Question 2.6: The Cross Section for Pion–Kaon Scattering Derive the explic......
The Cross Section for Pion-Kaon Scattering
Derive the explicit form of the differential cross section for electromagnetic \pi^{+} \mathrm{K}^{+}scattering in the center-of-momentum system (cm system).
According to (2.120) the cross section for \pi^{+} \mathrm{K}^{+}scattering is
\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), (1)
where
\begin{aligned} F & =-\frac{e^{2}}{q^{2}}\left(p_{1}+p_{3}\right) \cdot\left(p_{2}+p_{4}\right), \\ q^{\mu} & =\left(p_{3}-p_{1}\right)^{\mu}=-\left(p_{4}-p_{2}\right)^{\mu}, \qquad (2) \end{aligned}
and
\mathrm{d} \operatorname{Lips}\left(s ; p_{3} p_{4}\right)=(2 \pi)^{4} \delta^{4}\left(p_{3}+p_{4}-p_{1}-p_{2}\right) \frac{1}{(2 \pi)^{3}} \frac{\mathrm{d}^{3} \mathrm{p}_{3}}{2 \mathrm{E}_{3}} \frac{1}{(2 \pi)^{3}} \frac{\mathrm{d}^{3} \mathrm{p}_{4}}{2 \mathrm{E}_{4}} (3)
denote the invariant scattering amplitude and the Lorentz-invariant phase-space factor, respectively. The \mathrm{cm} system is defined by
{p}_{1}+{p}_{2}={p}_{3}+{p}_{4}=\mathrm{0} . (4)
In this system the scattering process is described by the scattering angle \theta_{\mathrm{cms}} (see Fig. 2.14). Now we transform the four-momenta to the \mathrm{cm} system,
\begin{array}{ll} p_{1}^{\mu}=\left(E_{1}, {p}\right), & p_{2}^{\mu}=\left(E_{2},-{p}\right) \\ p_{3}^{\mu}=\left(E_{3}, {p}^{\prime}\right), & p_{4}^{\mu}=\left(E_{4},-{p}^{\prime}\right) \qquad (5) \end{array}
which leads to the total energy
\begin{aligned} E_{\mathrm{cms}} & =E_{1}+E_{2}=E_{3}+E_{4} \\ & =\sqrt{{p}^{2}+m_{1}^{2}}+\sqrt{{p}^{2}+m_{2}^{2}} \\ & =\sqrt{{p}^{\prime 2}+m_{1}^{2}}+\sqrt{{p}^{2}+m_{2}^{2}} \\ & =\sqrt{p^{2}+m_{1}^{2}}+\sqrt{p^{2}+m_{2}^{2}} . \qquad (6) \end{aligned}
Here and in the following we denote the absolute value of the spatial momentum by
|{p}|=\left|{p}^{\prime}\right| \equiv p . (7)
By integrating the invariant phase-space factor \mathrm{d} \operatorname{Lips}\left(s ; p_{3} p_{4}\right) over the spatial momenta \mathrm{d}^{3} p_{4} we obtain
\int \frac{\mathrm{d}^{3} p_{4}}{E_{4}} \delta^{4}\left(p_{3}+p_{4}-p_{1}-p_{2}\right)=\frac{1}{E_{4}} \delta\left(E_{3}+E_{4}-E_{1}-E_{2}\right). (8)
The right-hand side of (8) contains E_{4} as well as \left|{p}_{4}\right|. These two variables, however, are not independent of each other: they are connected by
{p}_{4}={p}_{1}+{p}_{2}-{p}_{3}, \quad E_{4}=\sqrt{{p}_{4}^{2}+m_{2}^{2}}, \quad\left(m_{2}=m_{4}\right) . (9)
Next we transform \mathrm{d}^{3} p_{3} into spherical coordinates
\mathrm{d}^{3} p_{3}=p_{3}^{2} \mathrm{~d} p_{3} \mathrm{~d} \Omega, (10)
