Just a quick one

[RichIsnang]

If an anti photon and a photon collide do they release one or two photons? I'm guessing one right?

[Moontanman]

If an anti photon and a photon collide do they release one or two photons? I'm guessing one right?

 

 

Neither, a photon is it's own anti particle... there is no anti photon only photons...

[timo]

An anti-photon has no features distinguishing it from a photon (or alternatively: there is no such thing as an anti-photon). In principle, the reaction of two photons into N photons (and nothing else) should be possible for all N>=2 at one-loop level. In practice, two electromagnetic wave packets meeting each other (which is what you probably meant by "colliding photons") will just ignore each other - at least in vacuum.

[juanrga]

If an anti photon and a photon collide do they release one or two photons? I'm guessing one right?

 

The photon is its own antiparticle (photon has no charge and the inverse of 0 is -0 which is zero again). Therefore you are asking about photon-photon collision.

 

In QED, two photons can annihilate to give an electron plus and anti-electron (positron)

 

[math]\gamma + \gamma \rightarrow e^{-} + e^{+}[/math]

 

Two photons can also collide to give two photons

 

[math]\gamma + \gamma \rightarrow \gamma + \gamma[/math]

 

Both photon-photon collisions and annihilations are observed in current high energy particle accelerators, although in indirect ways. There are plans to build photon-photon colliders (also named gamma-gamma colliders).

Edited July 1, 2012 by juanrga

[RichIsnang]

>< sorry guys, don't know what I thinking when I wrote photon, meant just and anti particle/particle pair electron positron etc.

[timo]

An electron-positron annihilation must release at least two photons (assuming nothing else than photons comes out of the reaction). A single photon cannot satisfy conservation of momentum and energy simultaneously (*). Any number of resulting photons >=2 is possible for the same reasons as in the case of two incoming photons. But if I remember correctly, then at least for accessible collision energies 2 photons is the most probable outcome, with the probability rapidly dropping for each additional photon to be produced.

 

(*): In the center-of-mass frame, the total momentum is zero, whereas the energy is at least two times the mass of an electron. Since for a single photon, energy and magnitude of momentum are directly proportional, a single photon cannot satisfy the two constraints of zero momentum and non-zero energy at the same time.

[swansont]

If an anti photon and a photon collide do they release one or two photons? I'm guessing one right?

 

You can dismiss that scenario immediately as it would violate conservation of momentum.

[RichIsnang]

Edit: accidental repeat

 

Ooohh ok, good point, didn't think about that lol

 

How many solutions will there be to the equation? Like I know the numbers can't be exact, but will there be 1 solution of many? Looking at momentum before/after, energy etc.

Edited July 1, 2012 by RichIsnang

[timo]

How many solutions will there be to the equation? Like I know the numbers can't be exact, but will there be 1 solution of many? Looking at momentum before/after, energy etc.

No one posted an equation so far, and it is not clear what "the numbers" (which apparently cannot be exact) are supposed to be. Try rephrasing your issue with proper, grammatically correct sentences, and perhaps also proof-read your post before submitting. No offense intended, but some people trying to answer you put some effort into that, so I think you should at least put a fraction of that effort into the formulation of your question in the first place.

 

To answer what may have been your question: There is formally an infinite number of possibilities for an electron and a positron to annihilate into two photons. In the center-of-mass frame, the two resulting photons have to have opposite momenta, such that the total momentum is zero. However, this anti-parallel pair of photons can still point into every possible direction in space.

Edited July 1, 2012 by timo

[RichIsnang]

OK thanks timo, sorry about my poor grammar, one of the downsides of being dyslexic

 

all i meant by 'the numbers' was the initial momentum of the electron and positron

and all i meant by 'the equation' was the equation you may use to determine the energy of the photons and potentially directions.

 

 

but yeah you answered my question