Talk:Proton–proton chain

Comment

It appears to me the the prep-reaction at the end of the artical is redunent. The infomation is already mentioned in the first part of the artical. But I'm not that confident with my quantum mechanics to just remove it. Lotu 04:23, 6 April 2006 (UTC)

I'm trying to understand the first step in the proton-proton chain reaction a little better. I have a question, perhaps one of you knows the answer. Where do the positron and the neutrino come from? The two protons are made up of quarks, held together by gluons. One of the protons becomes a neutron ... end result is an up quark has become a down quark. The article says that the conversion of the proton to a neutron results from the weak interaction, and the weak interaction article says something about weak interactions allowing quarks and leptons (and their antiparticles) "effectively to change into each other." This is kind of vague. Can anyone out there explain this in a little more detail. The only thing that I can think of is that some of the excess energy has been converted into matter (the positron and the neutrino) because before the reaction we have 6 quarks (two protons) and after the reaction we have 6 quarks (a proton and a neutron) PLUS a positron and a neutrino. Thanks in advance to anyone able to answer my question or confirm my guess that the positron and the neutrino effectively come from some of the excess energy being converted into matter. Sincerely, Dino.

I haven't taken a class in quantum mechanics, so you might want to take my answer with a grain of salt. I think the exces of energy comes from the fact that two protons repel each other, while a proton and a neutron in the deuterium don't. --R.Koot 04:53, 11 August 2005 (UTC)

Yes, the positron and neutrino come from the excess binding energy in 2 protons vs. a deuteron. I suspect R.Koot is right about this coming from the lack of electrostatic repulsion between the proton and neutron (not neutrino) in the deuteron, but I'm not sure. The excess energy must take the form of a positron and a neutrino, not something else, in order to satisfy conservation laws. Specifically, the 2 protons have a charge of +2, but the deuteron has a charge of +1, so a particle with positive charge (the positron) must be emitted. But the positron has a lepton number of -1, so an electron-neutrino (with a lepton number of +1 and charge of 0) must be created to balance that out. ~~ N (t/c) 05:04, 11 August 2005 (UTC)

Fusion is when an unbound state of two particles decays into a bound state by releasing energy, usually as photons. Two protons cannot fuse into helium 2 as there is no such nucleus - because of the repulsion between the protons it would have negative binding energy (i.e. more energy overall than the free particles). However the unbound state of two protons can decay into a deuterium nucleus by emitting a W+ boson, which then decays into a positron and a neutrino. This is like beta+ decay - in terms of quarks, an up-quark turns into a down-quark and a W+. This is a weak interaction, hence the low probability

Another question: "To overcome the electromagnetic repulsion between two hydrogen nuclei requires a large amount of energy, and this reaction takes an average of 10^9 years to complete. " Read as is, it suggests that two protons have to remain in each other's (weak force?) viscinity for 10^9 years before they react. This seems a bit far fetched. What was really meant? Brendan 25 April 2006

I'm also confused by this. Perhaps what is meant is that the reaction takes 10^9 years to exhaust our sun's hydrogen?

In answer to the two above posts, one way to measure reaction rates is the number of reactions per unit volume, per unit time. For example, suppose you look at one cubic metre inside the sun, for one second, and count all the p-p reactions that happen. Chances are, you won't see any because they are very rare. An alternative way to think about the same rate is to assume that you have counted one reaction in your cubic metre, and then ask the question, "How long, on average, did I have to wait to see that reaction?" The answer, in the case of the p-p reaction in the sun's core is about a billion (10^9) years. The actual collision of the protons happens very quickly.

Apetre 21:17, 5 November 2006 (UTC)

Alternate Pathway

Is there a reason why two deuterium particles cannot merge into one helium particle?

