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Particles and Deep Inelastic Scattering

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Particles and Deep Inelastic Scattering Heidi Schellman Northwestern Particles and Deep Inelastic Scattering Heidi Schellman Northwestern University HUGS - JLab - June 2010 June 2010 HUGS 1 Heidi Schellman Northwestern k’ k q P P’ A generic scatter of a lepton off of some target. kµ and k0µ are the 4-momenta of the lepton and P µ and P 0µ indicate the target and the final state of the target, which may consist of many particles. qµ = kµ − k0µ is the 4-momentum transfer to the target. June 2010 HUGS 2 Heidi Schellman Northwestern Lorentz invariants k’ k q P P’ 2 2 02 2 2 2 02 2 2 2 The 5 invariant masses k = m` , k = m`0, P = M , P ≡ W , q ≡ −Q are invariants. In addition you can define 3 Mandelstam variables: s = (k + P )2, t = (k − k0)2 and u = (P − k0)2. 2 2 2 2 s + t + u = m` + M + m`0 + W . There are also handy variables ν = (p · q)=M , x = Q2=2Mµ and y = (p · q)=(p · k). June 2010 HUGS 3 Heidi Schellman Northwestern In the lab frame k’ k θ q M P’ The beam k is going in the z direction. Confine the scatter to the x − z plane. µ k = (Ek; 0; 0; k) P µ = (M; 0; 0; 0) 0µ 0 0 0 k = (Ek; k sin θ; 0; k cos θ) qµ = kµ − k0µ June 2010 HUGS 4 Heidi Schellman Northwestern In the lab frame k’ k θ q M P’ 2 2 2 s = ECM = 2EkM + M − m ! 2EkM 2 0 0 2 02 0 t = −Q = −2EkEk + 2kk cos θ + mk + mk ! −2kk (1 − cos θ) 0 ν = (p · q)=M = Ek − Ek energy transfer to target 0 y = (p · q)=(p · k) = (Ek − Ek)=Ek the inelasticity P 02 = W 2 = 2Mν + M 2 − Q2 invariant mass of P 0µ June 2010 HUGS 5 Heidi Schellman Northwestern In the CM frame k’ k q P P’ The beam k is going in the z direction. Confine the scatter to the x − z plane. µ k = (Ek; 0; 0; k) µ P = (EM ; 0; 0; −k) 0µ 0 0 0 k = (Ek; k sin θ; 0; k cos θ) qµ = kµ − k0µ June 2010 HUGS 6 Heidi Schellman Northwestern In the CM frame k’ k q P P’ 2 2 2 s = ECM = (Ek + EM ) ! 4k 2 2 02 0 0 0 t = −Q = mk + mk − 2(EkEk − kk cos θCM ) ! −2kk (1 − cos θCM ) (E − E0 )E + k(k − k0 cos θ) k2 ν = (p · q)=M = k k M ! (1 − cos θ ) M M CM 0 0 (Ek − Ek)EM + k(k − k cos θ) 1 − cos θCM y = (p · q)=(p · k) = 2 ! EkEm + k 2 June 2010 HUGS 7 Heidi Schellman Northwestern For such a two body process the cross section looks like dσ 1 1 1 2 = 2 jMj dt 64π s jkCM j kM p k = p ! s=2 CM s dσ 1 1 1 1 = jMj2 ! jMj2 dt 64π k2M 2 16π s2 June 2010 HUGS 8 Heidi Schellman Northwestern e− + µ+ ! e− + µ+ e e γ µ µ electron muon scattering occurs through the t channel (ie the exchanged γ has momentum qµ.). dσ s2 + u2 dσ s 1 = 2πα2 = 4πα2 1 + (1 − y)2] dt s2t2 dy Q4 2 June 2010 HUGS 9 Heidi Schellman Northwestern Angular dependence µ µ µ e µ e J = 1 e J = 0 e What angular dependence do we expect? These are spin 1/2 particles scattering through the exchange of a spin 1 particle so one expects to have final state with: no angular dependence dσ (Jz = 0), dy ! 1 dσ 2 1 2 (Jz = 1), dy ! (1 + cos θcm) ! 4 (1 − y) June 2010 HUGS 10 Heidi Schellman Northwestern − − νµ + e + ! µ + νe µ νµ − W e νe muon neutrino scattering also occurs through the t channel (ie the exchanged W has momentum qµ.). June 2010 HUGS 11 Heidi Schellman Northwestern If you do a substitution of weak interaction variables into the µe ! µe process p 2 α ! G M 2 π F W 1 1 2 ! 2 2 Q Q + MW 2 4 dσ 2 s 1 2 GF MW s = 4πα 4 1 + (1 − y) ! 2 2 2 dy Q 2 (MW − Q ) π Note that the weak interaction only allows one angular momentum combination (Jz = 0) in this case. June 2010 HUGS 12 Heidi Schellman Northwestern Now we can start studying more complex objects k’ k p=xP q p’ P’ If you scatter an electron off of something - you get an electron - or possibly an electron neutrino. But when you scatter off a proton, you can either get a proton out (elastic scattering) or a neutron (quasi-elastic scattering) or a bunch of hadrons (inelastic scattering). Since the proton breaks up, it must be composite. So maybe we can probe the stuff inside using electron, muons and