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
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For such a two body process the cross section looks like
dσ 1 1 1 2 = 2 |M| dt 64π s |kCM | kM √ k = √ → s/2 CM s dσ 1 1 1 1 = |M|2 → |M|2 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)
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− − 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
√ 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.
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A quasi-elastic neutrino reaction in the Minerva neutrino detector
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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 p=xP q p’ P’
in a frame where EP >> MP
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.
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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.
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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 → sˆ = 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
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What we expect to see
σ
x 1/3
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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
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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)
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n/p
Observed n/p ratio as a function of x.
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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 dσˆ (e + p → e0 + X) = (e + i → e + i;s, ˆ t,ˆ uˆ)f (x) dx dx i i
dσ By measuring dx on different targets and using the simple predictions for the σˆ(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 sˆ = (k + xp) = k + 2xkp + x M ' 2xk pµ = xs (1) tˆ = t = (k − k0)2 (2) uˆ = = (p − k0)2 ' xu (3)
In the cm frame
sˆ = xs (4) θ tˆ = −xQ2 = −xs sin2 cm (5) 2 θ uˆ = −xs cos2 cm (6) 2 (7)
June 2010 HUGS 27 Heidi Schellman Northwestern
If we substitute s/x and Q2 for sˆ and tˆ 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 sˆ = xs.
June 2010 HUGS 28 Heidi Schellman Northwestern
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 un(x)
should be the same as the d content of the proton dp(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 Heidi Schellman Northwestern
But it does mean that scattering from neutrons should be like:
dσ 4πα2s 1 (e + n) = × [1 + (1 − y)2] × dydx Q4 2 4 1 x[ (u (x) + u (x)) + (d (x) + d (x)) + 9 n n 9 n n 1 (s (x) + s (x))] + ... 9 n n 4πα2s 1 = × [1 + (1 − y)2] × Q4 2 4 1 x[ (d (x) + d (x)) + (u (x) + u (x)) + 9 p p 9 p p 1 (s (x) + s (x))] + ... 9 n n
June 2010 HUGS 30 Heidi Schellman Northwestern
And if you assume that the s quarks are the same for protons and neutrons you can look at the difference.
1 dσ dσ 1 (e + p) − (e + n) ∝ [u(x) + u(x) − d(x) − d(x)] x dx dx 3
You can use this to check to see if there really is one more u than d quark in the proton.
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