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1
THE RADIUS OF THE PROTON
IN THE SELF-CONSISTENT MODEL
Sergey G. Fedosin
e-mail: intelli@list.ru
Perm, Perm Region, Russia
Sviazeva Str. 22-79
Abstract
Based on the notion of strong gravitation, acting at the level of elementary
particles, and on the equality of the magnetic moment of the proton and the limiting
magnetic moment of the rotating non-uniformly charged ball, the radius of the proton
is found, which conforms to the experimental data. At the same time the dependence is
derived of distribution of the mass and charge density inside the proton. The ratio of
the density in the center of the proton to the average density is found, which equals
1.57.
Keywords: strong gravitation; de Broglie waves; magnetic moment; proton radius.
2
1 Introduction
Since the discovery of the proton in 1917 the question arose how to
determine the radius of this elementary particle. There are many theoretical
models to estimate the radius of the proton. Most of these models is associated
with the concept of the electromagnetic form factors as the amendment by
which the scattering amplitude of particles by proton is different from the
scattering amplitude by a point particle. The calculation of the form factors is
complex and requires taking into account many factors, including the radial
density distribution of charge and magnetic moment, the dynamics of quarks,
partons and virtual particles. There may be a variety of approaches – scattering
theory, chiral perturbation theory, lattice QCD, etc., description of which can
be found in [1], [2]. Form factors are determined from scattering experiments,
depend on the energy of the interacting particles, and allow us to find the root
mean square of the charge distribution and magnetic moment as a measure of
particle’s size. Information on the radius of the proton can be extracted from
the analysis of the Lamb shift in hydrogen and in a coupled system of a proton
and a negative muon [3].
2 Other estimates of proton radius
Consider some simple methods for determining the radius of the proton.
One of them is based on the fact that in the particles, when they are excited,
standing electromagnetic waves emerge. The maximum energy of these
standing waves does not exceed the rest energy in order to avoid the decay of
particles. From this it can be derived that the de Broglie waves are
electromagnetic oscillations, detectable in the laboratory frame in the
interaction of moving particles. To describe these oscillations it is necessary to
apply the Lorentz transformations to the standing waves inside the particles and
to find their form in the laboratory reference frame [4], [5].
In the simplest case the spherical standing waves are modeled by two
waves, one of which runs from the center to the surface of the particle and the
other at the same time is moving backwards. We can assume that in the
direction of a specified axis, for example OX , there are two counter-
propagating waves of the following form:
)sin( 101
ϕ
ω
+
′′
−
′′
=xKtUU , )sin( 202
ϕ
ω
+
′′
+
′′
=xKtUU ,
3
⎟
⎠
⎞
⎜
⎝
⎛+
+
′′
⎟
⎠
⎞
⎜
⎝
⎛−
−
′′
=+= 2
sin
2
cos2 2121
021
ϕϕ
ω
ϕϕ
txKUUUU , (1)
here 1
ϕ
, 2
ϕ
are the initial phases of the oscillations with 0tx
′′
==
, 0
U is
the amplitude of the periodic function,
ω
′ and
K
′ denote the angular
frequency and wave number and the primes over the variables mean that they
are considered in the rest frame of the particle.
As U any periodic function can be used, which satisfies the wave equation.
For example, it can be the strength or the field potential of the wave. The
phases of the waves in (1) must be shifted to
π
for emerging of the standing
wave. If 1
ϕ
π
=, 20
ϕ
=, then in the center of the particle with 0x′= there will
be always a node as the absence of visible oscillations, and (1) becomes as
follows:
()()
txKUU ′′′′
=
ω
cossin2 0. (2)
As a result of oscillations (2) velocities of charges of the particle substance
and the field potentials can periodically change inside the particle. This leads
inevitably to periodic oscillations of the field potentials also outside the particle
in the surrounding space.
Now we shall assume that the particle moves together with its standing
wave along the axis OX in the laboratory reference frame at the velocity u.
