Talk:Nuclear force

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Dr Chris Illert[edit]

Dr Chris Illert has worked on nuclear models and was awarded his Ph.D. on 26 September 2013 at the University of Western Sydney[1]. Enquiries about his work can be directed to the University of Wollongong via Michael Organ [2]


Ahhhhh, you know you're on Wikipedia when you need a physics(also known as madar chod) degree to understand a relatively straight forward topic like the Strong Nuclear Force.Diafel 06:44, 11 March 2006 (UTC)

I agree completely. Two sentences of introduction then BAM: Yukawa potentials, pions, quantum chromodynamics... while I've no doubt this information belongs SOMEWHERE in the article, it has no place in the introduction. Bear in mind the audience of this article: IIRC, the nuclear force is introduced as a concept at high school level. The material should start simple and general and ease into the details. As things stand, no-one without the requisite physics degree-level jargon will make it past the intro. And if you do have a physics degree, you're unlikely to be reading an encyclopedia article on such a basic concept. Think of the children, dammit!

It's looking better now, but one more suggestion: the paragraph in the intro discussing the recent change in terminology is too detailed and uses jargon which has no place in the intro. There should be a brief, simple sentence or two pointing out that the old "strong" usage has been dropped - this is important to get across in the intro - and the details of the confusion, involving QCD etc. should be in their own section near the end of the article. That would make it heaps clearer, while keeping all the present content.

Residual strong force[edit]

I made Residual strong force REDIRECT to Nuclear force. This IS correct, right? I hardly find this stuff easy going. :-) - Writtenonsand 13:25, 17 January 2006 (UTC)

Yes that is correct. linas 18:19, 17 January 2006 (UTC)

Rename article[edit]

Maybe we should rename this article to Stronge Nuclear Force instead. This helps to distinguish this force from the weak nuclear force and is more well-known nowadays. (unsigned, anonymous, 8 Feb 2006)

Yes, good idea. Maybe an admin can do this for us. linas 05:21, 9 February 2006 (UTC) Retract per below. linas 00:49, 10 February 2006 (UTC)
I disagree. The nuclear force is not confused with the weak interaction and is not the same as the strong residual force (though it is caused by residue of the strong interaction); it is specifically the force felt between nucleons. Compare modern usage: an arXiv search for "nuclear force" in modern literature (136 uses), "strong nuclear force" (3 uses), "weak nuclear force" (6 uses). Clearly, the preferred term is "nuclear force". -- Xerxes 15:08, 9 February 2006 (UTC)
Oh, ok. (If I was good at naming things, I would have majored in biology). linas 00:49, 10 February 2006 (UTC)

I would change it as I searched for both the strong and weak nuclear forces and your's came up only about 1. (anon) —Preceding unsigned comment added by (talk) 17:58, 13 May 2010 (UTC)

Nuclear stability[edit]

I'm not an expert on this stuff, but a back of the envelope calculation shows that the repulsive Coulomb force between two protons at a distance of 1.3fm is about 140N. This being the case, is the figure of 104N for the nuclear force a typo?

Pions or Gluons[edit]

Which particles are exchanged as part of the nuclear force? I always thought it was gluons, but the article says pions. VxP 15:38, 5 December 2006 (UTC)

Read the first paragraph instead. If by "nuclear force" you mean the force between nucleons, it's best understood as an exchange of virtual mesons (pi's and rho's). Of course these things are made of quark-antiquark pairs plus gluons, so in that sense the quarks serve as mashed potatoes to carry a residual bit of gluon gravy (the QCD or color force or honest-to-god strong force) between nucleons. You can think of forces between nucleons as a little bit like the van der Waals forces between neutral atoms like helium. The atoms are neutral overall, but self-polarization causes a little bit of electromagnetic interaction between them anyway. In the same way, nucleons are "colorless" or QCD white, but when they are close together, the gluons find a way to make a little interaction anyway, via the tricky path of getting the vacuum to polarize into virtual quark-antiquark pairs (those virtual mesons, the pi's and rho's), and when those go out, they can take virtual gluons with them. These gluons, in a sense, underlie the inter-nuclear force in the same way that electron-proton forces (i.e., electrostatic forces mediated by virtual photons) underlie van der Waals interactions between neutral atoms.