In principle they can merge, and actually in the Big Bang nucleosynthesis they do. In the Sun there are simply not enough deuterium. The deuterium is created at a rate about one per 7E9 years. And the rate of the D+p is about one per 2.4E5 years. So practically there is (7E9*7E9/2.4E5=) 2E14 times higher probability to do D+p than D+D in the Sun. — Preceding unsigned comment added by 90.147.118.180 (talk) 11:38, 2 February 2018 (UTC)

Minervamoon 23:56, 9 May 2006 (UTC)

Negative Mass

The proton-proton chain reaction needs to obey the law of conservation of mass. Does it?--67.10.200.101 00:51, 1 July 2007 (UTC)

Of course it obeys the Law of Conservation of Mass-Energy. Mass is not conserved in nuclear reactions because when an energy ΔE is released, then the mass goes down by Δm = ΔE/c^2. This is the whole idea behind this article, behind the proton-proton reaction, behind nuclear fusion, and behind E = mc^2. Did you lose something in the first minute?47.215.188.197 (talk) 07:15, 10 February 2017 (UTC)

The triple-alpha process

The first sentence read as follows:

The proton-proton chain reaction is one of several fusion reactions by which stars convert hydrogen to helium, the others being the CNO cycle, and the triple-alpha process.

The triple-alpha process is not a reaction for converting H to He; it's a reaction for converting He to C. I changed this to say:

The proton-proton chain reaction is one of several fusion reactions by which stars convert hydrogen to helium, the primary alternative being the CNO cycle.

However, I think something like this would be more accurate:

The proton-proton chain reaction is one of two primary fusion reactions by which stars convert hydrogen to helium, the other being the CNO cycle.

I don't think there are "several"--other than PP, PEP, and CNO, there are no other significant reactions. But I'm not a physicist; I could be wrong, so I only went as far as my knowledge. --69.107.75.113 07:49, 16 March 2007 (UTC)

If a fraction of the proton's mass is converted to energy, will the positive charge of that partial mass go with it as well? For example, if 0.7% of the mass gets converted, would the charge decrease by 0.7%? I am asking because when matter becomes excited (in the case of quarks) the charges all remain constant but the masses do not.

I can only reply "silly question". In our Universe, electric charge is conserved above all. There is no conceivable theory in which it might not be.47.215.188.197 (talk) 07:37, 10 February 2017 (UTC)

Many more nuclear fusion reactions happen in VERY hot stars. The important ones of these are as follows: C-12 + He-4 = O-16 + energy; O-16 + He-4 = Ne-20 + energy; Ne-20 + He-4 = Mg-24 + energy; Mg-24 + He-4 = Si-28 + energy; Si-28 + energy = S-32 + energy; and so on until Fe-56 (iron-56) is produced. Along the way, some additional neutrons have to be picked up from other reactions. The heaviest element with an equal number of protons and neutrons is Ca-40, and this one exists because 20 is a "magic number" in "nuclear shell theory". Thus with 20 protons and 20 neutrons, calcium-40 is "doubly magic". Its neighbor in the Periodic Table, potassium-40 is radioactive, but with a long half-life.47.215.188.197 (talk) 07:37, 10 February 2017 (UTC)

Energy/ charge release

If a fraction of the proton's mass is converted to energy, will the positive charge of that partial mass go with it as well? For example , if 0.7% of the mass gets converted ,would the charge decrease by 0.7%? I am asking because when matter becomes excited (in the case of quarks) the charges all remain constant but the masses do not.

Electric charge is always conserved. If a fraction of the proton's mass is converted to energy, the result won't be a proton any more (e.g. when two protons fuse you get a deuteron, a neutrino, and a positron, none of which is just a proton). --Rogermw (talk) 16:28, 25 August 2011 (UTC)

P-P balancing CNO

I read in Kaler's text that the solar fusion is 93% P-P and 7% CNO. Extrapolating, I figure an F0V star would be half and half. Interesting! But am I correct? (I suppose I'm correct in that there's SOME point where the two reactions are equally present.) 68Kustom (talk) 10:32, 10 February 2008 (UTC)

Yes, but that point depends on the metallicity of the star. Dauto (talk) 19:23, 22 August 2011 (UTC)

References?

