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In the standard description of proton-proton fusion, the first step of the interaction proceeds through the unbound diproton $\rm^2He$: $$ \begin{aligned} \rm p + p &\to \rm {}^2He^* \\ \rm ^2He^* &\to \rm{} ^2H + e^+ + \nu_e \\ \rm e^+ + e^- &\to \gamma + \gamma \end{aligned} \tag{p-p} $$ In nuclear positron emitters there is always some competition with electron capture. In the standard picture the electron capture branch is $$\begin{aligned} \rm p + e^- + p &\to \rm {}^2H + \nu_e \end{aligned} \tag{p-e-p} $$ The chance to distinguish between these apparently arises because p-e-p fusion gives neutrinos with a particular energy, which is higher than the endpoint of the p-p neutrino energy spectrum. If I have the experimental situation correct there is a claimed observation of solar neutrinos consistent with the p-e-p energy in 2012 or 2014, but the solar neutrino experiments are still blind to the p-p neutrinos.

I would expect some contribution to the p-p process from formation of free neutrons: $$ \begin{aligned} \rm p + e^- &\to \rm n + \nu_e \\ \rm p + n &\to \rm{}^2H + \gamma \end{aligned} \tag{n} $$ The first step in this process is endothermic: left alone, the neutron will decay to proton, electron, and antineutrino, releasing about 780 keV. However the sun's core temperature is 10–15 MK, which means the particles in the core have typical kinetic energies $kT \approx 1\rm\,MeV$: plenty hot to drive inverse beta decay.

Is there a gas of free neutrons in secular equilibrium at the Sun's core? If so, why is it omitted from discussions of solar fusion? If not, why not?

Edited to observe that I'm an idiot. I fat-fingered my value for $k$, and actually $kT\approx\rm1\,keV$. If there are neutrons in secular equilibrium, then, they'll mostly come from photodissociation of deuterium.

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  • $\begingroup$ The density of the solar core is on the order of 150g/cm^3, which brings nucleons only about a factor of 5.3 times closer to each other than in water and the thermal energy is non-relativistic, so it sounds like a pretty conventional plasma state. I haven't seen numbers for the equilibrium concentration, though. $\endgroup$
    – CuriousOne
    Jun 19, 2015 at 8:02
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    $\begingroup$ Average KE is nowhere near 1 MeV. $kT \simeq$ 1 keV. $\endgroup$
    – ProfRob
    Jun 19, 2015 at 11:55

2 Answers 2

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The sun's core has a density of

of 150 g/cm³ (150 times the density of liquid water) at the center, and a temperature of close to 15,700,000 kelvin, or about 15,700,000 degrees Celsius; by contrast, the surface of the Sun is close to 6,000 kelvin. The core is made of hot, dense gas in the plasmic state, at a pressure estimated at 265 billion bar (26.5 petapascals (PPa) or 3.84 trillion psi) at the center.

Neutron stars have a density

Neutron stars have overall densities of 3.7×10^17 to 5.9×10^17 kg/m3 (2.6×1014 to 4.1×10^14 times the density of the Sun), which is comparable to the approximate density of an atomic nucleus of 3×10^17 kg/m3.

The neutron is an unstable particle when free. Mean lifetime 881.5(15) s (free)

Its stability in the nucleus ( and not for all nuclei, there exists beta decay) is due to the strong force which creates a potential that binds the neutron below a free energy level. Thus the density of matter where the neutron appears has to be of the order of nuclear densities so as to have the equivalent binding energies and suppression of decays.

The densities of the plasma in the sun's core are order of magnitudes smaller of the nuclear densities necessary for the neutron to participate in the plasma. Once produced it decays into a proton an electron neutrino and an electron antineutrino. No free neutrons survive more thatn a few minutes to have a significant effect on the fusion processes.

To delve a bit into the energies available in the sun plasma , a temperature of ~1.5x10^7 degrees kelvin, using a conversion factor of 8.6x10^-5eV , comes out as

~ 1.3 keV.

