Electromagnetic nature of the nuclear forces and toroid structure of the deuteron and triton ()
1. INTRODUCTION
In order for a complete theory of the atomic nucleus to be created, that theory has to be able to explain the structure of the nucleus and the available experimental data about its behavior [1-4]. The main difficulty here is the incomplete knowledge of the forces of interaction between nucleons inside the nuclei and their models.
In the field of elementary particles there are two leading models: standard and helicon.
The standard model [5-7] presents elementary particles as quantum objects that can be both wave and particle. As particles they have mass, charge, spin, magnetic moment, quadrupole moment. The elementary particles mainly are modelling with spheres.
The helicon model [8-11], or ring model is less wellknown. According to that model the particles have ring spiral structure of charged filaments, one or more. The filaments are superconductors coiled around an imaginary ring. The helicon model is consistent with all the widely accepted and experimentally verified properties of elementary particles. This model implies the assumption that all known types of interactions (strong, electroweak, gravitational) should be electromagnetic in nature.
In [12] we introduce a toroid model of the nucleons that is to a certain degree in contradiction with the standard model, but it is in full agreement with the helicon model.
We consider the nucleons as tori, rotating with a constant angular velocity around an axis z, passing through their mass (geometrical) centre O and perpendicular to the plane of rotation of their central circle. From quantum mechanical point of view the nucleon is not a localized object in the three-dimensional physical space and therefore it cannot be consider a sphere or a torus [13]. We feel such a model is appropriate in the formal-heuristic sense of Niels Bohr. It is similar to the old quantum theory model of the electron in the hydrogen atom for which Bohr obtains good results for the description of its spectrums. Modern quantum theory confirms Bohr’s results based on other concepts.
Formal approaches are widely used in physics. In classical mechanics for example a variety of formulations have been proposed by Lagrange, Hamilton and others. The best example in that sense is quantum mechanics, where one introduces the so-called wave function which allows for the theoretical derivation of a number of physical properties confirmed extremely well by experiment.
We consider nucleons within nuclei to be space dimensional objects—tori, within which the electrical charges can be redistributed. This assumption does not contradict the quark model. The latter enables us to determine the electrostatic interaction between them. Based on this model, we obtained that the strong interactions are electromagnetic in nature. To achieve this we used exact analytical formulas first obtained by us about the electrostatic interaction between two charged conducting spheres with arbitrary charges and radii [14].
In this paper we apply this method in order to obtain the electrostatic interaction between nucleons in the nuclei of deuterium and tritium.
As noted by Feynman [15], at distances under 10–15 m either Coulomb’s law is not in force or the electrons and protons are not point charges. We consider the proton-neutron couples at distances under 10–15 m. Thus in this paper we determine that the known binding energy between the nucleons in the deuteron and triton is obtained through electromagnetic interactions! We also explain the other basic experimental data—spin, magnetic moment, stability of the nuclei of deuterium and tritium. Doing that, we use the experimentally obtained values of the radii and masses of the nucleons and the nuclei.
We obtain the volumes and mass densities of the nucleons; we also obtain the force of interaction within the nuclei under consideration—results obtained for the first time in nuclear physics.
2. METHOD FOR FINDING OUT ELECTROSTATIC INTERACTION BETWEEN TWO CHARGED CONDUSTIVE SPHERES
We will represent the part of the method presented by us in [10], necessary for performing the calculations for the nuclei of deuterium and tritium.
Let S1 and S2 be two isolated charged conductive spheres, with charges Q1, Q2 and radii r1, r2 respectively. Let’s denote with R the distance between their centers O1, O2 in an inertial system J. Since charges Q1 and Q2 are evenly distributed on the surfaces of S1 and S2, it is assumed that before the interaction between the spheres they are concentrated in the centers O1 and O2 respectively.
As a result of the electrostatic interaction between S1 and S2, on their surfaces appear induced charges 
and
, which are interrelated. Formally, we can consider that these charges are located on line segment
. On the surfaces of S1 and S2 appear uniformly distributed charges 
and
, we can assume that they are concentrated in their canters O1 and O2.