There are no external links on the page. If I add one, will it be taken off if it wasn't a source for the information? ZtObOr 00:37, 27 May 2008 (UTC)

fixed mistake in "The pp chain reaction"

I fixed a mistake in the section titled "The pp chain reaction." The text read like this:

This first step is extremely slow, because it depends on an endoergic beta positive decay, which requires energy to be absorbed, to convert one proton into a neutron. In fact this is the limiting step, with a proton waiting an average of 10⁹ years before fusing into deuterium

This is incorrect. As shown by the equation, the reaction gives a net release of energy. I've put in the following corrected explanation:

This first step is extremely slow, both because the protons have to tunnel through the Coulomb barrier and because it depends on weak interactions. —Preceding unsigned comment added by 76.93.42.50 (talk) 17:15, 30 May 2008 (UTC)

The above correction was 'undone' by an addition in Jan 2016 (by JGS952). I have deleted the offending phrase, inserted without citation for support, because it is incorrect. I have not chosen to repeat the Coulomb barrier story because it is mentioned earlier. The competition with the pp breakup is more significant here. I certainly am open to improvements in the text. Aqm2241 (talk) 23:39, 18 November 2016 (UTC)

Net neutrino production

Since both CNO and proton-proton chain produce 2 electron neutrinos per 32 MeV, as does any hydrogen to helium 4 process, it seems fair to say that the total neutrino flux from a hydrogen-burning star does not depend at all on the solar model, just on the total luminosity. But I remember that before neutrino oscillations were well established, some people thought that they could fix up a solar model to produce fewer neutrinos per MeV. Does anyone know what were they thinking? Were they just being silly?Likebox (talk) 20:23, 8 October 2009 (UTC)

Ok, this observation was made in the literature. The cite is John Bahcall "The Luminosity Constraint on Total Neutrino Fluxes" Physical Review C, vol 65, 25801 (2002).Likebox (talk) 20:23, 8 October 2009 (UTC)

They were not being silly. The observations cover only a fraction of the spectrum of neutrinos emitted by the sum, not the total neutrino flux which is indeed fixed by the luminosity of the star. Dauto (talk) 19:18, 22 August 2011 (UTC)

Deuterium + 2 MeV > proton + neutron and neutrons decay emitting antineutrinos (neutron source Wiki). We detect neutrinos on Earth by rare conversion of proton to neutron (with electron capture). With stellar density and neutrino flux there would be proton to neutron reactions competing with neutron decay. Shjacks45 (talk) 04:20, 3 March 2020 (UTC)

Conversion of protons to neutrons on a massive scale occurs when red dwarf collapses to a neutron star. See neutron star Wiki. Shjacks45 (talk) 04:25, 3 March 2020 (UTC)

Charge-balance

Any reason the nuclear reactions are not balanced with respect to charge? I know they're nuclear, so only the nuclei (not electrons) are the primary interactions to consider, but consider in The pp chain reaction:

1 1H

+

1 1H

→

2 1D

+

e+

+

ν e

+

0.42 MeV

A high-school chemistry student should recognize "hey, not a balanced reaction because charge is not conserved!" And the presence of electrons really does matter because this reaction is followed immediately by one that involves an electron.

1 1H+

+

1 1H+

→

2 1D+

+

e+

+

ν e

+

0.42 MeV

I see that the equations are autogenerated, so I don't know if these changes are best made in the wikisources or in the source that is used to (re)generate them. DMacks (talk) 16:25, 16 October 2009 (UTC)

In these nuclear reactions, the symbols refer to the nuclide only and not the atom or ion. The electrons that form an atom with the nuclide are irrelevant to the equations; they are not part of the reaction. Charge is not normally specified because the number of protons (bottom left) is equal to the charge, so you can calculate the charge of the nuclides if you want to check if the charge is balanced. Just because the charge is not explicitly mentioned does not mean it's not balanced. I hope this answers your question - btw. I created the equations using a script I wrote because I found it easier to manage any changes and such, if they need changes and you think it's easier to do it manually, feel free to do so.     — SkyLined (talk) 21:51, 19 October 2009 (UTC)

