Solution of polarised singlet DGLAP evolution equations by the method of characteristics

0.23 MB·English

by  D. K. Choudhury

Solution of polarised singlet DGLAP evolution
equations by the method of characteristics
6
0
0 D. K. Choudhury
∗
2
Department of Physics, Gauhati University, Guwahati-781014, India
n
a
J P. K. Sahariah
†
9
Department of Physics, Cotton College, Guwahati-781001, India

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v
9
5
0
Abstract
1
0
The polarised singlet coupled DGLAP equations (LO) are trans-
6
0 formed by a Taylor series expansion at low x into a pair of partial
h/ differential equations in x and t(t = lnQ2/Λ2). The pair of coupled
p partial differential equations is then reduced to canonical form and
-
p the resultant system solved by applying the method of characteristics
e under small x and t approximations. The result is compared with
h
some exact solutions available in the literature.
:
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1 Introduction
r
a
The measurement of the polarized structure functions of the proton by the
European Muon Collaboration in 1988 revealed that the structure of the pro-
ton is more profound to be understood only from the unpolarized DIS. Their
result[1, 2], which is often referred as the ’proton spin problem’ indicated
that the contribution of the quarks to the nucleon spin is very small. The
rest of the spin then must be carried by the gluon and/or by the angular
momentum of quarks and gluons. The spin distributions among the different
∗e-mail: [email protected]
†e-mail: [email protected]
1
contributors is expressed by a sum rule [3, 4]
1 1
= ∆Σ+∆G+L +L , (1)
q g
2 2
where ∆Σ is the contribution from the quarks, ∆G from the gluons and L
q
, L are the orbital angular momentum of quarks and gluons to the total
g
spin of the nucleon. In the static quark model with SU(6) symmetry, the
spin is entirely carried by the quarks i.e ∆Σ = 1. In the naive quark-parton
model, ∆Σ is predicted to be 0.7. These are in contradiction to the EMC
∼
result. Based on the next-to-leading order QCD analysis of the deep inelastic
polarized data, there are several model parametrization [5, 6, 7, 8, 9, 10] of
the polarized parton distribution functions. All of these models predict that
the value of ∆Σ is around 0.3 or less; that is, the quark contribution to the
nucleon spin is very small contrary to our expectation. In view of this, the
polarized gluon distribution has attracted in recent years special interest in
polarized DIS, because, a rather large contribution of the first moment of
the gluon distribution is essential to compensate for the low value of the
quark contribution to the total helicity of the nucleon. Recent data [11]
and analyses [6, 7, 8, 9, 10] suggest that the contribution of the polarized
gluon momentum is quite large, though the error involved is too large to
significantly constrain its contribution. It is therefore imperative to know
how the gluon is polarized inside the nucleon. In this paper we investigate
the polarized gluon and the quark singlet contribution to the nucleon spin
with the help of the DGLAP evolution equations. We adopt the method of
characteristics to solve the coupled polarized DGLAP evolution equations.
In 2 we discuss the formalism and in 3 we discuss our results.
§ §
2 Formalism
For Q2 evolution it is convenient to introduce a flavour singlet combination
of polarized quark and anti-quarks by
∆Σ(x,Q2) = (∆q(x,Q2)+∆q¯(x,Q2)), (2)
q
X
where the sum is over all quark and anti-quark flavours. Here ∆q(x,Q2) =
(q+(x,Q2) q−(x,Q2)), where q+(x,Q2) ,( q−(x,Q2)) refers to the quark
−
2
distribution with helicity parallel (anti-parallel) to the parent proton. The
polarized gluon distribution is similarly defined as
∆g(x,Q2) = g(+)(x,Q2) g(−)(x,Q2). (3)
−
The Q2 evolution of the polarized singlet and the gluon distribution mix
non-trivially and is given by the DGLAP [12, 13, 14] equations
∂ ∆Σ(x,t) α (t) ∆P 2n ∆P ∆Σ(x,t)
= s qq f qg , (4)
∂t ∆g(x,t) ! 2π ∆Pgq ∆Pgg !⊗ ∆g(x,t) !
where t = lnQ2 and ∆P are polarized splitting functions which are known
Λ2 ij
at LO [14] and NLO [15, 16, 17]. The symbol stands for the usual Mellin
⊗
convolution in the first variable defined as
1 dy x
(∆P ∆q)(x,t) = ∆P ∆q(y,t). (5)
× x y y!
