# Collisions of particles

1. Sep 12, 2009

### soupdejour

I have never done relativistic collisions when a photon is involved, and it's messing wtih me.

For example, I have a photon colliding with a pion moving at a speed 3c/4. Is it possible to use a frame so that the pion is at rest? I think not, because the momentum of the photon is the same in all frames because it travels at c.

Or does the momentum or energy of the photon change in different frames?

2. Sep 12, 2009

### humanino

Yes, the pion has a finite (non-vanishing) mass of approximately 135 or 139 MeV.
Certainly yes. Energy and momentum of a photon depend on the frame, and as usual for the components of a 4-vector, their transformation is such that the corresponding mass is conserved (here zero mass). The Lorentz transformation of a photon 4-vector is how the Doppler red(blue)shift formula is derived for instance.

3. Sep 12, 2009

### Bob_for_short

Yes, of course, it is well possible.
No, your conclusion is wrong. The energy-momentum is a four-vector that transforms from one reference frame to another as any four-vector.

4. Sep 12, 2009

### Bob S

5. Sep 12, 2009

### Bob_for_short

It is only in case when the momenta of involved particles have opposite signs.

6. Sep 12, 2009

### soupdejour

OK, I think I can visualize the wavelength of light changing in different frames.

I was confused that the momentum was different in different frames even though the speed was the same (c). But it makes sense that a photon just has a special four vector with "length" zero.

Thanks to both of you.

7. Sep 12, 2009

### soupdejour

A closely related follow-up question:

If I have a photon colliding with this pion, and I want to try to find the minimum energy of the pion to create some particle N. I know the mass of N and the wavelength of the photon.

$$\gamma + \pi \rightarrow N$$

For the minimum energy of pion to produce N, I just make N at rest.

I do energy conservation:

$$E_{\pi} = m_{N} c^2 - E_{\gamma}$$

If I do 3-momentum conservation:

$$E_{\pi} = \sqrt{ m_{p}^2 c^4 + E_{\gamma}^2 }$$

These two equations give me different answers! Is it possible that the N cannot be produced at rest? Thats weird...

Last edited: Sep 13, 2009
8. Sep 13, 2009

### Parlyne

You have two equations in two unknowns. You can certainly solve them for $E_\pi$ and $E_\gamma$. This just means that, for a given $m_N$, there is only one specific value for each of the energies that will produce and N at rest.

9. Sep 13, 2009

### Meir Achuz

Your second equation is wrong. It should be E_gamma+e_pi=m_N.

10. Sep 13, 2009

### soupdejour

Right. So for a given frequency of light, it isn't necessarily possible to produce N at rest. Only one frequency can produce N at rest.

Thanks!