I'm fine either way...student of mine mentioned it and I didn't (but now do...thanks!) see a self-consistent scheme here. DMacks (talk) 21:58, 19 October 2009 (UTC)

People have missed the point that all of these reactions take place in a nuclear plasma in which ALL of the electrons have been stripped off by the extreme heats and velocities involved. Remember PLASMA: all of the electrons are floating around freely, and they are not involved in such reactions as He-4 + He-3 --> Be-7 + energy and He-3 + D --> He-4 + n 47.215.188.197 (talk) 21:28, 12 February 2017 (UTC)

VERY POOR NOTATION:

1H + 1H → 2H + e+ + νe + 0.42 MeV

You have mass-energy on the left. Then you have mass-energy on the right:

2H + e+ + νe, that equals all of the mass-energy on the left. Then you have + "0.42 MeV" on the right as an additional term. You have counted the 0.43 MeV twice. That 430 KeV is actually contained in the diproton, the positron, and the neutron. These are not like (nonnuclear) chemical equations where you can write 2H2 + O2 ---> 2H2O + 100 kilojoules, or whatever the amount it.47.215.188.197 (talk) 08:34, 10 February 2017 (UTC)

Is the below statement in the The pp chain reaction section completely correct?

Statement: The positron immediately annihilates with an electron, and their mass energy is carried off by two gamma ray photons.

Is it not true that if the positron and electron are in motion, the energy of the photons would be slightly larger to take into account the two particle's energy of motion? Dave3457 (talk) 22:21, 20 August 2011 (UTC)

Yes that is true. Dauto (talk) 04:12, 21 August 2011 (UTC)

Thanks. I added "...as well as their kinetic energy...". Strictly speaking I should have said "..as well as their energy of motion..." but since they are low speed I didn't bother. Of course, if that's an issue for anyone, change it. Dave3457 (talk) 22:05, 21 August 2011 (UTC)

Kinetic energy and energy of motion are the same thing. Dauto (talk) 22:19, 21 August 2011 (UTC)

No they're not, Kinetic energy is the non-relativistic approximation of the true energy of motion. 1/2mv^2 is the first term of an infinity series. Maybe I'm using the wrong term. I'm using it in the context of... Total relativistic energy = rest mass X c^2 + energy of motion. Dave3457 (talk) 17:37, 22 August 2011 (UTC)

Yes, 1/2mv^2 is an approximate formula for the kinetic energy. The relativistically correct formula is also called kinetic energy. Dauto (talk) 19:14, 22 August 2011 (UTC)

How does this situation look to observers A, B, and C, all three present at the moment of annihilation but moving at very different speeds? A is in a frame of reference in which the positron and electron are close to stationary. B and C are each in a frame of reference moving at half the velocity (as a vector) of one of the departing photons (a different photon for each).

According to the above, B and C would see both photons at higher energy than A would. However it seems to me that B ought to see the photon it is following as having lower energy than C's, and vice versa for C.

The reason I ask is that it seems to me that relativity makes the original question ill-posed. Vaughan Pratt (talk) 04:10, 6 January 2017 (UTC)

The electron-positron annihilation step

The second reaction shown is the annihilation of the electron and the positron. On the right side of that equation, we see this:

2γ + 1.02 MeV

Is this the correct notation? All the energy released in the annihilation (1.02 MeV) is contained in the two gamma-ray photons. To me, this notation implies that there are two gamma ray photons released, plus an additional 1.02 MeV in thermal energy, which is not the case. --Rogermw (talk) 16:33, 25 August 2011 (UTC)

