# 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?

• Does this answer your question? Why is it that the Callan-Gross relation predicts that quark has spin 1/2? Jan 31 '20 at 12:08
• No, because I want to derive it specifically via the scattering coefficients $\hat{W}^{\mu\nu}$, not via Rosenbluth and Mott. It should be a simple error when executing the integration over the delta but it is eluding me. Jan 31 '20 at 12:21

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,

$$\hat{B}^{\mu\nu} = \left(p_{i}^{\mu}-\frac{p_{i} \cdot q}{q^{2}} q^{\mu}\right)\left(p_{i}^{\nu}-\frac{p_{i} \cdot q}{q^{2}} q^{\nu}\right)$$

$$B^{\mu\nu} = \left(\xi P^{\mu}-\frac{\xi P \cdot q}{q^{2}} q^{\mu}\right)\left(\xi P^{\nu}-\frac{\xi P \cdot q}{q^{2}} q^{\nu}\right)$$
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).