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Where is the mass coming from when neutrons are produced from protons in the Sun?

If a positron is made, will it possibly annihilate with an nearby electron?

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    $\begingroup$ Yes, the positrons annihilate with nearby electrons. See astronomy.stackexchange.com/a/33277/16685 $\endgroup$ – PM 2Ring Jan 29 at 12:15
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    $\begingroup$ Your top-line question is asking whether the Sun increases in mass, but the body of your question asks where does the extra mass come from in a neutron produced from a proton. These seem to be two different questions. $\endgroup$ – Wossname Jan 29 at 23:27
  • $\begingroup$ Does it even make sense to speak of mass instead of mass-energy for things like stars? $\endgroup$ – rubenvb Jan 30 at 15:13
  • $\begingroup$ Where is the mass coming from - Do you mean the mass of the neutrons? It is a bit unclear what you mean. Maybe you should ask where the neutrons come from. $\endgroup$ – descheleschilder Feb 1 at 11:20
  • $\begingroup$ It seems pretty clear that the Sun's mass doesn't increase. It sends particles into space (solar wind) and emits e.m. radiation. So her mass decreases. Her mass is increased by the absorption of cosmic particles but this increase is overwhelmed by the decrease. $\endgroup$ – descheleschilder Feb 1 at 11:26
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While a free neutron does have more mass than a free proton, a bound helium-4 nucleus has less mass than two free protons and two free neutrons. In fact, the helium-4 nucleus has less mass than four free protons. The difference goes into the binding energy of the nucleus. Therefore, as the other answers state correctly, stars are constantly losing mass, not gaining it, through their fusion reactions.

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    $\begingroup$ @jbradvi9: No, that is not true. As this answer and the one by Codename 47 clearly state, the neutrons and protons in a helium-4 nucleus mass less than 2 free protons and 2 free neutrons. Which is basically why fusion is a exothermic process, no? $\endgroup$ – jamesqf Jan 30 at 4:41
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    $\begingroup$ @jbradvi9: The intermediate steps along the way from 4x free protons to 1x He nucleus do not include free neutrons, so the mass of a free neutron isn't directly relevant. $\endgroup$ – Peter Cordes Jan 30 at 11:59
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    $\begingroup$ This answer should be edited to point out that a helium-4 nucleus has less mass than four free protons. While it is true that it also has less mass than two free protons and two free neutrons, that does not prove the point. $\endgroup$ – Paul Young Jan 30 at 18:27
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    $\begingroup$ Apparently, the loss rate is about 4 million tons per second. Plus 1.5 or so to the solar wind... $\endgroup$ – Peter - Reinstate Monica Jan 30 at 20:16
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    $\begingroup$ I edited on the back of a couple of up votes $\endgroup$ – Paul Young Jan 30 at 23:51
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In fact, the Sun is losing mass all the time. It radiates large amounts of energy, and through the energy-mass relationship $E = m c^2$, radiating energy means radiating mass.

Since the mass of a helium atom is less than the mass of the four free protons which enter the fusion process, one can consider fusion the process of converting mass into energy.

Enter image description here

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    $\begingroup$ To put a number on it; it's about 4 million tonnes per second. $\endgroup$ – Oscar Bravo Jan 29 at 12:01
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    $\begingroup$ Not to mention solar winds, which can (depending on the star) off gas entire solar masses worth of material in a relatively short time interval. $\endgroup$ – BooleanDesigns Jan 29 at 21:01
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    $\begingroup$ @MPW, consider that the solar wind and solar radiation pressure exert a rather significant outward force on anything trying to fall into the star. It's actually quite difficult. $\endgroup$ – CarlF Jan 29 at 22:17
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    $\begingroup$ @BooleanDesigns Indeed, I did not to mention solar wind, because the question was about nuclear fusion. Interestingly, solar wind loss is less than half of the mass loss due to fusion. I'd have thought it would be more... $\endgroup$ – Oscar Bravo Jan 30 at 6:46
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    $\begingroup$ @OscarBravo Well, the solar wind is powered by the same nuclear furnace, and most of the energy produced by the fusion escapes as radiation. There isn't that much left over to accelerate the solar wind against the massive gravity of the Sun - the solar wind isn't exactly slow, but it's still just hundreds of km/s. $\endgroup$ – Luaan Jan 30 at 7:35
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Yes, the conversion from hydrogen to helium by nuclear fusion releases energy at the expense of the products having less mass than the reactants. The linking equation between energy and loss of mass was proposed by Einstein, $E= \Delta m\,c^2$.

The numbers are quite astronomical.

Due to nuclear fusion the Sun loses about 4 million tonnes ($4\,000\,000\,000 \ \mathrm{kg}$) of mass every second and in the process produces $3.6 \times 10 ^{26} \ \mathrm J$ of energy per second.
This may seem a lot but if the Sun kept shining for another $5$ billion years it would have lost about $\frac {3}{10\,000}^{\text{th}}$ of its total mass.