where p_{3}=\left|{p}_{3}\right| and \mathrm{d} \Omega denotes the spherical angle into which the \pi^{+}is scattered. Because of
E_{3}^{2}=p_{3}^{2}+m_{1}^{2} (11)
we have
E_{3} \mathrm{~d} E_{3}=p_{3} \mathrm{~d} p_{3} (12)
and
\mathrm{d} \operatorname{Lips}\left(s ; p_{3} p_{4}\right)=\frac{1}{(4 \pi)^{2}} \delta\left(E_{3}+E_{4}-E_{1}-E_{2}\right) \frac{p_{3} \mathrm{~d} E_{3}}{E_{4}} \mathrm{~d} \Omega . (13)
This formula is valid in any Lorentz system. Now we go into the \mathrm{cm} system by making use of the relations
E_{3}^{2}=p^{2}+m_{1}^{2}, \quad E_{4}^{2}=p^{2}+m_{2}^{2}, (14)
and
E_{3} \mathrm{~d} E_{3}=E_{4} \mathrm{~d} E_{4}=p \mathrm{~d} p . (15)
Introducing the free variable
E^{\prime}=E_{3}+E_{4}, (16)
we also have, according to (15),
\mathrm{d} E^{\prime}=\frac{p}{E_{3}} \mathrm{~d} p+\frac{p}{E_{4}} \mathrm{~d} p=\frac{E^{\prime}}{E_{3} E_{4}} p \mathrm{~d} p=\frac{E^{\prime}}{E_{4}} \mathrm{~d} E_{3} . (17)
Now (13) assumes the form
\left.\mathrm{d} \operatorname{Lips}\left(s ; p_{3} p_{4}\right)\right|_{\mathrm{cms}}=\frac{1}{(4 \pi)^{2}} \delta\left(E_{\mathrm{cms}}-E^{\prime}\right) \frac{p}{E^{\prime}} \mathrm{d} E^{\prime} \mathrm{d} \Omega . (18)
An integration over E^{\prime} then yields
\left.\mathrm{d} \operatorname{Lips}\left(s ; p_{3} p_{4}\right)\right|_{\mathrm{cms}}=\frac{1}{(4 \pi)^{2}} \frac{p}{E_{\mathrm{cms}}} \mathrm{d} \Omega (19)
for the two-particle phase-space factor in the \mathrm{cm} system. Finally the flux factor has to be rewritten in terms of \mathrm{cm} variables. With the help of (5) and (6) we immediately find that
\begin{aligned} \sqrt{\left(p_{1} \cdot p_{2}\right)^{2}-m_{1}^{2} m_{2}^{2}} & =\sqrt{\left(E_{1} E_{2}+p^{2}\right)^{2}-\left(E_{1}^{2}-p^{2}\right)\left(E_{2}^{2}-p^{2}\right)} \\ & =p E_{\mathrm{cms}}. & (20) \end{aligned}
The result for the differential cross section is then
\left.\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}\right|_{\mathrm{cms}}=\frac{1}{\left(8 \pi E_{\mathrm{cms}}\right)^{2}}|F|^{2}. (21)
By introducing the second Mandelstam variable t the result (21) can be brought into an invariant form. t is simply defined as the square of the four-momentum transfer:
t=q^{2}=\left(p_{1}-p_{3}\right)^{2}=\left(p_{2}-p_{4}\right)^{2}. (22)
In the cm system we consequently have
\begin{aligned} t & =\left(0, {p}-{p}^{\prime}\right)^{2}=-\left({p}-{p}^{\prime}\right)^{2} \\ & =-\left({p}^{2}-2 {p} \cdot {p}^{\prime}+{p}^{\prime 2}\right) \\ & =-2 p^{2}\left(1-\cos \theta_{\mathrm{cms}}\right), & (23) \end{aligned}
which leads to
\mathrm{d} t=2 p^{2} \mathrm{~d} \cos \left(\theta_{\mathrm{cms}}\right). (24)
Since the cross section for spinless particles is cylindrically symmetric around the beam axis, i.e.