From the law for preservation of electric charges are in power the equations:
and
(1)
We will determine the charges
, 
and hence the charges
,
. Let as a consequence of 
be generated image charges 
(j = 1, 2, 3, ∙∙∙). Because each charge 
generates
, the charges with an odd index 
(m = 1, 2, 3, ∙∙∙) are located in the sphere
, and charges with an even index
—in the sphere
. Similarly are determined the image charges 
(j = 1, 2, 3, ∙∙∙), arising as a consequence from charge
. The charges with an odd index 
(m = 1, 2, 3 ∙∙∙) are located in the sphere
, and charges with an even index
—in the sphere
.
Let’s denote 
and
. We introduce for j = 1, 2, 3, ∙∙∙ the following denotations:
(2)
If 
are the distances of imagecharges
, respectively to the centers of the spheres
, in [14] we obtain that:
(3)
We find also that:
(4)
Lets
(5)
where 
at 
(
).
Since charges 
and 
are sums of all image charges, located respectively in the spheres 
and
, then
and
From here and from (4) and (5) it follows that
and 
Then, substituting these equations in (1), we get:
(6)
Lets denote the charges from formulas (4) and (6), which are located in the sphere 
as
, and those located in the sphere 
as 
(
). Thus 
and
, and for
, 
, 
and
,
. Their corresponding distances to the centers of the spheres, where they are situated, we denote with 
and 
(
), where
.
If
, and
, then, according to Coulomb’s law, for the magnitude F of the projection of the force of interaction on
, acting on the spheres 
and 
, we obtain
(7)
The potential energy of interaction between the spheres 
and
, according to [16], is
(8)
Let us point out that in (7) and (8) we do not take into consideration the interactions between the charges inside the spheres 
and 
as actually the interaction is external—between the charges on the surface of 
with the charges on the surface of
.
Let M be an arbitrary point in the electric field created by charges 
and 
(
). If M is at distances 
and 
from charges 
and 
respectively, then, using the metric relationships in a triangle, we can determine
and
Then based on the principle of linear superposition of states, the potential at point M will be the sum of the potentials of all charges in M [16]. Therefore
(9)
It is worthwhile to mention that using Eqs.7-9 one can determine the interaction between two charged spheres for arbitrary small distances between them, which represents a result obtained for the first time.
3. TOROID MODEL OF NUCLEONS
Using the results from [14] in [12] we first consider the nucleons as spheres, as they are viewed in the standard model. In this case we show that at distances 
m we can calculate with good approximation the binding energy and force of interaction between nucleons by modeling protons as point charges. But at distances 
m the proton should not be considered as a point charge. We also found out that using the standard model the binding energy of triton cannot be obtained. Using this model it is difficult to explain the experimentally obtained magnetic moments of the nuclei relative to the magnetic moments of the comprising nucleons; it is also hard to explain the experimental results for the radii and stability of the nuclei.
For this reason we remodel the nucleons as tori [12]. At that they are rotating with a constant angular velocity around a straight axis z, passing through their mass (geometrical) center O and perpendicular to the plane of rotation of their central circle (Figure 1).
After that we study a system of proton and neutron in order to determine the electrostatic interaction between them.
The two tori—of the proton and neutron we denote correspondingly as 
and 
and their centers as 
and
. We also assume that the central circles of 
and 
lie in parallel or coincident planes and rotate in the same or opposite directions with constant angular velocity around a the straight line z passing through 
and
, and perpendicular to the plane of their rotation. Thus if
, then 
(Figure 2).
Let us denote by 
and 
the centers of the forming circles of the tori 
and
, and with 
and
—the radii of the central circles of 
and
. We assume that 
and 
are at a distance 
from each other.
According to experimental data the radius of the proton 
is smaller than the radius of the neutron
. Because of that we consider
.
Figure 2. Cross section of a proton-neutron system.
Let 
and 
be the radii of the forming circles 
and
. It is clear that
, and
. Besides that
(10)
where 
and 
are the corresponding radii of the proton and neutron in this configuration. It should be noted that for different nuclei
, 
, 
and
, v, 
may have different values.
From geometrical considerations for 
will be fulfilled the equation
(11)
It is clear that
.
We assume that the volume mass densities of the proton and neutron are equal, i.e.
