# Particle momenta and cross section

1. Oct 4, 2015

### Schwarzschild90

Will someone explain this step to me?

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2. Oct 5, 2015

### DrDu

Some more background would be helpful. I suppose this is for massless particles where p^2=0?

3. Oct 5, 2015

### Schwarzschild90

Yes. It's for electron muon scattering

4. Oct 5, 2015

### DrDu

But neither electrons nor myons are massless. Or is this some approximation?

5. Oct 5, 2015

### Schwarzschild90

The event is viewed in a CM frame.

6. Oct 5, 2015

### ChrisVer

it has nothing to do with CM frame ($p^2= m^2$ in any frame -frame independent minkowski product).
Probably they consider electron/muon with high enough momenta/Energy so that they can be considered massless (i.e. the last $=$ in 2nd line should be $\approx$ instead)

7. Oct 5, 2015

### Staff: Mentor

They neglect the particle masses, right.

8. Oct 8, 2015

### Schwarzschild90

I was in a rush. CM has nothing to do with it.

Indeed, the particles considered are massless.

I figured it out.

Last edited: Oct 8, 2015
9. Oct 8, 2015

### Schwarzschild90

Why can we set E/c = p?

10. Oct 8, 2015

### Staff: Mentor

It is the same approximation of massless particles.

11. Oct 9, 2015

### ChrisVer

$E = \sqrt{p^2 + m^2} \Rightarrow E = p + \mathcal{O}\big(\frac{m^2}{2p}\big)$
(taylor expanding the square root for m<<p)