Question

The origin of the scaling violation of $q\left(x, Q^2\right)$ given by (10.37) can be traced back to (10.32). There, we assumed that $\alpha_s$ is a constant. Show that $(10.37)$ is also obtained for a running coupling constant. Assume that $\alpha_s$ in (10.32) is $\alpha_s\left(p_T^2\right)$ as given by (7.65).

   The origin of the scaling violation of $q\left(x, Q^2\right)$ given by (10.37) can be traced back to (10.32). There, we assumed that $\alpha_s$ is a constant. Show that $(10.37)$ is also obtained for a running coupling constant. Assume that $\alpha_s$ in (10.32) is $\alpha_s\left(p_T^2\right)$ as given by (7.65).

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Quarks and leptons: introductory course in modern particle physics

Quarks and leptons: introductory course in modern particle physics

Francis Halzen, Alan… 1st Edition

Chapter 10, Problem 7

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Equation (10.37) likely describes the scaling violation of the parton distribution function $q(x, Q^2)$, which means that $q(x, Q^2)$ does not solely depend on the variable $x$ but also varies with the scale $Q^2$. Equation (10.32) might be an expression or an Show more…

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The origin of the scaling violation of $q\left(x, Q^2\right)$ given by (10.37) can be traced back to (10.32). There, we assumed that $\alpha_s$ is a constant. Show that $(10.37)$ is also obtained for a running coupling constant. Assume that $\alpha_s$ in (10.32) is $\alpha_s\left(p_T^2\right)$ as given by (7.65).

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