. The volumes of the tori 
and
, according to [17], are correspondingly 
and
. Then if 
and 
are the corresponding masses of 
and
then from 
it follows that
(12)
In (12) we can substitute the experimentally measured masses of the proton and the neutron [18] 
kg, 
kg.
Lets denote with 
the radius of the empty part of the circle with a radius
. Then
(13)
Further let 
and 
from Figure 2 be a protonneutron couple bound in the nucleus of deuterium or tritium. If we denote with 
the radius of the nucleus under consideration, then
.
In order to apply the results from Section 2 for spheres we will remodel the tori.
We will emphasize that the potential energy and the force of interaction between two spheres depend on the distance between image-charges, i.e. from the lengths of the line segments 
from formula (3). These lengths in (3) are determined from the squares of the radii 
or 
of the spheres and from the length of their central line
, where 
is the smallest distance between their surfaces. For the square of the radii of each of the two spheres is fulfilled
, 
, where 
is the surface area of the corresponding sphere. Therefore, when we remodel the tori of the proton and the neutron we have to keep both their surface areas 
and 
and the distance 
between them.
Due to the central symmetry of the charge of the proton we can consider all of its charge 
to be concentrated in the geometrical center 
of the torus
.
Therefore we remodel the proton as a sphere
, with radius 
and the same centre 
on the straight line z and a surface area equivalent to the torus surface area, i.e. it has the same surface area as the torus
. At that 
is charged with a charge 
that is centrally symmetrical and can be redistributed.
The surface area 
of a torus 
according to [17] is
(14)
Then, as the surface areas of the torus 
and the sphere 
are equal, then from (14) it follows that the radius of 
is
(15)
We remodel the neutron 
with a torus 
with an equivalent surface area. At that 
has the same center 
on z and its surface is at the same distance 
from the surface of 
as in the case of 
and 
(Figure 3).
Let 
be a sphere. Its central circle, is forming for the torus
. We denote the centre of 
with
, at that 
and with
—the radius of
. Thus if 
then 
and from 
it follows that
, i.e.
(16)
It is clear that 
and
.
From the equality of the surface areas 
and 
of the tori 
and
, it follows that
(17)
For the number 
of spheres 
which have total surface area equal to the surface area of the torus 
is fulfilled
. Therefore
(18)
We assume that the centre 
is motionless relative to the inertial reference system J. We introduce a solid non-inertial reference system G that rotates with the constant angular velocity 
of rotation of 
relative to J. At that, the point 
is the centre of the coordinate system
, stationary connected with G and relative to which the spheres 
and 
are motionless to each other (Figure 3).
Then for the experimentally obtained value of
, giving values to q, from Eqs.10-13 we calculate
, 
, 
, 
and
. With those and formula (15) we obtain the radius 
of the surface-equivalent sphere
. Varying 
m and
, from formulas (16) and (17) wecalculate the radius 
of the forming sphere
.
Thus at
, through formulas (7) and (8), we find the binding energy 
and force of interaction
. According to (18), the binding energy 
and force of interaction 
between a the proton and the neutron will be
(19)
The Eqs.19 are valid because the forming spheres of the torus 
are situated symmetrically relative to the centre of the sphere
.
Moving on we will structure the nuclei of deuterium and tritium. The positioning of the nucleons in them must correspond to the principle of minimum potential energy. Due to the mass defect in the nucleus, the potential energy of the interaction in the atomic nuclei is calculated by the following formula [2]:
(20)
Here 
is the number of the protons, 
– number of the neutrons; 
kg and 
kg are the masses of the proton and of the neutron, respectively [18] and 
is the mass of the considered nucleus; 
m·s–1 is the velocity of the light in flat vacuum [18]. We will find the values of 
according to our method with formula (8), comparing them with the values obtained through formula (20).
Figure 3. Cross section of reduced model of the proton-neutron system.
Let us consider the proton in a unbound state as a torus
. Its radius according to [19] is 
m. If 
is the center of the forming circle of
, then we denote as 
the radius of this circle and with
.
Using formulas (10) and (13) for different values of the radius of the circle of the empty part q we find
, 
, 
and
. We also find the volume 
of the neutron in a free state since we know its mass 
and we have assumed that
. All the data are given in Table 1" target="_self"> Table 1.
