= Fixed-x electron scattering kinematics
{title2=$E'=E-Q^2/(2Mx)$}
Work in <natural units> with $c=1$, neglect the <electron mass>, and put the initial <proton> at rest with <mass> $M$. Let $k$ and $k'$ be the initial and final <Electron> <four-momenta>, $q=k-k'$, and $Q^2=-q^2>0$. Define the laboratory <energy> transfer $\nu_{\rm lab}=E-E'=P\cdot q/M$. This $\nu_{\rm lab}$ has units of <energy>; the invariant $P\cdot q$ has units of <energy> squared. The <Bjorken scaling variable> and the final <hadron> <invariant mass> obey
$$
Q^2=4EE'\sin^2(\theta/2),\qquad x=\frac{Q^2}{2M\nu_{\rm lab}},\qquad W^2=(P+q)^2=M^2+Q^2\left(\frac1x-1\right).
$$
These follow by squaring $k-k'$ and $P+q$, using $k^2=k'^2=0$ and $P=(M,\mathbf0)$. Solving the <Bjorken x> definition gives the outgoing <energy> and then the <scattering angle>:
$$
E'=E-\frac{Q^2}{2Mx},\qquad \sin^2\frac\theta2=\frac{Q^2}{4E(E-Q^2/(2Mx))}.
$$
Alternatively, substituting $Q^2=4EE'\sin^2(\theta/2)$ before solving gives $E'=E/[1+2E\sin^2(\theta/2)/(Mx)]$. A chosen $x,E,Q^2$ must satisfy $E'>0$ and $0\leq\sin^2(\theta/2)\leq1$. In addition, $W$ must exceed the lightest allowed inelastic <hadron> threshold; positive $W^2$ alone is insufficient.
For an original numerical example, take $M=0.938\,\mathrm{GeV}$, $x=0.3$, $E=15\,\mathrm{GeV}$ and $Q^2=3\,\mathrm{GeV}^2$. Then $\nu_{\rm lab}\simeq5.330\,\mathrm{GeV}$, $E'\simeq9.670\,\mathrm{GeV}$, $\theta\simeq8.25^\circ$, and $W\simeq2.807\,\mathrm{GeV}$. Varying $2\leq Q^2/\mathrm{GeV}^2\leq5$ and $10\leq E/\mathrm{GeV}\leq20$ at this same <Bjorken x> keeps every setting physical and gives $W^2>5.54\,\mathrm{GeV}^2$, above the <proton>-<pion> threshold and the stricter chosen continuum selection $W>2\,\mathrm{GeV}$. These are calculated settings, not historical measurements.
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