Particles and Deep Inelastic Scattering

Transcription

1 Particles and Deep Inelastic Scattering University HUGS - JLab - June 2010 June 2010 HUGS 1

2 k q k P P A generic scatter of a lepton off of some target. k µ and k µ are the 4-momenta of the lepton and P µ and P µ indicate the target and the final state of the target, which may consist of many particles. q µ = k µ k µ is the 4-momentum transfer to the target. June 2010 HUGS 2

3 Lorentz invariants k q k P P The 5 invariant masses k 2 = m 2 l, k 2 = m 2 l, P 2 = M 2, P 2 W 2, q 2 Q 2 are invariants. In addition you can define 3 Mandelstam variables: s = (k + P ) 2, t = (k k ) 2 and u = (P k ) 2. s + t + u = m 2 l + M 2 + m 2 l + W 2. There are also handy variables ν = (p q)/m, x = Q 2 /2Mµ and y = (p q)/(p k). June 2010 HUGS 3

4 In the lab frame k M q k θ P The beam k is going in the z direction. Confine the scatter to the x z plane. k µ = (E k, 0, 0, k) P µ = (M, 0, 0, 0) k µ = (E k, k sin θ, 0, k cos θ) q µ = k µ k µ June 2010 HUGS 4

5 In the lab frame k M q k θ P s = E 2 CM = 2E k M + M 2 m 2 2E k M t = Q 2 = 2E k E k + 2kk cos θ + m 2 k + m 2 k 2kk (1 cos θ) ν = (p q)/m = E k E k energy transfer to target y = (p q)/(p k) = (E k E k)/e k the inelasticity P 2 = W 2 = 2Mν + M 2 Q 2 invariant mass of P µ June 2010 HUGS 5

6 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. k µ = (E k, 0, 0, k) P µ = (E M, 0, 0, k) k µ = (E k, k sin θ, 0, k cos θ) q µ = k µ k µ June 2010 HUGS 6

7 In the CM frame k k q P P s = E 2 CM = (E k + E M ) 2 4k 2 t = Q 2 = m 2 k + m 2 k 2(E k E k kk cos θ CM ) 2kk (1 cos θ CM ) ν = (p q)/m = (E k E k )E M + k(k k cos θ) M y = (p q)/(p k) = (E k E k )E M + k(k k cos θ) E k E m + k 2 k2 M (1 cos θ CM ) 1 cos θ CM 2 June 2010 HUGS 7

8 For such a two body process the cross section looks like dσ dt = 1 64π 1 s 1 k CM 2 M 2 k CM = km s s/2 dσ dt = 1 64π 1 k 2 M 2 M π 1 s 2 M 2 June 2010 HUGS 8

9 e + µ + e + µ + e µ γ e µ electron muon scattering occurs through the t channel (ie the exchanged γ has momentum q µ.). dσ dt = 2πα2 s2 + u 2 s 2 t 2 dσ dy = s 1 4πα2 Q 4 2 [ 1 + (1 y) 2 ] ] June 2010 HUGS 9

10 Angular dependence µ µ µ e e J = 0 µ e e J = 1 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 (J z = 0), dσ dy 1 (J z = 1), dσ dy (1 + cos θ cm) (1 y)2 June 2010 HUGS 10

11 ν µ + e + µ + ν e µ ν µ W e ν e muon neutrino scattering also occurs through the t channel (ie the exchanged W has momentum q µ.). June 2010 HUGS 11

12 If you do a substitution of weak interaction variables into the µe µe process dσ dy = s 1 4πα2 Q 4 2 α 1 Q 2 [ 1 + (1 y) 2 ] 2 π G F MW 2 1 Q 2 + MW 2 G 2 F M W 4 s (MW 2 Q2 ) 2 π Note that the weak interaction only allows one angular momentum combination (J z = 0) in this case. June 2010 HUGS 12

13 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

14 A quasi-elastic neutrino reaction in the Minerva neutrino detector June 2010 HUGS 14

15 k k p=xp q p P Imagine a quark carrying momentum fraction x of the proton momentum P µ P 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

16 k k p=xp q p P in a frame where E P >> M P k µ = (E, 0, 0, E), k µ = (E, E sin θ, 0, E cos θ) P µ P = (E P, 0, 0, E P ) p µ = xp = (xe P, 0, 0, xe P ) p µ = k µ + p µ k µ = q µ + p µ p µ p µ = p µ p µ = x 2 MP 2 The last expression assume the quark is still the same after the scatter. June 2010 HUGS 16

