Question 2.11: Electron–Pion Scattering (II) Evaluate in detail the nonpola......
Electron-Pion Scattering (II)
Evaluate in detail the nonpolarized \pi^{+} \mathrm{e}^{-}cross section. Start with the expressions given in (2.116) and (2.120)
\mathrm{d} \sigma=P_{\mathrm{fi}} \frac{V^2}{2 E_1 2 E_2|\boldsymbol{v}|} \frac{V}{(2 \pi)^3} \frac{\mathrm{d}^3 p_3}{2 E_3} \frac{V}{(2 \pi)^3} \frac{\mathrm{d}^3 p_4}{2 E_4} (2.116)
\mathrm{d} \sigma=\frac{|F|^2}{4 \sqrt{\left(p_1 \cdot p_2\right)^2-m_1^2 m_2^2}} \mathrm{~d} \operatorname{Lips}\left(s ; p_3, p_4\right) . (2.120)
and use
\mathrm{d} \bar{\sigma}=\frac{1}{4 E \omega|{v}|} \frac{1}{2} \sum\limits_{s, s^{\prime}}\left|F_{s s^{\prime}}\right|^{2} \mathrm{~d} \operatorname{Lips}\left(s ; k^{\prime} p^{\prime}\right).
Determine \mathrm{d} \bar{\sigma} / \mathrm{d} \Omega in the rest system of the pion \left(p^{\mu}=(M, \mathrm{0})\right).
We denote the four-momenta of the pion before and after the collision by p^{\mu}=(M, \mathrm{0}) and p^{\prime \mu}=\left(E^{\prime}, {p}^{\prime}\right), respectively. k^{\mu}=(\omega, {k}) and k^{\prime \mu}=\left(\omega^{\prime}, {k}^{\prime}\right) are the corresponding four-momenta of the incoming and outgoing electrons.
The scattering angle \theta is the angle between the directions {k} and {k}^{\prime} and q=k^{\prime}-k denotes the momentum transfer. In the following we only consider high electron energies, i.e., since \omega, \omega^{\prime} \gg m, the electron rest mass can be neglected. Therefore we have \omega=|{k}| and \omega^{\prime}=\left|{k}^{\prime}\right| and the invariant flux factor is simply
4 \sqrt{(k \cdot p)^{2}-m^{2} M^{2}} \simeq 4 M \omega. (1)
According to the Feynman rules the spin average of the squared scattering amplitude describing the exchange of one photon is given by the contraction of lepton and hadron tensor:
\frac{1}{2} \sum\limits_{s, s^{\prime}}\left|F_{s s^{\prime}}\right|^{2}=\left(\frac{4 \pi \alpha}{q^{2}}\right)^{2} L_{\mu \nu} T^{\mu \nu}. (2)
Employing (2.163) and (2.172)
T^{\mu v}=\left(p+p^{\prime}\right)^\mu\left(p+p^{\prime}\right)^v (2.163)
L_{\mu \nu}=2\left(k_\mu^{\prime} k_\nu+k_v^{\prime} k_\mu+\frac{1}{2} q^2 g_{\mu \nu}\right) (2.172)
we obtain
L_{\mu \nu} T^{\mu \nu}=8\left(2(k \cdot p)\left(k^{\prime} \cdot p\right)+\frac{q^{2}}{2 M^{2}}\right). (3)
In the ultrarelativistic limit, q^{2} becomes
\begin{aligned} q^{2} & =\left(k-k^{\prime}\right)^{2}=\left(\omega-\omega^{\prime}\right)^{2}-\left({k}-{k}^{\prime}\right)^{2} \\ & \approx-2 \omega \omega^{\prime}(1-\cos \theta)=-4 \omega \omega^{\prime} \sin ^{2}\left(\frac{\theta}{2}\right) . \end{aligned}
Hence
\left(\frac{4 \pi \alpha}{q^{2}}\right)^{2}=\left(\frac{\pi \alpha}{\omega \omega^{\prime}}\right)^{2} \frac{1}{\sin ^{4}\left(\frac{\theta}{2}\right)} (4)
and
L_{\mu \nu} T^{\mu \nu}=16 M^{2} \omega \omega^{\prime} \cos ^{2}\left(\frac{\theta}{2}\right) (5)
Now we have to evaluate the invariant phase-space factor:
\begin{aligned} \mathrm{d} \operatorname{Lips}\left(s ; k^{\prime} p^{\prime}\right) & =(2 \pi)^{4} \delta^{4}\left(k^{\prime}+p^{\prime}-k-p\right) \frac{1}{(2 \pi)^{3}} \frac{\mathrm{d}^{3} p^{\prime}}{2 E^{\prime}} \frac{1}{(2 \pi)^{3}} \frac{\mathrm{d}^{3} k^{\prime}}{2 \omega^{\prime}} \\ & =\frac{1}{(4 \pi)^{2}} \delta^{3}\left({k}^{\prime}+{p}^{\prime}-{k}\right) \delta\left(E^{\prime}-M+\omega^{\prime}-\omega\right) \frac{\mathrm{d}^{3} p^{\prime}}{E^{\prime}} \frac{\mathrm{d}^{3} k^{\prime}}{\omega^{\prime}}. \qquad (6) \end{aligned}
