2-2 elastic scattering¶

We have reviewed the 2-2 elastic scattering with one massless neutrino in BDM Physics. To accommodate a more general case, we relax the massless assumption and do not identify the incoming or outgoing particles as any specific species. This allows us to build a general-purpose class suitable for arbitrary 2-2 scattering processes involving non-zero masses.

General expressions¶

The scheme for the scattering of two particles, labeled 1 and 2, is shown in Fig. 1. Each particle carries non-zero mass $m_1$ and $m_2$, respectively. The four-momenta of all particles are indicated in the diagram. The scattering angle quantifies the deflection relative to the incoming direction of particle 1. After the scattering event, particles 1 and 2 are deflected by angles $\vartheta$ and $\psi$, respectively.

Figure 1. The 2-2 particle scattering in lab frame.

We can write down the 4-momenta in lab frame,

\[ \begin{align*} p_1 = (E_1,\mathbf{p}_1),&\quad p_2 = (m_2,\mathbf{0}),\\ p_1^{\prime} = (E_1^\prime,\mathbf{p}_1^\prime),&\quad p_2^{\prime} = (E_2^\prime,\mathbf{p}_2^\prime), \end{align*} \]

and the corresponding $u$-channel

\[ \begin{gather*} (p_1-p_2^\prime)^2 = m_1^2+m_2^2-2E_1E_2^\prime +2 |\mathbf{p}_1| |\mathbf{p}_2^\prime| \cos\psi,\\ (p_2-p_1^\prime)^2 = m_1^2+m_2^2 - 2 E_1^\prime m_2. \end{gather*} \]

The two identities are equivalent due to the Lorentz-invariant nature of the process. Similar to the BDM case, suppose we know $E_2^\prime$ and its kinetic energy, such that $$ T_2 = E^\prime_2 - m_2 = E_1 - E_1^\prime. $$ We thus have $$ \begin{equation} E_1 (T_2+m_2) -|\mathbf{p}_1||\mathbf{p}_2^\prime| x = (E_1-T_2)m_2, \end{equation} $$ where $x= \cos\psi$. By letting $|\mathbf{p}_1|=\sqrt{E_1^2-m_1^2}$ and $|\mathbf{p}_2^\prime| =\sqrt{T_2(T_2+2m_2)}$, the only unknown in the above equation is $E_1$, which can be solved analytically,

\[ \begin{equation}\label{eq:E1} E_1=\frac{T_2^2 m_2 + |\mathbf{p}_2^\prime|x\sqrt{m_1^2 |\mathbf{p}_2^\prime|^2 x^2 + T_2^2 (m_2^2-m_1^2)}}{|\mathbf{p}_2^\prime|^2 x^2 - T_2^2}. \end{equation} \]

Note that $E_1$ gives the total energy of particle 1, thus $T_1 = E_1 - m_1$. It is true that $E_1 = T_1$ only in the special case where $m_1 = 0$.

Moreover, one can differentiate $E_1$ with respect to $T_2$, which yields $$ \begin{equation}\label{eq:dE1/dT2} \frac{dE_1}{dT_2}=m_2x^2\times \frac{\alpha +\beta + \gamma}{\eta} \end{equation} $$ where

\[ \begin{align*} \alpha &= m_1^2 \delta, \\ \beta & = m_2^2 (2T_2+\delta),\\ \gamma &= 2m_2 x \kappa,\\ \eta &= \delta^2 x \kappa, \end{align*} \]

with $\delta=-T_2 + (T_2 + 2m_2)x^2$ and $\kappa=\sqrt{(T_2+2m_2)(\alpha+ T_2 m_2^2 )}$. In most cases, elastic scattering does not change the mass of the particles, thus

\[\begin{equation} \frac{dE_1}{dT_2} = \frac{d}{dT_2}(T_1 + m_1) = \frac{dT_1}{dT_2} \end{equation}\]

and this allows all relevant quantities to be expressed in terms of the kinetic energies $T_i$. When constructing the corresponding class, the namespace will consistently refer to $T_i$ instead of $E_i$.