I agree completely! There is an obvious problem because the two gamma photons CONTAIN the released mass-energy, and so the mass-energy should NOT be listed separately with a + sign in between. The RHS of the equation lists the same amount of mass-energy TWICE, and I don't understand why nobody else has seen this since 2011. I don't even have a Ph.D. - I just have two master's degrees and one bachelor's degrees in the technical fields of electrical engineering and mathematics. I can SMELL equations that are not balanced, and that + sign unbalances everything! The system of notation sucks!47.215.188.197 (talk) 07:48, 10 February 2017 (UTC)

Incorrect image

This image shows beryllium-7 changing into lithium-7 with only the emission of a neutrino. It should also be shown emitting a positron. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 20:09, 28 December 2011 (UTC)

Bitching Betty, you are 100% correct, and I can see that nobody has bothered to make a correction in the five+ years since December 2011.47.215.188.197 (talk) 08:43, 10 February 2017 (UTC)

Deuterium events are rare?

I am confused by the third paragraph of this article. I feel that it should be changed, (or even omitted). It reads:

“In the Sun, deuterium-producing events are so rare (diprotons being the much more common result of nuclear reactions within the star) that a complete conversion of the star's hydrogen would take more than [ten billion] years at the prevailing conditions of its core.”

Regarding this first sentence; there seems to be conflicting "facts", that prevent me to conclude that the article is accurate. First of all it states that “deuterium-producing” events are "so rare in our sun, that it would take 10 billion years to convert all the sun’s hydrogen into helium," - yet from all sources, I’ve learned that our sun is 5 billion years old, and will last at least another 5 billion years (a total of 10^10 years).

The article then goes on to explain this very process (deuterium = one proton, one neutron) - shown in the adjacent illustration, and in the subtitle: “The proton-proton chain reaction” - without expanding on the exerted ‘much more “common” dipoton synthesis’ of converting hydrogen into helium.

In parenthesis, the sentence infers that the sun’s MAIN fusions product is “diproton” - or, helium-2. The article for helium-2 (http://en.wiki.x.io/wiki/Diproton#Helium-2_.28diproton.29) explains that diproton is only a “hypothetical” helium isotope. It appears to me that helium-2 “burning“ would convert the sun’s hydrogen exponentially faster than 10 billion years!

As for the second sentence, which reads: “The fact that the Sun is still shining is due to the slow nature of this reaction; if it went more quickly, the Sun would have exhausted its hydrogen long ago.” Whether or not the first sentence has merit, the sentence should unambiguously define which “reaction” (deuterium or diproton) is responsible for the “slow nature” of the sun’s actual proton-proton fusion process.

To me, it seems the contributor may have cut-and-pasted some information from another stellar nucleosynthesis web-page into this article, without making the necessary changes to the text to fit the context of the actual subject.

P.S. This is my first contribution to a science related wiki. I would also like to know; and see in the article, what boson binds the nuclei that the chain reaction actually produces - He-2 or He-3 (and how the [mesons] are created during the fusion process}. But clarification of the introduction is paramount at this time. I’m "below" a novice’s status on any subatomic subject, and wouldn’t dare make any changes myself.--Manixx2a (talk) 23:40, 1 June 2012 (UTC)

The problem here is that the words "rare" - and "uncommon" or "of low probability" are being confused. What we have here are events of low probability (atom-by-atom), but a stupendously-large number of atoms, sagans of them ("billions & billions"). The upshot is that huge numbers of the events "of low probability" happen every millisecond in stars, and there for, those nuclear events are NOT rare.47.215.188.197 (talk) 21:18, 12 February 2017 (UTC)

Not mentioned in this article is that Deuterium splits into proton and neutron at stellar temps (2 MeV per neutron source wiki). Also He3+D reacts to He4 at a lower temp than He3+He3, and He3 conc lower so rarer reaction. Shjacks45 (talk) 03:31, 3 March 2020 (UTC)

Misleading image

I think the first image (upper right corner) in this article (or its image text) is misleading, it seems to show that this is the only possible branch in the PP chain (provided the reader reads quickly). — Preceding unsigned comment added by 137.132.250.13 (talk) 09:41, 21 April 2013 (UTC)