Or so I've been told. Caveat is I have to give you my metaphorical picture because the math is quite beyond me. However, I have it on authority of people who can do the math that this is sometimes the mental picture they use. SBHarris 19:38, 5 December 2006 (UTC)

Thanks, that's the most accessible explanation I've read anywhere! VxP 20:18, 5 December 2006 (UTC)

charge independence and dineutrons[edit]

this article says that the residual nuclear force is nearly independent of charge. the article on dineutrons says that there are not bound pairs of neutrons. It seems to me that something should be added to this nuclear force article about the residual nuclear attraction between a pair of neutrons, and between a pair of protons, as opposed to the attraction between a proton and a neutron. I have not found information about whether a pair of protons (Helium-2, no neutrons) is a bound state with short lifetime, or not a bound state. I assume that the short lifetime of the neutral pi meson relative to the charged pi meson gives reason to why a pair of neutrons are not a bound state (and maybe a pair of protons is not a bound state). at any rate, it isn't simply that NN, NP, and PPs are strongly attracted to one another, nearly equally, is it? isn't the neutron-proton attraction much stronger than neutron-neutron and proton-proton?

(I have found the Wikipedia physics articles very helpful and interesting. I am not a physicist.) PhysicsStudent 17:10, 6 January 2007 (UTC)

Quoting excerpt from book, Amit Goswami [of University of Oregon], The Concepts of Physics, published by D. C. Heath, 1979, no edition number specified, ISBN 0-669-01897-X -- "But we may wonder why nuclei are not just bundles of neutrons alone. Why have the proton at all with its problem of electrical repulsion? The answer is that the attractive nuclear force between neutrons and protons is the strongest (stronger than the nuclear attraction between a pair of protons or a pair of neutrons). Because of this there is a tendency for nuclei to have an equal number of protons and neutrons." Mike Lepore, Stanfordville, New York 16:57, 6 March 2007 (UTC)
It might be helpful (here and on dineutron) to state the relative strengths of the N-P, P-P, N-N nuclear force. Rod57 03:26, 9 November 2007 (UTC)
Just to set the record straight. there is no difference in the strengths of the N-N, N-P, P-P nuclear interactions. (One can only assume that the Amit Goswami book is either mistaken or misquoted). To Understand the reason why there are no bi-neutron or bi-proton states one has to go a little deeper into the theory. Without going into any details, I will only say here that Pauli's exclusion principle plays a role here (protons and neutrons are fermions and must obey Pauli's principle). In a di-proton or di-neutron one would have the biding of two identical particles. By Pauli's principle these particles could not be in a symmetric state. one of them would have to be forced into a higher energy level. this extra energy is just enough to render these states unbound. Dauto (talk) 17:45, 21 January 2008 (UTC)

Mesons or Quarks and Gluons?[edit]

When I researched the nuclear force on the Internet, I found a source that said that the residual strong force actually occurred via exchange of quarks and gluons, not mesons.-- 02:00, 7 July 2007 (UTC)

I think that it is fair to say that at nuclear distances, it is a grey region as to whether there are quarks and gluons or mesons and baryons. What is written seems like a fair representation of current knowledge, although it is a bit technical. The description doesn't really say what is mediating the force (which is subjective) and merely gives the properties of the nuclear potential. jay 02:39, 7 July 2007 (UTC)

1/r^7 Potential[edit]

I posted that this fact needed a reference several months back and no one has tracked it down. I do not believe that it is correct because in order to have a power-law potential you need to have a long ranged field; however, QCD has a mass gap. With no long ranged fields, the potential has to fall off exponentially at long distances. If this is only valid at short distances, which distances and what makes up the units (m_N or m_pi)? This fact has been quoted in other wikipedia articles in contexts where it isn't right. jay 15:02, 7 July 2007 (UTC)

Non-conservation of orbital angular momentum[edit]

Article says:

The NN force has a noncentral or tensor component. This part of the force does not conserve orbital angular momentum, which is a constant of motion under central forces.

Can somebody explain why is that so? And can nuclear force realy violate law of conservation of angular momentum?

And linking to tensor article does not help the reader to understand the subject any better. It should be either explainded, or linked in better way. -- (talk) 18:06, 27 December 2007 (UTC)

It's talking about orbital angular momentum of the object in orbit only, not the total of the system, which of course is conserved. Think of an object stuck in a whirlpool. Not only would it feel a central force pulling it toward the center of the whirlpool, but also a radial force which pushes it faster the closer in it goes. That's a good example of a force which doesn't conserve angular momentum for the orbiting object-- it goes much faster than it would from just angular momentum conservation laws, because it gains angular momentum from the fluid.

