Electromagnetic nature of nuclear forces and the toroid structure of the helion and the alpha particle ()
1. INTRODUCTION
There are two basic models in the theory of elementary particles: the standard model [1-3] and the helicon model [4-7].
In [8] and [9] the toroid model of nucleons has been proposed, which is in certain contradiction with the standard model, however in perfect agreement with the helicon model.
Every nucleon is modelled with an imaginary torus, which rotates with constant angular velocity around an axis z passing through its mass centre (the geometric centre) o and perpendicular to the rotation plane of its central circle. From quantum mechanical point of view, the nucleon is not a localized object in configuration space. Therefore, our model is valid in a formal heuristic sense a la Niels Bohr similar to his model for the electron in the hydrogen atom, with which he obtained good results for its spectrum within the framework of the old quantum mechanics.
Additional reasoning to consider the nucleons as spatially extended objects is the fact that they are located close to each other in the atomic nuclei. Thus, we assume that they are tori and their electric charge can be redistributed. Essentially, this idea does not contradict to the quark substructure model of nucleons, and it enables us to determine the electrostatic interaction between them. Based on the exact analytical expressions for the electrostatic interaction between two charged conducting spheres with arbitrary radii and charges derived for the first time in [10], we conclude in [8] that strong interactions are electromagnetic in nature.
In [9], we ascertain that the experimentally determined potential binding energy between nucleons in the deuteron and triton can be obtained taking into account the electrostatic interaction only. To do that, the values of the experimentally known radii, and masses of the nucleons and the corresponding nuclei have been utilized. In [9], the volume and the mass density of nucleons, as well as their interaction force in the deuteron and the triton have been calculated—important results in nuclear physics presented for the first time.
Applying the method developed in [8,9], we calculate in the present paper the potential energy and the interaction force between the nucleons in the helium-3 and helium-4 nuclei. Basic experimentally measured characteristics such as stability, spin, radius and magnetic moment of the helion and the alpha particle are also explained.
According to R. Feynman’s conjecture at distances less than 10−15 m either Coulomb’s law is not valid, or electrons and protons cannot be considered as point-like particles [11]. In accordance with this conjecture, we assume that the distance within the corresponding protonneutron pairs is less 10−15 m.
2. TOROID MODEL OF NUCLEONS
In [8,9] the nucleons are modeled as tori. Moreover, we assume that they rotate with constant angular velocity around an axis z passing through their mass (geometric) centre o and perpendicular to the rotation plane of their central circle (Figure 1).
Next, we study the system consisting of a proton and a neutron and determine the electrostatic interaction between them.
The tori corresponding to the proton and the neutron are denoted by
and
, while their centres are denoted by
and
, respectively. We assume that the central circles of
and
lie in parallel or merging planes and rotate in the same or opposite direction with the same angular velocity
around the line z, which passes through
and
being perpendicular to the rotation plane. Then, if
, it follows that
(Figure 2).
Let
and
be the centres of the forming circles of the tori
and
, respectively, and
,
be the radii of the central circles of
and
. We assume that the distance between
and
is
m.
According to the available experimental data the proton radius
is not greater than the neutron radius
. That is why we must have
.
Let
and
be the radii of the forming circles
and
, respectively. It is clear that
and
. Moreover,
and
,(1)
where
and
are the radii of the proton and the neutron in the configuration shown in Figure 2. It is worthwhile to note that in some nuclei
,
,
and
,
,
may take different values.
Since
, simple geometric consideration yields

Figure 2. Cross section of the system containing a proton and a neutron.
. (2)
Obviously,
.
We assume that the mass and number densities of the proton and the neutron are equal, that is
. If
and
are the masses of
and
, then
and
.(3)
According to [12], the volumes of the tori
and
are
and
, respectively. Therefore, from
and from Eq.3 it follows that
. s are experimentally (4)
Eqs.3 and 4 contain the proton and the neutron mass whose values are experimentally known and can be readily substituted.
Let he radius of the of the empty part of the circle with radius equal to that of the proton
. Then
. (5)
In order to apply the results from [10] derived for the case of spheres, we remodel the tori as in [8,9].
Due to the spherical symmetry of the proton charge [13], we can assume that all its charge
is concentrated in the geometric centre
of the torus
.
We remodel the proton with a sphere
having the same centre
lying on the axis
, such that its area is equal to that of the torus
. Moreover, the charge
of the sphere is spherically symmetric and it can be redistributed.
According to [12] the area
of the torus
is
. (6)
Since the areas of the torus
and the sphere
are equal, from Eq.6 it follows that the radius
of
is
. src="10-8301676\cf42bf70-0c1d-4516-8777-95359cc6d547.jpg" width="22.8" height="28.5 " /> whose torus
has equivalent area and the same centre
. The distance
between its surface and that of
is the same as the distance between
and
(Figure 3).
Let
be the sphere, whose central circle is forming the torus
. We denote the centre of
with
so that
, and let
be the radius of
. If
, then
and
implies
, that is
. (8)

