Question 2.2: The Flux Factor Problem.Verify that in the center-of-mass sy......
The Flux Factor
Verify that in the center-of-mass system as well as in the laboratory system the flux factor 4E_{1}E_{2}\vert v\vert is given by the invariant expression
In the center-of-mass system the momenta of the projectile and target point in opposite directions, i.e.,
E_{1}=\sqrt{m_{1}^{2}+p^{2}}\ ,\quad p_{1}=+p\\E_{2}=\sqrt{m_{2}^{2}+p^{2}}\ ,\quad p_{2}=-p (1)
Therefore p_{1}\;p_{2}=E_{1}E_{2}+p^{2} , and we obtain
(p_{1}\,p_{2})^{2}-m_{1}^{2}m_{2}^{2}=(E_{1}E_{2}+p^{2})^{2}-m_{1}^{2}m_{2}^{2}\\=E_{1}^{2}(E_{2}^{2}-m_{2}^{2})+2E_{1}E_{2}p^{2}+p^{4}+(E_{1}^{2}-m_{1}^{2})m_{2}^{2}\\
=p^{2}(E_{1}^{2}+2E_{1}E_{2}+E_{2}^{2})\\
=p^{2}(E_{1}+E_{2})^{2}\\
=(E_{1}E_{2})^{2}\left|\frac{p_{1}}{E_{1}}-\frac{p_{2}}{E_{2}}\right|^{2}\;, (2)
which immediately leads to
4\sqrt{(p_{1}\,p_{2})^{2}-m_{1}^{2}m_{2}^{2}}=4E_{1}E_{2}\left|\frac{p_{1}}{E_{1}}-\frac{p_{2}}{E_{2}}\right|\\=4E_{1}E_{2}\left|v_{1}-v_{2}\right|\ . (3)
In the laboratory system the target particle (particle 2) is at rest and we have
E_{1}=\frac{m_{1}}{\sqrt{1-v^{2}}}\quad,\quad p_{1}=\frac{m_{1}v}{\sqrt{1-v^{2}}}\\E_{2}=m_{2}\ \ \ ,\ \ \ p_{2}=0\ . (4)
Then the scalar product p_1 \cdot p_2 is simply equal to m_{1}m_{2}/{\sqrt{1-v^{2}}} and we obtain
4\sqrt{(p_{1}\cdot p_{2})^{2}-m_{1}^{2}m_{2}^{2}}=4m_{1}m_{2}\sqrt{\frac{1}{1-v^{2}}-1}=4\frac{m_1m_{2}}{\sqrt{1-v^{2}}}|v|
=4E_1E_2|v|\ . (5)