If you want a gravity example, consider that the moon doesn't perfectly conserve angular momentum of orbit, because it spirals outward with time. In compensation, the rotation of the Earth decreases in rate. Total angular momentum is conserved, but orbital angular momentum isn't. The total force which acts on the moon is not a perfectly central force because such a force would have no place for a component that (like a rocket pushing in the direction of the moon's motion) increases the total energy of the moon in its orbit via this mechanism of angular momentum transfer, and would do so even if the orbit started out circular. Spiral orbits (inward or outward) can't be produced by simple central forces (think of the inward spiral of a satellite produced by air friction).

For nucleons, the spins of the nucleons contribute hugely to the forces produced (they need to be parallel to give the best force and for two neutrons that forces one into a higher energy state as noted above), so nuclear forces are a bit like (pairs of) whirlpools, with a vengeance. SBHarris 19:03, 21 January 2008 (UTC)


No. The nuclear force is indeed a residuum of the strong interaction, but they are still very seperate forces. Thγmφ (talk) 21:59, 24 May 2009 (UTC)

nuclear force at very short distances[edit]

It was my understanding that the nuclear force at short distances became proportional to r (in other words, it becomes the strong force) and that the nucleons within the nucleus (therefore?) moved independently of one another. Em3ryguy (talk) 17:19, 25 August 2008 (UTC) Em3ryguy (talk) 17:50, 25 August 2008 (UTC)

What does the nuclear force conserve?[edit]

Associated with every interaction is a conserved charge or Noether current." Gravity conserves mass-energy, electromagnetism electric charge, weak conserves weak isospin, and strong conserves color charge. What, if anything, does the nuclear force conserve? Isospin? If you know the answer, feel free to edit the article accordingly. (talk) 06:23, 9 September 2009 (UTC)

Forgive me as I venture an answer to my own question. My reading of pp. 191-92 of Schumm (2004) is that the current conserved by the nuclear force is indeed isospin. (talk) 22:04, 13 September 2009 (UTC)


The Introduction to this article is too technical. It should be comprehensible to a newcomer to science or a young person with a genuine interest in the subject. See Make technical articles accessible.

For example;

  • the third sentence in the Introduction says To a large extent, this force can be understood in terms of the exchange of virtual light mesons, such as the pions. This implies that to begin to understand nuclear force a reader must first understand the terms meson and pion!
  • The fourth sentence in the Introduction relies on the expression quantum chromodynamics!

See Introductory text. Young people and newcomers to science are likely to seek to understand why the isotopic mass of oxygen-16 is not exactly 16.000, despite the fact that the isotopic mass of carbon-12 is exactly 12.000. An elementary understanding of nuclear force is essential to resolving this paradox. In my view, the present Introduction to this article does not offer even an elementary understanding because it is too technical. Dolphin (t) 03:33, 21 May 2010 (UTC)

A month has passed since I started this thread about the Introduction to Nuclear force. In that time no User has responded in any way. I will go ahead and take sentences that are inappropriate in the Introduction, and paste them into other sections in the article. This will leave a very short Introduction but hopefully it will inspire an expert on the subject to develope a new Introduction. Any new Introduction should conform to the principle of Make technical articles accessible. Dolphin (t) 12:27, 21 June 2010 (UTC)

Lowest Order Feynman Diagram[edit]

Is the Feynman Diagram presented here the lowest order? I believe it should be possible for the pion exchange to take place only using two gluons. If so, should we not present the lowest order Feynman Diagram in the image? —Preceding unsigned comment added by (talk) 16:32, 3 June 2010 (UTC)

Nuclear force simplified[edit]

Hi Sbharris. Thank you taking an interest in Nuclear force and working on a new Introduction. Any article on a scientific phenomenon needs to address both the what and the why. What is it, and why does it occur. With an article on a complex phenomenon like nuclear force I think it is sufficient for the Introduction to deal only with an explanation of what it is. Why it occurs is usually best left for later in the article.

Your new introduction mixes both what and why. The why is reliant on an understanding of quarks, mesons, gluons, nucleons and van der Waal’s forces. Your new introduction implies that someone wishing to obtain an elementary understanding of nuclear force must first obtain an understanding of quarks, mesons etc. I disagree with that implication because I believe nuclear force is more fundamental than quarks and mesons. Teenagers in basic chemistry class are introduced to nuclear force as a means of understanding the lost mass that is evident when noting that the isotopic mass of carbon-12 is exactly 12.0000 whereas the isotopic mass of all other elements and isotopes is not exactly the same as the number of constituent protons and neutrons. This is years, if not decades, before these teenagers need to comprehend quarks, mesons etc.