Missing a few reaction pathways

The weak interactions in a star are as follows:

An up quark (+2/3) emits a W+ boson, changing into a down quark (-1/3). Protons are up+up+down, neutrons are up+down+down. So this emission from a single quark changes the identity of the hadron from a proton to a neutron. The W+ boson subsequently decays very quickly to a positron and neutrino. Energy is always conserved, and mass is the densest form of energy. The difference in mass of an up quark and down quark is about 2.5 MeV. However, the mass of the W+ is 80.4 GeV, and the mass of its decay products are 511 keV for the positron and less than 2.2 eV for the neutrino. While there's no way enough energy could be in one place to make a W+ boson (the mass of the proton including energy of the gluons is 938 MeV), quantum tunneling lets them be produced, as long as they decay fast enough. The change in energy, the difference in mass of the up and down quark, is carried as a very high mass W+ for a short time, then the decay products subtract from that difference. 2.5 MeV - 511 keV = about 2 MeV for the gamma photon produced for the difference in mass to allow energy to be conserved.

However, this is not the only weak interaction that can occur. Rather than having the W+ decay, it can also be passed from one up quark to another, simultaneously changing both up quarks into down quarks (change in energy is about 5 MeV). For example, three protons come together and form tritium and a gamma ray, or four protons come together and form helium-4. Both processes release 5 MeV gamma photons but no positron and neutrino like the proton + proton -> deuterium reaction.

A few other isomeric reactions that don't use weak changes but still produce gamma photons (though not as energetic):

deuterium + deuterium --> helium-4 + gamma

deuterium + tritium --> helium-4 + neutron + gamma

tritium + tritium --> helium-4 + 2 neutrons + gamma

tritium + proton --> helium-4 + gamma

helium-3 + neutron --> helium-4 + gamma

helium-3 + deuterium --> helium-4 + proton + gamma

helium-3 + tritium --> helium-4 + deuterium + gamma

Reactions that involve W+ emission and decay (2 MeV gamma photons):

proton + proton --> deuterium + positron + neutrino + gamma

proton + deuterium --> tritium + positron + neutrino + gamma

proton + helium-3 --> helium-4 + positron + neutrino + gamma

deuterium + deuterium --> tritium + neutron + positron + neutrino + gamma

deuterium + neutron --> helium-3 + positron + neutrino + gamma

tritium + neutron --> helium-4 + positron + neutrino + gamma

Reactions that involve W+ exchange (5 MeV gamma photons):

3 protons --> tritium + gamma

3 protons --> deuterium + neutron + gamma

4 protons --> helium-4 + gamma

4 protons --> helium-3 + neutron + gamma

5 protons --> deuterium + helium-3 + gamma

deuterium + helium-3 --> tritium + 2 neutrons + gamma

deuterium + helium-4 --> tritium + 3 neutrons + gamma

helium-3 + helium-3 --> helium-4 + 2 neutrons + gamma

Neutrinos and antineutrinos may be equivalent if neutrinos are majorana fermions, but until we know for sure we should make sure to understand that stars produce neutrinos, and a star made of antimatter would produce antineutrinos. Why is this important? Neutrinos cause beta- decay, while antineutrinos cause beta+ decay. Both by the weak force. For example, Carbon-14 beta decays to Nitrogen if a neutrino exchanges a W+ with a down quark, changing it into an up quark. while changing the neutrino into an electron. Similarly, in beta+ decays like Sodium-22 into Neon, an antineutrino exchanges a W- with a up quark, changing it into a down quark, while changing the antineutrino into a positron.

A star made of antimatter would also undergo an antiproton-antiproton chain reaction, where an antiup antiquark (-2/3) emits a W-, changing into an antidown antiquark (+1/3), while the W- decays to an electron and antineutrino. The same reaction patheays are available, and the gamma photons produced have the same energy as in normal stars.