Z
The polarized parton distributions at an initial scale enter as boundary val-
ues in the solution of the above equations. However, with exact splitting
functions, analytic solutions in the entire range of x is not possible. The
splitting functions can be simplified by taking their moments and expanding
around the rightmost singularity at N = 0 where N is the moment variable.
In this procedure the splitting functions[9] at LO get simplified to
∆P(0)(x) = 4[1+ 1δ(1 x)], ∆P(0)(x) = 1 2n [ 1+2δ(1 x)],
∆Pq(q0)(x) = 43[2 δ2(1 −x)],∆P(0)(xq)g = 3[4 2 13δf(−1 x)] n−fδ(1 x) . (6)
gq 3 − − gg − 6 − − 3 − )
With these simplified splitting functions, Eqs.(4) can be solved analytically
with some approximations about small x behaviour of these to be discussed
later. The two equations Eqs.(4) can be written separately in LO as
∂ α (t)
∆Σ(x,t) = s ∆P(0) ∆Σ(x,t)+2n ∆P(0)(x) ∆g(x,t) (7)
∂t 2π qq ⊗ f qg ⊗
h i
and
∂ α (t)
∆g(x,t) = s ∆P(0) ∆Σ(x,t)+∆P(0)(x) ∆g(x,t) . (8)
∂t 2π gq ⊗ gg ⊗
h i
Using the convolution integral(Eq.5), we write Eq.(7) and Eq.(8) as
∂ α (t) 1 x 1 x
∆Σ(x,t) = s dz∆P(0)(z)∆Σ( ,t)+2n dz∆P(0)(z)∆g( ,t) ,
∂t 2π (cid:20)Zx qq z f Zx qg z (cid:21)
(9)
3
∂ α (t) 1 x 1 x
∆g(x,t) = s dz∆P(0)(z)∆Σ( ,t)+ dz∆P(0)(z)∆g( ,t) .
∂t 2π (cid:20)Zx gq z Zx gg z (cid:21)
(10)
x
Introducing a variable u as u = 1 z, we can write as a series
− z
x x ∞
= = x uk. (11)
z 1 u
− kX=0
x x
Now we expand ∆Σ( ,t) and ∆g( ,t) in Eq.(9) and Eq.(10) in Taylor series
z z
using Eq.(11) as
x ∞ ∞ ∂∆Σ(x,t)
∆Σ( ,t) = ∆Σ(x+x uk,t) = ∆Σ(x,t)+(x uk) +... (12)
z ∂x
k=1 k=1
X X
and
x ∞ ∞ ∂∆g(x,t)
∆g( ,t) = ∆g(x+x uk,t) = ∆g(x,t)+(x uk) +..... (13)
z ∂x
k=1 k=1
X X
At low x the above series Eq.(12) and Eq.(13) are convergent[18]. Hence,
neglecting higher order terms, the series can be approximated as
x ∞ ∂∆Σ(x,t)
∆Σ( ,t) ∆Σ(x,t)+(x uk) (14)
z ≈ ∂x
k=1
X
and
x ∞ ∂∆g(x,t)
∆g( ,t) ∆g(x,t)+(x uk) . (15)
z ≈ ∂x
k=1
X
Using Eq.(14) and Eq.(15) and the splitting functions given by Eqs.(6), we
can write Eq.(9) and Eq.(10) as
β t∂∆Σ(x,t) ∂∆Σ(x,t) ∂∆g(x,t)
0 = I(1)(x)∆Σ(x,t)+I(2) +I(1)(x)∆g(x,t)+I(2)
2 ∂t q q ∂x g g ∂x
(16)
and
β t∂∆g(x,t) ∂∆Σ(x,t) ∂∆g(x,t)
0 = I(3)(x)∆Σ(x,t)+I(4) +I(3)(x)∆g(x,t)+I(4)
2 ∂t q q ∂x g g ∂x
(17)
4
4π
j
where we have used α (t) = in LO. The quantities I (i = q,g;j =
s β t i
0
1,2,3,4) in Eq.(16) and Eq.(17) are given by
4 1 dz 1
I(1)(x) = 1+ δ(1 z) , (18)
q 3 Zx z (cid:18) 2 − (cid:19)
4 1 dz 1 ∞
I(2)(x) = 1+ δ(1 z) x uk , (19)
q 3 Zx z (cid:18) 2 − (cid:19) k=1 !
