particles and deep inelastic scattering...these are spin 1/2 particles scattering through the...

Heidi Schellman Northwestern

Particles and Deep Inelastic Scattering

Heidi Schellman

Northwestern University

HUGS - JLab - June 2010

June 2010 HUGS 1


k

P

k’

P’

q

A generic scatter of a lepton off of some target. kµ and k′µ are the

4-momenta of the lepton and Pµ and P ′µ indicate the target and the final

state of the target, which may consist of many particles. qµ = kµ − k′µ is the

4-momentum transfer to the target.

June 2010 HUGS 2


Lorentz invariants

k

P

k’

P’

q

The 5 invariant masses k2 = m2` , k′2 = m2

`′, P2 = M2, P ′2 ≡W 2, q2 ≡ −Q2

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 = m2` +M2 +m2

`′ +W 2. There are also handy variables

ν = (p · q)/M , x = Q2/2Mµ and y = (p · q)/(p · k).

June 2010 HUGS 3


In the lab frame

k

M

k’

q

θ

P’

The beam k is going in the z direction. Confine the scatter to the x− z plane.

kµ = (Ek, 0, 0, k)

Pµ = (M, 0, 0, 0)

k′µ = (E′k, k

′ sin θ, 0, k′ cos θ)

qµ = kµ − k′µ

June 2010 HUGS 4


In the lab frame

k

M

k’

q

θ

P’

s = E2CM = 2EkM +M2 −m2 → 2EkM

t = −Q2 = −2EkE′k + 2kk′ cos θ +m2

k +m′2k → −2kk′(1− cos θ)

ν = (p · q)/M = Ek − E′k energy transfer to target

y = (p · q)/(p · k) = (Ek − E′k)/Ek the inelasticity

P ′2 = W 2 = 2Mν +M2 −Q2 invariant mass of P ′µ

June 2010 HUGS 5


In the CM frame

k

P

k’

q

P’

The beam k is going in the z direction. Confine the scatter to the x− z plane.

kµ = (Ek, 0, 0, k)

Pµ = (EM , 0, 0,−k)

k′µ = (E′k, k

′ sin θ, 0, k′ cos θ)

qµ = kµ − k′µ

June 2010 HUGS 6


In the CM frame

k

P

k’

q

P’

s = E2CM = (Ek + EM )2 → 4k2

t = −Q2 = m2k +m′2

k − 2(EkE′k − kk′ cos θCM )→ −2kk′(1− cos θCM )

ν = (p · q)/M =(Ek − E′

k)EM + k(k − k′ cos θ)M

→ k2

M(1− cos θCM )

y = (p · q)/(p · k) =(Ek − E′

k)EM + k(k − k′ cos θ)EkEm + k2

→ 1− cos θCM2

June 2010 HUGS 7


For such a two body process the cross section looks like

dσ

dt=

164π

1s

1|kCM |2

|M|2

kCM =kM√s→√s/2

dσ

dt=

164π

1k2M2

|M|2 → 116π

1s2|M|2

June 2010 HUGS 8


e− + µ+ → e− + µ+

e

µ

e

µγ

electron muon scattering occurs through the t channel (ie the exchanged γ

has momentum qµ.).

dσ

dt= 2πα2 s

2 + u2

s2t2dσ

dy= 4πα2 s

Q4

12[1 + (1− y)2]

]

June 2010 HUGS 9


Angular dependence

µ e

µ

eJ = 1

µ e

µ

e J = 0

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

(Jz = 0) , dσdy → 1

(Jz = 1) , dσdy → (1 + cos θcm)2 → 1

4 (1− y)2

June 2010 HUGS 10


νµ + e−+→ µ− + νe

νµ

e

µ

W

νe

−

muon neutrino scattering also occurs through the t channel (ie the exchanged

W has momentum qµ.).

June 2010 HUGS 11


If you do a substitution of weak interaction variables into the µe→ µe process

α →√

2πGFM

2W

1Q2

→ 1Q2 +M2

W

dσ

dy= 4πα2 s

Q4

12[1 + (1− y)2

]→ G2

FM4W

(M2W −Q2)2

s

π

Note that the weak interaction only allows one angular momentum

combination (Jz = 0) in this case.

