Relativistic Particle Collisions with Photons: Understanding Momentum and Energy

[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?

 

[humanino]

Is it possible to use a frame so that the pion is at rest?

Yes, the pion has a finite (non-vanishing) mass of approximately 135 or 139 MeV.

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

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.

 

[Bob_for_short]

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?

Yes, of course, it is well possible.

I think not, because the momentum of the photon is the same in all frames because it travels at c.

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

 

[Bob S]

Transfer to the pion rest frame. The photon energy increases. See
http://pdg.lbl.gov/2002/kinemarpp.pdf
for 4-vector kinematic Lorentz transormation.

 

[Bob_for_short]

Transfer to the pion rest frame. The photon energy increases.

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

 

[soupdejour]

The Lorentz transformation of a photon 4-vector is how the Doppler red(blue)shift formula is derived for instance.

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.

 

[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...

 

[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.

 

[Meir Achuz]

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

 

[soupdejour]

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.

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!