X
1 1 dz
I(1)(x) = 2n ( 1+2δ(1 z)) , (20)
g f2 x z − −
Z
1 1 dz ∞
I(2)(x) = 2n ( 1+2δ(1 z)) x uk , (21)
g f2 x z − − !
Z k=1
X
4 1 dz
I(3)(x) = (2 δ(1 z)) , (22)
q 3 x z − −
Z
4 1 dz ∞
I(4)(x) = (2 δ(1 z)) x uk , (23)
q 3 x z − − !
Z k=1
X
1 dz 13 n
I(3)(x) = 3 4 δ(1 z) fδ(1 z) (24)
g Zx z (cid:20) (cid:18) − 6 − (cid:19)− 3 − (cid:21)
and
1 dz 13 n ∞
I(4)(x) = 3 4 δ(1 z) fδ(1 z) x uk . (25)
g Zx z (cid:20) (cid:18) − 6 − (cid:19)− 3 − (cid:21) k=1 !
X
Carrying out the integrations in Eqs.(18-25), we recast Eq.(16) and Eq.(17)
as two coupled partial differential equations in two variables x and t as :
∂∆Σ(x,t) ∂∆g(x,t) ∂∆Σ(x,t) ∂∆g(x,t)
a′ +a′ +b′ +b′
11 ∂t 12 ∂t 11 ∂x 12 ∂x
= R′ ∆Σ(x,t)+R′ ∆g(x,t) (26)
11 12
and
∂∆Σ(x,t) ∂∆g(x,t) ∂∆Σ(x,t) ∂∆g(x,t)
a′ +a′ +b′ +b′
21 ∂t 22 ∂t 21 ∂x 22 ∂x
= R′ ∆Σ(x,t)+R′ ∆g(x,t), (27)
21 22
5
where
a′ = t, a′ = 0
11 12 , (28)
a′ = 0, a′ = t
21 22 )
b′ = 4 1 x xln1 , b′ = 2n 1 1+x+xln1
11 −3 − − x 12 − f 2 − x (29)
b′ = 4 (cid:16)1 2x xln1(cid:17) , b′ = 12 1 (cid:16) x xln1 (cid:17) 
21 −3 − − x 22 − − − x 
(cid:16) (cid:17) (cid:16) (cid:17)
and 
R′ = 4 1 +ln1 , R′ = 2n 1 2 ln1
11 3 2 x 12 f 2 − x . (30)
R′ = 4 (cid:16)2ln1 1(cid:17) , R′ = 12ln(cid:16)1 13 +(cid:17) nf 
21 3 x − 22 x − 2 3 
(cid:16) (cid:17) h (cid:16) (cid:17)i
Eq.(26) and Eq.(27) represent a system of first order coupled partial differ-
ential equations. We can reduce these equations to canonical form and then
solve by the method of characteristics. To do that, we introduce a vector
∆~u(x,t) defined by
∆Σ(x,t)
∆~u(x,t) = (31)
∆g(x,t)
!
