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Proton-Neutron Electric Potential Energy

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Proton-Neutron Electric Potential Energy 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 magnets (µαγνήτης, magnetis). • Only the Coulomb repulsion between protons is taken into account in mainstream nuclear physics. 2 The nuclear shell model The binding energy 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 binding energy 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 suffira d’ajuster le coeneutronfficient 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 coefficients 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 potential energy. 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.
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