Deriving Callan-Gross from Parton Model I want to derive the Callan-Gross relation from the parton model, not fom 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? 
 A: I also found
$W_{1}\left(x, Q\right)=2 \pi \sum_{i} f_{i}(x) Q_{i}^{2}=\frac{Q^{2}}{4} W_{2}\left(x, Q\right)$
from your starting point:
$W^{\mu \nu}\left(x, Q\right)=2 \pi \int_{x}^{1} \frac{d \xi}{\xi} \sum_{i} f_{i}(\xi) Q_{i}^{2} \delta\left(1-\frac{x}{\xi}\right)\left[A^{\mu \nu}+\frac{4 x}{Q^{2} \xi} B^{\mu \nu}\right]$

*

*I do not see a mistake in your equations.  I wonder if there is a mistake either in the orginal equation (32.32) or if the original equation was incorrectly combined with the convolution formula for the hadronic tensor (i.e. the starting point was incorrect).  I am curious about this.

Edit I: I think the textbook result is correct. I checked Peskin & Schroeder and found the same result at least for the imaginary part of $W_{1}$ and $W_{2}$. From p.626,
$$\operatorname{Im} W_{1}=\frac{y s}{4 x} \operatorname{Im} W_{2}$$
where p. 558 gives $Q^{2}=x y s$ which yields
$$\operatorname{Im} W_{1}=\frac{Q^2}{4 x^2} \operatorname{Im} W_{2}$$
as needed. So, you and I are doing something wrong and/or the Schwartz text has an error not in the result but in the preceding equations.
I checked the errata for both texts and no previously reported typos in these equations (although there can be mistakes that aren't yet caught).  I want to figure this out.
Edit II:  It was our mistake but not in the actual calculations. I talked to my friend Kora who pointed this out to me:
We need to distinguish between the partonic quantities and the hadronic quantities. Schwartz does this with the hats as we know, so that $\hat{W}^{\mu \nu}(z, Q)$ is the partonic tensor and $W^{\mu \nu}(x, Q)$ is the hadronic tensor.
The relationship between Bjorken variable $x$ (a measure of the fraction of hadron momentum carried by the struck parton) and the partonic version of $x$ (called $z$) is given in Eq. 32.30 as
$$z \equiv \frac{Q^{2}}{2 p_{i} \cdot q}$$
where the parton's momentum $p_i$ is given as a fraction of the hadron  momentum $P$,
$$p_{i}^{\mu}=\xi P^{\mu},$$
and since $x \equiv \frac{Q^{2}}{2 P \cdot q}$, we have $x=\xi z$.  All of this is clear in the book I think.
The part that we missed is that $B^{\mu \nu}$ also comes in partonic and hadronic versions, because it has a $p_i$ dependence.  This means that your $B^{\mu\nu}$ equation needs a hat,
$$\hat{B}^{\mu\nu} = \left(p_{i}^{\mu}-\frac{p_{i} \cdot q}{q^{2}} q^{\mu}\right)\left(p_{i}^{\nu}-\frac{p_{i} \cdot q}{q^{2}} q^{\nu}\right)$$
and its hadronic counterpart becomes
$$B^{\mu\nu} = \left(\xi P^{\mu}-\frac{\xi P \cdot q}{q^{2}} q^{\mu}\right)\left(\xi P^{\nu}-\frac{\xi P \cdot q}{q^{2}} q^{\nu}\right)$$
and we factor out $\xi^2$ to find $B^{\mu\nu}$ picking up a factor of $x^2$ (because $\xi \rightarrow x$ in the $d\xi$ integral).
Nice, right?
