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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 $0 \leq x$?

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  • $\begingroup$ Could it be that $q^{\mu}=(0,\textbf{p}_2-\textbf{p}_1)$, and therefore $p_2\cdot q$ is also negative? $\endgroup$
    – schris38
    Oct 28, 2023 at 13:27

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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.

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