neutrinos. June 2010 HUGS 13 Heidi Schellman Northwestern A quasi-elastic neutrino reaction in the Minerva neutrino detector June 2010 HUGS 14 Heidi Schellman Northwestern k’ k p=xP q p’ P’ µ Imagine a quark carrying momentum fraction x of the proton momentum PP and scatter an electron off of it. This is a t channel process. Imagine that you only detect the incoming and outgoing scattered electrons. Assume the electron mass is zero. June 2010 HUGS 15 Heidi Schellman Northwestern k’ k in a frame where EP >> Mp=P xP q p’ P’ kµ = (E; 0; 0;E); k0µ = (E0;E0 sin θ; 0;E0 cos θ) µ PP = (EP ; 0; 0; −EP ) µ p = xP = (xEP ; 0; 0; −xEP ) p0µ = kµ + pµ − k0µ = qµ + pµ µ 0µ 0 2 2 p pµ = p pµ = x MP The last expression assume the quark is still the same after the scatter. June 2010 HUGS 16 Heidi Schellman Northwestern Observables - charged lepton scattering You know the target mass M You can measure the incoming and outgoing electron or muon momenta kµ and k0µ. From these you can calculate the invariants s, ν, Q2. June 2010 HUGS 17 Heidi Schellman Northwestern June 2010 HUGS 18 Heidi Schellman Northwestern One can solve for x! (p0)2 = (q + p)2 = q2 + p2 + 2(q · p) 2 2 2 2 2 x MP = −Q + x MP + 2x(q · P ) Q2 = 2xνM Q2 x = 2Mν You can measure x, the fraction of the total momemtum carried by an individual parton just by measuring the incoming proton momentum and the incoming and outgoing electron 4-vectors. June 2010 HUGS 19 Heidi Schellman Northwestern So what would we expect to see? u d u d d u u Proton A proton has 2 u quarks and a d quark. So we'd expect the electron-quark electron center of mass energy s ! s^ = xs and the charge factor 2 2 qi 2 α ! α ( e ) . Naively we'd expect each quark to carry 1/3 of the momentum, so the x probability densities would be δ(x − 1=3). 2 d σ X sxi 1 = 4πα2q2 δ(x − 1=3) 1 + (1 − y)2] dydx i Q4 i 2 i=1;3 June 2010 HUGS 20 Heidi Schellman Northwestern What we expect to see σ x 1/3 June 2010 HUGS 21 Heidi Schellman Northwestern 2 1 The u quarks have charge 3 e while the d quarks have charge − 3 e. We'd also predict that a neutron would have a scattering cross section which is smaller by a factor of: σ (− 1 )2 + (− 1 )2 + ( 2 )2 2 n = 3 3 3 = 2 2 1 2 2 2 σp ( 3 ) + (− 3 ) + ( 3 ) 3 June 2010 HUGS 22 Heidi Schellman Northwestern What we actually see What we actually see when we plot cross sections vs x. There are no little spikes at (1/3)! (this is e + p ! e + X data from the Zeus experiment at HERA) June 2010 HUGS 23 Heidi Schellman Northwestern n/p Observed n=p ratio as a function of x. June 2010 HUGS 24 Heidi Schellman Northwestern New model of the proton u d u d d u u Proton Our new model has the proton containing the 3 "valence" u and d quarks but a whole sea of quark anti-quark pairs. June 2010 HUGS 25 Heidi Schellman Northwestern We can measure the momentum fraction x carried by these quarks! At each x the total cross section will be depend on the probability of there being a quark of fraction x around and on the rate for hard electron-quark scattering. These probability density functions are called "Parton Distribution Functions" or PDF's. dσ X dσ^ (e + p ! e0 + X) = (e + i ! e + i;s; ^ t;^ u^)f (x) dx dx i i dσ By measuring dx on different targets and using the simple predictions for the σ^(e + qi ! e + qi; xs) one can measure the different fi(x). June 2010 HUGS 26 Heidi Schellman Northwestern quark kinematics compared to proton kinematics In all frames: 2 2 2 2 µ s^ = (k + xp) = k + 2xkp + x M ' 2xk pµ = xs (1) t^ = t = (k − k0)2 (2) u^ = = (p − k0)2 ' xu (3) In the cm frame s^ = xs (4) θ t^ = −xQ2 = −xs sin2 cm (5) 2 θ u^ = −xs cos2 cm (6) 2 (7) June 2010 HUGS 27 Heidi Schellman Northwestern If we substitute s=x and Q2 for s^ and t^ we end up with: dσ 4πα2s (e + P ! e + X) = × dxdy Q4 1 [1 + (1 − y)2] × 2 4 1 x[ (u(x) + u(x)) + (d(x) + d(x)) 9 9 1 + (s(x) + s(x))] + ::: 9 Note: this is an approximation - we've neglected quark and lepton masses in assuming s^ = xs.
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