I suggest the Introduction to Nuclear force should address nothing more that what it is. Why it occurs should be left until later in the article. Could you do some more work on your Introduction to simplifly it in this way? Many thanks. Dolphin (t) 03:14, 22 June 2010 (UTC)

Well, I can always cut more "explanation" out of the LEDE and make it as short as you like. But ledes are supposed to provide a summary of the article, which includes some "why." The "what" is in the first paragraph-- it's a strong force between nucleons. The "why" must be expained in terms of what the force consists of, which is a fields of virtual pions, which themselves are virtual pairs of quarks and their associated virtual gluons. It is a residual of the gluon forces that hold the quarks together in nucleons (and of course all hadrons in general).

As for the binding mass-loss, it's not helpful as a "why," because it has nothing to do with "why." I'm well aware that in chemistry and physics courses it's used as though it was somehow explanatory, but it isn't. See the lede for mass-energy equivalence. Binding mass loss is simply due to binding energy loss (when this is removed from the system), and it happens in chemical reactions (when a proton binds to an electron to make hydrogen) just as well as nuclear reactions (when a proton binds to a neutron to make a deuteron). It's a side effect of energy loss from the system, since energy-loss always means mass-loss. The only reason the mass/energy loss is higher in nuclear reactions is because the forces involved are much higher than EM forces, so more energy is liberated in the interaction, and thus more lost (gamma ray vs. lower energy photon). But that's just another way of saying that the strong force is strong. It doesn't add anything. SBHarris 18:21, 22 June 2010 (UTC)


In File:Pn scatter quarks.png, shouldn't the central pion be color neutral? By working out the gluon colors, it seems that the central pion is made up of two blue quarks, whereas, for the pion to be color neutral, it has to be a blue-antiblue pair.

Note:I have never studied this subject, but just read it as a curiosity, so I might be missing some principles here.

Here is my interpretation of the diagram (top-down). From the proton, a blue up quark releases a blue-antired gluon (i'll represent as br*), and turns red. The br* hits the red down quark of the proton, and becomes blue. (Color A of the pion) Now, this blue quark emits a blue-antigreen gluon (bg*), and becomes green. The bg* is absorbed by a green up quark in the neutron, which becomes blue.

Now, from the neutron (happening simultaneously with the above): The blue down emits a br* gluon,and becomes red. This gluon hits the red down, which becomes blue. (Color B of the pion) The now blue quark emits a bg* gluon and becomes green. The bg* is absorbed by the green up in the neutron, which turns blue.

Everywhere in this exchange, charge is conserved. All particles are color-neutral, except for the pion formed in the exchange. Mesons are always made up of a pair of quarks with opposite colors (red-antired, etc.). But, this pion is made up of two blues. Could someone tell me why?

Also, could someone tell me if any quark changes into another during this reaction? The pion also has to be made up of a quark-antiquark pair, but (without changing a quark into another), it seems to be a down-down quark.

Thanks, ManishEarthTalkStalk 10:40, 8 July 2010 (UTC)

The pion is neutral so it has to be made of the two quarks of opposite charge-- either a [down + anti-down] or [up + anti-up]. Actually it's a weird quantum mechanical mix of both of these states, or else we'd have two kinds of neutral pions. You can look at it either way, but there's only one kind of neutral pion. You can see that up/anti-down and down/anti-up pions will be charged +1 and -1 respectively. In THIS diagram the pi-0 is drawn as a down/anti-down (they could as easily have drawn it as an up/antiup), and furthermore drawn as a blue/anti-blue (it could have been either of the other two color/anticolor pairs). That makes it both charge and color neutral. You may ask why the pion quarks made of matter and antimatter don't annihilate, and the answer is that they do! Neutral pions can disintigrate into nothing but photons (2 or 3 or whatever you like), just like an electron/positron pair, or a charmonium "J-psi". In this diagram, which (please note) is a diagram of a repulsion not attraction, a proton scatters off a neutron (or vice versa). The mediator is the virtual neutral pion. This is a little like a neutrino scattering off an electron, mediated by the neutral Z-zero. The same particles go in as go out.

So which particle is matter and which antimatter in the pion? In these diagrams the particles are "matter" if they are moving in time in the direction of their arrows. So in this pion, if the pion is going upward, the line represented by the blue upward arrow is the "particle" (the blue down quark) and the one forced to go the opposite way to the pion flight and thus opposite to its arrow, is the "antiparticle" (the antiblue antidown quark). BUT since you can picture the pion going in either direction (from proton to neutron or from neutron to proton) your choice of which of these is the quark and which the antiquark (the left one or the right one) is dependent on which direction you pick for the pion to move as a whole. It works out the same either way, which I suppose is why it's not shown one way or the other in the diagram. In the diagram you can see that an up quark gets changed to a down, and a down to an up, in each nucleon. The net result is no change in each particle. For charged pion interactions, a down is changed to an up in one particle and vice versa in the other, and neutron changes to proton and vice versa, in the interaction. Each of the 3 pions (+, -, or uncharged) come in blue-antiblue, red-antired, or green-antigreen varieties. We see only one of them here. SBHarris 02:01, 9 July 2010 (UTC)

