Talk:Nuclear force

fm

In nuclear physics, the fm unit is the Fermi. Conveniently fm sounds like Fermi, and also comes out femtometer. (Or femtometre, depending on where you are.) I suspect also in high energy physics. Gah4 (talk) 07:01, 22 December 2023 (UTC)

Quark Motion and quark distances

Quarks are likely to be separated by far smaller distances than a neutron diameter. Because they move 0.8 C and higher the Electromagnetic field intensity will have an intensity pulse in the direction of motion each time a quark cycles past. Let the quarks cycle in a roughly fixed pattern of two like quarks separated by the opposite sign quark with one like quark leading and one trailing. The magnetic field intensity wakes they will produce will cause magnetic curving forces that are balanced with one of the other quarks causing a curving force and the remaining causing a straightening force, true for each of the three quarks. Speed of light causes an additional barrier stabilizing the structure. Does this type of thinking open the door to electromagnetic forces being involved in quark attraction?

Also note that while a trio of quarks in a second proton sits inside the intensity wake of a first proton. The repulsion from highest intensity is towards the other proton. This would explain why at particular short distance the electromagnetic force is strong and attractive. Does this kind of thinking suggest a way that relativistic electromagnetic field forces could hold two protons together?

On a distantly related note when two wires attract an alternate theory is that the electrons seek the less dense fields behind other electrons. (To me this is far more likely than the favored theory of more dense packed protons due to relativistic shortening.) Bill field pulse (talk) 20:27, 20 January 2024 (UTC)

Force Curve diagram - possibly misleading due to electrostatic curve proximity

Dear community, Æ is my handle for a non-profit, non-industry youTube channel, on which has been produced 970 videos in 3 years to date. Æ is presenting geometric interpretations of the proton-proton "strong" binding nuclear force relation. Here is my problematic situation: Æ refers heavily to the force-curve graph, which is simple. However, the binding range of two protons is inside the electrostatic range, but the Wikipedia curve shows the diminution of the force on increasing distance to be asymptotic with the horizontal space line. That asymptosis, as depicted, must intersect with the more distant electrostatic curve. Æ seeks advice on this, because the graph as depicted may indeed be a proper representation, based on some convention with which Æ may not be familiar. In short, Æ is not citing a "mistake" in the graph (necessarily). What Æ is apparently now able to prove, is a relation between the outer "EM" relation and the inner "nuclear" relation. If this relation can be proved ( Æ can prove it ), then the asymptosis is misleading. It has misled me twice, although that is due to my seeming dyslexia more than anything else. However, in my most popular video release, about a year ago, which received over 1,750 views in 3 days, Æ caused a small furor when several of my viewers correctly pointed out that Æ cited the force relation invertly to that stated (correctly) in the Wikipedia diagram caption. My question is this: should Æ be able to publish a proof showing relation between the two force inversions, should Æ recommend modifying the [Nuclear Force] graph? Thank you. s.a.miller "anagalactic" "Drake Sterling" "Æ"

on youTube channel [ÆIOU] • 

27-May_2025 6pm PDT • • Anagalactic (talk) 00:27, 28 May 2025 (UTC)

The direct answer is No. If you have some proof, publish it in a reputable journal.

If you believe any content in this article on Nuclear force is incorrect, all we need is a reliable source. You can challenge any content that is not sourced, including graphs. I gather your talking about "Nuclear_Force.png"? Ideally you would find an alternative graph in a textbook or journal article. As drawn, the asymptote of the attractive residual strong force can never intersect the repulsive Coulomb force. As far as I know, the residual strong force falls rapidly to zero so it has no asymptote really. Johnjbarton (talk) 01:03, 28 May 2025 (UTC)

Thank you. That makes perfect sense. Anagalactic (talk) 01:27, 28 May 2025 (UTC)

Only For Reference Computing. (Generated by ChatGPT.)

Approximate expression of Newtonian mechanics

Proton

The strong nuclear force between protons.

("Phenomenological Approximation")

More sophisticated nuclear forces (such as OPEP/multi-meson exchange, tensor terms, spin-isospin dependence, Argonne v18, etc.) all have clear analytical expressions. However, these require the introduction of the spin operator ${\vec {\sigma }}$, the isospin operator ${\vec {\tau }}$, and the tensor operator $S_{12}$, making it difficult to simply draw a scalar curve.

Distance potential energy function

$V_{pp}(r)={\frac {V_{\rm {rep}}e^{\frac {-r}{a_{r}}}}{r}}-{\frac {V_{\rm {att}}e^{\frac {-r}{a_{a}}}}{r}}+{\frac {\alpha \hbar c}{r}}$

The first term is the short-range repulsive nuclear force, the second term is the medium-range attractive nuclear force, and the third term is the Coulomb repulsion between protons.