And it goes deeper than that. So called virtual W+ and W- bosons cannot actually be created and have energy be conserved. The real W bosons only come from black holes, where electroweak symmetry is restored and the masses of W and Z bosons and the masses of all quarks and leptons return to zero and all particles move at the speed of light. Real W bosons from the black hole at the center of our galaxy (and at the center of every galaxy) power the proton-proton chain reaction in the stars around it. Move closer to the black hole, and the reaction becomes easier to sustain. Move farther away, and less W bosons reach the star, making the weak interactions less likely to occur. Past a certain distance, stars can't form because they don't receive a large enough flux of real W bosons to sustain the chain reaction. W and Z bosons produced at the event horizon of a black hole produce all weak interactions far from the black hole, and our distance from the black hole is what makes the apparent mass of the W and Z so large.

Electroweak symmetry breaking when temperature drops below the electroweak symmetry scale, and electroweak symmetry being restored above that temperature, like at the event horizon of a black hole, is a well-described phenomenon central to the Standard Model. The constant vacuum expectation value of the Higgs field breaks the symmetry of the weak force, giving the W and Z bosons a large mass that makes them appear to have a short range, even though before the symmetry was broken, they were massless had infinite range. The electroweak mixing angle theta varies with momentum transfer (temperature), and the mass of the Z is defined as the mass of the W/cos(theta), causing the W and Z to have the same mass (0) when the mixing angle is zero. However, the absolute mass of the W is determined by distance from the supermassive black hole at the center of the galaxy. Increasing distance means a greater mass, and consequently a shorter range for the weak force. — Preceding unsigned comment added by 70.113.33.189 (talk) 03:17, 4 November 2014 (UTC)

Deuterium, Tritium, Lithium (6/7), Beryllium9, He3, Boron(10/11) have a very low cosmic abundance, much less than if actually a key stellar component. These isotopes (cept He3) break apart at energies less than solar interior(D at 2MeV into proton and neutron.) Although two body reactions are common, three body reactions (chance of 3 particles being in the same place) becomes mathematically rare, and 4 body reactions orders of magnitude rarer. Proton > Neutron reaction is single reactant with electron capture by proton. Shjacks45 (talk) 05:06, 3 March 2020 (UTC)

The sun and hydrostatic equilibrium

I removed "The fact that the Sun is still shining is due to the slow nature of this reaction; if it went more quickly, the Sun would have exhausted its hydrogen long ago." This is not so. why? The rate of fusion is determined by the pressure of the gravitational inward pressure, which, in the form of a gravitational collapse, is what provokes the fusion in the first place. If the rate of fusion increases it pushes the gravitational collapse outwatds, reducing the temperature in the star's core and reducing the rate of fusion. And it is this hydrostatic equilibrium and not any factor in the proton-proton chain reaction which determines the rate of fusion. This is a standard explanation for how stars including the sun function. ♫ SqueakBox talk contribs 17:41, 6 June 2015 (UTC)

I disagree with your end conclusion because you don't understand the scales of the factors. IF the weak nuclear force were not the bottleneck that slows everything down, the sun never would have become as large as it is. A much smaller sun would have ignited earlier, blown itself to smithereens, and blown away most of the cloud of hydrogen surrounding it. In other words, there never would have been any "hydrostatic equilibrium", our kind of a sun never would have existed. You made the assumption of equilibrium much too soon, and it never would have happened. "When you change the rules so much, you blow away the whole ball game!"47.215.188.197 (talk) 06:51, 10 February 2017 (UTC)

1971 offline/not real source?

