Total Energy in CM Frame

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Total Energy in CM Frame

Setting the lengths of the 4-momenta equal to each other,

${\mathbf P^*}^2={\mathbf P}^2$

we can use this for the collision of two particles of mass m. Since the total momentum is zero in the Center of Mass frame, we can express total energy in the center of mass frame as

$(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2=s=(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2$

$(E^*)^2-(\vec p\ ^*)^2=(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2$

$(E^*)^2=(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2$

$E^*=\sqrt{(E_{1}+E_{2})^2-(\vec{p_{1}}+\vec p_{2})^2}$

$E^*=\sqrt{E_{1}^2+2E_{1}E_{2}+E_{2}^2-\vec p_{1} . \vec p_{2} -\vec p_{1} . \vec p_{1} -\vec p_{2} . \vec p_{1} -\vec p_{2} . \vec p_{2} }$

$E^*=\sqrt{(E_{1}^2- p_{1}^2 )+(E_{2}^2-p_{2}^2 )+2E_{1}E_{2}-\vec p_{1} . \vec p_{2} -\vec p_{2} . \vec p_{1} }$

$E^*=\sqrt{(E_{1}^2- p_{1}^2 )+(E_{2}^2-p_{2}^2 )+2E_{1}E_{2}-p_{1} p_{2}\cos(\theta) - p_{2} p_{1}\cos(\theta) }$

$E^*=\sqrt{m_{1}^2+m_{2}^2+2E_{1}E_{2}-p_{1} p_{2}\cos(\theta) - p_{2} p_{1}\cos(\theta) }$

$E^*=\sqrt{m_{1}^2+m_{2}^2+2E_{1}E_{2}-2p_{1} p_{2}\cos(\theta) }$

$E^*=\sqrt{m_{1}^2+m_{2}^2+2E_{1}E_{2}-2p_{1} p_{2}\cos(\theta) }$

Using the relations $\beta\equiv \vec p/E\Longrightarrow \vec p=\beta E$

$E^*=\sqrt{2m^2+2E_{1}E_{2}(1-\beta_{1}\beta_{2}\cos(\theta))}$

where $\theta$ is the angle between the particles in the Lab frame.

In the frame where one particle (p₂) is at rest

$\Longrightarrow \beta_{2}=0$

$\Longrightarrow p_{2}=0$

which implies,

$E_{2}=\sqrt{p_{2}^2+m^2}=m$

$E^*=(2m(m+E_{1})^{1/2}=(2(.511MeV)(.511MeV+\sqrt{(11000 MeV)^2+(.511 MeV)^2})^{1/2}\approx 106.030760886 MeV$

where $E_{1}=\sqrt{p_{1}^2+m^2}\approx 11000 MeV$