# Formula COM energy [closed]

what is the formula that help me to calculate the center of mass energy for collision of proton and lead ion??.

Can this formula use for calculate COM energy for any collision between particles?

## closed as off-topic by stafusa, Jon Custer, peterh, Michael Seifert, John RennieOct 15 '17 at 6:17

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We can use the Mandelstam variable: $$s=(p_{1}+p_{2})^2$$ where $p_{1}$ is the 4-vector of one particle, and $p_{2}$ is the 4-vector of the other. $$s=p_{1}^2+p_{2}^2+2p_{1}p_{2}$$ For head-on collisions: $$p_{1}=(E_{1},\vec{p_{1}})$$ $$p_{2}=(E_{2},-\vec{p_{2}})$$ For proton-proton collisions: $$s=E_{p}^2+E_{p}^2+2E_{p}^2=4E_{p}^2$$ So the energy in the center of mass is: $$\sqrt{s}=2E_{p}$$

## Edit

In general:

$$s=(p_{1}+p_{2})^2= \left(E_{1}^2-\vec{p_{1}}^2\right)+ \left(E_{2}^2-\vec{p_{2}}^2\right)+ 2\left(E_{1} E_{2}-\vec{p_{1}}\vec{p_{2}}\cos(\theta)\right)$$ i.e., $$s=m_{1}^2+m_{2}^2+2\left[E_{1}E_{2}-|\vec{p_{1}}||\vec{p_{2}}|\cos(\theta)\right]=m_{1}^2+m_{2}^2+2E_{1}E_{2}\left[1-\beta_{1}\beta_{2}\cos(\theta)\right]$$ where $\theta$ is the angle between the colliding particles.

• OK that is clear but WHAT is the formula that can use to determine energy for muon if it traveling in the direction of the electron and traveling perpendicular to the direction of the electron. – FOFO Oct 7 '17 at 14:07
• I've edited my answer. See if this helps you. – Andrei Geanta Oct 7 '17 at 15:14
• IS my steps correct if not what should I do ??i need to find alpha,gamma,energy,and momentum for Electron which emitted from moun decay why is Pμ is negative ?? and the momentum under lorentz transformation is big=935.343....is it good?? or there is some mistake in my steps?? also when i try to calculate Eμ=Ee+Ev the result is not equal for Eμ=10.... [![enter image description here]] : i.stack.imgur.com/yHqzt.jpg – FOFO Oct 14 '17 at 20:16

I assume that you are speaking in terms of classical mechanics, as your second question indicates so.

For a system of particles, the position of the centre of mass, $$\vec {R_{CM}} = \frac {1}{M} \sum_i m_i \vec {r_i}$$

Taking X, Y and Z components,

$X = \frac {1}{M} \sum_{i} m_i x_i$, $Y = \frac {1}{M} \sum_{i} m_i y_i$ and $Z = \frac {1}{M}\sum_{i} m_i z_i$.

Here, $M$ is the total mass of the system, and $\vec {R_{CM}}$ is the position vector of the centre of mass.

In case of classical mechanics, this formula can be used everywhere there is a mention of Centre of Mass.