Chapter 4, Electrodynamics of Spinless Particles Video Solutions, Quarks and leptons: introductory course in modern particle physics

Problem 1

Working in a box of volume $V=L^3$, show that the number of allowed states of momentum with $x$ component in the range $p_x$ to $p_x+d p_x$ is $(L / 2 \pi) d p_x$. Convince yourself that you need to impose periodic boundary conditions on the wavefunction and its derivative to ensure no net particle flow out of the volume.

Lottie Adams

Numerade Educator

07:26

Problem 2

In the center-of-mass frame for the process $A B \rightarrow C D$, show that
$$
\begin{gathered}
d Q=\frac{1}{4 \pi^2} \frac{p_f}{4 \sqrt{s}} d \Omega \\
F=4 p_l \sqrt{s},
\end{gathered}
$$
and hence that the differential cross section is
$$
\left.\frac{d \sigma}{d \Omega}\right|_{c m}=\frac{1}{64 \pi^2 s} \frac{p_f}{p_j}|: \Omega|^2,
$$
where $d \Omega$ is the element of solid angle about $\mathbf{p}_C, s=\left(E_A+E_B\right)^2,\left|\mathbf{p}_A\right|=$ $\left|\mathbf{p}_B\right|=p_1$ and $\left|\mathbf{p}_C\right|=\left|\mathbf{p}_D\right|=p_f$.

Michael Talbot

Numerade Educator

02:29

Problem 3

Use (4.18) to show that for very high-energy "spinless" electron-muon scattering.
$$
\left.\frac{d \sigma}{d \Omega}\right|_{c m}=\frac{\alpha^2}{4 s}\left(\frac{3+\cos \theta}{1-\cos \theta}\right)^2
$$
where $\theta$ is the scattering angle and $\alpha=e^2 / 4 \pi$. Neglect the particle masses.

Mayukh Banik

Numerade Educator

04:55

Problem 4

Use the Feynman rules to obtain the invariant amplitudes for the "spinless" processes $\mathrm{e}^{-} \mu^{+} \rightarrow \mathrm{e}^{-} \mu^{+}$and $\mathrm{e}^{-} \mathrm{e}^{+} \rightarrow \mu^{-} \mu^{+}$. Check your answers by appropriately crossing the $\mathrm{e}^{-} \mu^{-} \rightarrow \mathrm{e}^{-} \mu^{-}$amplitude of Section 4.2.

Ameer Said

Numerade Educator

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Problem 5

Show that
$$
s+t+u=m_A^2+m_B^2+m_C^2+m_D^2,
$$
where $m_i$ is the rest mass of particle $i$.
To display the kinematic (or physical) regions of processes related by crossing, we construct a two-dimensional plot which maintains the symmetry of $s, t, u$. The three axes $s, t, u=0$ are drawn (Fig. 4.7) to form an equilateral triangle of height $\sum m_i^2$. From any point inside and also outside (if attention is paid to the signs of $s, t, u)$ the triangle, the sum of the perpendicular distances to the axes is equal to the height of the triangle [see (4.44)].

It is easy to show that $s$ is the square of the total center-of-mass energy of the process $\mathrm{AB} \rightarrow \mathrm{CD}$ [see (3.12) or Exercise 4.6]. It is conventional to take the reaction under study to be the $s$ channel process. In our last example, this was $\mathrm{e}^{-} \mathrm{e}^{+} \rightarrow \mathrm{e}^{-} \mathrm{e}^{+}$scattering. The crossed reactions $\mathrm{A} \overline{\mathrm{D}} \rightarrow \mathrm{C} \overline{\mathrm{B}}$ and $\overline{\mathrm{D}} \mathrm{B} \rightarrow \mathrm{C} \overline{\mathrm{A}}$ are called the $u$ and $t$ channels, respectively, since $u$ and $t$ are equal to the square of the total center-of-mass energy in the respective channels (see Exercise 4.7).

Lainey Roebuck

Numerade Educator

Problem 6

Taking $\mathrm{e}^{-} \mathrm{e}^{+} \rightarrow \mathrm{e}^{-} \mathrm{e}^{+}$to be the $s$ channel process, verify that
$$
\begin{aligned}
& s=4\left(k^2+m^2\right) \\
& t=-2 k^2(1-\cos \theta) \\
& u=-2 k^2(1+\cos \theta)
\end{aligned}
$$
where $\theta$ is the center-of-mass scattering angle and $k=\left|\mathbf{k}_i\right|=\left|\mathbf{k}_j\right|$, where $\mathbf{k}_i$ and $\mathbf{k}_f$ are, respectively, the momenta of the incident and scattered electrons in the center-of-mass frame. Show that the process is physically allowed provided $s \geq 4 m^2, t \leq 0$, and $u \leq 0$. The physical region is shown shaded on Fig. 4.7. Note that $t=0(u=0)$ corresponds to forward (backward) scattering.

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02:16

Problem 7

For the crossed reaction $\mathrm{A} \overline{\mathrm{D}} \rightarrow \mathrm{C} \overline{\mathrm{B}}\left(\mathrm{e}^{-} \mathrm{e}^{-} \rightarrow \mathrm{e}^{-} \mathrm{e}^{-}\right)$, show that $u$ becomes the square of the total center-of-mass energy and that this process would become physical in a different kinematic region: $u \geq 4 \mathrm{~m}^2$, $t \leq 0$, and $s \leq 0$. (Note that, for example, $-p_D=(E, \mathbf{p})$, where $E$ and $\mathbf{p}$ refer to the incoming $\overline{\mathrm{D}}$ ).

Zulfiqar Ali

Numerade Educator

01:38

Problem 8

If the $s$ channel process is $\mathrm{e}^{-} \mu^{-} \rightarrow \mathrm{e}^{-} \mu^{-}$, show that the boundaries of the physical regions of this and the crossed channel reactions are given by
$$
t=0, \quad s u=\left(M^2-m^2\right)^2
$$
where $m$ and $M$ are the electron and muon masses, respectively. Construct the Mandelstam plot.

Nicole Krahulik

Numerade Educator

Problem 9

Verify that crossing relation (4.42) is of the form
$$
\Re_{e^{-} e^{+}}(s, t, u)=\Re_{e^{-} e^{-}}(u, t, s)
$$

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Problem 10

Show that the invariant amplitude, (4.41), for "spinless" electron-positron scattering can be written as
$$
\pi_{\mathrm{e}^{-} \mathrm{e}^{+}}(s, t, u)=e^2\left(\frac{s-u}{t}+\frac{t-u}{s}\right) .
$$

Comment on the symmetry of 9 under $s \leftrightarrow t$.
Let us look back at the amplitude for "spinless" electron-electron scattering. The amplitude (4.40) is derived taking, of course, the process to be $A B \rightarrow C D$,
that is, the $s$ channel process. In terms of invariant variables, (4.40) becomes
$$
\pi_{\mathrm{e}^{-} \mathrm{e}^{-}}=e^2\left(\frac{u-s}{t}+\frac{t-s}{u}\right) .
$$

The resulting cross section is sketched in Fig. 4.8, and the origin of the forward and backward peaks is identified; $-t$ and $-u$ are the squares of the three-momentum transferred in Figs. 4.4a and 4.4c, respectively, that is, of the momentum carried by the virtual photon. When the photon has a very small momentum squared $\left(-q^2\right)$, that is, almost on its mass shell, then by the uncertainty principle the range of the interaction is very large. Interactions with small deflections therefore occur with large cross sections.

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