I think I understand... So then (Out of the different possibilities) can this be viewed like this?
  • The blue and red quarks in the proton undergo the same reaction which I wrote above and create the blue quark in the meson.
  • The green quark in the proton releases an antiblue-green gluon and becomes blue.
  • This gluon becomes a pair of (virtual?) quarks, one green (goes to the proton), and one antiblue (becomes part of the pion)
No, when the green up quark releases an antiblue-green gluon, it becomes a blue up-quark (same proton, now on the right). That antiblue-green converts the blue down-quark from the pion into a green down-quark, which then becomes part of the outgoing proton. Remember the gluons are the little chain-like things (always two colors-- one color the other an anticolor), the colored straight lines are quarks and they stay quarks (even the same kind of quark) in strong interactions.
Asterisk.pngThen where can the anti down come from? As you said, quarks cannot change into other quarks. To form an antidown, thus, Some sort of decay/reaction must occur. The only one I can think of is the conversion of a gluon into a down-anti down pair. So, if the pion is going from the proton to the neutron, the red down will become blue and go off with the pion, while the green up will emit a green-anti blue gluon which will form a green down and a blue antidown, which goes off with the pion. If the pion goes from neutron to proton, the red down becomes blue and enters the pion, while the green up emits a gluon which forms a down and an anti down. The anti down goes off with the pion. Is this correct? ManishEarthTalkStalk 12:21, 13 July 2010 (UTC)
Yes. Which ever way the pion goes, the blue down going THAT way has, in some ways, a simpler history-- it's a red down that interacts with two gluons to change to first blue (in the pion) and then to a green down at the end. It stays a down the whole time and changes color twice. BUT for the OTHER blue-down that you pick as the one that is doing the opposite way the pion is going (and is thus an anti-blue antidown) the history is more complicated. Each one of these has to be made as a new anti-down from a new [green down PLUS antiblue-antidown] pair. This comes from a green antiblue gluon that originates from a green particle that changes to blue, and thus emits a green-antiblue gluon that undergoes pair production to become an antiblue antidown (in the pion) and a new green down that stays with the original particle. The antiblue antidown in the pion anihilates a red down in the other particle, and the resulting red-antiblue gluon is available to turn some blue quark in the other particle (either an up or a down) into a red version of itself. SBHarris 06:40, 18 July 2010 (UTC)
So it happens like I said? Production of a short-lived pair from a gluon of which the antiparticle annihilates? It's much clearer now... Thanks a million! ManishEarthTalkStalk 09:08, 18 July 2010 (UTC)
P.S. I'll try to make the animation soon. ManishEarthTalkStalk 09:14, 18 July 2010 (UTC)
  • The virtual antiblue annihilates with the red (in the neutron) and releases an antiblue-red gluon. This annihilation makes the virtual green quark in the proton real. (One question: Can quarks of different colors annihilate/form virtual pairs if the appropriate gluons are present?)
  • The gluon turns the blue quark in the neutron red.
No, the green quark line crosses the antiblue-red gluon line, but there's no interaction there-- it just happens to go over the top of it in 2-D. But there's no connection. Ignore the crossing. This is best seen as a blue up-quark emitting a blue-antired gluon to become a red up-quark. The blue-antired gluon combines with the red down-quark to turn it into a blue down, which goes off with the pion. It's a blue down or blue anti-down depending on which direction the pion is coming in. The same is true of the OTHER down quark in the pion-- one of them is down, the other antidown, depending on which direction the pion goes. In gluon interactions the quarks can only change color-- they can't change type (up cannot change to down, etc). If you see an interaction where an up and a down go in, an up and a down have to go OUT (although they will have switched colors). So if it looks like up turns to down, that's actually not true. All that happens is colors have switched. It takes weak force interactions to change an up to a down quark and vice-versa (like beta decay).SBHarris 00:46, 10 July 2010 (UTC)

Thanks, ManishEarthTalkStalk 04:41, 9 July 2010 (UTC)

Could you also answer this question? When a body experiences increase in mass because of its speed, is the increase in mass due to particles being created by the energy, or is it just increase in apparent mass? In other words, if you accelerated a particle to near lightspeed, would smaller particles appear to accompany it, or would it just seem to be more massive. I tend to believe the second possibility, because the new particles (from the first one) would have to exist when seen from certain reference frames only. Thanks, ManishEarthTalkStalk 05:02, 9 July 2010 (UTC)