Figure 3. Cross section of the reduced model of the proton-neutron system.
Clearly,
and
.
The equality of the areas
and
of the tori
and
, together with Eq.6 yields
Eq.6 yields
. d6c53ced-871f-4123-9fb1-02cf79648fda.jpg" width="12.825" height="22.8 " /> of the spheres
whose total area is equal to that of the torus
, the following relation
holds. Therefore
. (10)
Let us assume that the centre
is fixed with respect to some inertial reference frame
. Next, we introduce a rigid inertial coordinate system
, which rotates with the constant angular velocity
of the torus
rotation with respect to
. The point
is the centre of the coordinate system
, which is firmly fixed with
and the spheres
and
are at rest one against the other (Figure 3).
Thus, taking into account the experimentally found value for
(
is the radius of the considering nucleus), assigning different values for
and varying
m and
, from Eqs.1-5 we can find
,
and
,
,
. Combining with Eq.7 we determine the radius
of the sphere
with equivalent area. From Eqs.8 and 9 we determine the radius
of the forming sphere
.15.425" height="29.925 " /> and taking into account ([10], Eqs.10 and 11 therein), we can determine the interaction force and taking into account ([10], Eqs.10 and 11 therein), we can determine the interaction force
and the binding energy
. According to Eq.10 the binding energy
between the proton and the neutron become
(11)
Eq.11 hold because the forming spheres of the torus
are located symmetrically with respect to the centre of the sphere
.oton in a free state as the torus
According to [14] its radius is
m. Let
is the centre of forming circle of the torus
and let
is the radius of this circle. Moreover,
.
Utilizing expressions (1) and (5) for different values of the radius
of the empty part of the circle with radius
, we have calculated in [9] the quantities
, 
and the volume of the proton torus 
[12]. The mass of the proton in a free state is
kg [15]. This can be used to calculate
.
These results are depicted in Table 1 for various values of
.
Since, the mass of the neutron
kg [15] is also known, its volume
in a free state can be determined, where
has been taken into account as already mentioned in Section 2.
Due to the mass defect in the atomic nuclei, the volumes of the nucleons change. We conjecture however, that the mass densities of the proton and the neutron in all nuclei do not change
. Based on the values presented in Table 1, we ascertain the potential binding energy and calculate the interaction force in the deuteron and the triton in [9].
We model the deuteron as two immersed concentric tori. The inner one
corresponds to the proton, while the outer
corresponds to the neutron, such that the distance between
and
is of the order of
m. In addition, we assume that the two tori rotate in the same direction with constant angular velocity
around a straight line
passing through their common centre and perpendicular to the rotation plane.
We model the triton with three tori whose central circles

Table 1. Dimensions, volume and mass density of the proton.
lie in three parallel planes. The tori
and
representing the neutrons are located symmetrically on both sides of the torus
representing the proton. The distance between the neutron tori
,
and the proton one
is the same and of the order of
m. The tori
and
rotate around an axis
, which passes through their centres and is perpendicular to the rotation plane. The tori
and
rotate with constant angular velocity
in the same direction (clockwise, for example), while
rotates with the same angular velocity in the opposite direction.
In the subsequent exposition we will model the helium-3 and the helium-4 nuclei. The nucleon disposition must comply with the principle of the minimum of potential energy. Taking into account the mass defect in atomic nuclei, the potential energy of interaction can be calculated according to the following formula [16]
. (12)
Here
is the number of protons,
is the number of neutrons, while
kg and
kg are the proton and the neutron mass, respectively [15],
is the mass of the nucleus and
ms−1 is the speed of light in vacuum [15].
Based on the values for mass densities in Table 1, we calculate the corresponding values of
for the helion and the alpha particle, we follow the procedure described in Section 2. Finally, our results will be compared with the values obtained by virtue of Eq.12.
4. MODEL OF THE HELION
The helion is a mirror nucleus of the triton and therefore its structure is analogous to that of the latter [9]. It consists of one neutron
and two protons
and
. Since the proton tori
and
repel due to the electrostatic interaction between them, their central circles will lie in two parallel planes symmetrically located on both sides of the plane in which the central circle of the neutron torus
is located. In addition, the distance between the surfaces of
,
and that of
will be the same
m (Figure 4).
This configuration ensures symmetry with respect to the mass centre (the geometric centre) of the helion. This also implies the helion stability
.
The centres
and
of the tori
,
and
, respectively lie on one axis
perpendicular to their plane of rotation with
being the mass centre of the helion. The tori
and
rotate with constant angular velocity
in a certain direction, while
rotates with the same angular velocity but in the opposite direction. From here it follows directly that the

