Proton-neutron electromagnetic interaction.

ISINN-22, Dubna (27-30 may 2014)

Bernard Schaeffer, Paris, France [email protected] 1 Electromagnetic interactions in a nucleus

• A proton attracts a not so neutral neutron as amber (ἤλεκτρον, elektron) attracts small neutral pieces of paper. • Nucleons repulse themselves as collinear and opposite (µαγνήτης, magnetis). • Only the Coulomb repulsion between protons is taken into account in mainstream nuclear .

2

The nuclear shell model The binding of even the simplest bound nucleus, the deuteron 2H cannot be calculated with the nuclear shell model, the fundamental laws of the strong force being unknown.

The force center of the nucleus being undefined, we may assume that, in contrast with electrons, nucleons do not move, as suggested by the drawings of the atom and its nucleus.3 Estimation of 2H

Applying electric Coulomb’s law with the measured radius of the proton, rp = 0.88 fm, one obtains,

not far from 2.2 MeV, the measured value of the deuteron binding energy, indicating a possible electric origin of the nuclear interaction. 4 Proton electric charge

• It is well known that the proton contains one positive elementary electric charge, assumed here to be punctual: +e = +1.6 × 10-19 Coulomb

• The proton-proton repulsion, inexistent in the deuteron, is not the only Coulomb force in a nucleus.

5 Neutron electric charges

It is less known that the not so neutral neutron contains electric charges with no net charge, assumed for the sake of simplicity to be punctual charges +e and –e (+2e/3, -2e/3 would give a 30% error).

A proton attracts the negative charge of a nearby neutron and repulses its positive charge farther away, resulting in a net attraction, according to the 1/r Coulomb’s law.

6 Constantes

e2 =1.6 MeV 4⇡✏0rp Noyau Nuclear magnetic energy constant: •

µ0 µnµp 2 gngp | 3 | = ↵mpc | | =9.147 871 896 MeV 4⇡RP 16

3 2 eiejRP gigj RP Uem = ↵mpc 2 + | | Sij (18) 2 e rij 16 rij 3 i i=j ✓ ◆ i i=j ✓ ◆ X X6 X X6 where 4 5 S = cos (µ~ , µ~ ) 3 cos (µ~ , ~r ) cos (µ~ , ~r ) (19) ij i j i ij j ij

R g g R 3 U ↵m c2 P + | n p| P S (20) em ⇡ p r 16 r np " np ✓ np ◆ # It is more complicated for the magnetic moments: Electromagnetic contents of 2H Snp = cos (µ~ n, µ~ p) 3 cos (µ~ n, ~rnp) cos (µ~ p, ~rnp)= (21) Spin 3 electric= charges1 3 (1/r1 law):( 1) = 2 The proton repulses the +e⇥ charge⇥ of the Il sura d’ajuster le coeneutroncient and attracts magn´etique, its -e charge, (toujoursresulting in ?)+ comprise entre 2 pour le deuton a net attraction. -e et 3). Il faut aussi tenir compte des g qui varient entre gn =+e 3.826 and gp =5.585. Formule 2 magnetic moments (1/r3 law): Proton: μ > 0 p µ > µ | p| | n| Neutron: μn < 0 Cette formule, approximativeThus, for collinear (30% and opposite d’erreur), magnetic moments, peut ´etre the g´en´eralis´ee avec des coecients 7 ad hoc. Cette approximationresulting magnetic vient moment de ce is qu’on positive, thus a n´eglig´elarepulsive. r´epulsion entre le proton et la charge `electrique du neutron. En fait seule la partie magn´etique change. Cela veut dire que l’´energie de liaison par nucl´eon d’un noyau varie assez peu. Le principal param`etre est le nombre de liaisons neutron-proton par nucl´eon qui varie de 0,5 pour le deuton `a3 soit un rapport de 6 pour un noyau de masse infinie. La particule ↵ serait 6 fois plus ´energ´etique par nucl´eon. En fait les moments magn´etiques y sont inclin´es `a600 ce qui divise l’´energie magn´etique r´epulsive par 2, donnant 12 MeV (AVERIFIER). En fait elle est diminu´ee par la r´epulsion entre protons et neutrons n´eglig´ee provisoirement. Le probl`eme se r´eduit donc au calcul de Snp et au nombre de liaisons (neutron-proton en 1`ere approximation) par nucl´eon ou, de pr´ef´erence, par proton.

p-3 Constantes

2 e e e e U H = (1) e 4⇡✏ r + a r a 2a 0 ✓ np np ◆ Exact electric2 dipole formula 2H e 1 1 1 U2e = (2) The 2a/r approximation4⇡✏0 rnp of+ thea dipolernp isa invalid2a in the deuteron 2H where the✓ proton touches the neutron◆ (a ≈ r). 2 2 e 1 1 2a 4a Therefore,U H = the exact electric dipole formula [1] has to be (3) e 4⇡✏ r + a r a r2 ⇡r2 used instead of0 the✓ usualnp simplifiednp formula:◆ ou e2 1 1 e2 2a (4) 4⇡✏ r + a r a ⇡4⇡✏ r2 0 ✓ np np ◆ 0 4He avec nn et pp • rnp : separation distance between the proton and the neutron. • 2a : separation distance between the neutron positive and negative charges 2 e e e e of the deuteron.U H = (5) e 4⇡✏ r + a r a 2a [1] B. Schaeffer, Advanced0 ✓ Electromagnetics,np np Vol. 2, No. 1,◆ September 8 2013. Potentiel ´electromagn´etique:

2 2 3 4 4R 4R 1 R 3 g g 1 g + g R U He = ↵m c2 P P + P + | n p| + p n P em p r + a r a 4 ⇥ r 4 ⇥ 16 4 ⇥ 16 r " np np pp ! ✓ np ◆ # The electromagnetic potential, for one bond (or one nucleon of the ↵ particle), is:

2 2 3 4 4R 4R 1 R 3 g g 1 g + g R U He = ↵m c2 P P + P + | n p| + p n P em p r + a r a 4 ⇥ r 4 ⇥ 16 4 ⇥ 16 r " ! ✓ ◆ # 2 Convertir ↵mpc : Fundamental constants of the nuclear energy potential • Electrostatic attraction:

2 2 938 e ↵mpc = = =6.85 MeV (6) 137 4⇡✏0RP Numerically,

2 2 3 4 3 0.210 3 0.210 3 3.83 5.59 + 5.59 +3.83 0.210 U He =6.85 ⇥ ⇥ + ⇥ em r + a r a 4 ⇥ 16 r "✓ np np ◆ ✓ np ◆ # (7) The minimum of the potential is the binding energy per nucleon, the number of neutron- proton bonds being equal to the number of nucleons.

p-1 Proton-neutron electric .

Electrostatic energy of 3 aligned punctual electric charges ep, en-, en+:

The third term is unphysically infinite for an isolated neutron where a ⋍ 0. The intuitive solution is to replace 1/2a by the exact electric dipole potential energy 1/(r+a) - 1/(r-a). The potential is thus doubbled and zero for a = 0:

The potential, being negative, is attractive. No empirical polarizability, only fundamental electric Coulomb’s law. Proton-neutron

B. Schaemagnetic↵er potential energy Magnetic dipole-dipole interaction potential energy (interactionFormule g´en´erale between du potentieltwo magnetic moments after Maxwell [2]):

µ 3(µ~ ~r )(µ~ ~r ) µ µ µ µ µ µ U = 0 µ~ µ~ n • np p • np = 0| n p| [ 1 3( 1)(1)] = 2 0| n p| (9) m 4⇡r3 n • p r2 4⇡r3 4⇡r3 np  np np np The potential energy of the 2H interacting magnetic moments, eiej µ0 3(µ~i µ~j)(µ~i µ~j) µ0 3(µ~ n ~rnp)(µ~ p ~rnp) Uem = Ue+Uem = + 3 [µ~i µ~j • 2 • + 3 µ~ n µ~ p • 2 • ] collinear and opposite,4⇡✏ is0r ijpositive,4⇡r thus• repulsiver : 4⇡r • r i i=j ij ij np  np ! X X6 (10)

qiqj Uem = 4⇡✏0rij µ and µ are the neutron and proton magnetic moments. i i=j n p 6 X X 10 [2] J. C. Maxwell, Aµ treatise on electricity3(µ~ andµ~ )(magnetismµ~ µ~ ) . Vol.3(µ~ II. 1995,~r )( µ~art. 384-387~r ) + 0 µ~ µ~ i • j i • j i • ij j • ij (11) 4⇡r3 i • j r2 r2 i i=j ij " ij ij # X X6

µ 3(µ~ µ~ )(µ~ µ~ ) µ µ µ U = 0 µ~ µ~ n • p n • p = 0| n p| [ 1 3 ( 1)] (12) m 4⇡r3 n • p r2 4⇡r3 ⇥ np  np np

p-2 B. Schae↵er

Corrections 5-5-14 , 27-4-14 Potentiel total:

2 2 e 2 2 µ µ µ U H = +2 0| n p| em 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ ✓ np ◆ 2 2 e 1 1 µ µ µ U H =2 + 0| n p| em 4⇡✏ r + a r a 4⇡r3  0 ✓ np np ◆ ✓ np ◆ Potentiel par nucl´eon: litt´eral 2 2 e 1 1 µ µ µ U H /A = + 0| n p| em 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ np B. Schae↵er 2 4 e 1 1 U He/A =4 e ⇥ 4⇡✏ r + a r a 0 ✓ np np ◆