The question of mass increase is complicated and you should read mass in special relativity to get a feel for it. Basically, there are two types of mass, one of which goes up with speed and the other of which does not. Generally in the Dirac view, the increase in relativistic mass is actually associated with the appearance of virtual particle-antiparticle pairs, which can become real particles if enough mass-energy (kinetic energy) is supplied, and there is something to transfer the momentum to when the new real pair is created. Otherwise, the virtual pair sort of go along for the ride, supplying the extra mass (or momentum, if you like) but not extra charge (since that does not change with velocity-- so the virtual particle antiparticle pair insure that mass can go up without charge increasing). SBHarris 00:46, 10 July 2010 (UTC)
Hmmm... If the kinetic energy is supplied, will both the occupant of a spaceship and an observer see the new particles? Because the virtual particles only exist for the observer, while the occupant detects nothing wrong. As long as they are virtual, they are unobservable anyways, but then, if they are made real, will everyone see them? I find it hard to digest that a particle can exist selectively for different observers... ManishEarthTalkStalk 12:21, 13 July 2010 (UTC)
No, for a single particle, only the observer that sees kinetic energy for that particle (which is frame dependent) sees the new virtual particles that travel with it (and only that observer see the extra momentum, energy, and so on). Since they never become real, there's no problem with one observer seeing them and another not. To make them real, you need a second particle to provide a reference, and then all observers see the new particles as capable of being made by the invariant mass that represents the kinetic energy (the invariant mass is the same for all observers). All observers see the same kinetic energy available to make new particles, but each observer locates it in a different place. For example, in a two-electron collision with a center-of-mass energy to make an electron/positron pair (1.022 MeV), the observer on one electron sees the other electron with 1.022 MeV kinetic energy, but an observer on the other electron sees it all on the first electron. An observer in the COM frame sees it divided between the electrons. SBHarris 03:23, 30 July 2010 (UTC)
Aah, I see. Thanks a million... Could you also answer the wuark question above (It's hard to see, so I added a star)? I'm thinking of creating an animation of one of the possible ways it can happen. Thx, ManishEarthTalkStalk 03:38, 18 July 2010 (UTC)

Animation preview[edit]

Well, I've finished the animation for File:Pn_scatter_quarks.png, and I've uploaded the .swf animation and the animated GIF file to Google Docs SWF , GIF(Click "Open" to see it animated). I have a few questions for finishing touches:

  • How should I represent anticolors? I used Cyan, Magenta, and Yellow, by File:Quark_Anticolours.png. Or I could keep File:Quark_Anticolours.png and File:Quark Colours.png on the side...
  • What should the speed of the animation be? I can make it slower, but it might get chunky. Alternatively, I can insert extra frames, but that takes time, so I want to know how much slower before I do it.
  • So that the animation can loop, I've let the proton/neutron exchange gluons normally (What happens before and after File:Pn_scatter_quarks.png, i.e., when this happens). When this internal gluon exchange is happening, the nucleons are bounded by a circle and are labelled. Should I have this stage?
  • The quarks look a bit plain, being just brightly colored circles. If you want, I can replace them with better looking circles, but I would prefer it if you gave me a template image which I can color and put in place of the quarks, as I'm not too good at creating cool-looking images.
  • To show particle-antiparticle annihilation/pair production, I have used a little explosion symbol. If you can think of anything better, tell me.

The above changes are what I've thought of and which I think would be easy to implement (Each one is a 5 minute tops change). If you think of any more, just mention them here. ManishEarthTalkStalk 05:05, 23 July 2010 (UTC)

Basic properties section[edit]

This section covers the gamut of the concepts about the nature of the the strong interactive force of attraction between the atomic nucleons in very short order. In doing so, it brings up without commentary the essential paradox involved in such a concept, which is how an attractive force can overpower a repulsive force at short distances, and then at shorter distances in mid-air change first to zero and then to a repulsive force? The charge independence concept of the force precludes the use of the coulombic repulsion concept to explain the short distance repulsive force (in the case of neutrons). So shouldn't there be an explanation of this problem in this article?WFPM (talk) 15:15, 18 August 2010 (UTC)

