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I have read about particle decay, a process in which one particle becomes several other particles. However, I have not been able to find much information about its opposite: several particles combining into one particle. Is such a process possible, and if so, under what conditions? For example, a free neutron may decay into a proton, electron, and electron antineutrino. Could a proton, electron, and electron antineutrino somehow be joined into a neutron?

Edit: Everyone, thank you for your help, but let me try to make what I'm looking for clearer. I want to know whether several particles can join into ONE particle, in an exact reverse of that one particle decaying into several particles. As far as I know, I don't think an atomic nucleus counts as one particle. Please correct me if I'm wrong.

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  • $\begingroup$ Particle combining takes place in stars every time. The most effective producers of elements are exploding stars including Black holes. $\endgroup$
    – HolgerFiedler
    Commented Apr 11, 2017 at 4:34
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    $\begingroup$ @HolgerFiedler True, but successful proton-proton fusion has a very low probability. Wikipedia says "The half-life of a proton in the core of the Sun before it is involved in a successful proton-proton fusion is estimated to be about one billion years, even at the extreme pressures and temperatures found there." This is actually rather fortunate: if the probability were higher, stars would consume their fuel faster, and die sooner. $\endgroup$
    – PM 2Ring
    Commented Apr 11, 2017 at 7:18

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Even if the reaction is not prohibited by energy/momentum or other conservation laws, it would have a tiny cross section: shooting one kind of particle at a target is easy, but shooting two at the same time, not so much (particularly if one is an antineutrino). However there is such a thing as electron capture by a proton in the nucleus, producing a neutron and an electron neutrino: $p + e^- \rightarrow n + \nu_e$.

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    $\begingroup$ And so-called "inverse beta-decay" which is not what the OP asked about, but $\bar{\nu}_e + p \to n + e^+$, which is used as a tool in reactor neutrino experiments. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Apr 11, 2017 at 2:35
  • $\begingroup$ You don't have to shoot a nutrino at the other two particles — given the two, it can conjur a nutrino pair from the vacuum. That's why you get a nutrino out rather than shooting an antinutrino in. $\endgroup$
    – JDługosz
    Commented Apr 12, 2017 at 4:05
  • $\begingroup$ Cross-sections aren't well-defined for 2->1 processes. $\endgroup$
    – innisfree
    Commented Apr 12, 2017 at 9:43
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    $\begingroup$ You don't have enough phase-space to integrate over the Dirac for energy-momentum conservation, you have $\delta^4 (p_i -p_f) d^3 p$ $\endgroup$
    – innisfree
    Commented Apr 12, 2017 at 9:45
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    $\begingroup$ @andy 'finding the same set of particles somewhere with exactly the required momenta for that perfect balance to create a perfectly on-shell single particle is vanishingly unlikely.' I don't agree with that. E.g. $\sigma (pp \to h) $ is nonvanishing. You kill the remaining delta-function with an integral overs pdfs. Whether it's likely/unlikely/vanishingly unlikely depends on distribution of momenta of initial state. $\endgroup$
    – innisfree
    Commented Apr 16, 2017 at 8:48
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As Nick says, in laboratory conditions the crossections are very very small. In a way two body resonances are the only reversible example with high probability to study in the laboratory.

In the grand cosmological laboratory of the universe, in its history at the hadron epoch:

In physical cosmology, the hadron epoch was the period in the evolution of the early universe during which the mass of the universe was dominated by hadrons. It started approximately 10-6 seconds after the Big Bang, when the temperature of the universe had fallen sufficiently to allow the quarks from the preceding quark epoch to bind together into hadron

In this model the inverse processes exist , due to the very high average energies of the quarks and gluons the crossections for generating protons and neutrons out of three quarks is high and reversible, until the universe cools enough that quarks are no longer asymptotically free. So there is a three to one process.

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    $\begingroup$ Lol, only a physicist would call a millionth of a second "an epoch" :) $\endgroup$
    – Jasmine
    Commented Apr 11, 2017 at 16:23
  • $\begingroup$ Cross-sections aren't well-defined for 2->1 processes. $\endgroup$
    – innisfree
    Commented Apr 12, 2017 at 9:43
  • $\begingroup$ @innisfree counterexample e+e- --> Z $\endgroup$
    – anna v
    Commented Apr 12, 2017 at 10:39
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    $\begingroup$ $\sigma (ee\to Z) $ isn't well defined. I don't see how just saying it is a counter-example. You don't have enough phase space integrals (you have 3) to integrate 4 Dirac deltas. $\endgroup$
    – innisfree
    Commented Apr 12, 2017 at 10:47
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    $\begingroup$ Now, you can have a distribution of initial state momenta, e.g. parton distribution function for quarks, and integrate over that distribution to kill the remaining delta. $\endgroup$
    – innisfree
    Commented Apr 12, 2017 at 10:51
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The triple-alpha process is a way $C^{12}$ is formed in stars from three $He^4$ nuclei once (most of) the hydrogen in the core has been burned. It is really two reactions,
$He^4+He^4 \to Be^8, Be^8+He^4 \to C^{12}.$
$ Be^8$ is unstable, but lasts long enough for this to happen if the helium is dense and hot enough.

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Yes, and way more!

Every process (we know so far) is CPT-invariant. This stands for Charge (so the charge of the particle, if it is a particle/anti-particle), Parity (in which direction it rotates, a little bit like left-/right-handedness or chirality) and Time (so forward/backward in time, if a decay is forward -> the recombination, what you asked for, is backward; or vice-versa). Invariant means that if you invert all of those quantities simultaneously, your process is still the same, e.g. can still happen.

Or in other words, for every decay, there exists a recombination of the same particles (only with every charge and parity inverted).

This happens all the time. Otherwise, no heavier particles would even exist! But it is way more difficult do achieve it artificially because all particles have to hit one spot at the same time. But, for example inside the sun, this is quite "common".

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  • $\begingroup$ I actually was thinking about this angle, because I heard that the fundamental laws of physics are time-reversible. $\endgroup$
    – Inflationary_Bubble
    Commented Apr 12, 2017 at 15:50
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Yes, the reverse of the "particle decay" process is possible. The proof of this is the existence of the universe (and us in it). If it were not possible, there would not be molecules, elements, compounds, etc.. The best example of this, is hydrogen (the most abundant element), which is formed by "uniting" protons and electrons,etc.. The conditions required, obviously, are the ones that existed right after the Big Bang.

The reason there is more "literature" on the decay process, is that it is a lot easier and less expensive to observe and measure the decay processes, than to combine particles. It takes a lot of energy and money to perform the combination process.

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