Corrections 5-5-14 , 27-4-14 4 3 µ µ µ Proton-neutronU He/A = Coulomb-Maxwell0| n p| Potentiel total: m 2 4⇡r3 ✓ np ◆ electromagnetic2 potential energy 2 e 2 2 2 µ µ µ U 4HHe= e 1 1 +2 3 0|µ0nµnpµ|p Uemem /A =4 + |3 3 | With electric4⇡✏0⇥ attraction4⇡✏rnp0 +ranp and+ ar magneticnp rnpa arepulsion2 4 ⇡4r⇡nprnp combined, the✓ total✓ binding energy ◆potential◆ ✓ ✓of 2H is: ◆◆ num´erique 2 2 e 1 1 µ µ µ U H =2 + 0| n p| em 2H 2 2 0.08493 U 4=1⇡✏0.442rnp + a rnp a +2 4⇡rnp em ✓ ◆ ✓ 3 ◆ rnp + a rnp a ⇥ rnp Potentiel par nucl´eon:or, numerically:✓ ◆ litt´eral 2H 1 1 0.0849 Uem =2 21.442 + 3 2 e r1np + a rnp1 a µrnpµ µ H  ✓ ◆ 0 n p Uem /A = + | 3 | Only4 4fundamental⇡✏0 rnp + lawsa randnp constants.a 4⇡rnp He ✓ 1 1 ◆ 0.1274 Uem No/A fit,=5 empirical.76 parameter or cutoff.+ 3 11 rnp2 + a rnp a rnp 4 ✓e 1 ◆ 1 U He/A =4 Energie de liaisone d’un noyau⇥ 4 en⇡✏ g´en´eralr + a r a 0 ✓ np np ◆ Atome d’hydrog`ene: 4He 3 µ0 µnµp Um H /A = 1| 2 | 2 B = 13.62eV 4⇡↵r3mec (12) ✓ ⇡ 2 np ◆ Atome d’hydrog`ene: 2 4He e 1 1 3 µ0 µnµp U /A =4 1 2 2 + | | em ⇥ 4⇡✏ 13r .6 eV+ a ↵r mec a 2 4⇡r3 (13) 0 ✓ np ⇡ 2 np ◆ ✓ np ◆ num´erique 2 1 B H /A =1.11 MeV ↵m c2 (14) ⇡ 6 p 2 2 2 0.0849 U H =1.442 +2 em 1 2 3 r1np.11+MeVa rnp↵mac ⇥ rnp (15) ✓ ⇡ 6 p ◆ 4 1 1 0.1274 He 4He 2 Uem /AB=5/A.76=7.07 6.85 MeV = ↵m+pc (16) r +⇡ a r a r3 ✓ np np ◆ np Energie de liaison d’un noyau7.07 MeV en g´en´eral6.85 MeV = ↵m c2 (17) ⇡ p

Atome d’hydrog`ene: 2 e 2 H = ↵mpc =6.8461 9012 652MeV 4⇡✏0BRP = 13.6 eV ↵ m c (12) ⇡ 2 e Atome d’hydrog`ene: p-2 1 13.6 eV ↵2m c2 (13) ⇡ 2 e

2 1 B H /A =1.11 MeV ↵m c2 (14) ⇡ 6 p 1 1.11 MeV ↵m c2 (15) ⇡ 6 p

4 B He/A =7.07 6.85 MeV = ↵m c2 (16) ⇡ p

7.07 MeV 6.85 MeV = ↵m c2 (17) ⇡ p

2 e 2 = ↵mpc =6.846 901 65 MeV 4⇡✏0RP

e2 =1.6 MeV 4⇡✏0rp

p-2 Theoretical proton-neutron electromagnetic potentials

-2.2 MeV The equilibrium between attractive and repulsive forces is obtained at a minimum or at a horizontal saddle point.

12 Best result with the exact electric dipole formula (continuous curve). Constantes

2 2 ¯h 2 e 2 2 µ µ µ 2 + U H + +2 0| n p| = 0 (32) 2mr em 4⇡✏ r + a r a 4⇡r3  0 ✓ np np ◆ ✓ np ◆ 2 2 2 ¯h e 2 2 µ µ µ U H = 2 + +2 0| n p| (33) em 2mr 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ ✓ np ◆ For the spherically symmetrical electrostatic potential, one has:

2 2 2 ¯h 1 2 e 2 2 µ µ µ B. Schae↵Eer H =+ + 2 0| n p| em 2m b2 rb 4⇡✏ r + a r a 4⇡r3 ✓ ◆  0 ✓ np np ◆ ✓ np ◆

