1
$\begingroup$

Consider a process $$ \tag 1 N+N\to X+\text{all}, $$ where $N$ are nucleons and $X$ is some massive particle. Assuming the parton model is valid and there is vertex $PP'X$, where $P,P'$ are partons inside the nucleons, the process $(1)$ can be described as the $2 \to 1$ Drell-Yan like process $$ \tag 2 P+P'\to X $$ Following Drell and Yan, let's assume that the parton model is applicable if the lifetime of the virtual state with the partons inside the nucleon is much larger than the time of the interaction $(2)$, $$ \tag 3 t_{\text{lifetime}}\gg t_{\text{int}} $$ If it holds, parton can be assumed as free particle carrying amount $x$ of the nucleon momentum, and probabilistic interpretation of the process $(2)$ becomes valid (this approximation is known as impulse approximation).

Assuming that nucleons have sufficiently large momenta $|\mathbf p|$ at their CM frame, $t_{\text{lifetime}}$ can be estimated as (using time-energy uncertainty principle) $$ \tag 4 t_{\text{lifetime}} \simeq 2|\mathbf p|/m_{N}^{2} $$ As for the interaction time, naively I think that it can be estimated using Fermi Golden rule, $$ \tag 5 t_{\text{int}}^{-1}\simeq \frac{1}{V}\frac{1}{2E_{P}2E_{P'}}\int |M_{\text{hard}}|^{2}\delta(p_{X} - p_{P}-p_{P'})\frac{d^{3}\mathbf p_{X}}{(2\pi)^{3}2E_{X}}, $$ where $V\simeq m_{N}^{-3}\gamma_{N}^{-3}$ is the proper volume of the nucleon and $M_{\text{hard}}$ is the matrix element of the process $(2)$.

Assuming that this is correct, the validity criterion $(3)$ depends on the mass of $X$ particle and on its coupling to partons. However, in the literature (see this overview article for example, expression $(33)$ and corresponding discussion) the criterion $(3)$ is replaced by very simple relation $$ \tag 6 s_{PP'} \gg \Lambda_{\text{QCD}}^{2}, $$ where $s_{PP'} = m_{X}^{2}$ is partonic invariant mass. Therefore we replaced probability criterion by more special QCD perturbativity criterion.

My question is the following. Whereas $(3)$ is model dependent criterion (it depends on the type of nucleon, the coupling of $X$ to partons and $X$ mass, as is demonstrated by $(5)$-$(6)$), $(6)$ depends only on $X$ mass. I don't clearly understand why it is reasonable to replace $(3)$ by $(6)$, since naively $(3)$ can be violated while $(6)$ holds, and vice versa. Could you please comment on that?

P.S. In principle, my question is related to general process $2\to n$, but for clarity let's concentrate on $2 \to 1$ process.

$\endgroup$
  • $\begingroup$ Few comments since no reply yet: For DIS, the energy of the probe in the nucleon rest frame is $q^0 \sim Q^2/2xm_N$. The reciprocal of that is representative of the timescale of scattering event which need be $\gg$ than the typical partonic interaction time $\sim 1/m_N$ for justification of parton model. So we get the equalities $\tau_{scattering} \sim 1/q^0 \gg \tau_{int} \sim 1/m_N$ which is also implied if $q^0 \gg m_N \sim \Lambda_{QCD}$, not dissimilar to the 'perturbativity' condition you mention at the close of your post. $\endgroup$ – CAF Jul 5 '18 at 17:54
  • $\begingroup$ @CAF : thank you for the answer; however, some aspects are not clear for me. First of all, what is $x$ in nucleon rest frame? Then, how to obtain the expression for $q_{0}$? And also why the scattering time is related to $q_{0}$ by the relation $\tau_{\text{scattering}} \sim q^{-1}_{0}$? Assuming that this is due to uncertainty principle, $q_{0}$ has to be the energy of the excited state, while I don't clearly understand why it is so. $\endgroup$ – Name YYY Jul 5 '18 at 21:09
  • $\begingroup$ In the IMF frame of the proton, $x$ has the interpretation of momentum fraction carried by the struck parton. $x$ is written in terms of Lorentz invariants, e.g in DIS $x=Q^2/2P\cdot q$ but I think in rest frame of proton this interpretation simply no longer holds. In proton rest frame, $P^{\mu} = (m_P, 0,0,0)$ with $q^{\mu} = (q^0, 0,0,q_z)$ e.g. then the product $P \cdot q = m_P q^0$ so that $q^0 = P \cdot q/m_P$. Use the expression for $x$ as given above gives the result. Yes, it's all order of magnitude estimates via uncertainty principle using characteristic scales in the problem. $\endgroup$ – CAF Jul 6 '18 at 9:05

Your Answer

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

Browse other questions tagged or ask your own question.