In Table 1 we have denoted with 
the radius of the forming circle of the torus of the proton and with 
the radius of the central circle of the torus of the proton.
Due to the mass defect in atomic nuclei the volumes of the nucleons within them change. We will assume however that the volume mass densities of the proton and neutron do not change, i.e.
.
4. MODEL OF THE DEUTERON
We model the nucleus of deuterium as two concentric toroids inlaid one in the other. The internal torus 
corresponds to the proton and the external one 
corresponds to the neutron with 
and 
being at a distance 
m. This configuration provides symmetry relative to the center of masses O of the deuteron placed in its geometrical center (Figure 4). From that follows the stability of the deuteron
.
We assume that both tori spin in the same direction with a constant angular velocity 
around the line z going through their common center 
and perpendicular to their plane of rotation.
Table 1. Size, volume and mass density of the proton and neutron.
Figure 4. Cross-section of the model of the deuteron.
From this structure follows that the spin 
of the deuteron will be a sum of the spins 
of the proton and neutron, i.e. the spin of the deuteron is
, which has been experimentally obtained.
The nucleons, binding to each other within the deuteron almost double their total mass relative to their separate masses and as a consequence the centrifugal force increases. From this follows that the radius of the deuteron is larger than the radii of the nucleons and as has been experimentally determined it is 
m [18].
Let us assume that the charge of each of the nucleons is distributed parallel along a circle with a center O. Then a circular current appears and the magnetic moments of the nucleons are proportional to the magnitude of the charges and their angular velocity but also to the square of their distance from the center of rotation. The inlaying of the proton within the neutron increases the radius of the neutron more relative to the increased radius of the proton. This explains why the sum of the magnetic moments of the proton 
JT–1 and neutron 
JT–1 is larger than the magnetic moment of the deuteron
JT–1 [18].
Let 
and 
be the centers of the forming circles of 
and 
and let the radii of those circles be 
and 
correspondingly. We denote as 
and 
the radii of the central circles of the tori 
and 
(Figure 4).
In this case 
and
, also 
and
. Based on the method we described in Section 3 using mass volume density 
of the nucleons from Table 1 we find
, 
, 
and
. In formula (12) we consider the mass defect, proportionally for the masses of the proton and neutron. The experimentally obtained mass of the deuteron is 
kg [18]. Then, according to formula (20), the binding energy of the deuteron is 
J.
This value we confirm at different values of the distance 
between the tori (Table 2) for 
with the corresponding formula from (19). From (19) we also obtain the force of interaction 
for the deuteron.
In Table 2 we have denoted with 
and 
respectively the radius of the forming circle of the torus of the neutron and the proton, with 
and 
respecttively the radius of the central circle of the torus of the neutron and the proton and with 
the force of interaction between the nucleons in the deuteron.
5. MODEL OF THE TRITON
The triton is obtained structurally from the deuteron by adding one more neutron. The second neutron tries to take over the place of the first one.
The neutron, although it can be assumed electrically neutral, i.e. with a common charge n = 0, has an internal electric structure, its negative charge, as opposed to the positive one is distributed primarily at its surface [20,21].
Then because of the repulsion between the tori 
and 
the central circles of the nucleons will be situated symmetrically in two planes parallel to the plane in which the central circle of proton torus 
is situated. At that 
and 
will be at the same distance 
m from
. This configuration provides symmetry relative to the center of masses (geometrical center) O of the triton (Figure 5). From this follows the relative stability of the triton 
y.
Let us denote as 
the centers of the tori 
and as
—the center of
. Then the center 
and the points 
are on the same line z, perpendicular to their plane of rotation.
The tori 
and 
rotate around z with a constant angular velocity 
in the same direction (e.g. clockwise) and 
will rotate with the same velocity in the opposite direction (counterclockwise), i.e. with a velocity
. Thus we obtain that the spin of the triton is 
which is experimentally confirmed.
Table 2. Size of the nucleons and force of interaction in the deuteron.
Figure 5. Cross-section of the model of the triton.
The neutron that rotates in an opposite direction will decrease the centrifugal force caused by the nucleons rotating in the same direction. Therefore the radius of the triton
. According to experiments 
m [22].