2 2 2 2µ0 µnµp 2H ¯h @ 2 @ U H =2e 2 2 Constantesµ0 µnµp E = + m+ | 3 | 2 | | em 2m @r2 r @r 4⇡✏ 4⇡rrnp + a r a 4⇡r3 ✓ ◆  ✓ 0 ✓ np ◆ np ◆ ✓ np ◆ 2H µ0 2m K Um 2 = 3 2 µnµp +4⇡2r E +⇥ | = 0| (33) 2 2 r 2 ¯h np r ¯h @ 2 @ e  2 2 µ0 µnµp 2 2 +¯h +2 2 E e 2 2 µ µ µ 2 | | =0 2 2H 2 +e U H + 2 2 +2 µ00| n p| = 03 (34) 2m @r U r @=r em 4⇡✏0 rnp + a +rnp a 3 2 µ µ 4⇡rnp ✓ em2mr ◆  4⇡✏0 rnp✓+ a rnp a 34⇡r◆np n✓ p ◆ 4⇡✏ 0 rnp +✓a rnp a ◆ 4⇡✓rnp ⇥ ◆| | 2✓ 2 ◆ 2 ¯h e 2 2 µ µ µ U H = 2 + +2 0| n p| (35) Potentiel ELECTRIQUE:em 3 2 2mr 24⇡✏0 rnp + a rnp a 4⇡rnp @ 2 @ 2m e ✓ 2 ◆2 ✓ ◆µ0 µnµp For the spherically symmetrical electrostatic potential, one has: 2 + + 2 E + 2 +2 | 3 | =0 @r r @r ¯h 2H 4⇡✏e 0 rnp +2 a rnp 2 a 4⇡r 2 U = 2 np (18) 2H ¯h ✓e1 2 ✓e 2 2 ◆ µ0✓µnµp ◆◆ Eem =+ 2 4+⇡✏0 rnp + a rnp a 2 | 3 | 2m b rb 4✓⇡✏0 rnp + a rnp a ◆ 4⇡rnp ✓ ◆  ✓ ◆ ✓ ◆ 2 2m K Potentiel total: + 2 E + = 0 (34) 2 2 r 2 r 2H ¯h @ 2 @ ¯he 2 2 µ0 µnµp Eem = 2 + + ✓ ◆ 2 | 3 | 2 2m @er r @r 2 4⇡✏0 r2np + a rnp aµ µ µ4⇡r For the sphericallyU symmetricalH =✓ electrostatic◆  ✓ potential,+2 one◆0 has:| n✓ p| np ◆ em 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ ✓ np ◆ ¯h2 @2 2 @ 2 e2 2 2 µ µ µ + @ + E 2 @ 2m K 2 0| n p| =0 2m @r2 r @r2 + 4⇡✏ +r + a rE +a =4⇡ 0r3 (35) ✓ 2 e ◆ 2 2 0 ✓ np 22 np ◆ µ✓ µ npµ ◆ U H = @r r @r ¯h +2r 0| n p| em 4⇡✏ r + a r ✓a ◆ 4⇡r3 2 0 np 2 np np @ 2 @ 2m ✓ e 2 2◆ ✓ µ0 µnµp ◆ 2 + + 2 E + +2 | 3 | =0 Formules approximatives@r r @r par¯h nucl´eon:4⇡✏0 rnp + a rnp a 4⇡rnp ¯h2 ✓ 1 1✓ ◆ ✓ ◆◆ Schrodinger repris HyperPhysics2 8-5-14m K (36) 2 + E + = 0 (36) 2mpb br r ¯h2 r 2 2 ✓ ◆ ✓ ◆ ¯h 2 e2 2 2 µ0 µnµp For2 the spherically+ E H symmetricale electrostatic2 potential,2 one has:µ20 µnµ|p | = 0 (19) 2mr em + 4⇡✏ r + a r +2a | 4⇡|r3 =0 @02 np2 @ 2m np K 3 np  4⇡✏0 ✓r+ + a + r Ea+ ◆ = 04✓⇡rnp ◆ (37) @r✓2 r @r ¯h2 ◆r ✓ ◆ =✓ 1◆ (20) 2 2 2H ¯h 1 1 e 2 2 µ0 µnµp E = ¯h2 1 +1 +2 | | em (38)3 2mpb 2bmb rb r 4⇡✏0 r + a r a 4⇡rnp ✓ p ✓ ◆ ◆ =✓ 0 ◆ ✓ ◆ (21) e2 2 2 µ µ µ + 2 +2 0| n p| =0 2H 1 4⇡✏e r + a 2r a 2 4⇡r3 µ0 µnµp Schrodinger repris Wiki 8-5-140 ✓ ◆ ✓ np ◆ Uem /A = +2 | 3 | 22 4⇡✏0 rnp +2 a rnp a 4⇡rnp 2H ¯h  1 1✓ e2 2 2 ◆ µ✓0 µnµp ◆ Eem = + ¯h 2 +2 | 3 | 2mEp b (rb )=2 r 4⇡✏0 r ++ aV(rr) a (r)4⇡rnp (22) 4He ✓ e ◆ 2 ✓ 2 ◆ ✓ 3 µ0◆µnµp 2 2m r Uem /A2 =2 1 e  2 2 µ0 +µnµp | | Proton-neutronU H /A = Schrödinger+2 equation3 em ⇥ 4⇡✏0 rnp + a rnp a | 3 4| 4⇡rnp Schrodinger 2 4⇡✏0 ✓rnp + a rnp a ◆4⇡rnp ✓ ◆ With the Coulomb-Maxwell potential✓ V, the 2H Schrödinger◆ ✓ equation◆ writes: 2 2 4 e 2 2 2 3 µ µ µ ¯h 1 2 He 22 e 2 2 0 nµ0p 2 µnµp Uem /A =2H 2 + | 3 | + ¯hU ⇥ 4⇡✏0 rnpH+ a rnp a 4 4⇡rnp | | =0 2 em ✓ ◆ ✓ ◆ 3 2m b rb +4⇡✏U0em r +Va(r, ar) a= 0 4⇡ r (23) 2 2m 2  ✓ ¯h 1 ◆ 2 2H e ✓ 2 2 µ0◆ 2 µnµp ✓ ◆ With the simplest2 exponential+ Uem h well andi the Coulomb| 3 |-Maxwell =0 2m b rb 4⇡✏0 r + a r a 4⇡ r electromagnetic ✓ ◆potential,2 we obtain:✓ ◆ ✓ ◆ 2 ¯h 2 2H ¯h 1 2 2 +e U 2 V (r, a,2 b) = 0µ 2 µ µ (24) 2 H 2 em 0 n p r/b ¯h 1 2 2m 2 e 2 2 µ 2 µ µ 2 + Uem H 0 n p | r/b3 | e =0 2 + Uem | 3 | e =0 2m b 2m rbb rb 4⇡✏4⇡✏h00 rr++a a r ar ia4⇡ 4r ⇡ r  ✓  ✓ ◆ ◆ 2 ✓ ◆ ◆✓ ✓◆ ◆ B. Schae↵er ¯h +[E p-5V (r/br, a, b)] = 0 (25) Dividing the Schrödinger2m equation by e , we obtain the classical electromagnetic 2H Coulomb-Maxwell potential except for the first term. 2 2 ¯h 2 2 ¯h 1 2 2 e 2 2 µ 2 µ µ + U H +[E U] = 00 | n p| =0 (26) 2m b2 rb 2emmr4⇡✏ r +p-5a r a 4⇡ r3 ✓ ◆ 0 ✓ ◆ ✓ 13◆ r/b 2m e 2 + [E V (r)] = 0 (27) r ¯h2 r/b = e =1 2 2m K 2 2 ¯h 1 2 2 + e 2 E2+ 2 = 0µ0 2 µnµp (28) +rU H ¯h r | | =0 2m b2 rb em 4⇡✏ r + a r a 4⇡ r3 ✓ ◆ 0 ✓ ◆ ✓ ◆ Formules exactes ENERGIE MAGNETIQUE: p-4 4 3 µ µ µ U He/A = 0| n p| m 4 4⇡r3 ✓ np ◆ Formules dipˆole a<

