11
$\begingroup$

Electrons annihilate with positrons produced through fusion processes in stars. Which particle interaction produces new electrons so that the sun isn't deplete of electrons? Or is something else happening altogether?

Regular fusion cycles in stars produce neutrinos and positrons as by products. Those positrons annihilate with electrons that are already there in the plasma of the star to produce the light that we eventually see. How are those electrons replaced?

Improve this question
$\endgroup$
  • $\begingroup$ You might be confusing regular fusion (which happens in regular stars like the sun) with pair-production stars -- enormous stars with such high energy density in their cores that they spontaneously produce electron-positron pairs. $\endgroup$ – antlersoft Sep 6 '19 at 15:21
  • 1
    $\begingroup$ Regular fusion cycles in stars produce neutrinos and positrons as by products. Those positrons annihilate with electrons that are already there in the plasma of the star to produce the light that we eventually see. How are those electrons replaced? $\endgroup$ – Josh Bilak Sep 6 '19 at 15:40
  • $\begingroup$ I inserted your clarification in comment into your question. I don't want to remove possibly important meanings, but I think the text should be made now somehow more clear. Feel free to edit it, to make it saying exactly what you want to know. $\endgroup$ – peterh - Reinstate Monica Sep 6 '19 at 16:23
12
$\begingroup$

The proton-proton chain ultimately converts four protons into one helium nucleus. The charge of the 4 protons was balanced by 4 electrons, but helium contains 2 protons (and 2 neutrons), so it only needs 2 electrons to be balanced.

As you point out, the process of converting a proton to a neutron releases a positron (and an electron neutrino), and that positron quickly annihilates with an electron.

Here's the diagram from that Wikipedia page of the main p-p chain.

p-p chain

So the process actually consumes 6 protons, and emits 2 protons, a helium nucleus, and 2 positrons (plus a couple of neutrinos), and a couple of gamma photons. The positrons annihilate with 2 electrons, releasing more gamma photons (usually 2 or 3 apiece, depending on the spin alignments of the positron & electron).

If you add everything up, you'll see that the electromagnetic charge balance is unchanged.

We started with 4 protons, which are balanced by 4 electrons nearby in the stellar core plasma. (We can ignore the intermediate pair of hydrogens that are eventually re-emitted). We end up with a helium nucleus that only needs 2 electrons to be electrically balanced, so if those other 2 electrons weren't annihilated then the star would build up an excess of negative charge.

Improve this answer
$\endgroup$
  • $\begingroup$ That clarifies how charge conservation is not violated and provides more detail to the overall process, but if we are constantly losing those pairs of electrons that you mentioned, how does the star have electrons after billions of years of burning? are they brought back into the cycle from a neutron to proton/electron/neutrino reaction? If so, what triggers this? If not, is there something other reaction going on that yields electrons? $\endgroup$ – Josh Bilak Sep 6 '19 at 17:10
  • $\begingroup$ @Josh No, the electrons essentially get consumed by the process of creating neutrons. But why is that a problem? A star generally burns less than 50% of its original supply of hydrogen over its lifetime. $\endgroup$ – PM 2Ring Sep 6 '19 at 17:14
  • 1
    $\begingroup$ @Josh Also bear in mind that p-p fusion is a very slow process. Most of the time when 2 protons fuse into diprotium it immediately fissions back into 2 protons rather than transmuting to a deuteron. The odds of the transmutation happening are on the order of 1 in $10^{26}$, and a typical proton in the solar core has a mean lifetime of a billion years before it is successfully fused to helium. This is a good thing, since it gives the Sun a long lifespan. It also means that per cubic metre, the energy production of the solar core is roughly equal to the energy production of a compost heap. :) $\endgroup$ – PM 2Ring Sep 6 '19 at 17:23
  • 1
    $\begingroup$ So stellar nucleosynthesis is gradually reducing the number of electrons & protons in the universe, but increasing the number of neutrons. When a neutron star forms, a whole bunch of protons + electrons are rapidly converted to neutrons (plus neutrinos). In some very large stars, high energy gamma rays do create electron + positron pairs, but they soon annihilate, creating more gammas, and that process doesn't last for long, since such stars soon explode in a pair-instability supernova, which totally blows them apart. $\endgroup$ – PM 2Ring Sep 6 '19 at 17:31
  • 6
    $\begingroup$ @JoshBilak I think the point is just that yes the star is depleting its electrons, but at exactly the same rate that it's depleting its protons; they remain in balance. So there's no possibility of ending up with the star "running out" of electrons; to do so it would also have had to have converted every single proton into a neutron, which obviously doesn't happen. $\endgroup$ – Ben Sep 7 '19 at 9:23
5
$\begingroup$

They are not replaced.

Fusion in ordinary stars means actually many processes, neutrinos are involved most commonly in these:

  • $p + p \rightarrow D + \nu_e + e^+$
  • $T \rightarrow He_3 + \nu_e + e^+$

The created positrons (very) quickly find an electron to annihillate into two (sometimes 3) gamma photons: $e^- + e^+ \rightarrow 2 \gamma$. As you can see, both the electric charge and both the lepton number (detailed below) remains. Very rarely also neutrinos (or some more exotic particles) can be created, but even these reactions keep the conservation laws.