Although the angle $\vartheta$ for particle 1 is irrelevant to our study, it can still be determined via 3-momentum conservation \begin{equation} \sin\vartheta = \frac{|\mathbf{p}_2^\prime|}{|\mathbf{p}_1^\prime|} \sin\psi \end{equation} where $|\mathbf{p}_1^\prime| = \sqrt{E_1^{\prime 2} - m_1^2}$ and $E_1^\prime = E_1 - T_2$.

Validation¶

Now recall the $\nu\chi$ scattering with $E_1 = E_\nu$, $T_2 = T_\chi$, $m_1 = m_\nu = 0$, and $m_2 = m_\chi$. With $|\mathbf{p}_2^\prime| = |\mathbf{p}_\chi| = \sqrt{T_\chi (T_\chi + m_\chi)}$, Eq. \eqref{eq:E1} becomes

\[\begin{equation}\label{eq:Ev} E_\nu = T_\chi m_\chi \frac{T_\chi + |\mathbf{p}_\chi| x}{(|\mathbf{p}_\chi| x - T_\chi)(|\mathbf{p}_\chi| x + T_\chi)} = \frac{T_\chi m_\chi}{|\mathbf{p}_\chi| x - T_\chi}. \end{equation}\]

Additionally,

\[\begin{align*} \alpha &= 0, \\ \beta &= m_\chi^2 \left(T_\chi + (T_\chi + 2m_\chi)x^2\right), \\ \gamma &= 2m_\chi^2 |\mathbf{p}_\chi| x, \\ \eta &= \delta^2 m_\chi |\mathbf{p}_\chi| x, \end{align*}\]

and after some tedious algebra, we obtain

\[\begin{align*} m_\chi x^2 \frac{\alpha + \beta + \gamma}{\eta} &= \frac{x}{\delta^2} \frac{m_\chi^2}{|\mathbf{p}_\chi|} \left(T_\chi + (T_\chi + 2m_\chi)x^2 + 2|\mathbf{p}_\chi| x \right) \\ &= \frac{x}{\delta^2} \frac{m_\chi^2}{|\mathbf{p}_\chi|} \left( T_\chi + \frac{|\mathbf{p}_\chi|^2}{T_\chi} x^2 + 2|\mathbf{p}_\chi| x \right) \\ &= \frac{x}{\delta^2} \frac{m_\chi^2}{|\mathbf{p}_\chi|} \frac{(T_\chi + |\mathbf{p}_\chi| x)^2}{T_\chi}. \end{align*}\]

We also use

\[\begin{align*} \delta^2 &= \left(-T_\chi + \frac{|\mathbf{p}_\chi|^2}{T_\chi} x^2 \right)^2 = \frac{(|\mathbf{p}_\chi|^2 x^2 - T_\chi^2)^2}{T_\chi^2} \\ &= \frac{(|\mathbf{p}_\chi| x - T_\chi)^2 (|\mathbf{p}_\chi| x + T_\chi)^2}{T_\chi^2}, \end{align*}\]

which leads to the final result

\[\begin{equation}\label{eq:dEv/dTx} m_\chi x^2 \frac{\alpha + \beta + \gamma}{\eta} = \left( \frac{m_\chi}{|\mathbf{p}_\chi| x - T_\chi} \right)^2 \frac{T_\chi}{|\mathbf{p}_\chi|} x = \frac{dE_\nu}{dT_\chi}. \end{equation}\]

We observe that Eqs. \eqref{eq:Ev} and \eqref{eq:dEv/dTx} are exactly Eqs. (5) and (8) in BDM Physics, respectively. We therefore conclude that Eqs. \eqref{eq:E1} and \eqref{eq:dE1/dT2} are the general expressions for 2-2 elastic scattering involving non-zero masses.