$\hbar c\simeq 197.326{\text{MeV·fm}}$，$\alpha \simeq 1/137$

$a_{a}\sim {\frac {1}{\mu _{\pi }}}$， $\mu _{\pi }={\frac {m_{\pi }c}{\hbar }}\approx {\frac {135{\text{MeV}}}{197.326{\text{MeV·fm}}}}\approx 0.684{\text{fm}}^{-1}$$\Rightarrow a_{a}\approx 1.46{\text{fm}}.$

$a_{r}$ takes a shorter length to indicate a “hard core”, and $V_{rep},V_{att}>0$ is the intensity (unit: MeV·fm).

Commonly used parameters: $a_{r}\approx 0.3{\text{fm}},a_{a}\approx 1.4-1.5{\text{fm}},V_{rep}\approx 2000{\text{MeV·fm}},V_{att}\approx 300{\text{MeV·fm}}.$

$V_{pp}(r)$: $r\lesssim 0.5{\text{fm}}$ strongly repulsive, $r\sim 1-2{\text{fm}}$ slightly attractive, and at a distance the repulsive force is dominated by the Coulomb term.

Force formula

$F_{pp}(r)=-{\frac {dV_{pp}}{dr}}={\frac {V_{\rm {rep}}e^{-{\frac {r}{a_{r}}}}}{r^{2}}}(1+{\frac {r}{a_{r}}})-{\frac {V_{\rm {att}}e^{-{\frac {r}{a_{a}}}}}{r^{2}}}(1+{\frac {r}{a_{a}}})+{\frac {\alpha \hbar c}{r^{2}}}$

The first term is the short-range outward repulsion, the second term is the medium-range inward attraction, and the third term is the Coulomb outward repulsion.

More precise writing

Analytical writing (operator form and coordinate space expansion)

$V_{\pi }({\boldsymbol {r}})={\frac {f_{\pi NN}^{2}}{m_{\pi }^{2}}}({\boldsymbol {\tau }}_{1}\cdot {\boldsymbol {\tau }}_{2})({\boldsymbol {\sigma }}_{1}\cdot \nabla )({\boldsymbol {\sigma }}_{2}\cdot \nabla ){\frac {e^{-m_{\pi }r}}{4\pi r}}$

Expanded into center term + tensor term (coordinate space)

$V_{\pi }({\boldsymbol {r}})={\frac {f_{\pi NN}^{2}}{4\pi }}({\boldsymbol {\tau }}_{1}\cdot {\boldsymbol {\tau }}_{2})\left[({\boldsymbol {\sigma }}_{1}\cdot {\boldsymbol {\sigma }}_{2})W_{S}(r)+S_{12}W_{T}(r)\right]$

Proton-Neutron

There is no Coulomb term for proton-neutron.

$V_{pn}({\boldsymbol {r}})=V_{\rm {strong}}({\boldsymbol {r}})+{\frac {\mu _{0}}{4\pi }}{\frac {1}{r^{3}}}\left[{\boldsymbol {\mu }}_{p}\cdot {\boldsymbol {\mu }}+{n}-3({\boldsymbol {\mu }}_{p}\cdot {\hat {r}})({\boldsymbol {\mu }}_{n}\cdot {\hat {r}})\right]+V_{\rm {Breit}}(r)+V_{\rm {other}}^{\rm {EM}}(r)$

The second term is $V_{\rm {mm}}$.

$F_{pn}(r)=-{\frac {dV_{pn}}{dr}}={\frac {V_{\rm {rep}}e^{-{\frac {r}{a_{r}}}}}{r^{2}}}(1+{\frac {r}{a_{r}}})-{\frac {V_{\rm {att}}e^{-{\frac {r}{a_{a}}}}}{r^{2}}}(1+{\frac {r}{a_{a}}})$

Neutron

Neutrons have no Coulomb term.

$V_{nn}(r)=V_{\rm {strong}}(r)+V_{\rm {EM}}(r)$

$F_{NN}(r)=-{\frac {dV_{NN}}{dr}}={\frac {V_{\rm {rep}}e^{-{\frac {r}{a_{r}}}}}{r^{2}}}(1+{\frac {r}{a_{r}}})-{\frac {V_{\rm {att}}e^{-{\frac {r}{a_{a}}}}}{r^{2}}}(1+{\frac {r}{a_{a}}})$

Figure Reid 1968, the first nuclear force image, differs from this parameter, but the process is the same.

Reid 1968

Figure Reid 1968, the first nuclear force image, differs from this parameter, but the process is the same.

Parameters similar to Reid 1968: $V_{\rm {rep}}\approx 2600$, $a_{r}\approx 0.28$, $V_{\rm {att}}\approx 800$, $a_{a}\approx 1.05$.

Nuclear force Reid 1968 Approximately.

Socie-Leaner (talk) 06:17, 26 September 2025 (UTC)