The magnetic moments caused by the redistributed charges in the neutrons cancel each other out. Thus the magnetic moment of the triton is caused only by the charge of the proton. As a consequence of the increased radius of the proton relative to the free state the magnetic moment of the triton 
JT–1 is larger than the magnetic moment of the proton 
JT–1, as shown by experimental data [18].
Let 
and 
be the centers of the forming circles respectively of 
and 
with corresponding radii 
and 
. We denote 
and for
,
.
In this case 
and if 
is the radius of the triton, then
.
Due to the central symmetry of the charge of the proton we can assume that all of its charge 
is concentrated in the geometrical center O. Because of that we model the proton as a sphere 
with a center O and radius
, equivalent in surface are to
. At that 
is charged with a charge
, which is centrally symmetrical and can be redistributed.
With 
(
) we denote the tori that are equivalent in surface to 
with centers 
and centers of the forming circles 
(Figure 6). At that 
and 
have the same distance 
between them as 
and
. With 
we denoted the forming sphere of
.
We assume the point O to be stationary relative to the inertial reference frame J. We introduce two solid noninertial reference systems 
and
, that rotate with the constant angular velocities 
and 
respectively of 
and 
relative to J.
The point O is a center of the coordinate system 
stationary connected with the reference system 
relative to which the spheres 
and 
are stationary to each other (Figure 6).
The two tori 
and 
are symmetrical relative to the sphere
. Then it is enough to study the electrostatic interaction only between one of them and
. We assume with some approximation that between 
and 
there is no electrostatic interaction since their total charges are zero. Besides that, to simplify the calculations, we assume that the distance between the surfaces of each of the tori is the same and equal to
. Then the distance
.
Since
, using the model we revealed in Section 3, with the volume mass density 
of the nucleons from Table 1 we find
, 
and
,
. In formula (12) we consider the mass defect proportionally for the masses of the proton and two neutrons. The experimenttally obtained mass of the triton is 
kg [18].
Then, according to formula (20) the binding energy of the triton is 
J. This value we confirm for different values of the distance 
between the tori (Table 3) for
, 
or 
using the corresponding formula from (19). From (19) we also obtain the force of interaction for the triton 
for 
or
.
In Table 3 we have denoted with 
and 
respectively the radius of the forming circle of the torus of the neutron and the proton, with 
and 
respectively the radius of the central circle of the torus of the neutron and the proton and with 
the force of interaction between the nucleons in the triton.
We should note that by determining the value of
, we can also vary h, where in this case is fulfilled 
.
6. DISCUSSION
In [12] considering the nucleons as tori we theoreticcally determine the potential energy and the force of interaction in the systems: proton-neutron, proton-proton and proton-neutron-proton, which we derive using experimentally obtained results for the radii and the masses of the nucleons in unbound condition.
Figure 6. Cross-section of the reduced model of the triton.
Table 3. Size of the nucleons and force of interaction in the triton.
Using our method, we have shown that the electromagnetic forces for the proton-neutron pair are quite strong (in the order of the nuclear forces) an short-ranged. This suggests that the binding energy of the nucleons have electromagnetic nature. We can also explain other basic experimental data like stability, radius, magnetic moment and spin of the nuclei.
In this paper, we concretize the general results obtained in [12] for the nuclei of deuterium and tritium.
7. CONCLUSIONS
Nuclear physics bases its knowledge on experiments and has numerous different contradicting models. Considering nucleons as tori and modeling the deuteron and triton, we obtain and explain their basic experimentally obtained characteristics and also obtain new characteristics of these nuclei.
Our model can also be applied for more complicated atomic nuclei. Based on particular charge and current configurations to it can be considered the interaction betweennucleons in electrodynamic aspect; to be find analytical expressions for the magnetic moments, to determine the angular velocity of nucleons, the linear velocity at particular points on their surface, etc.; to explain excited states of the nuclei; to be find out the potential of the electromagnetic field generated by atomic nuclei and to calculate their quadrupole moments for the deuteron and the other nuclei.
8. ACKNOWLEDGEMENTS
The authors express their gratitude to the computer specialist Stefan Bozhkov, who with the help of Wolfram Mathematica 7.0 performs the calculations in this article. The results of the present studies are published with the financial support of the Fund for Scientific Research with the Ministry of Education and Science under contract DTC No. 02/35.