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

1. Apr 23, 2012

### Ai52487963

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

$$E^2 - p^2c^2 = m^2c^4$$

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:

$$P^{\mu}_{tot} = (\frac{2E}{c},0)$$ for the CM and
$$P^{\mu '}_{tot} = (\frac{2E'}{c},p'_{\mu^+} - p'_{\mu^-}cos\theta,-p'_{\mu^-}sin\theta)$$

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:

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

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:

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

and replace with the invariant:

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

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

$$\left( 1 - cos \theta -sin \theta \right)^2 \Rightarrow \left( \frac{\theta^2}{2} + \theta \right)^2$$

and squared

$$\left( \frac{\theta^4}{4} + \theta^3 + \theta^2 \right)$$

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:

$$\left( \frac{\theta^4}{4} + \theta^3 + \theta^2 \right) \Rightarrow \theta^2$$

Then solving for Ecm:

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

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:

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

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

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

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

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

Last edited: Apr 23, 2012
2. Apr 23, 2012

### BruceW

This bit isn't right:
$$\left( \frac{2E}{c}\right)^2 = \left( \frac{2E'}{c}\right)^2 - \left( p'_{\mu^+} - p'_{\mu^-}cos \theta - p'_{\mu^-}sin \theta \right)^2$$
Because the components of momenta should each be squared on their own, then added, but you've added the x and y components, then squared. (Easy mistake, I've done this before too).

3. Apr 24, 2012

### Ai52487963

So in that case, would this be a legitimate way of approaching the problem? I don't like the sines and cosines and I want to get rid of them, so

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

for small theta, cos -> 1 and sin -> 0,

$$\left( \frac{2E}{c}\right)^2 = \left( \frac{2E'}{c}\right)^2 - \left( p'_{\mu^+} - p'_{\mu^-} \right)^2$$

and because the magnitudes of the momenta are the same,

$$\left( \frac{2E}{c}\right)^2 = \left( \frac{2E'}{c}\right)^2 - |p'|\left( 1 - 1 \right)^2$$

just tells us that the energy in the CM frame is the same as in the lab frame (eg that the small angle deflection doesnt really matter at energies as high as 500 GeV). There is a part of the question that reminds you that the muon mass is much much less than the muon energy, but I don't see how that comes in to play here.

The question point-value isn't very high, so I should expect a pretty straightforward answer (versus the ninja-ing I did earlier), but this just seems like a "duh" answer. Or did I mess up the momentum squaring again?

4. Apr 24, 2012

### BruceW

This is right, but then when you use the approximation with the angles, you are effectively saying that the angle is zero (which is why the answer says that the energy is the same in the CM and lab frame - because there is no angle, so they are the same thing!)

I don't think you need to use any small-angle approximations, just use the angle you are given in the equation. You also said about the muon mass energy being much less than the total energy of the particles. You can make an approximation using this fact, which is similar to the approximation you used in the first post.