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Chapter 8 - Deep Inelastic Scattering - Coggle Diagram
Chapter 8 - Deep Inelastic Scattering
Kinematic variables: \(W, x, y, \nu, Q^2\)
\(Q^2 = -q^2\)
\(Q^2 = -(p_1 - p_3)^2 \sim 4E_1E_3\sin^2(\frac{\theta}{2})\) > 0
\(x = \frac{Q^2}{2p_2\cdot q}\)
\(x = \frac{Q^2}{Q^2 + W^2 - m_p^2}=\frac{Q^2}{2m_p(E_1-E_3)}\)
Therefore, \( 0 \leq x \leq 1\) and it represents the "
elasticity
": \(x=1\) means \(W^2 = m_p^2\), where the proton is intact.
\(W^2 = p_4^2\)
Invariant mass-squared of final state hadronic system. Since final state must involve at least one baryon, \(W^2 = p_4^2 = (q + p_2)^2 \geq m_p^2\)
\(y = \frac{p_2 \cdot q}{p_2 \cdot p_1}\)
In a frame where
target
(proton) is
at rest
, \(y = 1 - \frac{E_3}{E_1}\)
y: "
inelasticity
", representing the
fraction of energy lost by the electron
in a rest-proton frame. Therefore, \(0 \leq y \leq 1\) (electron
must lose energy
)
\(\nu = \frac{p_2\cdot q}{m_p}\)
In a frame where proton is at rest, \(\nu = E_1 - E_3\)
Relationships
Assuming \(s = (p_1+p_2)^2 = 2 p_1 \cdot p_2 + m_e^2 + m_p^2 \) and neglecting \(m_e^2\), we have
\(x = \frac{Q^2}{2m_p\nu}\)
\(y = \frac{2m_p}{s - m_p^2}\nu\)
\(Q^2 = (s - m_p^2)xy\)
Therefore, for a given CM energy \((\sqrt{s})\), the
kinematics
of inelastic scattering can be expressed by
any two
of Lorentz-invariant {\(x, y, \nu, Q^2\)} except {y, \(\nu\)}
Inelastic: proton breaks up. Need
2 kinematic variables
.
Deep inelastic scattering (DIS)
\(\frac{d\sigma}{dx dQ^2} = \frac{4\pi\alpha^2}{Q^4}\Big((1-y)\frac{F_2(x, Q^2)}{x} + y^2 F_1(x, Q^2) \Big)\)
Rewritten from Rosenbluth formula; first factor comes from photon propagator and vertex.
Structure functions
: \(F_1\) is purely magnetic; \(F_2\) is both electric and magnetic.
For each bin of \(x, Q^2\), measure the cross section. Vary y. Then determine \(F_2, F_1\).
Measuring structure functions
SLAC: 2-mile linear accelerator; \(e^-\) hits hydrogen target, scatters off into a movable spectrometer placed at varying angles. Elevated from ground to measured momentum.
Result 1:
Bjorken scaling
: structure functions are almost
independent
of \(Q^2\)! \(F_1(x, Q^2) \rightarrow F_1(x)\), \(F_2(x, Q^2) \rightarrow F_2(x)\)
Implication:
point-like
constituents
inside proton
.
Result 2: Gallan-Gross relation: \(F_2(x) = 2x F_1(X)\)
Implication:
elastic
scattering of \(e^-\) off \(spin-\frac{1}{2}\) particles inside proton (quarks!).
Parton model
Partons
are
point-like free particles
inside proton; proton is assumed in
infinite momentum frame
: ignore \(m_p\), and ignore transverse motion of "partons".
For a proton with momentum \(p=(E, E)\), suppose \(e^-\) hits a quark. The struck quark has momentum \((\xi E, \xi E)\). Therefore, \((\xi p + q)^2 = m_q^2\).
As it turns out, \(\xi = \frac{Q^2}{2p_2q} = x\).
Therefore, interpret \(x\) as the fraction of the proton momentum carried by the struck quark!
Start from the \(e^-q\) scattering cross section, it can be shown that the
electron-parton cross section
for a given Bjorken x is:
\(\frac{d\sigma}{dQ^2} = \frac{4\pi\alpha^2Q_q^2}{Q^4}\Big( (1-y) + \frac{y^2}{2}\Big)\)
x is implicit: \(y = \frac{Q^2}{(s-m_p^2)x}\)
Parton distribution function
(
PDF
)
\(q^p(x)\delta x\):
probability
of finding quark \(q\) (i.e.
all quarks of type \(\mathbf{q}\)
) carrying a fraction \(x\) of proton's momentum. \(0 < q^p(x) < 1\).
Follows a distribution, e.g. a
delta function
at 1/3 indicates proton has 3 quarks.
Quarks exchange gluons with each other,
smearing
the PDF.
If gluons produce virtual quarks, expect PDF to
rise at low x
due to \(\frac{1}{q^2}\) gluon propagator.
