# Deriving Callan-Gross from Parton Model

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I want to derive the Callan-Gross relation from the parton model, not from the Rosenbluth and Mott cross sections, but I am having some problems obtaining the textbook result. I am following *M.D. Schwartz: Quantum Field Theory and the Standard Model* (pp.672, 675, 678).

Starting from the hard scattering coefficient obtained from the partonic scattering amplitude for $\gamma^\ast q_i\rightarrow q_i$ (eq. 32.32),
$$\hat{W}^{\mu\nu}(z,Q^2)=2\pi Q^2_i\delta(1-z)\left[A^{\mu\nu}+\frac{4z}{Q^2}B^{\mu\nu}\right],$$
where $A^{\mu\nu}:=-g^{\mu\nu}+\frac{q^\mu q^\nu}{Q^2}$, $B^{\mu\nu}:=\left(p^\mu+\frac{pq}{Q^2}q^\mu\right)\left(p^\nu+\frac{pq}{Q^2}q^\nu\right)$, and the convolution formula for the hardonic tensor $W^{\mu\nu}(x,Q^2)$ obtained from factorisation, we arrive at
\begin{align*}
W^{\mu\nu}(x,Q^2)
&=2\pi\int^1_x\frac{d\xi}{\xi}\sum_if_i(\xi)Q^2_i\delta(1-\frac{x}{\xi})\left[A^{\mu\nu}+\frac{4x}{Q^2\xi}B^{\mu\nu}\right]\\
&=2\pi\int^1_xd\xi\sum_if_i(\xi)Q^2_i\delta(\xi-x)\left[A^{\mu\nu}+\frac{4x}{Q^2\xi}B^{\mu\nu}\right]\\
&=2\pi\sum_if_i(x)Q^2_i\left[A^{\mu\nu}+\frac{4}{Q^2}B^{\mu\nu}\right],\end{align*}
such that $W_1(x,Q^2)=2\pi\sum_if_i(x)Q^2_i=\frac{Q^2}{4}W_2(x,Q^2)$.

Now, the textbook says that the result should be $W_1(x,Q^2)=\frac{Q^2}{4x^2}W_2(x,Q^2)$ (eq. 32.23, 32.24). Did I make a mistake somewhere in my calculations?

recategorized Feb 4, 2020

The integration is done correctly. Make in the integral the substitution $\xi=x/u$ to replace the integration variable $\xi$ by $u$, and $d\xi/\xi$ by $-du/u$, reverse the integration bounds to get rid of the minus sign (since $u<1$) and you'll get the same expression. Thus the problem must lie elsewhere.

What happens to the converted z from line 2 to line 3?
Why to you use Q² among W arguments ? Similarily, for the missing square of Q²/4z ( instead of Q²/4z² ), it depends on notations.and arguments of W(), since the proportionality is sufficient to conclude on a -1/2 spin.

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