|
GR9677 #53
|
|
|
|
|
Alternate Solutions |
| There are no Alternate Solutions for this problem. Be the first to post one! |
|
|
Comments |
South92
2009-10-20 08:18:11 | So what happens if the positronium is in the 2s state (which has zero orbital angular momentum)? My understanding is that the positronium could decay into its ground state, emitting a single photon (there would be recoil, thus conserving momentum), and anihilate to produce two more photons. Is this incorrect?
The_Duck
2010-07-05 14:39:54 |
I don't really know what I'm talking about here, but I think if you're in the singlet 2s state you can't drop down to the singlet 1s state by emitting a single photon, because of this very reason: both states have J=L=M=0, so you can't conserve angular momentum if you emit a single photon.
|
|  | WoolfianOperator
2009-10-18 18:05:16 | Here's a go.
We know that energy has to be conserved. A is out. We know momentum must be conserved, a single photon cannot have 0 momentum. B is out. We know from Feynman diagrams that a process will occur with probability  , with each vertex of the diagram contributing a factor of  . Because  equals  , the diagram with the least amount of vertices is most likely to occur. See Griffiths Introduction to Particle Physics for more on this. The more photons the more complicated the diagram and therefore there will be more vertices. So choice C is best of the remain 3 options. | | phys2718
2008-10-12 08:42:45 | The comments on antiparticles are incorrect. The photon and antiphoton both carry spin 1, just as the electron and positron both carry spin 1/2 (not -1/2 for the positron). Spin is not a so-called "additive" quantum number in the sense that we simply add charge or baryon number as scalars, and in fact antiparticles carry the same spin as their particles. Two positive spins may be combined so as to produce 0 total spin (the singlet configuration) which is the case here for the positronium as well as the two emitted photons.
physicsisgod
2008-10-29 21:44:15 |
Maybe we are confusing the magnitude of a particle's spin, which is positive, and its direction, which can be up or down. Spin is a vector, really, so it's not "positive" or "negative".
|
| | backyard
2006-11-01 23:13:38 | why not zero photons?
barefoot0
2006-11-02 19:30:21 |
Good question...
I do not think you could have a decay without emiting some kind of radiation.
I would love to hear a better answer.
|
ben
2006-11-03 14:21:57 |
can't have zero photons because of conservation of energy. since positronium is unstable the electron will soon spiral into the positron and the two particles will annihilate, producing photons. i can't think of any other particle an electron and its antiparticle could decay to.
|
scottopoly
2006-11-03 19:32:10 |
It has to emit something, just to conserve energy. 2 photons are certainly the most likely. Anything else would probably take more energy. 1 photon is forbidden by both momentum conservation and angular momentum conservation.
|
wittensdog
2009-11-03 19:58:07 |
Aside from issues involving parity conservation in EM processes and so forth, the more photons you add, the more interaction vertices you add, and you bring along another factor of alpha. So even if 3 and 4 photons were allowed, they're going to be less likely.
|
wittensdog
2009-11-03 20:33:33 |
I mean, if you want to get really anal about it and consider other forces and interactions,
1.) electrons and positrons don't feel the strong force, so rule that out
2.) forget anything to do with gravitons and quantum gravity, those decays, if they even exist, are going to be so ridiculously suppressed it's not even funny
3.) Weak interactions are also very suppressed, and really the only neutral end product of interest is a Z boson, which is much too heavy and would need to decay in a very small time to something lighter, and the cross section to go to something like a neutrino anti-neutrino pair is going to be really low, and may even violate some conservation law I'm not thinking of right now
So really the EM interaction is what we have to work with, so we're going to be working with some non-zero number of photons, since in order to conserve energy, zero photons would mean some other force is at play, since there are no particles involved in the EM interaction that are lighter than electrons and positrons. 1 photon violates momentum conservation and also I believe parity conservation, but 2 is okay. Anymore photons is going to be suppressed in the diagrams.
|
|  | comorado
2006-10-30 02:06:52 | We have conseravtion of momentum here, before the decay  , after decay must be  , so the most simple answer is two fotons with opposite momentum.
barefoot0
2006-11-02 19:35:30 |
I dont think anything is said reguarding the momentum only angular momentum.
|
scottopoly
2006-11-03 19:28:45 |
You could always transform to the rest frame of the positronium, so WLOG, we know it's momentum is zero.
|
scottopoly
2006-11-03 19:29:29 |
You could always transform to the rest frame of the positronium, so WLOG, we know it's momentum is zero.
|
scottopoly
2006-11-03 19:29:39 |
You could always transform to the rest frame of the positronium, so WLOG, we know it's momentum is zero.
|
| |
|
| Post A Comment! |
|
|
Bare Basic LaTeX Rosetta Stone
|
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces  .
|
| type this... |
to get... |
| $\int_0^\infty$ |
 |
| $\partial$ |
 |
| $\Rightarrow$ |
 |
| $\ddot{x},\dot{x}$ |
 |
| $\sqrt{z}$ |
 |
| $\langle my \rangle$ |
 |
| $\left( abacadabra \right)_{me}$ |
_{me}) |
| $\vec{E}$ |
 |
| $\frac{a}{b}$ |
 |
|
|
|
|
|