In order for a scattering of an electron in such a plasma on a proton to produce a neutron only the tail end of the distribution of electrons can have energies needed for the creation of a neutron: the proton neutron mass difference ~1.4 MeV and in addition the neutrino will take away part of the energy.

The small number of neutrons produced from the tail of the electron proton energy distributions will be further diminished by the decay of the neutron . Thus there will not be any stable subset of neutron in the plasma, even ignoring that when they hit a proton there is a good probability of binding into a deuteron and thus be out of the hypothetical "gas".

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    $\begingroup$ Hi Anna, I'm not sure this answers the question. As long as the neutron lifeime is not zero there should be a non-zero equilibrium density of neutrons. Rob is asking what this concentration is. $\endgroup$ Jun 19, 2015 at 10:16
  • $\begingroup$ @JohnRennie I elaborated a bit. $\endgroup$
    – anna v
    Jun 19, 2015 at 11:40
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    $\begingroup$ @annav I'm aware that the neutron is unstable; I was thinking of the population in secular equilibrium. $\endgroup$
    – rob
    Jun 19, 2015 at 14:59
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    $\begingroup$ to get numbers one would have to take the Maxwell distribution for the electrons and the protons and estimate the numbers with enough energy to have a probability of creating a neutron neutrino pair. This from the tail of the distribution and extra orders of magnitude are lost due to the weak interaction for this reaction. (physicsnet.co.uk/a-level-physics-as-a2/particles-radiation/… feynman diagram) So the number of neutrons generated will be very small. (10^-24) to the 1 of strong interaction (fusion) just by coupling constants. $\endgroup$
    – anna v
    Jun 19, 2015 at 20:17
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[Edit after helpful comment]: The process referred to in the answer is ok (it does happen) but any neutrons thereby produced are rapidly captured (see comment from ProfRob) so the neutron density is a lot lower than I estimated.

Here is a very rough guess just to give you something to play with, and invite others to improve the guess.

The start of the pp chain is when two protons come together and make a deuteron via a weak decay. But in this process there is some non-zero probability that instead of a deuteron you get a proton and a free neutron. The ratio between the rates of these two processes is very hard to guess. It will depend on details of the diproton (unbound) state which I don't know. But suppose we say a ratio of $10^{3 \pm 1}$ just to have something to play with. That is, by a wild guess, I am saying that one in every 100 to 10000 weak decays during a collision results in a free neutron instead of a deuteron. There are $3.7 \times 10^{38}$ protons fused per second, so this suggests the rate of production of free neutrons is $R = 0.001 \times 2 \times 10^{38}$ per second. The neutron lifetime is $\tau = 878\,$s, so we get an equilibrium ($N = R \tau$) when there are $2 \times 10^{38 \pm 1}$ neutrons in the Sun. This is about $10^{-19}$ times the number of protons.

Since this is so small, one might also consider the electron capture reaction $$ p + e \rightarrow n + \nu_e $$ I don't have a cross section to hand but since it's a weak decay I expect it is like the one for fusion but without the Coulomb barrier, hence $10^5$ times larger. However, with an energy deficit of $1.4\,$MeV there is no way to get the required energy from the tail of the thermal distribution. It would have to involve fast electrons or protons from some other reaction before they have thermalized. Therefore I expect it is small compared to the other contribution, but I would like to re-emphasize that this answer has a wild guess in it.

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    $\begingroup$ Why wouldn't a free neutron immediately be captured by a proton to form a stable deuteron before the neutron decays? Cross-section is about 1 barn,. proton density $10^{32}$ m$^{-3}$, mean free path $10^{-4}$m, neutron speed $>3\times 10^5$ m/s, lifetime to capture $< 3\times 10^{-10}$ s? $\endgroup$
    – ProfRob
    Nov 11, 2023 at 13:26
  • $\begingroup$ @ProfRob Yes you're surely right! Thanks. $\endgroup$ Nov 11, 2023 at 15:34

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