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 m 1 & m 2 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 m 2 >> m 1 = m = μ so that the recoil of m 2 due to m 1 scattering from it can be neglected. Effectively the same as assuming infinitely massive m 2. Now, transform back to the lab frame.

Recall: 2 body problem in Center of Mass coordinates: Center of Mass Coordinate: ( M  (m 1 +m 2 )) R  (m 1 r 1 +m 2 r 2 )  (M) Relative Coordinate: r  r 1 - r 2 Define: Reduced Mass: μ  (m 1 m 2 )  (m 1 +m 2 ) A useful relation: μ -1  (m 1 ) -1 +(m 2 ) -1 Algebra  Inverse coordinate relations: r 1 = R + (μ/m 1 )r; r 2 = R - (μ/m 2 )r Velocities related by: v 1 = V + (μ/m 1 )v; v 2 = V - (μ/m 2 )v

To get from the 1 body CM frame scattering problem just discussed to the 2 body lab frame problem, just replacing m  μ in 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)[r 2 {1- V(r)/E} - s 2 ] -½  Angle between the initial & final directions of the relative coordinate r between m 1 & m 2 in the CM coordinate system. θ = Θ only if m 2 is stationary (or infinitely massive) throughout the scattering. –NOTE: θ  θ the angle describing the orbit r(θ)!

Kinematics of the Transformation Assume m 2 is initially at rest in the lab frame. –Clearly, after m 1 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

In the lab frame, the situation looks like:  m 2 is initially at rest In the CM frame, the situation looks like:   Looks like this to an observer moving with the Center of Mass.

In the lab frame:  m 2 initially 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.

Terminology, notation, changed slightly: r 1, v 1 = position, velocity of the incident particle, m 1 AFTER scattering in the LAB system. (r 1 )´, (v 1 )´, = position, velocity of m 1 AFTER 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) r 1 = R + (r 1 )´ & v 1 = V + (v 1 )´ See figure (after scattering!):

r 1 = R + (r 1 )´ & v 1 = V + (v 1 )´ Figure (after scattering!): v 1 & (v 1 )´ make angles  & Θ, respectively with direction of V. Initial velocity of m 1 in lab system = v 0. m 2 is initially at rest in the lab system  v 0 = initial relative velocity (= initial v in the general formalism). Linear momentum conservation:  (m 1 + m 2 )V = m 1 v 0  V = (μ/m 2 )v 0 (1) From the figure: v 1 cos  = (v 1 )´cosΘ + V (2) Also: v 1 sin  = (v 1 )´sinΘ (3) Divide (2) by (3) & use (1) (ρ  (μv 0 )/[m 2 (v 1 )´]):  tan  = (sinΘ)/(cosΘ + ρ) (4) Note: if m 2 is infinite, ρ = 0 &  = Θ

r 1 = R + (r 1 )´ & v 1 = V + (v 1 )´ Figure (after scattering!): Alternative relation from the Law of Cosines. From the figure: (v 1 ) 2 = [(v 1 )´] 2 + V 2 + 2(v 1 )´VcosΘ Also: v 1 sin  = (v 1 )´sinΘ & V = (μ/m 2 )v 0 Combine & get (ρ  (μv 0 )/[m 2 (v 1 )´]):  cos  = (cosΘ + ρ)/[1+2ρcosΘ + ρ 2 ] ½ (4´)

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

Transforming  To analyze scattering cross sections in the lab frame, its not sufficient to do simple kinematics! Also need to transform the cross section σ itself from a function of Θ to 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 )sin Θ |d Θ | = 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 + ρ 2 ] ½ Take derivative & get: (ρ = (m 1 /m 2 )[1 +(M/m 2 )(Q/E)] ] -½ ) σ´(  ) = σ(Θ)[1+2ρcosΘ + ρ 2 ] ½ (cos Θ + ρ) -1 (6)

Note: σ´(  ) & σ(Θ) are both measured in the lab frame! They’re expressed in terms of different coordinates. Special Case #1: Elastic scattering with m 1 = m 2 :  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 m 1 << m 2 (effectively, m 2 is infinite)  ρ  0  σ´(  )  σ(Θ)

More Details Obviously, scattering slows down the incident particle! More kinematics: We had (v 1 ) 2 = [(v 1 )´] 2 + V 2 + 2(v 1 )´VcosΘ Also, ρ = (μv 0 )/[m 2 (v 1 )´] and V = (μ/m 2 )v 0 Combine these to get (algebra): [(v 1 ) 2 /(v 0 ) 2 ] = [μ 2 /(m 2 ρ) 2 ][1+ 2ρcosΘ + ρ 2 ] (a ) Special case: Elastic scattering  ρ = (m 1 /m 2 ) –Let E 0  (½)m 1 (v 0 ) 2 = initial KE of m 1 before scattering –Let E 1  (½)m 1 (v 1 ) 2 = final KE of m 1 after scattering (a)  (E 1 /E 0 ) = [1+2 ρcosΘ + ρ 2 ]/(1+ ρ) 2 If m 1 = m 2, (E 1 /E 0 ) = (½)(1+ cosΘ) = cos 2  (Typo in text, forgot the square!). For max Θ = π,  = (½)π  (E 1 /E 0 ) = 0. The incident particle stops in the lab system!! Principle behind “moderator” in neutron scattering.

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 of MOST of 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!