(reposting the above which was mistakenly posted on article page) If we can find a reliable source that discusses or lays out the problem and the solution, then sure. However, the paradox you mention is not unique to the nuclear (nucleon-nucleon) force, but also happens with van der Waals forces. According to the page on the Lennard-Jones potential, in that case the shorter distance repulsion is attributed to Pauli repulsion. My guess would be, since the article notes the analogy with van der Waals forces, that something similar happens for the nuclear force (quarks being fermions). That's my guess, but I'm sure there is a reliable source out there that we can reference if we want to give more description of the cause of the repulsive-attractive duality of the force. Still, without sources I don't think it's up to us to point out problems or paradoxes in physics that we on our own discover (per WP:OR) - we can only report problems or paradoxes that appear in reliable sources. --FyzixFighter (talk) 19:27, 18 August 2010 (UTC)

Well, I thank you for the transfer and subsequent discussion. I would note that in James Clerk Maxwell's discussion about the "Atom" in the 9th edition of the EB he discusses this paradox at length. So the topic has had a long history of contemplation and discussion. In the "atom" article he also points out that nobody has ever really seen a physical force, but only deduces its existence from ensuing activity. And he makes the case that the direction of a force is determined by the directional change of the energy content of the system.WFPM (talk) 16:13, 19 August 2010 (UTC)

Yes, and there are literally thousands of high energy scattering experiments in the literature, in which protons are projected against nuclei at various energies. Their behavior is elastic (Rutherfordian) until the energy exceeds a certain threshhold, and then the curve changes, showing that they're now subject to a different type of force that is non-coulombic, inside the radius that corresponds with that energy. Rutherford actually did these experiments first, with alphas scattered at hydrogen (rather than his previous gold), You can back-calculate the critical radius using more or less simply E = 2ke^2/r electromagnetic potentials. Inside that radius, the force changes, and becomes more repulsive than simple electrostatics. That's the nuclear force. SBHarris 23:15, 19 August 2010 (UTC)

Do you think that the discussion about the closeness of packing of the nucleons within the nucleus is consistent with the idea that a cloverleaf orbital indicates that the electron passes through the nucleus evidently 4 times in the course of the path through each orbital. Since there's no angular momentum factor involved in these paths, I don't understand what is controlling them.WFPM (talk) 14:15, 20 August 2010 (UTC)

Animation Added[edit]

I've added an animation to the page. I'm not sure If I've sized it or placed it correctly (Three images on top of each other look ugly.) Feel free to move it around. ManishEarthTalkStalk 01:47, 1 February 2011 (UTC)

If anyone wants any similar animations, just ask me. It shouldn't take me too long now. ManishEarthTalkStalk 01:48, 1 February 2011 (UTC)
Now THAT is pretty cool. Probably even nuclear physicists have never seen such a thing, even though their diagrams show it. How much work would it be to add the reverse process (just as valid) that has the pion go back from neutron to proton, after going from proton to neutron? I think of pions as ping-pong balls going back and forth, after all. SBHarris 06:01, 1 February 2011 (UTC)
Discussion continued at User_Talk:Manishearth ManishEarthTalkStalk 13:19, 1 February 2011 (UTC)

Repulsive? Where exactly does it become repulsive?[edit]

The article states: "the nuclear force is only felt among hadrons. At small separations between nucleons (less than ~ 0.5 fm between their centers) the force is very powerfully repulsive, which keeps the nucleons at a certain average separation, even if they are of different types." This is confusing. Shouldn't one state explicitly that at higher distances the force becomes attractive?

Fixed. I'm looking for a nice illustrative graph. There's one referenced, but I don't trust the one on commons, which has significantly different distances for attraction than others I've seen in the literature. Still thinking about this. SBHarris 18:19, 9 September 2011 (UTC)
Even now (Oct. 2014) the text of the article is not in agreement with the figure. The text says the force becomes repulsive at "distances less than 0.7 fm," but the figure (Nuclear_force.png) shows the force transitioning from positive to negative at about 0.86 fm. (Am I misinterpreting the figure?) What mathematical expression is used to generate that data? I could make my own darn plot, if that expression were provided. (talk) 19:40, 21 October 2014 (UTC)

Third intro sentence, binding energy[edit]

"The energy released causes the masses of nuclei to be less than the total mass of the protons and neutrons which form them." I could be wrong but I think that's only true for light elements. Nuclei heavier than iron have more mass than constituent nucleons. LieAfterLie (talk) 23:41, 18 May 2012 (UTC)

Actually, no they do not. Nuclei heavier than iron have more mass PER NUCLEON than iron does, but they are all still less massive than the protons and neutrons that make them up. You can verify that yourself by taking the isotopic mass of some heavy nuclide like U-238, which is 238.05078826 u, and subtracting the mass of 92 electrons from it (0.0050469277 u) to get 238.0003189 u. That's the mass of a U-238 nucleus. So what is the mass of the 92 free protons and 146 free neutrons that make it up? It's 92*(1.007276466812 u) + 146 (1.00866491600 u) = 239.934513 u. This is a lot more massive. The diffence is the binding energy and it's equal to 1.9341938 u. The particles are about 0.8 % heavier than the nucleus. This is typical: atomic binding energies never exceed 1%, but they may come close. SBHarris 00:00, 19 May 2012 (UTC)

No "Van der Waals" forces are not "between neutral atoms"[edit]

and even more notably, the range of van der Waals forces is NOT shorter than "the electrical forces that hold the atoms themselves together". Since atoms are not generally described as "held together by electrical forces."

