Elastic scattering $A+B \rightarrow A+B$: How to express the scattering angle in terms of $s, t$? There is an elastic reaction $A + B \rightarrow A + B$ and we need to express the scattering angle in terms of Mandelstam variables $s, t$.
From what I know: $s=(p_A + p_B)^2$ and $t=(p_A - p_C)^2$.
In the CM frame $p_A + p_B = 0$, so $p_A = - p_B$. Would it be correct now to write the energy as $E_A = - E_B$ or $E_A + E_B = 0$? So now we have:
$E= \sqrt{p^2 c^2 + m^2 c^4}$ but c=1
$E = \sqrt{p^2 + m^2}$
$E_A + E_B = \sqrt{p_A^2 + m_A^2} + \sqrt{p_B^2 + m_B^2}$
Therefore
$ s = (p_A + p_B)^2 = (E_A + E_B)^2 - (p_A + p_B)^2 = p_A^2 + m_A^2 + p_B^2 + m_B^2 - p^2$
But I'm kinda stuck and actually I'm not sure whether my thinking is correct or not and what should I do. Could anyone help me with clarifying the steps?
 A: I wouldn't do your homework for you, but I'll give you a trail map, since this is an elementary exercise in relativistic kinematics you should be clear on.
Assuming your $p_C=p_{A'}$ the outgoing momentum of particle A, and you are working in the center of momentum frame, as the PDG and presumably your text specify, you must first convince yourself from the conservation of energy that all four legs of the amplitude have a momentum of the same magnitude,
$$
q\equiv |\vec p_A|= |\vec p_B|=|\vec p_{A'}|=|\vec p_{B'}|,
$$
hence $E_A=E_{A'}$.
Since, in the center of momentum,
$$
s=(E_A+E_B)^2=\left (\sqrt{m_A^2+q^2}+\sqrt{m_B^2+q^2}\right  )^2,
$$
you may readily solve for q in terms of s and the two masses. Make sure you do it efficiently/nicely!
(Hint:
$4s~q^2= s^2-2s(m_A^2+m_B^2) +(m_A^2-m_B^2)^2= (s-(m_A+m_B)^2)(s-(m_A-m_B)^2)$, essentially Heron's triangle area formula.)
But you know
$$t= (p_A-p_{A'})^2= (E_A-E_{A'})^2- (\vec p_A-\vec p_{A'})^2\\
=-2 q^2 +2\vec p_A \cdot \vec p_{A'}= 2q^2 (-1+\cos\theta),
$$
where θ is the scattering angle you are to solve for,
in terms of t and q, hence s and the two masses.
A: In a collision in the center of mass system which conserves both momentum and energy, we can say that both masses rebound from the center of mass with the same speed that they had on approach (in directions which are opposite to each other). But in two or three dimensions, the initial energy and momentum tell us only that about the final directions. The directions will depend on the nature of the impact. Consider the collision of two billiard balls.