e2 1 1 e2 2a 4⇡✏ r + a r a ⇡ 4⇡✏ r2 0 ✓ np np ◆ 0 ✓ np ◆ e2

⇡ 4⇡✏0rnp VERIFICATION 2H

3 2 2 0.210 2 0.210 3.83 5.59 0.210 U H =6.85 ⇥ ⇥ +6.85 2 ⇥ (39) em r + a r a ⇥ ⇥ 16 r ✓ np np ◆ ✓ np ◆ 2 2 2 0.0849 U H =1.44 + (40) em r + a r a r3 ✓ np np ◆ np 4He The electromagnetic potential, for one bond of the ↵ particle, is: 4He 4He 4He Uem = Ue + Um +↵m c2 4RP 4RP p rnp+a rnp a (41) 3 2 ⇣ 3 gngp R⌘P +↵mpc | | 4 ⇥ 16 rnp Numerically, ⇣ ⌘⇣ ⌘ 4He 4 0.210 4 0.210 U =6.85 ⇥ ⇥ em rnp+a rnp a 3 (42) 3 ⇣3.83 5.59 0.210 ⌘ +6.85 ⇥ ⇥ 4 ⇥ 16 rnp ERREUR 3 corrig´epar 4 ⇣ ⌘

4 2 2 0.0637 U He =2.88 + (43) em r + a r a r3 ✓ np np ◆ np FORMULE MAGNETIQUE

4 2 2 3 He 2 3 gngp 1 gp+gn RP U = ↵mpc | | + (44) m 4 ⇥ 16 4 ⇥ 16 rnp ⇣ ⌘⇣ ⌘ 4He 3 0.210 3 0.210 U =6.85 ⇥ ⇥ m rnp+a rnp a 3 (45) 3 ⇣3.83 5.59+5.592+3.83⌘2 0.210 +6.85 ⇥ ⇥ 4 ⇥ 16 rnp ⇣ ⌘

p-6 B. Schae↵er Proton-neutron Schrödinger equation 2 2 2 ¯h 1 2 e 2 2 µ 2 µ µ U H = + + 0 | n p| em 2m b2 rb 4⇡✏ r + a r a 4⇡ r3 ✓ ◆ 0 ✓ ◆ ✓ ◆ 2 2 ¯h 1 2 2 e 2 2 µ 2 µ µ + U H 0 | n p| =0 2m b2 rb em 4⇡✏ r + a r a 4⇡ r3 ✓ ◆ 0 ✓ ◆ ✓ ◆ r/b e Zero binding

r/b energy = e =1

B. Schae2 ↵er 2 ¯h 1 2 2 e 2 2 µ 2 µ µ + U H 0 | n p| =0 2m b2 rb em 4⇡✏ r + a r a 4⇡ r3 ✓ ◆ 0 ✓ ◆ ✓ ◆ Formules exactes ENERGIE MAGNETIQUE:2H: -2.2 MeV