Possibly the LDF is meant here for "van der Waals forces", and "molecules" for "atoms", but the whole thing is very confused. (talk) 16:12, 23 June 2012 (UTC)

LDF = London dispersion forces or London forces would be more specific, as van der Waals forces (which now include London forces) have become something of a catchbasin of various small effects, including LDFs. But Van der Waals discovered them first, and his first application was to describe LDFs between neutral gas molecules, including monatomic gas molecules of inert gases. These are forces between atoms that BEGIN as neutral, and the same is true of nuclear forces-- they induce their own diples and multipoles. Nuclear forces are far smaller than the strong forces between quarks inside nucleons, and have a different type of range description. LDFs and Van der Waals forces are shorter range than simple EM static forces from charges, inasmuch as they are dipole or induced higher pole expansions between particles that have no net charge (just as nucleons have no net color charge, and are "white" or neutral). And (finally) indeed atoms are described as "held together by electrical forces." What do YOU think holds the electrons and nucleus of an atom together-- elves and faeries? SBHarris 16:26, 23 June 2012 (UTC)

Introduction is sad[edit]

Everything else sounds like real physics but even I can disprove the introduction. If the strong force acted like LDF all nucleons would feel strong attraction, not just protons and neutrons with the right isospin, as is confirmed in experiments. The introduction is sad. — Preceding unsigned comment added by (talk) 03:12, 28 September 2012 (UTC)

Energy released[edit]

"[The nuclear force] is responsible for binding [sic] of protons and neutrons into atomic nuclei. The energy released causes the masses of nuclei to be less than the total mass of the protons and neutrons which form them; this is the energy used in nuclear power and nuclear weapons."

What "energy released"? Energy released by what? I think someone edited one of these two sentences and accidentally made them incoherent — Preceding unsigned comment added by (talk) 05:04, 19 March 2013 (UTC)

The energy released by the binding. It's an attractive process and it turns potential energy into some other sort , like EM energy, or kinetic energy, and that is then available for release from the system. I'll go ahead and add it so it 's clear but it's pretty well implied. SBHarris 06:11, 19 March 2013 (UTC)

Identification of nuclear force[edit]

The article does not specify how the info on nuclear force is obtained, through what measurements.-- (talk) 14:21, 10 December 2013 (UTC)

true. i think it is theory, there is also this: " with a force like that of electric charge, but of far greater power." what charge i wonder? sth like a pulsar? (talk) 07:13, 12 November 2014 (UTC)

Another plot, please[edit]

I'd like to see the nuclear force between two protons plotted as a function of separation distance, and also on the same plot, electrostatic force between two protons. This would allow readers to visualize the distance at which the nuclear force attraction exactly cancels out the electrostatic force repulsion. Pretty basic stuff that the article should certainly include. (talk) 19:53, 21 October 2014 (UTC)

Since protons are not bound into diprotons, such a graph could hardly exist, as there is no point at which proton attraction equals proton repulsion. The net force is always repulsion. SBHarris 01:53, 22 October 2014 (UTC)

" Its quantitative description relies on internucleon potentials with phenomenological constants determined from fitting experimental data." Isn't that really too technical?[edit]

I mean, pardon my french, but at first glance I have no fucking idea what that means at all. Wikipedia has a policy of comprehensibility: and having this as the second sentence of the article just basically says to me "Abandon all hope of comprehension all ye who read beyond here." I mean the key mouthfuls aren't even wikilinks.

I would appreciate if someone who understands what that sentence means could simplify it for us plebeians.

Citizen Premier (talk) 01:18, 7 February 2015 (UTC)

I had a go at the article and then some, hopefully that is all ok. I absolutely share the sentiment that the initial portion of any Wikipedia article should be as simple and clear to understand as possible. Many of the technical articles are absolutely ludicrous. Its a public encyclopedia; give those 8th graders a chance! (We need citations now people!) Bdushaw (talk) 09:00, 3 June 2015 (UTC)