1
$\begingroup$

Consider the leading-order (LO) DGLAP ((Dokshitzer–Gribov–Lipatov–Altarelli–Parisi) equation $$x \mu^2 \frac{d xg(x,\mu^2)}{d\mu^2}= \alpha_s \int_x^1 dz P_{gg}(z) \frac{x}{z} g(\frac{x}{z}, \mu^2) + \dots\tag1$$

Define an unintegrated parton density function (PDF) by $$F(x,\mu^2) \equiv x \frac{dg(x,\mu^2)}{d\mu^2},$$ with the initial condition $$\mu^2 F^{(0)}(x,\mu^2) = \theta(1-x) \theta \left(\frac{\mu^2}{Q_o^2}-1\right),\tag2$$ where $Q_o$ is some non perturbative scale. The first order contribution to $F$ is then given by inserting $(2)$ into $(1)$ so that $$\mu^2 F^{(1)}(x,\mu^2) \approx \int_x^1 dz P_{gg}(z) \int_{Q_o^2}^{\mu^2} dk^2 \alpha_s(k^2) F^{(0)}(\frac{x}{z}, k^2)$$

My questions are

1)From the definition of $F$ it follows that $$g(x,\mu^2) = x^{-1} \int^{\mu^2} dk^2 F(x,k^2)$$ so I can see how the integral in the last display arises but what I don't see is why does this correspond to the first order contribution to $F$? Doesn't equation (1) tell us that the evolution of the LO gluon PDF is given by the integral convolution of the LO splitting function kernels with the LO gluon PDF itself? So where does next-leading order (NLO) effects, i.e that suggested by 'first order in $F$' arise?

2)What are the physical meanings of the $F^{(i)}$? Are these defined through some expansion and, if so, what is this expansion in?

$\endgroup$
3
  • $\begingroup$ Interesting: DGLAP equations played a big role in my research when I was still in particle physics but I have never seen this method to solve DGLAP equations. We solved them numerically using Mellin transforms. Would you share a reference? $\endgroup$
    – user154997
    Aug 12 '17 at 20:30
  • $\begingroup$ @luc thanks for your comment - I'm reading this from a thesis published in 2013, I don't think it's on the arxiv however. $\endgroup$
    – CAF
    Aug 12 '17 at 21:19
  • $\begingroup$ It would probably be better to narrow this down to one question, and ask your other question separately. $\endgroup$
    – David Z
    Aug 13 '17 at 0:40
1
$\begingroup$

First, the DLAP equation at any order can be written

$$\mu^2x\frac{dg(x,\mu^2)}{d\mu^2} = \int_x^1 dz P_{gg}(z) \frac{x}{z} g\left(\frac{x}{z}, \mu^2\right) + \cdots$$

Then, the kernel $P_{gg}(z)$ is expanded in powers of $\alpha_S$:

$$P_{gg}(z) = P_{gg}^{(0)}(z)\frac{\alpha_S}{2\pi} + P_{gg}^{(1)}(z)\left(\frac{\alpha_S}{2\pi}\right)^2 + \cdots$$

Then LO means only the first term and NLO means the first two terms are included. Thus I think that the wording "first order contribution" used in the work you study for $F^{(1)}$ is a different notion. Basically, this is a scheme to solve the DGLAP equations by successive approximations: the idea would be that $F^{(n)}$ converges to the solution as $n$ increases, I think.

This should work as follow. After defining (beware I added a factor $\mu^2$ on the right-hand side so as to keep the equations more symmetric later on)

$$F(x,\mu^2) \equiv \mu^2 x \frac{dg(x,\mu^2)}{d\mu^2},\label{defF}\tag{II}$$

the DGLAP equation reads

$$F(x,\mu^2)=\int_x^1 dz P_{gg}(z) \frac{x}{z} g\left(\frac{x}{z}, \mu^2\right) + \cdots\label{dglapalt}\tag{I}$$

we start from a first approximation given by your equation (2), to which correspond an approximation $g^{(0)}$ of $g$ given by

$$xg^{(0)}(x,\mu^2) = \int_{Q_0^2}^{\mu^2} \frac{dk_0^2}{k_0^2} F^{(0)}(x, k_0^2).$$

