects 3.10 & 3.11 can be used in analyzing experiments for(almost)any kind of(low energy)scattering!Exception: At high enough energies, need to do all of this with Relativity! See Sect. 7.7!

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