Download Homework Set No. 4, Physics 880.02

Homework Set No. 4, Physics 880.02
Deadline – Thursday, June 4, 2009
1. (20 pts) Similar to what we did in class, solve the DGLAP equation for gluon distribution
Z 1 0
αs
dx
2
2 ∂
G(x,
Q
)
=
γGG (x/x0 ) G(x0 , Q2 )
Q
∂Q2
2 π x x0
with
γGG (z) =
2 Nc
z
in the small-x asymptotics, but now with fixed coupling constant αs (independent of Q2 ). In
particular show that, in the saddle point approximation, the small-x asymptotics for gluon
distribution is given by
s
!
2
1
Q
α
N
s
c
ln ln 2 .
xG(x, Q2 ) ∼ exp 2
π
x
Q0
This is called double logarithmic approximation (DLA) of DGLAP, since we resum both ln x1
2
Q2
in the new parameter αs ln x1 ln Q
and ln Q
2.
Q2
0
0
2. (30 pts) Calculate the cross section for
gluon + gluon → quark + antiquark
at the Born level shown in the figure below. The figure is for the amplitude, which needs to
be squared and multiplied by appropriate factors to get the cross section.
p q
k1
q
k2
q
p
q
k1
k2
p q
q
k1
k2
You should find
3 αs2 2
dσ̂gg→qq̄
=
(t̂ + û2 )
Ep
d3 p
8 ŝ
1
4
1
− 2
9 t̂ û ŝ
2 δ(ŝ + t̂ + û)
(1)
with the Mandelstam variables ŝ = (k1 + k2 )2 , t̂ = (k1 − p)2 , û = (k2 − p)2 . The factor of 2 in
front of the δ-function in Eq. (1) comes from the fact that either the quark or the antiquark
can carry momentum p. (q and q̄ in the figure denote the quark and the antiquark. Time
flows upward.) Assume that quarks are massless.
As a starting point you may take the formula derived in class:
Ep
1
dσ̂gg→qq̄
2 δ(ŝ + t̂ + û) h|M |2 i
=
3
dp
16 (2 π)2 E1 E2
(2)
where E1 = k10 , E2 = k20 , and M is the amplitude (sum of all three diagrams drawn above).
Angle brackets h. . .i denote summation over final state quantum numbers (polarizations and
colors of the quarks) and averaging over initial state quantum numbers (polarizations and
colors of the gluons). The factor of 2 mentioned above in Eq. (1) is already included in
Eq. (2).
Hints:
One quick and dirty way of arriving at the solution is to use the fact that incoming gluons
are physical,
k µ λµ (k) = 0,
(3)
in the amplitude to eliminate some of the terms. Then after squaring the amplitude you
may replace the polarization sums by
X
∗µ λ (k) λν (k) → −gµν .
(4)
λ=±1
Alternatively you may use Eq. (4) without Eq. (3). However, if you follow this path you
would have to subtract ghost loop contributions (amplitudes with ghost-antighost pair in
initial state instead of gluons, see Sterman pp. 233-237). This strategy is a more systematic
way of arriving at the right answer.
Other ways of solving the problem include the use of explicit parameterizations for polarization vectors in calculating amplitudes. One may also use polarization sum
X
∗µ λ (k) λν (k) = −gµν +
λ=±1
k̄µ kν + kµ k̄ν
,
k · k̄
where for k µ = (k 0 , ~k) we defined k̄ µ = (k 0 , −~k).
The following γ-matrix formulas may be useful:
γ µ γ ν γµ = −2 γ ν
γ µ γ ν γ ρ γµ = 4 g ν ρ
γ µ γ ν γ ρ γ σ γµ = −2 γ σ γ ρ γ ν
2
tr[γ µ γ ν ] = 4 g µν
tr[γ µ γ ν γ ρ γ σ ] = 4 (g µν g ρσ + g µσ g νρ − g µρ g νσ ).
For color traces the following expressions may come in handy:
T a T a = CF 1
with
CF
Nc2 − 1
,
=
2 Nc
tr[T a T b T a T b ] = −
CF
2
f abc f abd = Nc δ cd .
3