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I am considering DIS with an electron and a proton: $$ e^-(k) p^+(P) \rightarrow e^-(k') X(p_X) $$ I define $q = k-k'$ and $Q^2 = -q^2 > 0$. The variable $x_B = \frac{Q^2}{2P\cdot q}$ is should be kinematically only allowed to take values $$ 0 \leq x_B \leq 1. $$ I could not show this by only using $$ k - k' = P + p_X, \quad k^2 = k'^2 = m_e^2 \geq 0, \quad P^2 = m_p^2 \geq 0, \quad p_X^2 \geq 0. $$ It seems to me like one needs that $p_X^2 \geq m_p^2$, from which $0 \leq x_B \leq 1$ follows trivially.

In https://arxiv.org/abs/1106.5607 it is said $p_X^2 \geq m_p^2$ is due to baryon number conservation. Can someone please explain the reasoning?

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    $\begingroup$ Consider to spell out acronyms. $\endgroup$
    – Qmechanic
    Jan 17, 2020 at 15:12

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Please note that one of your equalities is wrong, it should be: $k-k'=p_X-P$.

A simple way to prove that $0\leq x_B \leq 1$ is by writing: $x_B=\frac{Q^2}{2P\cdot q}=\frac{Q^2}{(P+q)^2-m_p^2+Q^2}$. The factor $(P+q)^2$, often called $s$, is equal to the squared of the center-of-mass energy. This energy can be, in principle, as high as you want, hence $x_B\geq 0$. On the other hand, the energy in a collision involving a proton can, of course, néver be less than the energy stored in the proton mass, hence $x_B\leq 1$.

I'm not convinced about the argument relating $p_X^2\geq m_p^2$ to baryon number conservation. From baryon number conservation it follows that, should the proton be broken up in the reaction, other hadrons such as pions appear in the end state to conserve this number. Therefore, there is clearly some lower bound on $p_X$, because not àll the energy can be absorbed by the rescattered electron, and there should be something left to form baryons. However, since the mass of the proton is dynamical (i.e. it is generated from QCD interactions, not from intrinsic mass of its constituents), I don't see why this lower bound should be equal to the proton mass.

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