4 3 µ µ µ r/b U HeFor/A b= infinite,0| n p | = e =1 , the Schrödinger m 4 4⇡r3 term is ✓ zero, np simplified◆ to the electromagnetic Formules dipˆole 2 Coulomb-Maxwell2 potential (dark line), giving, at 2 a<

4He 4He 4He ⇡ 4⇡✏0rnp Uem = Ue + Um VERIFICATION+↵m c2 4RP 4RP p rnp+a rnp a (41) 3 2H 2 ⇣ 3 gngp R⌘P +↵mpc | | 4 ⇥ 16 rnp ⇣ ⌘⇣ ⌘ 3 Numerically, 2H 2 0.210 2 0.210 3.83 5.59 0.210 U =6.85 4He 4 0.210 4 0.210 +6.85 2 (37) em U ⇥=6.85 ⇥ ⇥ ⇥ ⇥ em rnp + a rnp+a rnprnp aa ⇥ ⇥ 16 rnp 3 (42) ✓ 3 ⇣3.83 5.59 0.210 ⌘◆ ✓ ◆ +6.85 ⇥ ⇥ 4 ⇥ 16 rnp 2H 2 2 0.0849 ERREUR 3 corrig´epar 4 U =1.44 ⇣ ⌘ + (38) em r + a r a r3 ✓ np np ◆ np 4 2 2 0.0637 U He =2.88 + (43) 4He em 3 rnp + a rnp a rnp The electromagnetic potential,✓ for one◆ bond of the ↵ particle, is: FORMULE MAGNETIQUE 4He 4He 4He Uem = Ue + Um p-6 +↵m c2 4RP 4RP p rnp+a rnp a (39) 3 2 ⇣ 3 gngp R⌘P +↵mpc | | 4 ⇥ 16 rnp ⇣ ⌘⇣ ⌘ Numerically, 4He 4 0.210 4 0.210 U =6.85 ⇥ ⇥ em rnp+a rnp a 3 (40) 3 ⇣3.83 5.59 0.210 ⌘ +6.85 ⇥ ⇥ 4 ⇥ 16 rnp ERREUR 3 corrig´epar 4 ⇣ ⌘

4 2 2 0.0637 U He =2.88 + (41) em r + a r a r3 ✓ np np ◆ np FORMULE MAGNETIQUE

4 2 2 3 He 2 3 gngp 1 gp+gn RP U = ↵mpc | | + (42) m 4 ⇥ 16 4 ⇥ 16 rnp ⇣ ⌘⇣ ⌘ 4He 3 0.210 3 0.210 U =6.85 ⇥ ⇥ m rnp+a rnp a 3 (43) 3 ⇣3.83 5.59+5.592+3.83⌘2 0.210 +6.85 ⇥ ⇥ 4 ⇥ 16 rnp ⇣ ⌘

p-6 α particle structure

4He, with 2 protons + 2 neutrons, may be r considered in a first approximation as a n nn regular tetrahedron with 60° angles. µ n µ n The of 4He being zero, n the magnetic moments of identical nucleons have to be collinear and rnp rnp rnp r oppositely paired. np p p µ p µ p rpp The other nucleon-nucleon interactions, being each ¼ of the total proton-neutron interactions are neglected provisionally.15 4He electric potential energy (only proton-neutron interactions considered) Per nucleon (= per bond in 4He) n • 2 protons interacting with each neutron, rnn the electric potential is twice that of 2H/A 4 • He having one proton-neutron bond, n twice the 2H binding energy per nucleon, the electric potential is again doubbled rnp rnp r r • The electric potential of 4He is thus np np 4 times larger than that of 2H p p

rpp

16 4He magnetic potential energy (only proton-neutron interactions considered) Per nucleon (= per bond in 4He) • 2 protons interacting with each neutron: the magnetic potential is twice that of 2H/A. • The magnetic potential of one bond is given by the Maxwell formula of the magnetic dipole-dipole interaction energy:

The proton and neutron magnetic moments being perpendicular, the first term is zero. Being inclined at 60° and 120° with respect to the edges of the tetrahedron, one has − (−½×½) = +¼. • The magnetic part of the 4He binding energy per nucleon is thus 2×3×¼ = 3/2 times larger than that of 2H/A: 17

2H and 4He potential : (only proton-neutron interactions considered, per nucleon, analytical formulas) The 2H potential energy per nucleon is ½ of its bond energy:

The 4He electric potential energy/A is 4 times that of 2H. The 4He magnetic potential energy/A is 3/2 times that of 2H:

18 2H and 4He potential energies: (only proton-neutron interactions, per nucleon, numerical formulas)

The 2H potential energy per nucleon is ½ of its bond energy:

The 4He electric potential energy per nucleon is 4 times that of 2H. The magnetic moments being perpendicular and inclined at 60°, the magnetic potential energy is 1.5 that of 2H:

19 2H and 4He potential energies.

