2
$\begingroup$

Consider the $s$-channel process. I use the metric convection $(+,-,-,-)$. The constraints are: Conservation of four momentum: $p_{1}^{\mu}+p_{2}^{\mu}=p^{\mu}$ and on mass shell condition $p_{i}^{2}=m_{i}^{2}$.

The calculation can be simplified using the center of mass frame. In this frame, the 3-momenta of colliding particles are equal in magnitude and opposite in direction: $$\vec{p}_{1}=-\vec{p}_{2}$$ and particles' 4-momenta obey: $$ p_{1}=(E_1,\vec{p}_1) \\ p_{2}=(E_2,-\vec{p}_{1}) \\ p=(E_1+E_2,0).$$ Define the Lorentz-invariant quantity $s$ as the square of four momentum of sum of colliding particles: $$s=(p_1+p_2)^{2}.$$ As the sum of energies and rest mass energy can be measured, it's advisable to expressed the momentum in terms of these parameters, now take the scalar product of conservation of four-momentum with the momentum of particle 1: $$p\cdot p_1=p_1 \cdot p_1+p_1 \cdot p_2 \rightarrow \sqrt{s}E_{1}=m_{1}^{2}+p_{1}\cdot p_{2}.$$ The last term can be eliminate using the squaring of conservation of four-momentum: $$p_{1}^{2}+p_{2}^{2}+2p_{1}\cdot p_{2}=s \rightarrow p_1 \cdot p_2=\frac{s-(m_{1}^{2}+m_{2}^{2})}{2}.$$ Substitute this expression to obtain $E_1$ in terms of $s,m_1,m_2$: $$E_1=\frac{1}{2\sqrt{s}}(s+m_{1}^{2}-m_{2}^{2}).$$ Therefore, the magnitude of 3-momentum of particle 1 can be found: $$|\vec{p}_1|=\sqrt{E_{1}^{2}-m_{1}^{2}}=\sqrt{\frac{1}{4s}(s+m_{1}^{2}-m_{2}^{2})^{2}-m_{1}^{2}}$$ A couple of steps of algebra: $$|\vec{p}_{1}|=\sqrt{\frac{1}{4s}[s^{2}+2(m_{1}^{2}-m_{2}^{2})s+(m_{1}^{2}-m_{2}^{2})^{2}}{-4sm_{1}^{2}]}$$ Finally, I obtain the magnitude: $$|\vec{p}_{1}|=\frac{1}{2\sqrt{s}}\sqrt{s^{2}-2(m_{1}^{2}+m_{2}^{2})s+(m_{1}^{2}-m_{2}^{2})^2}.$$ I am struggling on converting to laboratory frame to obtain the momentum of particle 1 in this frame where the particle 2 is initially at rest, I am not sure about how the inverse transformation can be done.

$\endgroup$

1 Answer 1

0
$\begingroup$

Evaluate $s=(p_1 + p_2)^2$ in the lab frame. Hint: $p_1^\mu = (\epsilon, \mathbf{k}), p_2^\mu = (m_2, 0)$.

Incidentally, your final equation can be simplified into a form that's somewhat more symmetric: $$|\mathbf{p}_1|^2 = \frac{1}{4s}(s-m_{12}^2)(s-\Delta_{12}^2)$$ where $m_{12} = m_1 + m_2$ and $\Delta_{12} = m_1-m_2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.