In order to evaluate the cross section for the electron to be scattered into a given final state, we have to integrate over all final states of the pion. This is readily achieved by means of the \delta^{3} function in (6) (owing to momentum conservation only one final pion state is possible for a given final state of the electron). In the remaining integrand {p}^{\prime} must then be replaced by {k}-{k}^{\prime}=-{q}. Because of the identity
\begin{aligned} E^{\prime} & =\sqrt{M^{2}+{p}^{\prime 2}}=\sqrt{M^{2}+{q}^{2}} \\ & =\sqrt{M^{2}+\omega^{2}+\omega^{\prime 2}-2 \omega \omega^{\prime} \cos \theta} \qquad (7) \end{aligned}
E^{\prime} is not an independent quantity either. Except for this factor there are no further dependences of the integrand on {p}^{\prime}, including the scattering amplitude and flux factor. Finally \omega^{\prime}=\left|{k}^{\prime}\right| leads to
\frac{\mathrm{d}^{3} k^{\prime}}{\omega^{\prime}}=\omega^{\prime} \mathrm{d} \omega^{\prime} \mathrm{d} \Omega, (8)
where \mathrm{d} \Omega denotes the spherical angle into which the electron is scattered. The \omega^{\prime} integration is performed by using the remaining energy \delta function, i.e., for a given scattering angle the absolute value of the electron momentum is fixed by kinematic arguments. Note that in the argument of the \delta function E^{\prime} depends on \omega^{\prime} as well (cf. (7)). Therefore we have to employ the relation
\delta(f(x))=\sum\limits_{i} \frac{\delta\left(x-x_{i}\right)}{\left.f^{\prime}(x)\right|_{x=x_{i}}}, (9)
where the x_{i} denote the zeros of the function f(x). According to (7) we have
\frac{\mathrm{d}\left(E^{\prime}-M+\omega^{\prime}-\omega\right)}{\mathrm{d} \omega^{\prime}}=\frac{\omega^{\prime}-\omega \cos \theta+E^{\prime}}{E^{\prime}}. (10)
Inserting now (7) into the identity E^{\prime}=M+\omega-\omega^{\prime}, which has been derived by integrating over the \delta function, we obtain
M\left(\omega-\omega^{\prime}\right)=\omega \omega^{\prime}(1-\cos \theta), (11)
and inserting E^{\prime}=M+\omega-\omega^{\prime} into the right-hand side of (10) yields
\frac{\mathrm{d}}{\mathrm{d} \omega^{\prime}}\left(E^{\prime}-M+\omega^{\prime}-\omega\right)=\frac{\omega(1-\cos \theta)+M}{E^{\prime}} , (12)
Combining equations (11) and (12), we obtain
\frac{\mathrm{d}}{\mathrm{d} \omega^{\prime}}\left(E^{\prime}+\omega^{\prime}\right)=\frac{M \omega}{E^{\prime} \omega^{\prime}}. (13)
The (partially integrated) Lorentz-invariant phase-space factor now assumes the form
\mathrm{dLips}=\frac{1}{(4 \pi)^{2}} \frac{\omega^{2}}{M \omega} \mathrm{d} \Omega, (14)
where \omega^{\prime} is fixed by (11). Summarizing equations (1), (4), (5), and (14) finally yields the nonpolarized cross section
\left(\frac{\mathrm{d} \bar{\sigma}}{\mathrm{d} \Omega}\right)_{\text {n.s. }}=\frac{\alpha^{2}}{4 \omega^{2} \sin ^{4}\left(\frac{\theta}{2}\right)} \frac{\omega^{\prime}}{\omega} \cos ^{2}\left(\frac{\theta}{2}\right) . (15)