Figure 4. Cross section of the helion model.
spin of the helion is
, which is an experimentally established fact [16].
The proton rotating in an opposite direction decreases the centrifugal force, which is originated by the two nucleons rotating in the same direction. That is why the helion radius
m [17] is smaller than that of the deuteron
m [15].
The magnetic moments due to the proton charges compensate each other. Thus, the magnetic moment of the helion is generated by the charge of the neutron only. As a consequence of the enlarged neutron radius as compared to that in a free state, the magnetic moment of the helion
JT−1 is greater in absolute value than the magnetic moment of the neutron
JT−1, which is in agreement with the experimental data [15].
Let
and
be the centres of the forming circles of
and
, respectively with radii
and
. We introduce the notations
and for
,
(Figure 4).
This means that
. Since the neutron radius is equal to the radius of the helion, it follows that 
and
.
Due to the spherical symmetry of the proton charges, we can assume that their charges
C are concentrated in their geometric centres
. On these grounds we can model the protons with spheres
having the same centres
and radii
with areas equal to the areas of
. Moreover, the charge of each
is
, which is spherically symmetric and can be redistributed.
With
we denote the torus which is equivalent in area to
with the same centre
. The distance between
and
is equal to the distance
between
and
. Let us denote: with
the sphere whose central circle is the forming circle of the torus
; with
the center of the sphere and with
the radius of the sphere (Figure 5).
We assume that the points
are at rest with respect to an inertial frame
. Let us also introduce two rigid non inertial reference frames
and
rotating with respect to
with angular velocity equal to that of the rotation of the torus
.
The point
is the centre of the coordinate system
and is fixed firmly with the reference frame
relative to which the spheres
and
are at rest one against the other (Figure 5).
The torus
is located symmetrically with respect to the two spheres
. It suffices to analyse the electrostatic interaction between
and any of the spheres. Based on the calculations presented in [8], we assume that the electrostatic interaction between the spheres
and
can be approximated with interaction between point-like charges
,
concentrated in their centres, i.e.
and
. For the sake of simplicity, we assume that the distance
between the surfaces of the protons
and
is the same as the distance between
and
, i.e.
. Therefore, we must have
(Figure 4).
Using Eqs.1-5 from Section 2, we calculate
,
and
,
,
for different values of the mass density
taken from Table 1 of Section 3 and different values of the distance
. The mass defect of the two protons and the neutron has been taken into account in Eqs.3 and 4 accordingly to the mass proportions of the nucleons. According to Eqs.7-9 we calculate
,
and the distance
between the centers of the spheres
and
.
The experimentally established mass of the helion is
kg [15]. According to Eq.12 the binding energy of the helion is

Figure 5. Cross section of the reduced model of the triton.
J. Using the corresponding Eq.
11, we confirm this value for
, where
or
. By virtue of the same Eq.11 we find the magnitude of the interaction force for the helion
, where
or
.
In Table 2,
and
denote the radii of the forming circles, while
and
are the radii of the central circles of the neutron and the proton tori in the helion, respectively. In addition,
is the interaction force between the nucleons in the helion.
It is important to note that in order to obtain the value of
, we can vary
as well for different values of the distance
, such that
holds. This implies that the distance
between the surfaces of the two protons will vary according to
.
5. MODEL OF THE ALPHA PARTICLE
Structurally, the alpha particle is obtained by adding one neutron to the helion. We assume that the centres of the protons
,
and the neutrons
,
are symmetrically located with respect to the mass centre of the nucleus. The central circles of the tori lie in parallel planes. The distance between the tori
,
(
) is the same
m, while the elongation between
,
is some small distance (Figure 6).
From this configuration it follows the stability of the alpha particle is
.
The centres
and
(
) of the tori
and
lie on one axis
perpendicular to the plane of rotation. The tori
and
rotate in the same direction around
with constant angular velocity
, while
and
rotate with the same velocity in the opposite direction. It follows that both the spin and the magnetic moment of the alpha particle are zero, which is an experimentally established fact [15,16].
Moreover, the decrease of the centrifugal force as compared to the helion implies that the radius of the

Figure 6. Cross section of the model of the alpha particle.

Table 2. Dimensions of nucleons and interaction force between them in the helion.

