PX156 - J4 - two body collisions

three common reference frames:
- centre of momentum (COM) frame
  - $C O M : {\vec{p}}_{1} = - {\vec{p}}_{2}$
- fixed target frame
- lab frame

mandelstam-s

if there are a set of particles, $i$ , in the initial or final state of a collision interaction, then $s$ is invariant:

s = {(\sum_{i} E_{i})}^{2} - {| \sum {\vec{p}}_{i} c |}^{2}

$s$ is conserved in the initial and final state
$s$ is invariant and does not depend on reference frame

interpretation of $s$

consider a collision in the lab frame

\begin{aligned} s & = (E_{1} + E_{2})^{2} - | {\vec{p}}_{1} c + {\vec{p}}_{2} c |^{2} \\ = (E_{1}^{2} + E_{2}^{2} + 2 E_{1} E_{2}) - (| p_{1} |^{2} c^{2} + | p_{2} |^{2} c^{2} + 2 c^{2} {\vec{p}}_{1} \cdot {\vec{p}}_{2}) \\ = m_{1}^{2} c^{4} + m_{2}^{2} c^{4} + 2 E_{1} E_{2} - 2 c^{2} {\vec{p}}_{1} \cdot {\vec{p}}_{2} \\ = m_{1}^{2} c^{4} + m_{2}^{2} c^{4} + 2 E_{1} E_{2} - 2 c^{2} | p_{1} | | p_{2} | \cos (π - θ) \end{aligned}

massless limit: $E_{i} >> m_{i} c^{2} :$

s ≃ 2 E_{1} E_{2} - 2 E_{1} E_{2} \cos (π - θ)

in $C O M$ frame: $θ = 0 :$

s = 4 E_{1} E_{2}

in the massless limit: $E_{1} = E_{2} = E$ in $C O M$ frame:

\begin{aligned} ∴ s & = 4 E^{2} \\ \sqrt{s} & = 2 E \end{aligned}

this is the total energy available to make new particles in the final state
$\sqrt{s_{C O M}} \equiv$ ' $C O M$ energy'
this tells if the final state can be made in $C O M$ frame
figure of merit for designing accelerators
why build a collider?
- $\sqrt{s_{C O M}} ≃ 2 E$
- for a fixed target:

\begin{aligned} s & = (E_{1} + m_{2} c^{2}) - | {\vec{p}}_{1} c |^{2} \\ = E_{1}^{2} + 2 E_{1} m_{2} c^{2} + m_{2}^{2} c^{4} - | {\vec{p}}_{1} c |^{2} \\ = m_{1}^{2} c^{4} + 2 E_{1} m_{2} c^{2} + m_{2} c^{4} \end{aligned}

- massless limit: $s ≃ 2 E_{1} m_{2} c^{2}$
- or, $\sqrt{s_{F T}} = \sqrt{2 E_{1} m_{2} c^{2}}$

collider: $\sqrt{s_{C O M}} \propto E$
fixed target: $\sqrt{s_{F T}} \propto \sqrt{E}$
$\sqrt{s_{c o l l i d e r}}$ scales with $E$ , so, colliders are far more efficient to probe high energy effects
eg: what energy must the initial proton have to produce the final state in fixed target frame considering the interaction:

p + p \to p + p + p + \bar{p}

- to find: $\sqrt{s}$ in the $C O M$ system, where the minimum energy of the final state configuration is clear
- in the $C O M$ system, all final state protons are stationary
- $\sqrt{s}$ of the final state in the $C O M$ frame is: $s_{C O M} = (\sum_{i} E_{i})^{2} = 16 m_{p}^{2} c^{4}$
$$\therefore \sqrt{s_{COM}} = 4 m_{p}c^{2}$$
- in the initial state of the fixed target system:

\begin{aligned} s_{F T} & = (E + m_{p} c^{2})^{2} - | p c |^{2} \\ = 2 E m_{p} c^{2} + 2 m_{p}^{2} c^{4} \end{aligned}

- if $s$ is invariant: $\sqrt{s_{C O M}} = \sqrt{s_{F T}}$
$$E = 7 m_{p}c^{2}$$

mandelstam-s

interpretation of s

interpretation of $s$