2H: -1.1 MeV

Equilibrium at the horizontal saddle point: not a real minimum, due to the Coulomb singularity 4He: -6.9 MeV

20 B. Schae↵er

FormulePairing N=Z effect of N=Z nuclei Calculated as proton-neutron bonds between α particles: B (A) B(4He) A 8 = Integer + B(2H) 1 MeV (96) A A 4 A ✓ ◆ ✓ ◆

B (A) 28.28 A 8 = Integer +2.225 1 MeV (97) A A 4 A ✓ ◆ ✓ ◆

B (Z) 14.14 Z 4 = Integer +2.225 1 MeV Z Z 2 Z ✓ ◆ ✓ ◆ Ne marche qu’`apartir de Z=4. Cette formule donne la mˆeme valeur quel que soit N. Comment introduire N? Chaque neutron ajout´eapporte une ´energie d’un 3 deutons qui ne peut d´epasser 6 par proton pour un noyau infiniment grand. Si on ajoute un deuton avec chaque neutron, l’´energie par proton augmente de 2,2 / Z MeV. Pour Z=50, cela fait 4%. De 10 en 10 Z cela fait 40%, bien trop fort. Pour Z=10, cela fait 22%. De 10 en 10 Z cela fait 40%, En fait, il faut ramener `aZ; on a alors:

B (Z) 14.14 Z 4 = Integer +2.225 1 +2.2(N/Z 1) MeV Z Z 2 Z ✓ ◆ ✓ ◆ 21 Essayer:

B (Z) 14.14 Z N 4 = Integer +2.225 MeV Z Z 2 Z ✓ ◆ ✓ ◆ On obtient une relation lin´eaire en N pour Z constant. Il faut donc modifier COURBES N/Z

N/Z E/Zmax E/Z(N/Z)=E/Zmax = cste N/Z + cste N/Z +1

p-32 Constantes Constantes

Energie de liaison d’un noyau en g´en´eral Energie de liaison d’un noyau en g´en´eral Atome d’hydrog`ene: H 1 2 2 Atome d’hydrog`ene: B = ↵ mec = 13.6 eV (1) H 1 22 2 B = ↵ mec = 13.6 eV (1) 2 2 1 B H ↵m c2 1.1 MeV (2) ⇡ 6 p ⇡ 2H 1 2 B ↵mpc 1.1 MeV (2) ⇡4 6 ⇡ B He =6.8 7.07 MeV (3) ⇡ 4 B He =6.8 7.07 MeV (3) Fundamental2 nuclear constants e 2⇡ Electric: = ↵mpc =6.846 901 65 MeV 42⇡✏0RP e 2 Noyau = ↵mpc =6.846 901 65 MeV 4⇡✏ R Nuclear magnetic energy0 constant:P Noyau• Magnetic: Nuclear magnetic energyµ0 constant:µnµp 2 gngp • | 3 | = ↵mpc | | =9.147 871 896 MeV 4⇡RP 16 µ0 µRnP µ: protonp Compton2 gradius,ngp 0.210 308 910 fm ≈ rp / 4 | | = ↵mpc | | =9.147 871 896 MeV rp :3 proton radius g : proton Landé factor 4⇡RP 16 p 3 m : proton2 eiejRP gigj RP p gn : neutron Landé factor Uem = ↵mpc 2 + | | 22 Sij (4) α : fine2 structure constante rij 16 rij 3 i i=j ✓ ◆ i i=j ✓ ◆ X X6 X X6 3 2 4 eiejRP gigj RP 5 where Uem = ↵mpc 2 + | | Sij (4) 2 e rij 16 rij 3 S =i cosi=j(µ~✓, µ~ ) 3◆cos (µ~i , ~ri=)j cos (µ~ ✓, ~r )◆ (5) ij X X6 i j Xi Xij6 j ij where 4 5 Sij = cos (µ~ i, µ~ j) 3 cos (µ~ i, ~rij) cos (µ~ j, ~rij) (5) R g g R 3 U ↵m c2 P + | n p| P S (6) em ⇡ p r 16 r np " np ✓ np ◆ # R g g R 3 It is more complicatedU for↵ them c magnetic2 P moments:+ | n p| P S (6) em ⇡ p r 16 r np " np ✓ np ◆ # S = cos (µ~ , µ~ ) 3 cos (µ~ , ~r ) cos (µ~ , ~r )= (7) np n p n np p np It is more complicated for the magnetic= 1 moments:3 1 ( 1) = 2 ⇥ ⇥ Il sura d’ajusterSnp = lecos coe(µ~cientn, µ~ p magn´etique,) 3 cos (µ~ n (toujours, ~rnp) cos ?)(µ~ p compris, ~rnp)= entre 2 pour le deuton(7) et 3). Il faut aussi tenir compte= des1 g qui3 varient1 ( entre1) = 2gn = 3.826 and gp =5.585. Cette formule, approximative (30% d’erreur),⇥ ⇥ peut ´etre g´en´eralis´ee avec des coecients Ilad su hoc.ra d’ajuster Cette approximation le coecient vient magn´etique, de ce qu’on (toujours a n´eglig´ela ?) compris r´epulsion entre entre 2 pour le proton le deuton et la et 3).charge Il faut `electrique aussi tenir du neutron. compte des En fait g qui seule varient la partie entre magn´etiquegn = 3.826 change. and g Celap =5 veut.585. dire que Cettel’´energie formule, de liaison approximative par nucl´eon (30% d’un d’erreur), noyau varie peut assez ´etre peu. g´en´eralis´ee Le principal avec param`etre des coecients est le ad hoc.nombre Cette de approximation liaisons neutron-proton vient de par ce nucl´eon qu’on a qui n´eglig´ela varie de r´epulsion 0,5 pour le entre deuton le proton `a3 soit et un la rapport de 6 pour un noyau de masse infinie. La particule ↵ serait 6 fois plus ´energ´etique charge `electrique du neutron. En fait seule la partie magn´etique change.0 Cela veut dire que l’´energiepar nucl´eon. de liaison En par fait nucl´eon les moments d’un magn´etiques noyau varie y assez sont peu. inclin´es Le `a60 principalce qui param`etre divise l’´energie est le magn´etique r´epulsive par 2, donnant 12 MeV (AVERIFIER). En fait elle est diminu´ee nombre de liaisons neutron-proton par nucl´eon qui varie de 0,5 pour le deuton `a3 soit un par la r´epulsion entre protons et neutrons n´eglig´ee provisoirement. rapport de 6 pour un noyau de masse infinie. La particule ↵ serait 6 fois plus ´energ´etique Le probl`eme se r´eduit donc au calcul de Snp et au nombre de0 liaisons (neutron-proton par nucl´eon.en 1`ere approximation) En fait les moments par nucl´eon magn´etiques ou, de pr´ef´erence, y sont inclin´es par proton. `a60 ce qui divise l’´energie magn´etique r´epulsive par 2, donnant 12 MeV (AVERIFIER). En fait elle est diminu´ee par la r´epulsion entre protons et neutrons n´eglig´ee provisoirement. Le probl`eme se r´eduit donc au calcul de Snp et au nombre de liaisons (neutron-proton en 1`ere approximation) par nucl´eon ou, de pr´ef´erence, par proton.

