Talk:Positronium

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From my understanding, positronium can either exist in para- or ortho- states. (There is no way to add two spin-1/2 particles so you get total spin other than 0 or 1.) If the lifetime of the para state is 10^-10, and the lifetime of the ortho state is 10^-8, then how can the (presumably average) lifetime of positronium be 10^-7? What's going on here? Ckerr 08:24, 11 November 2005 (UTC)

Could be a difference in meaning of "lifetime" the 10^-7 number is "at most". — Omegatron 06:28, 11 November 2005 (UTC)

If so, that's an unusual use of the term "lifetime", since there is no such thing as a maximum lifetime. It's "possible" to have any compound, no matter how unstable, living for an arbitrary amount of time, since radioactive decay is a stochastic process. Ckerr 08:24, 11 November 2005 (UTC)

I've added a clarifying bit of material on the lifetimes of the two states of Ps. The mean lifetime is the amount of time it takes for the population of an exponentially decaying population to be reduced by a factor of "e" (=2.71828...). The mean lifetime is a bit longer than the half-life by about 30%.

[edit] practical uses?

I came here hoping to find some solution to the energy crisis. Someone should add to the article how this could be useful? 4.230.102.132 05:32, 24 November 2005 (UTC)

Positronium would not be a solution to the energy crisis in any useful way. Positronium does not store energy for any useful length of time (minutes to days), nor can it be stabilized to do so. No energy is gained in the process of positron annihilation, since it requires energy to create anti-matter in the first place. There is no such thing as a "positron well" where one can go get usefully large amounts of antimatter to be stored. In principle, one could store energy in positrons (or other antimatter) by creating the antimatter in the first place and storing it in charged particle traps, then releasing it in a controlled way to use the annihilation radiation as energy. But it's not a solution to the rising cost of oil.

At best, anti-matter is an interesting tool of science-fiction authors as an energy source. In reality (with present technology) it is very inefficient and actually wastes energy as a result of the equipment required (typically, powerful accelerators). Nimur 01:01, 3 April 2006 (UTC)

--Mplskid 07:30, 12 July 2006 (UTC)--Mplskid 07:30, 12 July 2006 (UTC)--Mplskid 07:30, 12 July 2006 (UTC)--Mplskid 07:30, 12 July 2006 (UTC)== Anything emitted during the "spiral"? ==

The particles "spiral" closer to each other (although this actually takes place in quantized steps of decreasing radius), until their existence is terminated by electron-positron annihilation. At annihilation, gamma rays are produced.

Is anything produced during the "spiral"? And if not, what happens to that energy? Ewlyahoocom 12:19, 1 April 2006 (UTC)

I'm not 100% sure, but I do not believe there is electromagnetic radiation during this inward spiral (this would be a classical phenomenon). Instead, the energy is converted between orbital and spin angular momentum, so the system is not losing (or radiating) energy. After the particles annihilate, then they produce the gamma radiation. Nimur 00:58, 3 April 2006 (UTC)

Hmm. I'm not so sure about this explanation. Positrons and electrons have spin 1/2, so exactly how could they get more spin angular momentum? They can't magically turn into bosons, which is why they're called spin 1/2 particles. A decreasing radius of orbit would necessitate a change in orbital angular momentum, so something must be going on. From my background (one postgraduate class in particle physics) I would say that EM radiation is emitted. This would explain why positronium annihilation has a well-defined energy: all the extra energy is bled off in the form of photons before the two antiparticles annihilate.

There is no "spiral." This is a classical notion, implying the existence of trajectories, which is inconsistent with the tenets of quantum mechanics. Any atom in its ground state exists as a cloud of charge distribution that is static in the absence of external forces. Any two particles within an atom can be found within a prescribed distance from each other with a definite probability that can be calculated. Positronium is a genuine atom, and its two particles must be within a certain small distance in order to annihilate. The probability for this is calculable, and represents the fraction of time the particles are able to annihilate, which leads to the annihilation rate or lifetime.Mplskid 07:30, 12 July 2006 (UTC)

Ewlyahoocom asked the right question. Spiral is correct, only QM does not describe it. Also, it is probably not monotonic - and definitely not in quantized steps. Radiation is correct, only it is not far-field (propagating) radiation. The energy goes into kinetic energy and field energy. Since positronium does not have enough angular momentum to form a photon (needed for a quantized step), it cannot radiate far-field. If the conditions were right, it could radiate photon pairs prior to annihilation. However, until annihilation actually begins, the field-energy and photon-frequency relations are not correct. I'm trying to figure out the mechanics of the annihilation process right now. (That requires understanding what an electron really is - and not its average, or statistical, QM picture.)

-To anwser the first question asked why is the mean life time 10^-7 s. The 2 forms of positronium are not produced in equal quanties when its formed. para-Ps can only be formed in one quantum state where the spins are anti-parallel (↑↓-↓↑)/sqrt(2) S=0. and ortho-Ps can be formed in one of three quantum states where the spins are parallel ↑↑, ↓↓, or (↑↓+↓↑)/sqrt(2) S=1. If you add (1/4)*0.125 ns + (3/4)*143 ns = 1.07*10^-7 s