and express the two equations Eq.(26) and Eq.(27) in matrix form
a′∆~u (x,t)+b′∆~u (x,t) = R′∆~u(x,t), (32)
t x
where the matrices a′, b′ and R′ are
a′ a′
a′ = 11 12 , (33)
a′ a′
21 22 !
b′ b′
b′ = 11 12 (34)
b′ b′
21 22 !
and
R′ R′
R′ = 11 12 , (35)
R′ R′
21 22 !
the elements being given by Eqs.(28-30) respectively. The matrix a′ being
non-singular, we multiply Eq.(32) from left by a′−1 and get
∆~u (x,t)+A′∆~u (x,t) = B′∆~u(x,t), (36)
t x
6
where the new matrices A′ and B′ are:
A′ = a′−1b′ (37)
and
B′ = a′−1R′. (38)
Eq.(32) represents a system of two coupled first order partial differential
equationinthetwo variablesx andt forthevector∆~u prescribed byEq.(31).
Its principal part, i.e. ∆~u (x,t)+A′∆~u (x,t) is completely characterized by
t x
the coefficient matrix A′. Since the matrix A′ has n [here n=2] distinct
eigenvalues, the system Eq.(36 ) is a hyperbolic one and it is possible to
obtain its canonical form in the following way [19, 20]: Let λ′(i)(i = 1,2) be
the eigenvalues of the matrix A′ (Eq.37) and P′ be the matrix formed by
the corresponding eigenvectors. Now, if Λ is the diagonal matrix with the
eigenvalues λ and λ as the two elements, then we have
1 2
λ 0
P′−1A′P′ Λ = 1 . (39)
≡ 0 λ′2 !
Let ∆~v(x,t) be a new function defined in terms of ∆~u(x,t) by the equation
∆~u(x,t) = P′.∆~v(x,t). (40)
so that
∆~v = P′−1.∆~u. (41)
Differentiating Eq.(40) with respect to t and x respectively we get
∆~u = P′∆~v +P′∆~v, ∆~u = P′∆~v +P′∆~v. (42)
t t t x x x
Substituting Eqs.(40) and (42) in Eq.(32), we obtain
P′∆~v +P′∆~v +A′P′∆~v +A′P′∆~v = B′P′∆~v. (43)
t t x x
Multiplying Eq.(43) from left by P′−1 and using Eq.(39), we obtain
∆~v +Λ∆~v = ∆~e, (44)
t x
where
∆~e = P′−1(B′.P′ P′ A′.P′).∆~v. (45)
− t − x
7
Eq.(44) is in canonical form. In components, it is
∆v(i) +λ′(i)∆v(i) = ∆e(i) (i = 1,2), (46)
t x
∂∆v(i) ∂∆v(i)
Eq.(46) shows that the principal part viz. +λ′(i) involves only
∂t ∂x
thecomponent∆v(i) ofthevector∆~v anditsderivativesi.e. theequationsare
decoupled. These equations (Eq.46) can be reduced to ordinary differential
equations
d∆v(i)
= ∆e(i)(x,t,∆v(i)(x,t)) (i = 1,2) (47)
dt
along the characteristic curves defined by the equations
dx(i)(t)
λ′(i)(x,t) = . (48)
dt
In order to integrate the ordinary differential equations (Eq.47) to get an-
alytical forms for ∆~v, one has to express the right hand side in terms of t.
This is done from the solution of the characteristic equations Eq.(48), which
expresses x(i) as a function of t. However, the integration of Eq.(48) to get
such an expression depends on the nature of the eigenvalues λ′(i) of the ma-
trix A′. For simple forms of the eigenvalues, it is possible to integrate Eq.(47)
analytically. We discuss this below.