p-1

p-1 Constantes

Corrections 5-5-14 , 27-4-14 Potentiel total:

2 2 e 2 2 µ µ µ U H = +2 0| n p| em 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ ✓ np ◆ 2 2 e 1 1 µ µ µ U H =2 + 0| n p| em 4⇡✏ r + a r a 4⇡r3  0 ✓ np np ◆ ✓ np ◆ Potentiel par nucl´eon: litt´eral 2 2 e 1 1 µ µ µ U H /A = + 0| n p| em 4⇡✏ r + a r a 4⇡r3 0 ✓ np np ◆ np 2 4 e 1 1 U He/A =4 e ⇥ 4⇡✏ r + a r a 0 ✓ np np ◆

4 3 µ µ µ U He/A = 0| n p| m 2 4⇡r3 ✓ np ◆ 2 4 e 1 1 3 µ µ µ U He/A =4 + 0| n p| em ⇥ 4⇡✏ r + a r a 2 4⇡r3 0 ✓ np np ◆ ✓ np ◆ num´erique

2 2 2 0.0849 U H =1.442 +2 em r + a r a ⇥ r3 ✓ np np ◆ np

2 1 1 0.0849 U H =2 1.442 + em r + a r a r3  ✓ np np ◆ np

4 1 1 0.1274 U He/A =5.76 + em r + a r a r3 ✓ np np ◆ np Energie de liaison d’un noyau en g´en´eral

Atome d’hydrog`ene: 1 BH = 13.6 eV ↵2m c2 (13) ⇡ 2 e Atome d’hydrog`ene: 1 13.6 eV ↵2m c2 (14) ⇡ 2 e

2 1 B H /A =1.11 MeV ↵m c2 (15) H, 2H, 4He, 58Fe energies⇡ 6 p compared (absolute values) 1 2 1.11 MeV ↵mpc (16) • H atom binding energy:⇡ 6 1

4He 2 BindingB energy/A =7 per.07 nucleon:6.85 MeV = ↵mpc (17) ⇡ 2 80.000 • H7 . 07: MeV 6.85 MeV = ↵m c2 (18) ⇡ p • 4He : 500.000 58 2 • Fe: 8.79 MeV =1.28 ↵mpc 650.000 (19) 23 p-3 Nuclear to ratio order of magnitude:

mp : proton mass me : electron mass α : fine structure constant 24 Discussion • Strong force (unknown but strength 1!) • Magic numbers (numerology) Doubtful • Charge independence (neutron has +e,-e)} • Quark (virtual, unobservable, undetectable) • Schrödinger equation (no : useless) • With the bare application of electric and magnetic forces with fundamental constants only, the binding energies of 2H and 4He nuclei have been calculated without fitting, cutoff, input parameter…

l 25 The quantitative adequation (no fitting) between theory and experiment proves the electromagnetic nature of the nuclear energy.

Спасибо за внимание Thank you for your attention

26