June 2010 HUGS 12


Now we can start studying more complexobjects

kk’

q

P’p’xPp=

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


A quasi-elastic neutrino reaction in the Minerva neutrino detector

June 2010 HUGS 14


kk’

q

P’p’xPp=

Imagine a quark carrying momentum fraction x of the proton momentum PµPand 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


kk’

q

P’p’xPp=

in a frame where EP >> MP

kµ = (E, 0, 0, E), k′µ = (E′, E′ sin θ, 0, E′ cos θ)

PµP = (EP , 0, 0,−EP )

pµ = xP = (xEP , 0, 0,−xEP )

p′µ = kµ + pµ − k′µ = qµ + pµ

pµpµ = p′µp′µ = x2M2

P

The last expression assume the quark is still the same after the scatter.

June 2010 HUGS 16


Observables - charged lepton scatteringYou know the target mass M

You can measure the incoming and outgoing electron or muon momenta kµ

and k′µ.

From these you can calculate the invariants s, ν, Q2.

June 2010 HUGS 17


June 2010 HUGS 18


One can solve for x!

(p′)2 = (q + p)2 = q2 + p2 + 2(q · p)

x2M2P = −Q2 + x2M2

P + 2x(q · P )

Q2 = 2xνM

x =Q2

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


So what would we expect to see?

u

d

u

uu

d d

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

α2 → α2( qi

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).

d2σ

dydx=∑i=1,3

4πα2q2isxiQ4

δ(xi − 1/3)12[1 + (1− y)2]

]June 2010 HUGS 20


What we expect to see

1/3x

σ

June 2010 HUGS 21


The u quarks have charge 23e while the d quarks have charge − 1

3e.

We’d also predict that a neutron would have a scattering cross section which

is smaller by a factor of:

σnσp

=(− 1

3 )2 + (− 13 )2 + ( 2

3 )2

( 23 )2 + (− 1

3 )2 + ( 23 )2

=23

June 2010 HUGS 22


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


n/p

Observed n/p ratio as a function of x.

June 2010 HUGS 24


New model of the proton

u

d

u

uu

d d

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


We can measure the momentum fractionx 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

dσ

dx(e+ i→ e+ i; s, t, u)fi(x)

By measuring dσdx 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


quark kinematics compared to protonkinematics

In all frames:

s = (k + xp)2 = k2 + 2xkp+ x2M2 ' 2xkµpµ = xs (1)

t = t = (k − k′)2 (2)

u = = (p− k′)2 ' xu (3)

In the cm frame

s = xs (4)

t = −xQ2 = −xs sin2 θcm2

(5)

u = −xs cos2θcm2

(6)

(7)

June 2010 HUGS 27


If we substitute s/x and Q2 for s and t we end up with:

dσ

dxdy(e+ P → e+X) =

4πα2s

Q4×

12[1 + (1− y)2]×

x[49(u(x) + u(x)) +

19(d(x) + d(x))

+19(s(x) + s(x))] + ...

Note: this is an approximation - we’ve neglected quark and lepton masses in

assuming s = xs.

June 2010 HUGS 28


What can neutrons tell usEarly 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


But it does mean that scattering from neutrons should be like:

dσ

dydx(e+ n) =

4πα2s

Q4× 1

2[1 + (1− y)2]×

x[49(un(x) + un(x)) +

19(dn(x) + dn(x)) +

19(sn(x) + sn(x))] + ...

=4πα2s

Q4× 1

2[1 + (1− y)2]×

x[49(dp(x) + dp(x)) +

19(up(x) + up(x)) +

19(sn(x) + sn(x))] + ...

June 2010 HUGS 30


And if you assume that the s quarks are the same for protons and neutrons

you can look at the difference.

1x

[dσ

dx(e+ p)− dσ

dx(e+ n)

]∝ 1

3[u(x) + u(x)− d(x)− d(x)]

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

particles and deep inelastic scattering...these are spin 1/2 particles scattering through the...

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