In the second volume of The Quantum theory of Fields, Weinberg provides the inelastic cross-section for the scattering of an electron from a nucleon with four momentum $p$ based on the parton model:
$$ \tag{20.6.8} \frac{d^2\sigma}{d\Omega d\nu} = \frac{e^4}{4E_e^2} \frac{\cos^2(\theta/2)}{\sin^4(\theta/2)} \sum_i Q_i^2 \int_0^1 \left( 1 + \frac{q^2}{2m_N^2 x^2} \tan^2(\theta/2) \right) \delta\left(\nu - \frac{q^2}{2m_N x}\right)$$ (see section 20.6). Here:
- $Q_i$ is the charge of the parton $i$
- $E_e$ is the incident electron energy in the nucleon rest frame
- $q$ is the difference between the scattered and the incident electron four-momentum
- $m_N$ is the nucleon rest mass
- $\nu$ is the energy of the photon in the nucleon rest frame
- $d\Omega=\sin\theta d\theta d\phi$ is the solid angle into which the electron in scattered
I am familiar with the derivation of this formula from the electron-muon scattering cross section: the muon is identified with the parton and the electron mass is set to zero. However, Weinberg says that this can also be derived from the Fenyman amplitude for Compton scattering: $$ \tag{8.7.38} \sum_{\sigma, \sigma'}|M|^2 = \frac{e^4}{64(2\pi)^6 \nu \nu' p^0 p'^0} \left[ \frac{8(q\cdot q')^2}{(q\cdot p)(q'\cdot p)} + 32(e\cdot e')^2 \right]$$ Where (unprimed/primed quantities referring to incident/scattered particles):
- $\nu$, $\nu'$ are the photon energies in the charged particle initial rest frame
- $e$, $e'$ are the polarization vectors
- $q$, $q'$ are the photon four momenta
- $p^0$, $p'^0$ are the charged particle four momenta
He also remarks that in the gauge used to derive (8.7.38): $$ \sum_{e, e'} (e\cdot e')^2 = 2 + \frac{m^4 (q\cdot q')^2}{(q\cdot p)^2(q'\cdot p)^2} + 2 \frac{m^2 (q\cdot q')}{(q\cdot p)(q'\cdot p)} $$
The problem is that I don't see any way to fit a Compton process in the electron-parton scattering above; for instance, in the former process photons are always on-shell, while in the latter, the only photon is clearly off-shell. Similarly the on-shell/off-shell status is reversed for the charged particle.
Of course I could put the incident photon in (8.7.38) off-shell, integrate over $q'$ and $p'$, and obtain a tensor to combine with the leptonic tensor of the electron, but it seems a lot of work compared with assuming that the hadronic and the leptonic tensors are structurally the same from the start.
So the question is: how can I use (8.7.38) as a shortcut to derive (20.6.8)?