The other method of mass loss from the Sun is the solar wind, which are ionised particles "boiled off" from the surface of the Sun.
This process accounts for about $\frac 14$ the total loss of mass from the Sun.

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  • $\begingroup$ What is the difference between increasing of mass due to formation of neutrons and decrease of mass due to formation of denser element in this case helium? $\endgroup$ – jbradvi9 Jan 29 at 17:10
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    $\begingroup$ The difference is that by themselves protons and electrons never become a neutron without the input of a lot of energy, in chemical parlance it is an endothermic reaction whereas the production of a helium nucleus results in the emission of energy - it is an exothermic reaction. The "natural" direction of a reaction is to decrease the potential energy. $\endgroup$ – Farcher Jan 29 at 17:24
  • $\begingroup$ Please correct me if I am wrong when an upQ expel a +1charge in form of a positron the system loses one positron mass and a mass of the nearby anihileted electron but You get a mass increased due formation of a downQ in the former proton or actual neutron.If the mass of the neutron minus mass of proton is greater than the two masses of an electron plus binding energy the system increases in mass....is it correct?But this same equation could also give decrease in system mass if the binding energy takes away lot of mass... $\endgroup$ – jbradvi9 Jan 29 at 23:17
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    $\begingroup$ The numbers are astronomical? No kidding! ;-) $\endgroup$ – Peter - Reinstate Monica Jan 30 at 20:50
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Stars fuse hydrogen (bare protons) to helium (two protons and two neutrons). But the reaction has a number of intermediate steps. The first step fuses two protons to deuterium, and a positron and a neutrino:

$$ ^1_1H + ^1_1H ~ \rightarrow ~ ^2_1H + e^+ + \nu_e $$

You are correct that a neutron has higher rest mass than a proton, but this reaction never generates a free neutron. The neutron is bound to the other proton from the moment it is generated. The binding energy compensates for the extra rest mass energy.

https://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction

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  • $\begingroup$ Is this mass defect radiated out as photons or mass depends on frequency of the wave function so maybe nucleons tied together give some energy for the strong nuclear interactiion and that way have longer wavelenghts and less mass due to this fact? $\endgroup$ – jbradvi9 Jan 30 at 3:10
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    $\begingroup$ @jbradvi sorry I didn't understand your question. The energy will be divided between the rest mass, the kinetic energy, photon(s) and the binding energy. You can think of the binding energy as negative, because the particles have lower energy when they are bound than when they are separate. $\endgroup$ – craq Jan 30 at 4:13
  • $\begingroup$ Is there a mechanism that transfers the binding energy into the proton to create the neutron? $\endgroup$ – Wossname Jan 30 at 22:48
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    $\begingroup$ @Wossname I don't know if "mechanism" is the right word. As I understand it, the protons are forced extremely close together (aka collide) because of the massive pressure and temperature in a star. At that range the strong force becomes relevant, and you can imagine that they become momentarily "bound" as a "nucleus" with 2 protons and 0 neutrons. Because the strong force is acting, you could talk about or calculate a binding energy at that point. Almost immediately, that very unstable "nucleus" converts itself to a deuterium nucleus, a positron, a neutrino and energy. $\endgroup$ – craq Jan 31 at 2:34
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No it is decreasing in mass.

You must take into account binding energy in the atom core. The protons and neutrons are bound together really strongly by the strong nuclear force inside the core of a Helium atom.

It takes lots and lots of energy to try to pull them apart.

Firstly you must overcome electromagnetic force to push them together so closely they fusion.

But at the scale of an atom core, the strong nuclear force is very much stronger, so once they fusion the electromagnetic force has no chance to repel the protons from each other.

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The sun loses about 4 million tons per second in mass. This is largely in the form of light (which has a mass from E=mc^2), and charged particles such as helium nuclei.

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  • $\begingroup$ So photons take away mass or energy and temperature?...... $\endgroup$ – jbradvi9 Jan 30 at 20:19
  • $\begingroup$ mass is equivalent to energy, so as the sun is incredibly luminous, that amount of energy has a significant mass. Photons themselves are massless, it is only the wavelength of the photon that determines the mass loss. Temperature is a scalar measurement which relates to the average wavelength of the energy emission. $\endgroup$ – Dr Jonathan Kimmitt Feb 6 at 17:14
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Note that the question asked Where does the mass coming from when neutrons are produced from protons. A neutron is basically(*) a proton+electron. So the mass was already there as an electron and it basically merged with a proton to form the neutron.

As others correctly explained, when protons and neutrons bind into a nucleus they release energy. E=MC^2, the bound nucleus has lower mass than the free protons and neutrons. That energy comes out as sunlight.

(*)Footnote: For simplicity I deliberately ignored neutrinos.

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