Ishfaq Ahmad, The Nucleus, 1:42,59, (1971), The Proton type-nuclear fission reaction

What type of source is this? Is it book, video, article in journal? What is its DOI, ISSN, ISBN? Was it cited in other papers? I'm unable to find any real citations of the paper (before wikipedia used it - diff 13 March 2011 by User:Ironboy11). I think we can find better source that is easier to check (for example, some online pdf).`a5b (talk) 04:46, 9 July 2015 (UTC)

Ok, there is Pakistanian scientific journal "The Nucleus" (website www.thenucleuspak.org.pk), but google scholar has no such article indexed in it in 1965-1975 by Ishfaq Ahmad:

https://scholar.google.com/scholar?q=Proton++type-nuclear+fission+reaction&hl=en&as_publication=Nucleus&as_sdt=0%2C5&as_ylo=1965&as_yhi=1975 https://scholar.google.com/scholar?q=Ahmad+proton&btnG=&hl=en&as_publication=Nucleus&as_sdt=0%2C5&as_ylo=1965&as_yhi=1975 https://scholar.google.com/scholar?q=Ishfaq+proton&btnG=&hl=en&as_publication=Nucleus&as_sdt=0%2C5&as_ylo=1965&as_yhi=1975

This is a very poor title for a picture.

This is a very poor title for a picture: 678px-Proton-Proton_II_chain_reaction. The "678px" is practically an irrelevant number, and why would anyone think that it would be? What IS important is that the larger of the two dimensions is 800 pixels! Then whatever the smaller dimension "d" is, it is "trapped" such that 0 < d < 800.
In contrast, if you specify d = 678, then the larger dimension "D" can me anything at all from 679 to ∞. So, 678 tells us/you nothing about the overall size of the picture at all. Giving lower bounds on the smaller number is useless, but giving an upper bound on the larger number is very useful. In this case, we know that the upper bound in D = 800, and this tells us/you something about the size of the picture. 47.215.188.197 (talk) 06:59, 10 February 2017 (UTC)

Reaction + Energy Breakdown

• proton-proton fusion into deuterium accounts for 40% of the reactions by number
  • releases 1.44 MeV of energy for each reaction
    • 10.4% of the Sun's total energy
• deuterium-proton fusion into helium-3 accounts for 40% of the reactions by number
  • releases 5.49 MeV of energy for each reaction
    • 39.5% of the Sun's total energy
• helium-3/helium-3 fusion into helium-4 accounts for 17% of the reactions by number
  • releases 12.86 MeV of energy for each reaction
    • 39.3% of the Sun's total energy
• helium-3/helium-4 fusion into two helium-4s accounts for 3% of the reactions by number
  • releases 19.99 MeV of energy for each reaction
    • 10.8% of the Sun's total energy

from http://www.forbes.com/sites/startswithabang/2017/09/05/the-suns-energy-doesnt-come-from-fusing-hydrogen-into-helium-mostly/

Zeryphex (talk) 03:49, 21 November 2018 (UTC)

How does He3+He4 equal 2He4? Proton-proton fusion but how much energy when it falls apart (neutron source wiki notes 2MeV splits D)? Fusion energy comes from binding energy difference, so this means Proton-proton product has binding energy? You missed D+He3 which requires lower temp than He3-He3 (and He3 conc lower). Would think He3 would be more common in our universe if He3 was that critical to stellar fusion. Shjacks45 (talk) 03:41, 3 March 2020 (UTC)

Misleading energy addition in the "The proton–proton chain"

In this section, the steps of the p-p chain are summarized in equations with the "reactants" and the products of the fusion processes. Very conveniently, the energy yield of every step was also added. However, the way the energy yield is shown, as an addition to the products, is misleading. Take the example below:

2 1D

+

1 1H

→

3 2He

+

γ

+

5.493 MeV

The way this is written, it seems that, besides the ³
₂He
and the
γ
particle, there's more 5.493 MeV of energy as a product. In reality, that energy is distributed in the energy of the
γ
photon and the kinetic energy gained by the ³
₂He
nuclide. In summary: the energy isn't an additional product of the reaction, as might be interpreted from the equation.

I suggest changing removing the energy yield from inside the equations, and add a comment below explaining that the reaction generates x MeV of energy (and even explain where that energy is found). GoldSkulltulaHunter (talk) 06:04, 8 November 2022 (UTC)