GR9677 #53



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whereami 20180406 01:23:24  i think the answers are all wrong, or maybe I am making a mistake. \r\nThe problem states that orbital angular momentum is zero. That\'s not even spin. And everybody is saying that spin is zero. This really bugs me. \r\nI don\'t think L has anything to do with S. \r\nCan somebody help me ?
kronotsky 20181022 03:12:24 
Spin is indeed not conserved by itself; this is something neglected by previous answers. Total angular momentum is conserved  but in the zero momentum frame centered around the center of mass, the orbital angular momentum of emitted photons should be zero since r is parallel to p, so orbital angular momentum will be conserved and therefore so will spin. Since L and p do not commute, the resulting combined photon state must be in the kernel of the commutator (i\'m a little worried about this being true, but I really don\'t want to find out lol), which need not be true in general, so the fact that we are thinking about the center of momentum/masscentered frame is important (i.e. this logic is more difficult to apply for less simple complicated positronium diagrams).

kronotsky 20181023 02:14:26 
After thinking about this a little more, if the initial angular momentum is zero, then in the COM frame both the angular momentum and the total momentum is zero for the final multiplephoton state, so L and p do commute in this special case.

kronotsky 20181023 02:16:13 
Also, I meant to say that \"total angular momentum is conserved and therefore so is spin\" since j = l + s and l = 0.

  gman 20101110 17:51:00  There are two kinds of positronium: ORTHOpositronium and PARApositronium. The difference is in the way the positron and electron spins align.
Ortho is the spin triplet > total spin=1
Para is the spin singlet > total spin=0
Since para is unchanged under charge conjugation (due to antisymmetry of fermion wavefunctions in the singlet), and since photons are negative under charge conjugation, para must decay into an even number of photons. Zero is forbidden due to energy conservation, so it's 2, 4, 6, ... Successively higher numbers of photons are less likely because of the additional vertex factors. So to lowest order, para goes to 2 photons.
Ortho on the other hand is negative under charge conjugation, so it must go to an odd number of photons. 1 is out due to momentum conservation, so thats 3 5 7, etc... again decreasing in probability.
But that's technical... Just remember this. PARA is the singlet, spin 0. zero is small, so that's a small number of photons. You need TWO lines to be PARAllel. If you have two ORTHOgonal lines, cross them to get THREE mutually ORTHOgonal lines.
I know, it's dumb. But you'll remember it on the test....
mvgnzls 20110918 13:08:46 
actually there are 3 types of positronium
the third is Dipositronium

  pkamal 20101002 18:32:53  If emission of 2 photons conserve spin, whats wrong with emission of 4 photons? It will also conserve spin.
flyboy621 20101024 17:20:26 
I think you can assume that emission of 4 photons COULD occur, but is less likely.

  South92 20091020 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 20100705 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.

tmraotmh 20110407 19:50:19 
A photon is massless, so can't have no momentum. Thus in the center of mass frame, we can't have just one photon or momentum will not be conserved.

  WoolfianOperator 20091018 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 20081012 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 socalled "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 20081029 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 20061101 23:13:38  why not zero photons?
barefoot0 20061102 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 20061103 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 20061103 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 20091103 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 20091103 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 antineutrino 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 nonzero 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 20061030 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 20061102 19:35:30 
I dont think anything is said reguarding the momentum only angular momentum.

scottopoly 20061103 19:28:45 
You could always transform to the rest frame of the positronium, so WLOG, we know it's momentum is zero.

scottopoly 20061103 19:29:29 
You could always transform to the rest frame of the positronium, so WLOG, we know it's momentum is zero.

scottopoly 20061103 19:29:39 
You could always transform to the rest frame of the positronium, so WLOG, we know it's momentum is zero.

 

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