The eigenvalues of the matrix A′ are
1
λ′(1,2) = (b′ +b′ s), (49)
9t 11 22 ±
where
s = b′2 +4b′ b′ 2b′ b′ +b′2 . (50)
11 12 21 − 11 22 22
q
Taking the elements of the matrices a′ and b′ as given by Eqs.(28-29), the
eigenvalues are found to be
2 1
λ′(1,2)(x,t) = 10+10x+10xln
27t − x
(cid:20)(cid:18) (cid:19)
1 12
55 101x+46x2 +( 110x+101x2)ln +55x2(ln )2 ,
±s − − x x 
(51)
8
where the + ( ) sign corresponds to λ′(1)(λ′(2)). And the two components of
−
the vector ∆~e are
∆e(1,2) =
1 (−b′11+b′22+s) −b′21sR′11−(−b′11+b2′2s2∓s)R′21 2b′21R′12 (−b′11+b′22∓s)R′22 ∆v
t ∓ (cid:16) 9b′21 (cid:17) ∓ 9s ∓ 9s  1
+ (−b′11+b′22−s) −b′21sR′11−(−b′11+b2′2s2∓s)R′21 2b′21R′12 (−b′11+b′22∓s)R′22 ∆v , (52)
∓ (cid:16) 9b′21 (cid:17) ∓ 9s ∓ 9s  2
  
where the upper sign (-) corresponds to ∆e(1) and the lower sign (+) to
∆e(2). We note that both λ′(i) and ∆e(i) are factorisable in x and t, the
quantities b′ and R′ being functions of x only. As we have already observed,
ij ij
integration of Eq.(47) depends on the analytical solution of the characteristic
equation . But with the above eigenvalues (Eq.(51)), analytical solution of
the characteristic equations(Eq.(48)) cannot be found and so we cannot find
the necessary transformation equation to express x as a function of t for the
integration of Eq.(47). But under certain extreme situations, it is possible
to get analytical solutions which we discuss below.
2.1 Solution when x 0
→
The eigen values of the matrix A′ can be simplified by making certain ap-
proximations about the elements of the matrix b′. Before we do that, we
write the eigenvalues in simple form. The eigenvalues being factorisable in x
and t can be written as
λ′(i)(x,t) = λ′(ia)(x)λ′(ib)(t) (i = 1,2), (53)
where
2 1
λ′(ia)(x) = 10+10x+10xln
27 − x
(cid:20)(cid:18) (cid:19)
1 12
55 101x+46x2 +( 110x+101x2)ln +55x2ln (54)
±s − − x x 

and
1
λ′(ib)(t) = . (55)
t
9
In Eq.(54), +(-) sign corresponds to i=1 (2) respectively. Now in the limit
x 0, Eq.(54) takes the values
→
2
λ′(ia) ( 10 √55). (56)
→ 27 − ±
The characteristic equation Eq.(48) now takes the form
dx(i)(t)
= λ′(ib)(t)dt. (57)
λ′(ia)(x)
Integrating Eq.(57) we get
x(i)(t) = α(i)lnt+Ci, (58)
where
2
α(i) = ( 10 √55), i = 1,2 (59)
27 − ±
and C(i) are two constants of integration. To get the constants of integration,
let (x¯,t¯) be a fixed point [19] in the (x,t) plane through which the two
characteristic curves Eq.(58) pass through. That is, x(1)(t¯) = x¯ and x(2)(t¯) =
x¯. Then from the Eqs.(58) we get
t
x(i)(t) = x¯+α(i)ln . (i = 1,2) (60)
t¯
(cid:18) (cid:19)
These are the two characteristic curves that pass through a common point
(x¯,t¯) up to which we can evolve the two functions ∆v(1) and ∆v(2) defined
by Eq.(41). Furthermore, if the two characteristic curves cut the initial line
t = t (t = lnQ20) at x(1)(t ) = τ′ and x(2)(t ) = τ′ respectively, then
0 0 Λ2 0 1 0 2
t
τ′ = x¯+α(1)ln 0 (61)
1 t¯
(cid:18) (cid:19)
and
t
τ′ = x¯+α(2)ln 0 . (62)
2 t¯
(cid:18) (cid:19)
Hence the two characteristic equations corresponding to the two eigenvalues
are respectively
t
x(1)(t) = τ′ +α(1)ln (63)
1 t
(cid:18) 0(cid:19)
10