\mathrm{d} \Omega_{\mathrm{cms}}=2 \pi \mathrm{d} \cos \left(\theta_{\mathrm{cms}}\right), (25)
we obtain the relation
\frac{\mathrm{d}}{\mathrm{d} t}=\frac{\pi}{p^{2}} \frac{\mathrm{d}}{\mathrm{d} \Omega_{\mathrm{cms}}}. (26)
Therefore the two-particle cross section is, in invariant form
\begin{aligned} \frac{\mathrm{d} \sigma}{\mathrm{d} t} & =\frac{1}{64 \pi} \frac{1}{\left(p E_{\mathrm{cms}}\right)^{2}}|F|^{2} \\ & =\frac{1}{64 \pi} \frac{|F|^{2}}{\left(p_{1} \cdot p_{2}\right)^{2}-m_{1}^{2} m_{2}^{2}}, & (27) \end{aligned}
where we have inserted (20). Finally we introduce the Mandelstam variable s (see Exercise 2.3) and take p_{1}^{2}=m_{1}^{2} and p_{2}^{2}=m_{2}^{2} into account. Equation (27) then assumes the form
\frac{\mathrm{d} \sigma}{\mathrm{d} t}=\frac{1}{16 \pi} \frac{|F|^{2}}{\left[s-\left(m_{1}+m_{2}\right)^{2}\right]\left[s-\left(m_{1}-m_{2}\right)^{2}\right]}. (28)
As already mentioned these expressions are valid for any scattering reaction with two unpolarized particles in both initial and final states. Now we want to express |F|^{2} in the case of \pi^{+} \mathrm{K}^{+}scattering by invariant Mandelstam variables. From the definitions
\begin{aligned} & s=\left(p_{1}+p_{2}\right)^{2}=\left(p_{3}+p_{4}\right)^{2} \\ & u=\left(p_{1}-p_{4}\right)^{2}=\left(p_{2}-p_{3}\right)^{2} & (29) \end{aligned}
we derive the relations
\begin{aligned} & 2 p_{1} \cdot p_{2}=2 p_{3} \cdot p_{4}=s-m_{1}^{2}-m_{2}^{2}, \\ & 2 p_{1} \cdot p_{4}=m_{1}^{2}+m_{2}^{2}-u. & (30) \end{aligned}
With the help of these relations the invariant scattering amplitude (2) assumes the form
|F|^{2}=\left(\frac{4 \pi \alpha}{t}\right)(s-u)^{2}, (31)
where we have transformed the fine-structure constant
\alpha=\frac{e^{2}}{4 \pi} \simeq \frac{1}{137} (32)
to so-called Heaviside-Lorentz units. In these units Gauss’s law reads \nabla \cdot {E}=\varrho. It is particularly simple to transform (31) into the \mathrm{cm} system, since we have, according to (29) and (5),
\begin{aligned} s & =\left(E_{1}+E_{2}, \mathrm{0}\right)^{2}=E_{\mathrm{cms}}^{2}, \\ u & =\left(E_{1}-E_{4}, {p}+{p}^{\prime}\right)^{2}=\left(E_{1}-E_{4}\right)^{2}-\left({p}+{p}^{\prime}\right)^{2} \\ & =E_{1}^{2}+E_{4}^{2}-2 E_{1} E_{4}-\left(E_{1}^{2}-m_{1}^{2}\right)-\left(E_{4}^{2}-m_{2}^{2}\right)-2 {p} \cdot {p}^{\prime} \\ & =m_{1}^{2}+m_{2}^{2}-2 p^{2} \cos \left(\theta_{\mathrm{cms}}\right)-2 E_{1} E_{4}, \\ t & =-2 p^{2}\left(1-\cos \left(\theta_{\mathrm{cms}}\right)\right) . & (33) \end{aligned}
The variable t has already been given in (23).
