Say you have the distribution function $\rho_i$ of a quark $i$ (depending on the momentum fraction of that quark), and a structure function $\hat{F}_i$ (depending on the "Bjorken variable" $x$ and the energy scale $q^2$) of such quark. i.e. a term describing its contribution to the hadronic tensor which describes some process, such as deep inelastic scattering of an electron off the hadron. Many sources, such as this Scholarpedia article by Guido Altarelli, say that in order to calculate the structure function of the whole hadron you have to combine the previous two function as the convolution \begin{equation} F(x,q^2)=\sum_i\int_x^1\rho_i(\xi)F_i\Bigl(\frac{x}{\xi},q^2\Bigr)\frac{1}{\xi}\,\mathrm{d}\xi. \end{equation} I can't find why it should have this form.

I tried to prove it as follows. If the quark/parton has a moment equal to $\xi p$ where $p$ is the momentum of the hadron, then its $x$ variable, defined as \begin{equation} x=-\frac{q^2}{2\xi p\cdot q}, \end{equation} is $1/\xi$ times the hadron's $x$, which has just $p$ instead of $\xi p$. (The $q$ vector, in deep inelasting scattering problems, is the momentum of the internal particle exchanged between the electron and the hadron. I don't know if there is a general definition of such $x$ outside of DIS problems.)

Therefore, if we have assigned a value of $x$ to the hadron, a parton with momentum fraction $\xi$ will be described by a structure function of the form $\hat{F}(x/\xi,q^2)$. In order to obtain the structure function of the hadron, I weigh this with the parton density distribution, and sum over all types of quarks, obtaining \begin{equation} F(x,q^2)=\sum_i\int_0^1\rho_i(\xi)\hat{F}_i\Bigl(\frac{x}{\xi},q^2\Bigr)\,\mathrm{d}\xi. \end{equation} My proof doesn't explain the factor $1/\xi$ in the integral, or the fact that the lower limit is $x$ and not $0$. Why is that so?

  • $\begingroup$ the article says, "Note that the integral is from x to 1, because the energy can only be lost by radiation before interacting with the photon (which eventually has to find a fraction x)." $\endgroup$
    – JEB
    Commented Apr 17, 2020 at 13:24
  • $\begingroup$ I don't really understand what the author wants to convey with that sentence. The "energy can be only lost by radiation before interacting with the photon" bit may refer to the fact that only the diagram in which the gluon is emitted by the incoming quark contributes to the logarithmic correction to $F_2$. But why does the photon have to find a fraction $x$? Where? $\endgroup$
    – yellon
    Commented Apr 17, 2020 at 22:17

1 Answer 1


So when you scatter an electron of fixed beam energy off a nucleon ($M$) and into some lab angle and momentum, there are 2 kinematic variables, which are converted into the (positive) four momentum squared $Q^2$ and energy loss, $\nu$.

The former can be shown to be the spatial resolution of the exchanged virtual photon, and the latter is converted into:

$$ x = \frac{Q^2}{2M\nu} $$

In the naive picture, if the nucleon is a bag of non-interacting point particles, then the cross section becomes independent of $Q^2$ and is only a function of $x$.

In the Bjorken-limit (infinite-momentum frame), the scattering is off a non-interacting charged point particle that carries a fraction $x$ of the nucleon's total momentum, and the structure functions $u(x), d(x), s(x), ...$ are interpreted as the quark content of the nucleon, and that is what experiments aim to measure. That is, you factor out the well-known point particle scattering, and the variations in the cross section are due entirely to the structure functions.

The problem is renormalization, running $\alpha_s$, finite resolution and all that, for which an explanation is seriously technical. Qualitatively, the quark content of a nucleon is scale dependent. For instance, at low resolution, there are 3 valence quarks. As you crank it up, low-$x$ shows sea quarks. At higher $Q^2$, a single sea quark becomes a quark plus gluons + other vacuum fluctuations, and so on.

So, when you observe a scattering event at $x$ with finite $Q^2$, you know that the virtual photon has "found" a quark with momentum fraction $x$, but that doesn't mean you're probing $u(x)$; rather you have probed $u(x')$ with $x \le x' < 1 $. This is interpreted as follows: The photon has resolved the quark at length scale $\sqrt{Q^2}$, and what was quark with momentum fraction $x'$ is really said quark that has radiated a gluon with momentum fraction $(x'-x)$, leaving you with a quark at $x$, right where your virtual photon saw it.

  • $\begingroup$ I'm confused by tour $x$s: do you refer to the "Bjorken $x$" (for lack of a better name...) or the momentum fraction of the parton? In the last paragraph, I interpret the first sentence as "you observe a scattering event at $x=Q^2/2M\nu$ with finite $Q^2$", since these are kinematic variables you can measure; but then how is $x$ also the momentum fraction of the struck quark? As far as I know they coincide only in the "naive" parton model, but not in its refined version. $\endgroup$
    – yellon
    Commented Apr 18, 2020 at 22:15

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