# Proof that Bjorken-$x$ is positive

For a $$t$$-channel deep inelastic scattering process the Bjorken-$$x$$ is defined as:

$$x=\frac{Q^2}{2p_2\cdot q},$$

where $$Q^2:=-q^2$$ [in the $$(+,-,-,-)$$ Minkowski sign convention], $$q$$ is the transferred four-momentum, and $$p_2$$ is the four-momentum of one of the particles before the collision. $$0\leq x\leq1$$. What is the proof that $$0 \leq x$$?

• Could it be that $q^{\mu}=(0,\textbf{p}_2-\textbf{p}_1)$, and therefore $p_2\cdot q$ is also negative? Oct 28, 2023 at 13:27

In the Bjorken scattering we look at scattering of an electron beam at high velocities with a massive nucleus $$N$$, with the four-momentum $$p_2$$ and mass $$m$$. Denote the momentum of the electron before scattering as $$k$$ and after the scattering as $$k^\prime$$, with energies $$E$$ and $$E^\prime$$, therefore $$q = k - k^\prime \, .$$ Now we can choose the reference of frame in which the nucleus $$N$$ is at rest and the electron is moving, i.e. $$p_2=(m, 0, 0, 0) \; \text{and}\; k=(E, \vec{k}) \,$$ where we set the mass of the electron $$m_e=0$$, as it is negligible at high velocities, therefore $$|\vec{k}| = E$$. After the scattering we have $$k^\prime = (E^\prime, \vec{k^\prime}), \; |\vec{k^\prime}| = E^\prime.$$ Denote the scattering angle between $$k$$ and $$k^\prime$$ by $$\theta$$. We then have $$-q^2 = -k^2 + 2kk^\prime - k^{\prime 2} = -0 + 2kk^\prime - 0 = 2EE^\prime -2EE^\prime \cos(\theta) = 2EE^\prime(1-\cos(\theta)) \\= 4EE^\prime\sin^2(\theta/2) \geq0.$$ As for the denominator $$p_2 q = p_2k - p_2k^\prime = m(E-E^\prime) \geq 0,$$ because we set the nucleon to be at rest, so the incoming electron will lose some energy to the nucleus through the scattering process.