# Mass-energy equivalence

1. Aug 4, 2011

### kannank

Pair-production is the event when a particle and anti-particle is created from a single photon. We don't see 2 or more photons participating in a single pair-production event. Further, it seems in all the events of energy-mass conversion, photons act independently.

Two or more photons can never participate in a single event.

Is this a right statement?

Starting from this -

I know it doesn't make any sense to mix the fundamental equations below. But I'm going to try it nevertheless.

1) de-broglie wave-length of matter, $\lambda$ = h$/$mv
2) E = h$\mu$
3) c=$\mu\lambda$

The result is, E=mvc. [ This looks familiar to the mass-energy equivalence, E=mc2 ]

My questions,

1) We famously know with a particle with non-zero mass cannot have a a velocity equal to c. Can we conversely say a particle with zero mass cannot have a velocity not equal to c?
2) Is mass a 'slowed-down' energy quanta (photon)?
3) If an event reduce the velocity of a photon, does it get converted to mass?
4) Is this what happens in pair-production?
5) Is pair-production the only possible mechanism of creating mass? Can there be another mechanism of converting energy quanta to mass?

cheers!
KANNAN

2. Aug 4, 2011

### LostConjugate

In a vacuum, yes.

These are some good questions. I am looking forward to some answers.

A "slowed" down photon in my mind is the group velocity of many photons traveling at c with different frequencies.

3. Aug 4, 2011

### kannank

Thanks.

What I explained is clearly not a mathematical model. A 'slow photon' is like a 'frozen photon'. Analogues to water getting frozen to ice in a refrigerator, photon gets frozen to mass under certain conditions.

for instance, the energy quantum created during high energy collisions gets frozen to mass (pair-production).

The more important question is "could there be another way to freeze photon to mass?" If yes, may be it could take care of the Baryon asymmetry.

cheers!
KANNAN

4. Aug 4, 2011

### PAllen

Stop right here. This is all wrong. Pair production form a single photon (in isolation) can never occur - you can't conserve energy and momentum. Further pair production from two interacting photons *does* occur, though it not likely. What really happens with pair production is that the photon interacts with a charged particle (usually a nucleus) which is very likely interaction. This charged particle participates in conserving energy and momentum, enabling the pair production to occur.
No, this is wrong, as noted above. If you want to get into details, the interaction of two photons is mediated by virtual charged fermions (which is why it is rare, but not impossible).
This mixing is nonsensical, I choose not to comment in detail on it.
Correct, a massless particle must move with velocity c according to special relativity.
No.
Nothing can reduce vaccuum speed of a photon. In a refractive medium, light waves travel less than c, but this does not give photons any mass.
No.
Not sure what you mean by energy quanta. All particles have energy. All such energy can produce particle pairs in collisions. Collision of two high energy protons will produce a plethora of particle antiparticle pairs among other things.

5. Aug 4, 2011

### LostConjugate

Perhaps not frozen but locked into a discrete energy level, a discrete frequency with electron volts equivalent to the mass of an electron.

The problem is, it is not that simple. Electrons have an associated electrostatic field, photons don't. So an electron is not just a photon with a locked frequency.

Or is it?

A photon is obviously part of a field that originated somewhere previously, but if you look around the photon then you would see an EM field that appeared to be associated with the photon if you did not know better. If the photon changes it's position (perpendicular to its direction of propagation) the original field must have changed it's position and the photon would appear to emit a new photon (permutation) of the mistakenly associated EM field.

6. Aug 4, 2011

### kannank

Thanks for a good reply. I have some questions though.

First of all, we shouldn't have mixed usages of 'photon' & 'quantum of energy'. It's confusing. I was the one to make that mistake. Sorry.

My understanding about pair-production happens when fermions (hadrons specifically ) collide. Is pair-production possible in a boson-fermion collision? If yes, this thread is considered closed. Please give me a good reference.

These two statements are contradicting. There are some complex QM reasons of why the phase velocity of photons seems to be reduced in a refractive medium. It is not because of photons slowing down. Your first statement is right, photons never actually slow down anywhere.