I simply integrated with respect to $\mu^2$. This approximation can then be inserted in the right-hand side of the DGLAP equation (\ref{dglapalt}) to give

$$F^{(1)}(x,\mu^2) \approx \int_x^1 dz_0 P_{gg}(z_0) \int_{Q_o^2}^{\mu^2} \frac{dk_0^2}{k_0^2} F^{(0)}\left(\frac{x}{z_0}, k_0^2\right)+\cdots$$

Then from $F^{(1)}$, we can get $g^{(1)}$ by integrating with respect to $\mu^2$ again,

$$xg^{(1)}(x,\mu^2) = \int_{Q_0^2}^{\mu^2} \frac{dk_1^2}{k_1^2} F^{(1)}(x, k_1^2)=\int_x^1 dz_0 P_{gg}(z_0)\int_{Q_0^2}^{\mu^2}\frac{dk_1^2}{k_1^2}\int_{Q_0^2}^{k_1^2} \frac{dk_0^2}{k_0^2} F^{(0)}\left(\frac{x}{z_0},k_0^2\right) + \cdots$$

then insert $g^{(1)}$ in the right-hand side of (\ref{dglapalt}) to get you $F^{(2)}$,

$$\begin{aligned} F^{(2)}(x,\mu^2)&=\int_x^1 dz_1 P_{gg}(z_1) \frac{x}{z_1} g^{(1)}\left(\frac{x}{z_1}, \mu^2\right) + \cdots\\ &=\int_x^1 dz_1 \int_{z_1}^1 dz_0 P_{gg}(z_1) P_{gg}(z_0)\int_{Q_0^2}^{\mu^2}\frac{dk_1^2}{k_1^2}\int_{Q_0^2}^{k_1^2} \frac{dk_0^2}{k_0^2} F^{(0)}\left(\frac{z_1}{z_0},k_0^2\right) + \cdots \end{aligned}$$

We can then continue by computing $g^{(2)}$, etc. Each step of this procedure add one more splitting. This can probably be connected to the origin of the DGLAP equations, which appear as the result of the resummation of a class of amplitudes featuring 1, 2, 3, etc colinear gluon emissions.

$\endgroup$
9
  • $\begingroup$ Thanks for your answer! Since you've had some background in working with these things, perhaps I could ask you some more questions? What would the NLO DGLAP equation look like? I think it's correct to think of the evolution of a NLO gluon PDF as the integral conv. of a gluon PDF with NLO splitting functions rather than a NLO gluon as being a term in some kind of expansion of the gluon PDF. So maybe the NLO DGLAP equation would be an additional term with $\alpha_s^2P_{gg}^{(1)}$ convoluted with $xg$ also? $\endgroup$
    – CAF
    Aug 13 '17 at 10:33
  • $\begingroup$ What you wrote after "I think it's correct" is… correct indeed! As I tried to state in my answer, NLO means including the terms of order $\alpha_S^2$ in $P_{gg}$ and all the others. So the equation is still the same (1) but with different $P_{gg}$, just to be crystal clear. Note that, to state the obvious, when you plug PDF's into cross sections, you will also need NLO expressions for the parton-parton cross sections, otherwise it makes no sense. $\endgroup$
    – user154997
    Aug 14 '17 at 9:33
  • $\begingroup$ Thanks! So just to check I understand fully, equation (1) to NLO instead should read $$x \mu^2 \frac{dxg(x,\mu^2)}{d\mu^2} = \frac{\alpha_s}{2\pi} \int_x^1 dz P_{gg}^{(0)}(z) \frac{x}{z} g(\frac{x}{z}, \mu^2) + \frac{\alpha_s^2}{(2\pi)^2} \int_x^1 dz P_{gg}^{(1)}(z) \frac{x}{z} g(\frac{x}{z}, \mu^2)?$$ $\endgroup$
    – CAF
    Aug 14 '17 at 12:34
  • $\begingroup$ Yes, exactly. . $\endgroup$
    – user154997
    Aug 14 '17 at 12:59
  • $\begingroup$ Thanks - so just to reiterate the idea is that the thesis is solving (1) by successive approximations via the $F^{(i)}$ and the summation of more and more $F^{(i)}$ converges to (1)? So from your answer I think I can understand conceptually what is going on but do you know (and could possibly elaborate in your answer) how that equation for $\mu^2 F^{(1)}$ is obtained? It's just how using (1) along with the definition of $F$ that I don't see how the $F^{(i)}$ appear. +1 $\endgroup$
    – CAF
    Aug 15 '17 at 9:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.