Question 3.6: The Breit System Prove the existence of a Lorentz reference ......
The Breit System
Prove the existence of a Lorentz reference frame where (3.61) holds.
P_\mu=(\tilde{P}, 0,0,-\tilde{P}), \quad \tilde{P} \gg M_{\mathrm{N}}, \quad P^2=\tilde{P}^2-\tilde{P}^2=0 \approx M_{\mathrm{N}}^2 (3.61)
We start with the laboratory system. Here
\begin{aligned} & q_{\mu}=\left(q_{0}, {q}\right), \quad q_{0}>0, \\ & P_{\mu}=(M, \mathrm{0}) . & (1) \end{aligned}
First we perform a rotation in such a way that {q} is parallel to the z direction.
\begin{aligned} & q_{\mu}^{\prime}=\left(q_{0}^{\prime}, 0,0, q_{3}^{\prime}\right), \quad q_{0}^{\prime}, q_{3}^{\prime}>0 \\ & P_{\mu}^{\prime}=\left(M_{\mathrm{N}}, 0,0,0\right) . & (2) \end{aligned}
Now we boost the system in the z direction, i.e., we transform to a reference frame moving with the velocity
v_{z}=\frac{q_{0}^{\prime}}{q_{3}^{\prime}} c=\beta c . (3)
This is possible, because the momentum transfer q_{\mu} is spacelike, i.e.,
Q^{2}=q_{3}^{\prime 2}-q_{0}^{\prime 2}>0 \rightarrow q_{3}^{\prime}>q_{0}^{\prime} . (4)
This follows easily from (3.50a), according to which, Q^{2}=4 E E^{\prime} \sin ^{2} \theta / 2>0. Then in the new reference frame
\begin{aligned} q_{\mu}^{\prime \prime} & =\gamma\left(q_{0}^{\prime}-\beta q_{3}^{\prime}, 0,0, q_{3}^{\prime}-\beta q_{0}^{\prime}\right) \\ & =\gamma\left(0,0,0, q_{3}^{\prime}-\beta^{2} q_{3}^{\prime}\right), \beta=\frac{q_{0}^{\prime}}{q_{3}^{\prime}} \\ & =\left(0,0,0, q_{3}^{\prime} \sqrt{1-\frac{q_{0}^{\prime 2}}{q_{3}^{\prime 2}}}\right)=\left(0,0,0, \sqrt{Q^{2}}\right) , & (5) \end{aligned}
with
\beta=\frac{v_{z}}{c}=\frac{q_{0}^{\prime}}{q_{3}^{\prime}} \quad \text { and } \quad \gamma=\left(1-\beta^{2}\right)^{-1 / 2}. (6)
Under this transformation the nucleon momentum becomes
P_{\mu}^{\prime \prime}=\gamma\left(M_{\mathrm{N}}, 0,0,-\beta M_{\mathrm{N}}\right). (7)
Since we consider reactions where Q^{2} and
v=q^{\prime} \cdot P^{\prime}=q_{0}^{\prime}=q^{\prime \prime} \cdot P^{\prime \prime} (8)
get very large, but their ratio x=Q^{2} / 2 v remains constant,
\frac{v}{M_{\mathrm{N}}}=\frac{q^{\prime \prime} \cdot P^{\prime \prime}}{M_{\mathrm{N}}}=\beta \gamma \sqrt{Q^{2}}=\frac{Q^{2}}{2 M_{\mathrm{N}} x} \Rightarrow \beta \gamma=\frac{\nu / M_{\mathrm{N}}}{\sqrt{Q^{2}}}=\frac{\sqrt{Q^{2}}}{2 M_{\mathrm{N}} x} (9)
must hold. Since \sqrt{Q^{2}} \gg M_{\mathrm{N}} and x \leq 1, equation (9) can be fulfilled only for \beta \approx 1, \gamma \gg 1. Hence (7) becomes
P_{\mu}^{\prime \prime} \approx(\tilde{P}, 0,0,-\tilde{P}), \quad \text { where } \quad \tilde{P}=\beta \gamma M_{\mathrm{N}}=\frac{\sqrt{Q^{2}}}{2 x} (10)
This is (3.62).
P_{i,\mu}=\xi_{i}~(\tilde{P},0,0,-\tilde{P})~. (3.62)
In the Breit system the nucleon (and with it all partons) and the photon move in the z direction towards each other. In the final state the scattered parton carries the momentum ( x denotes the initial momentum fraction of the parton, i.e., \left.P_{\mu}^{\mathrm{i}}=x P_{\mu}^{\prime \prime}\right)
\begin{aligned} P_{\mu}^{\mathrm{f}} & =x P_{\mu}^{\prime \prime}+q_{\mu}^{\prime \prime}=x(\tilde{P}, 0,0,-\tilde{P})+q_{\mu}^{\prime \prime}=\left(\frac{\sqrt{Q^{2}}}{2}, 0,0,-\frac{\sqrt{Q^{2}}}{2}\right)+q_{\mu}^{\prime \prime} \\ & =\left(\frac{\sqrt{Q^{2}}}{2}, 0,0, \frac{\sqrt{Q^{2}}}{2}\right). & (11) \end{aligned}
Obviously the spatial momentum of the parton is flipped to its opposite direction by the reaction (see Fig. 3.9). In the Breit system the parton is simply reflected. We shall come back to this in Example 3.8.