'Energy Quanta' is the same as photon. Rest of it you are right.

KANNAN

7. Aug 4, 2011

### xts

$$\pi^0+p\rightarrow p + \hbox{large number of various pairs, e.g.: } (\pi^+ + \pi^-)$$
That is most common particle reaction ever observed: it happens billions time per second in every detector of high energy experiment, as the hadron cascades develop.
Or, if you start wondering maybe it is internal structure of hadrons involved:
$$\gamma + e^- \rightarrow e^-+e^-+e^+$$

Last edited: Aug 4, 2011
8. Aug 4, 2011

### Phrak

I don't see how you decided this when none of the terms have been defined.

9. Aug 4, 2011

### PAllen

Reference:

http://en.wikipedia.org/wiki/Pair_production

See specifically, the authoritative references at the bottom of the wiki article.

My two statments are not contradictory with sensible interpretation. The first assumes vaccuum. The second explicity says 'light waves' travel less than c in a refractive medium - light waves are a composite phenomenon.

10. Aug 4, 2011

### kannank

True. But this has nothing to do with the frozen-photon model we are talking about. This is an already existing question. During pair-production why do we always see matter from the list of standard model and its corresponding anti-particle? When there are possibilities of creating particles of any random mass (within the limits of mass-energy equivalence), why don't we see that?

Not a clue!

11. Aug 4, 2011

### kannank

No pair-production mechanisms is explained with bosons involved in collision experiments. If true pair production is possible with bosons, it would happen naturally. You don't need an accelerator for that. Bosons already travel at speed-of-light. You can't accelerate them, anyways.

Now I do know of a boson based collision.

http://en.wikipedia.org/wiki/Two-photon_physics

If you read it you'll know it's again the fermions collided in the accelerator and the photons take part in the event one-by-one, not together.

OK. So a photon does not slow down in any medium. Only the observed phase velocity of the wave slows down.

12. Aug 4, 2011

### xts

Not all bosons travel at speed of light. Actually - just a few of them. Helium atoms may travel veeeeery slowly. They are perfect bosons. And you may accelerate them.

13. Aug 4, 2011

### kannank

Here you are!

You do know I meant those 'pure' bosons in Standard Model, don't you?

14. Aug 4, 2011

### PAllen

This really silly. Pions are bosons. Z-bosons are bosons. Both have mass and of course travel at less than c, and can produce particle/antiparticle pairs in collision with each other or with fermions.

Note that photon + proton -> e+ + e- + proton

is a boson - fermion interaction. In one form or another, this is one of the most common pair production reactions in the universe.

15. Aug 4, 2011

### kannank

It was my mistake to drag bosons and boson-fermion collisions into this discussion. Let's just stick to photons and let's stick the the original problem.

"Does photons involve in an event collectively?"

Is there any 'single unbreakable' event (pair-production or otherwise) where 2 or more photons participate? Like I said - if there is one - this thread is not worth a discussion.

I know an electron jumps 2 states when 2 photons hit. These are 2 events. That's why i stress on 'single unbreakable' event.

16. Aug 4, 2011

### PAllen

I don't understand. In my first response I noted that two photons can interact to produce a particle pair (rare reaction). You actually provided a reference for this. If you are bothered that virtual fermions are involved, well we can make the following ironclad statement:

two photons cannot directly interact without involvement of virtual charged particles.

What exactly are you asking? It really is not clear.

17. Aug 4, 2011

### LostConjugate

Im with you brotha!

18. Aug 5, 2011

### kannank

Good we have sort of an agreement (though hazed by the virtual particles)

Now consider a proton-proton collision at 3TeV + 3TeV = 6TeV. During the event of collision, we know a 6TeV energy is involved ( plus ~ 2 GeV restmass-energy equivalent of the 2 protons).

From the quantum theory of photons (and energy), we know energy exist and transfers as quanta. The question is how many 'energy quanta' are involved in this 6TeV event? 1, 2, 100, 1 million or indefinite?

The idea is to find an (theoretical / logical) answer to this question and then see the what that means.

cheers!
KANNAN