Table 3. Dimensions of nucleons and the interaction force in the alpha particle.
alpha particle
m [17] is smaller than the radius of the helion
m.
Let
and
be the centres of the forming circles of
and
with radii
and
. For
, we denote
and
(Figure 6).
The radius of the neutron is equal to the radius of the alpha particle, that is
and
. From the isosceles trapezoid
it follows that the distance
between the surfaces of the tori
and
is
. (13)
We assume again that the proton charges are concentrated in their geometric centres
. Similar to Section 4, we model the protons as spheres
with centres
and radii
, whose areas are equal to that of
. Moreover, the charge of
is
, which is spherically symmetric and can be redistributed.
With
, we denote the tori, whose areas are equal to the areas of
with the same centres
. The distance between the objects
and
is the same as the distance between
and
. Let us denote: with
the sphere whose central circle is the forming circle of the torus
; with
the center of the sphere and with
the radius of the sphere (Figure 6).
We further assume that the points
and
are at rest with respect to an inertial reference frame
. In addition, four firm noninertial coordinate systems
are introduced rotating at constant angular velocity
(or
) equal to the angular velocity of rotation of the neutrons
and
with respect to
. Every point
is a centre of two coordinate systems
fixed firmly with the reference frames
with respect to which the spheres
and
are at rest.
Since the total charges of the tori
and
are zero it can be assumed with good approximation that there is no electrostatic interaction between them.
Due to the symmetry, it suffices to find the interaction between any of the spheres
and the spheres
. We assume as we noted in Section 4 that the interaction between the spheres
and
is taken as point charges
, i.e.
and
. For the sake of calculative simplicity it is assumed that the distance
between the surfaces of the proton tori
and
, as well as the distance
between the surfaces of the neutron tori
and
, is the same as the one between
and
, i.e.
.
We have
and
. Therefore,
. Taking into account the mass density
of the nucleons from Table 1 of Section 3 and using Eqs.1-5 from Section 2 for different values
all parameters
,
and
,
,
can be found. In Eqs.3 and 4 the corresponding mass defect for the two protons and the two neutrons has been taken into account. According to Eqs.7-9 we calculate
,
and the distances
and
between the centres of the spheres
,
and
,
, where
is given by Eq.13.
The experimentally established mass of the alpha particle is
kg [15]. According to Eq.12 the binding energy for the alpha particle
J. Using the corresponding Eq. 11 we confirm this value for
where
or
. By virtue of the same Eq.11 we find the magnitude of the interaction force for the alpha particle
, where 
or
.
The quantities
and
in Table 3 denote the radii of the forming circles of the tori,
and
are the radii of the central circles of the neutron and proton tori, respectively and
is the interaction force between the nucleons in the alpha particle.
We would like to note that in order to obtain the value of
, the quantity
for different values of the distance
can be varied as well, such that
. It follows that the distance
between the proton surfaces will vary
. According to Eq.13 the distance
will vary, too.
6. DISCUSSION
Describing the nucleons as tori, we calculate in [8] the potential energy and the interaction force between the nucleon couples. This approach is based on recently derived for the first time analytical expressions describing the electrostatic interaction between two charged spheres in the most general case [10]. According to the above mentioned formulae one can determine the electrostatic interaction between spheres at distances much less than their radii.
Based on this method, we find that strong interactions in atomic nuclei are electromagnetic in origin. Additional finding is the fact that the electrostatic interactions between the couples proton-neutron are short range (attracting forces), while the interactions between the couples proton-proton are long range ones (repelling forces) [18].
In [9] the basic experimental results for the deuteron and the triton such as binding energy, radii, spins and magnetic moments have been explained.
In the present paper, we extend this investigation for the nuclei of helium-3 and helium-4 and confirm that our model is capable to explain all essential experimental results for the basic simple nuclei. We find new results— the volumes and the mass densities of the nuclei, as well as the force of interaction in the considered nuclei.
7. CONCLUSIONS
The presented here model can be applied for the more complicated atomic nuclei. The electrostatic interaction between nucleons changes their electric structure. Thus, all nucleons enter to various extents in interactions, which compensate each other.
Let us also note that the electromagnetic forces between nucleons depend on whether they rotate in the same or different directions that is they depend on the orientation of their spins.
Based on our research, we are confident that all available experimental data about atomic nuclei can be explained by the model proposed. Other new properties of atomic nuclei can also be found.
It is essential that we obtain the basic nuclear characteristic—the binding energy in all of the considered nuclei, using only electromagnetic interactions. The most significant conclusion from our studies is that nuclear forces are electromagnetic in origin.
8. ACKNOWLEDGEMENTS
The author would like to thank Stefan Bozhkov, for performing the calculations in the present paper on Wolfram Mathematica 7.0.
The results of the present studies are published with the financial support from the Fund for Scientific Research at the Ministry of Education and Science of Bulgaria under contract DTC No. 02/35.