17 Observables - charged lepton scattering You know the target mass M You can measure the incoming and outgoing electron or muon momenta k µ and k µ. From these you can calculate the invariants s, ν, Q 2. June 2010 HUGS 17

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19 One can solve for x! (p ) 2 = (q + p) 2 = q 2 + p 2 + 2(q p) x 2 M 2 P = Q 2 + x 2 M 2 P + 2x(q P ) Q 2 = 2xνM x = Q 2 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

20 So what would we expect to see? d d d u u 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 ŝ = xs and the charge factor α 2 α 2 ( q i e ) 2. Naively we d expect each quark to carry 1/3 of the momentum, so the x probability densities would be δ(x 1/3). d 2 σ dydx = i=1,3 4πα 2 q 2 i sx i Q 4 δ(x i 1/3) 1 2 [ 1 + (1 y) 2 ] ] June 2010 HUGS 20

21 What we expect to see σ 1/3 x June 2010 HUGS 21

22 The u quarks have charge 2 3 e while the d quarks have charge 1 3 e. We d also predict that a neutron would have a scattering cross section which is smaller by a factor of: σ n σ p = ( 1 3 )2 + ( 1 3 )2 + ( 2 3 )2 ( 2 3 )2 + ( 1 3 )2 + ( 2 3 )2 = 2 3 June 2010 HUGS 22

23 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

24 n/p Observed n/p ratio as a function of x. June 2010 HUGS 24

25 New model of the proton d u 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

26 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 dx (e + p e + X) = i dˆσ dx (e + i e + i; ŝ, ˆt, û)f i (x) By measuring dσ dx on different targets and using the simple predictions for the ˆσ(e + q i e + q i ; xs) one can measure the different f i (x). June 2010 HUGS 26

27 quark kinematics compared to proton kinematics In all frames: In the cm frame ŝ = (k + xp) 2 = k 2 + 2xkp + x 2 M 2 2xk µ p µ = xs (1) ˆt = t = (k k ) 2 (2) û = = (p k ) 2 xu (3) ŝ = xs (4) ˆt = xq 2 = xs sin 2 θ cm 2 û = xs cos 2 θ cm 2 (5) (6) (7) June 2010 HUGS 27

28 If we substitute s/x and Q 2 for ŝ and ˆt we end up with: dσ dxdy (e + P e + X) = 4πα2 s Q [1 + (1 y)2 ] x[ 4 9 (u(x) + u(x)) + 1 (d(x) + d(x)) (s(x) + s(x))] Note: this is an approximation - we ve neglected quark and lepton masses in assuming ŝ = xs. June 2010 HUGS 28

29 What can neutrons tell us Early in the history of particle physics Heisenberg noticed that the strong interactions didn t care if particle was a neutron or a proton. In modern language this means that the strong interaction is flavor blind. This leads to an approximate symmetry called Isospin where the proton and neutron are the +1/2 and 1/2 eigenstates and the symmetry acts like spin. The u and d quarks have the same relation under isospin as the proton and neutron and it has been argued that the u content of the neutron u n (x) should be the same as the d content of the proton d p (x) ie applying an isospin rotation to a neutron changes it to a proton and also exchanges all the u and d quarks. We re not certain this is perfectly so (after all the neutron and proton do differ a bit). June 2010 HUGS 29

30 But it does mean that scattering from neutrons should be like: dσ dydx (e + n) = 4πα2 s Q [1 + (1 y)2 ] x[ 4 9 (u n(x) + u n (x)) (d n(x) + d n (x)) (s n(x) + s n (x))] +... = 4πα2 s Q [1 + (1 y)2 ] x[ 4 9 (d p(x) + d p (x)) (u p(x) + u p (x)) (s n(x) + s n (x))] +... June 2010 HUGS 30

31 And if you assume that the s quarks are the same for protons and neutrons you can look at the difference. 1 x [ ] dσ dσ (e + p) (e + n) dx dx 1 [u(x) + u(x) d(x) d(x)] 3 You can use this to check to see if there really is one more u than d quark in the proton. June 2010 HUGS 31