(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find CM energy of a mu+ mu- collider, with each beam having an energy E of 500 GeV. The beams cross at a small angle of 250 mrad.

2. Relevant equations

[tex]E^2 - p^2c^2 = m^2c^4[/tex]

3. The attempt at a solution

So I have a diagram for the lab frame which has the mu- coming in at a small angle to the mu+ which goes along the x-axis, say. In the CM frame, (I think) this would be a head-on collision if I'm not wrong.

So we look at the 4-momenta in each frame:

[tex]P^{\mu}_{tot} = (\frac{2E}{c},0)[/tex] for the CM and

[tex]P^{\mu '}_{tot} = (\frac{2E'}{c},p'_{\mu^+} - p'_{\mu^-}cos\theta,-p'_{\mu^-}sin\theta)[/tex]

for the lab frame. We have the x-components of the momenta subtracting and a y-component of momentum that isn't cancelled by the mu+. So the next step is to set the lab and CM frame momenta equal to each other:

[tex]\left( \frac{2E}{c}\right)^2 = \left( \frac{2E'}{c}\right)^2 - \left( p'_{\mu^+} - p'_{\mu^-}cos \theta - p'_{\mu^-}sin \theta \right)^2[/tex]

Now we know that the magnitudes of the momenta in the lab frame are equal, since the beams have equal energy. We can then pull out the momentum:

[tex]\left( \frac{2E}{c}\right)^2 = \left( \frac{2E'}{c}\right)^2 - \left|p_{\mu} \right|^2 \left( 1 - cos \theta -sin \theta \right)^2[/tex]

and replace with the invariant:

[tex]\left( \frac{2E}{c}\right)^2 = \left( \frac{2E'}{c}\right)^2 - \left( \frac{E'^2 - m_{\mu}^2c^4}{c^2} \right) \left( 1 - cos \theta -sin \theta \right)^2[/tex]

the angular part, through the small angle approximation reduces to:

[tex]\left( 1 - cos \theta -sin \theta \right)^2 \Rightarrow \left( \frac{\theta^2}{2} + \theta \right)^2[/tex]

and squared

[tex]\left( \frac{\theta^4}{4} + \theta^3 + \theta^2 \right)[/tex]

And because theta is small, theta to the 4th is even smaller, so we can get away with neglecting terms of order higher than 2:

[tex]\left( \frac{\theta^4}{4} + \theta^3 + \theta^2 \right) \Rightarrow \theta^2[/tex]

Then solving for Ecm:

[tex]E_{cm} = \left[ \left(\frac{2E'}{c}\right)^2 + \left( \frac{E'^2 - m_{\mu}^2c^4}{c^2} \right) \theta^2 \right]^{1/2} \cdot \frac{c}{2}[/tex]

The plus sign between the two terms on the right comes from the fact that when you small angle approximate the thetas, the 1's cancel from that term and you can factor out a minus sign to cancel the one in front of the momentum term.

Does this seem right? I feel a bit uneasy about it. I mean, if the angle is zero, it reduces nicely down to Ecm = Elab, but that doesn't really seem especially illuminating. Did my sequence of handwaviness work?

Incedentally, because the muon mass is so much smaller than the energy, it reduces down to:

[tex]E_{cm} = \left[ \left(\frac{2E'}{c}\right)^2 + \left( \frac{E'}{c} \right)^{2} \theta^2 \right]^{1/2} \cdot \frac{c}{2}[/tex]

[tex]\Rightarrow E_{cm} = \left[ 4\left(\frac{E'}{c}\right)^2 + \left( \frac{E'}{c} \right)^2\theta^2 \right]^{1/2} \cdot \frac{c}{2}[/tex]

[tex]\Rightarrow E_{cm} = \frac{E'}{c}(4+\theta^2)^{1/2} \frac{c}{2}[/tex]

[tex]\Rightarrow E_{cm} = \frac{E'}{2}(4+\theta^2)^{1/2}[/tex]

Does this seem right? Should I not have small angle'd the theta terms or what?

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# Homework Help: Relativistic collision in CM frame at small angle

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