How to determine particle energies in center of momentum frame?

[RicardoMP]

Homework Statement

I want to show for the following process that, except for the angle [tex]\theta[/tex], all momenta and energies are fixed by energy-momentum conservation.

Relevant Equations

[tex]p_A=\frac{1}{2\sqrt{s}}(s+m^2_A-m^2_B\space,\space 0\space,\space 0\space,\space\sqrt{\eta_i})[/tex]
[tex]p_B=\frac{1}{2\sqrt{s}}(s-m^2_A+m^2_B\space,\space 0\space,\space 0\space,\space -\sqrt{\eta_i})[/tex]
[tex]p_C=\frac{1}{2\sqrt{s}}(s+m^2_C-m^2_D\space,\space \sqrt{n_f}sin(\theta)\space,\space 0\space,\space\sqrt{n_f}cos(\theta))[/tex]
[tex]p_D=\frac{1}{2\sqrt{s}}(s-m^2_C+m^2_D\space,\space -\sqrt{n_f}sin(\theta)\space,\space 0\space,\space-\sqrt{n_f}cos(\theta))[/tex]

, where [tex]\eta_i=4s|\vec{p_i}|^2[/tex] and [tex]\eta_f=4s|\vec{p_f}|^2[/tex].

That said, my approach was to determine the energies and 3-momenta at the center of momentum reference frame for each particle, with a fixed s, and check it corresponds to each one of the above, but I'm having some trouble proving that, for example, [tex]E_A=\frac{s+m^2_A-m^2_B}{2\sqrt{s}}[/tex]. I've been playing around with [tex](p_A+p_B)^2=(E_A+E_B)^2=s[/tex] but I'm failing to determine [tex]E_A[/tex]. How should I do it?

 

[Nate810]

You can use 4-momentum conservation. In the CMS frame, we have $p_A + p_B = 0$. Thus, we have \begin{align}E_A + E_B &= \sqrt{(p_A + p_B)^2} = 0 \\\implies E_A &= -E_B\end{align}We also know that $E_B = \frac{s + m_B^2 - m_A^2}{2\sqrt{s}}$. So we have\begin{align}E_A &= - \frac{s + m_B^2 - m_A^2}{2\sqrt{s}} \\&= \frac{s + m_A^2 - m_B^2}{2\sqrt{s}}\end{align}