Not known a priori; must be probed by experiments measuring structure functions.
\(\frac{d\sigma^{ep}}{dxdQ^2} = \frac{4\pi\alpha^2}{Q^4} \Big[ (1-y) + \frac{y^2}{2} \Big] \sum_i Q_i^2 q_i^p(x) \)
This is the DIS \(e^-p\) scattering cross section under the parton model.
Comparing with the previous modified Rosenbuth formula, we have:
\(F_2(x, Q^2) = 2xF_1(x, Q^2) = x\sum_i Q_i^2 q_i^p(x)\)
Implication 1: electron scatters off
point-like quarks
; therefore, structure functions \(F_{1,2}(x)\) have no strong \(Q^2\) dependence.
Implication 2: quarks are spin-1/2 particles; therefore, quark's
magnetic
moment is
directly related
to
electric
charge, resulting in fixed relation between \(F_2(x)\) and \(F_1(x)\)
Structure functions for protons and neutrons
Structure function
Protons/neutrons are dynamical systems: quarks interact with gluons, creating
virtual sea quarks
.
Ignoring heavy quarks, we have \(F_2^{ep}(x) = x \Big( \frac{4}{9} u^p(x) + \frac{1}{9}d^p(x) + \frac{4}{9}\bar{u}^p(x) + \frac{1}{9} \bar{d}^p(x) \Big) \),
neutron expect to be
same as proton
with \(u\leftrightarrow d\) quarks exchanged (isospin symmetry).
Therefore, \(d^n(x) = u^p(x) \equiv u(x)\), \(u^n(x) = d^p(x) \equiv d(x)\); same for anti quarks.
Therefore, \(F_2^{ep}(x) = 2x F_1^{ep}(x) = x \Big( \frac{4}{9} u(x) + \frac{1}{9}d(x) + \frac{4}{9}\bar{u}(x) + \frac{1}{9} \bar{d}(x) \Big) \), \(F_2^{en}(x) = 2x F_1^{en}(x) = x \Big( \frac{4}{9} d(x) + \frac{1}{9}u(x) + \frac{4}{9}\bar{d}(x) + \frac{1}{9} \bar{u}(x) \Big) \)
Integrate over x: , \(\int_0^1 F_2^{ep} dx = \frac{4}{9} f_u + \frac{1}{9} f_d\), \(\int_0^1 F_2^{en} dx = \frac{4}{9} f_d + \frac{1}{9} f_n\)
where \(f_u = \int_0^1 x[u(x) + \bar{u}(x)]\), \(f_d = \int_0^1 x[d(x) + \bar{d}(x)]\)
e.g. \(f_u\) is the
fraction of the proton momentum carried by up + anti-up quarks.
Experimental results: \(f_u = 0.36, f_d = 0.18\).
Twice more
up-quark than down-quark in proton; remaining 50% of momentum is carried by
gluons
.
Valence/sea quarks
uud in proton:
valence
quarks.
Sea quarks
are also produced from \(g \rightarrow q\bar{q}\)
e.g., \(u(x)\) can be break down into \(u_V(x), u_S(x)\). Expect \(u_S(x) = d_S(x) = \bar{u}_S(x) = \bar{d}_S(x) = S(x)\)
Then \(F_2^{ep}(x) = x\Big(\frac{4}{9}u_V(x) + \frac{1}{9}d_V(x) + \frac{10}{9}S(x)\Big)\), \(F_2^{en}(x) = x\Big(\frac{4}{9}d_V(x) + \frac{1}{9}u_V(x) + \frac{10}{9}S(x)\Big)\)
Therefore, \(\frac{F_2^{en}}{F_2^{ep}} = \frac{4d_V(x) + u_V(x) + 10 S(x)}{4u_V(x) + d_V(x) + 10S(x)}\)
Since sea quark contribution is low at high x, and \(u_V = 2 d_V\), expect \(\frac{F_2^{en}}{F_2^{ep}} \rightarrow \frac{2}{3}\) as \(x \rightarrow 1\)
Experiment: ratio is ~ \(\frac{1}{4}\)!
Exclusion principle
: two u quark both at low x is disfavored -> one u is high-x, one is low-x. Then \(d_V/u_V \rightarrow 0\), so total ratio is
1/4
.
Since
sea quark dominates at low x
(small momentum due to \(\frac{1}{q^2}\) suppression, expect \(\frac{F_2^{ep}(x)}{F_2^{en}(x)} = 1\) as \(x \rightarrow 0\)
Confirmed by experiment
HERA
ep collider at DESY; study high-\(Q^2\) low-x inelastic scattering.
Overall
consistent with Bjorken scaling
(structure function independent of x). Implies proton radius \(r_p < 10^{-18}m\).
Scaling violation: at high \(Q^2\), a larger fraction of quarks are found to have low-x value.
Why? High-\(Q^2\) is sensitive to
quarks (having small x) radiating gluons
! \(q \rightarrow qg\).