Sometimes gamma photons can "decay" to electron-positron pairs (or, very rarely, to another particle-antiparticle pairs), this is called pair production. It can happen only near an electrically charged particle (because the photons are going with $c$, but the resulting particles aren't, so to preserve the impulse we need someone to take away the excess impulse - this requirement hugely decreases the probability of this reaction).

None of them destroys electrons. The only nuclear reaction which actually destroys electrons, is actually K-capture, what happens typically simultanously with the $\beta^+$-decay. If it happens, an electron disappears, instead we get an electron neutrino ($\nu_e$).

In the nuclear processes of stars, the net result of the reaction creating/destroying electrons looks like $n \rightarrow p + e + \nu_e$, or $p \rightarrow n + \overline{\nu_e} + e^+$, or their reverse. Note, these are only the net results, the actual processes are more complex (involving the quarks and intermediate bosons of the weak interaction ($W^+$, $W^-$, $Z^0$)). We can say as if neutrons would decay to a protons or electrons (or the reverse), or that protons decay to positron and neutron (or the reverse).

Any time if an electron is created, also an electron antineutrino is created with it. The important thing is, that both of them remain the same:

  • the lepton number (total count of electrons and electron neutrinos, antiparticles count negatively)
  • and the electric charge (electron: -1, positron: +1, proton: +1, neutron: 0, neutrinos: 0)

All the reactions in the stars keep these laws.

P.s. stars are fusing mostly hydrogen to heavier elements. Hydrogen has no neutrons, all the heavier elements have (typically, as the proton number of the nuclei grows, also the ratio of the neutrons grows with it). Thus, the long-term tendency is really that the count of the electrons and protons are decreasing in the stars, while the count of the neutrons grows. Nothing replaces them. The ultimate end, which is possible only in larger stars (much bigger than the Sun) is the neutron stars, which has only a very little electrons (and protons), and the star is mostly a big neutron ball.

Improve this answer
$\endgroup$
  • $\begingroup$ So, electrons already in the stellar plasma interact with positrons that come from the p→n+νe+e+ reaction. This annihilation into gamma radiation does not cause the electrons to be "destroyed?" If it does, than the electrons originally in the star from the nebula it formed from would run out eventually if some other common reaction in the star didn't provide more. does the n→p+ve+e reaction replenish them? I understand that the star does not violate conservation laws. Can you clarify, not how charge is conserved, but how actual electrons remain in the star. $\endgroup$ – Josh Bilak Sep 6 '19 at 17:05
  • $\begingroup$ @JoshBilak No, electron + positron creates two gamma photons. I did not explain this version in the post, but not I did. Yes, annihillation destroys electrons, but it destroys the same number of positrons, too. The stellar plasma is a soup of many particles, and the annihillation of positrons with electrons has a very high probability (compared to the other reactions). Thus, the few created positrons live only very little before annihillation (maybe nanoseconds or so). But not this is important, but that both the total lepton number and the total electric charge is preserved in the reactions. $\endgroup$ – peterh - Reinstate Monica Sep 6 '19 at 17:24
  • $\begingroup$ The charge is converved because as we detail the different possible reactions, all of them, we can't find a single one what would violate the charge conservation. That the electrons aren't depleted in the star, would not be a strict requirement. The strict requirement is that both the charge and the lepton number is conserved. The electrons remain only because it is their only way the keep the conservation laws. But there is an exception: there is a single way for a star to destroy (nearly) all of its electrons: if they "join" their protons into neutrons. This also destroys (nearly) all the $\endgroup$ – peterh - Reinstate Monica Sep 6 '19 at 17:28
  • $\begingroup$ protons in it (and creates such a huge mass of neutrinos that we can detect it billions of light years away). Note, as I wrote in the post, the actual reaction is more complex, only the net result is that electron + proton -> neutron + neutrino! This happens in supernova explosions. The result is that the soup of protons and electrons becomes a neutral ball of neutrons. That is the neutron star. The last one we could see happened in 1987 (for us, actually it happened many millennia ago). $\endgroup$ – peterh - Reinstate Monica Sep 6 '19 at 17:30
  • $\begingroup$ The Sun is too little to become a neutron star ever, but larger stars can. The problem with it that neutron has a little bit bigger mass than the proton, thus too many neutrons don't like to exist together in the star. The free neutron decays to a proton + electron + neutrino with around 20 minutes half life, in some neutron-rich nuclei they can exist longer (for example, tritium has 1 proton and 2 neutrons, decays with 12 years of half life), but only the nuclear processes can't create too much neutrons. A neutron star can be created only if there is something what "compresses" the protons $\endgroup$ – peterh - Reinstate Monica Sep 6 '19 at 17:33
1
$\begingroup$

Hydrogen fusion

I'm stealing a bit from other answers, just to clarify the point here. What follows is not exactly how it all happens, but should clarify how electrons and positrons are balanced.

The key to the answer is in this part of the reaction: two hydrogen atoms become one hydrogen atom. A hydrogen atom is made of one electron and one proton and zero or more neutrons. Now in this step, in one hydrogen atom proton flips into neutron, emits a positron, which in turn may annihilate the electron of said hydrogen atom. Thus results in hydrogen atom (with one proton and one neutron and one electron) and two gamma rays.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.