How are the field oscillations modified inside and outside the particle with
respect to its movement? We should express in (2) the primed coordinates and
the time inside the moving particle through the coordinates and the time in the
laboratory reference frame using the Lorentz transformations ( c refers to the
speed of light):
22
2
1
/
cu
cxut
t−
−
=
′, 22
1cu
tux
x
−
−
=
′, yy =
′, zz =
′,
4
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
′
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
′
=22
2
22
01
)/(
cos
1
)(
sin2
cu
cxut
cu
tuxK
UU
ω
. (3)
From (3) we see that as a result of displacement of the standing wave with
the particle for the external motionless observer in the laboratory frame the
wavelength and the frequency will change. More precisely, on the observed
wave additional antinodes appear, with a wavelength between them, differing
from the wavelength 2
K
π
λ
′=′ in the reference frame of the particle. We shall
stop the wave (3) for a moment with 0t= and shall find the wavelengths as
the spatial separation between the points of the wave in the same phase. When
0x= the sine in (3) will be zero, while when 1
x
λ
= the phase of the sine will
change from 0 to 2
π
. Hence we obtain:
()
π
λ
2sin
1
sin 22
1=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
′
cu
K, 22
22
11
12 cu
K
cu −
′
=
′
−
=
λ
π
λ
. (4)
Similarly for the wavelength of the cosine in (3) we find:
()
π
λω
2cos
1
/
cos 22
2
2−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
′
−
cu
cu , u
cuc
ω
π
λ
′
−
=
222
2
12 . (5)
We shall now estimate the temporal separation between the points of the
wave in one phase with 0x=, considering this separation as the corresponding
period of the wave:
()
π
2sin
1
sin 22
1−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
′
−
cu
TuK , uK
cu
T′
−
=
22
1
12
π
. (6)
5
()
π
ω
2cos
1
cos 22
2=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
′
cu
T, 22
21cuTT −
′
=. (7)
From (4) − (7) we obtain the following expressions for the velocities:
u
T=
1
1
λ
, b
u
c
T
υ
λ
==
2
2
2. (8)
As we see from (8) the oscillations of the wave (3) associated with the
cosine, are propagating at the phase velocity of de Broglie b
υ
. Besides, the
oscillations of the wave (3) associated with sine, move in space at the same
velocity u as the particle itself. The wavelength 2
λ
in (5) can be transformed
so as to bring it to the standard form for the de Broglie wavelength. We shall
associate the angular frequency of the oscillations inside the particle, similarly
to the electromagnetic wave, with the energy of oscillations: W
ω
′′
==, where
2
h
π
== is Dirac constant, h is Planck constant. This gives the following:
uW
cuch
′
−
=
222
2
1
λ
. (9)
Similarly from (4) we have:
W
cuch
′
−
=
22
1
1
λ
. (10)
In the limiting case when the oscillation energy is compared with the rest
energy of the particle, 2
Wmc
ω
′′
===, from (9) it follows:
22
2
1
f
huc
h
mu
λ
−
==
℘, (11)
6
where m is the mass of particle, ℘ is relativistic momentum of the particle.
The formula (11) defines de Broglie wavelength with the help of particle
momentum. We shall note that de Broglie wrote (11) on condition that the
energy of the particle
2
22
1
mc
W
uc
=
−
is equal to the energy of the wave
accompanying the particle. According to the obtained expression (9), the
wavelength 2
λ
must be present in the particle also at low excitation energy
.W′ In this case as the excitation energy decreases, the wavelength should
increase.
As a rule in the experiments only 2bf
h
λλ
==
℘ is found from (11), and not
the wavelength 2
λ
from (9). This can occur because among the number of
interacting particles at the same time there are particles with different
excitation energies W′ and different 2
λ
, so that the wave phenomena are
blurred. The same is true for the waves with wavelength 1
λ
in (10). Only for
the most actively interacting particles, the excitation energies W′ of which are
close to the rest energy of the particles, the limiting value of the wavelength is
reached equal to the de Broglie wavelength. Thus this wavelength is revealed
in the experiment. When 2
Wmc
ω
′′
=== we can also predict for the particles
the wave phenomena with the critical wavelength
22
1
1
f
huc
mc
λ
−
=. In
particular, c
h
mc
λ
= is the Compton wavelength, discovered in the Compton
effect. According to our point of view, emerging of de Broglie wave should be
treated as a purely relativistic effect, which arises as a consequence of the
Lorentz transformation of the standing wave, moving with the particle.
As a result, we have to assume that the wave-particle duality is realized in
full only in those particular particles, the excitation energies of which reach
their rest energies. In this case the difference of particles and field quanta, if
they are treated from the point of view of their wave properties, becomes
7
minimal. At low excitation energies the particles can not emit their energy
greatly, and the amplitudes of the oscillations of the field potentials near the
particles are small. Then the particles would interact with each other not in the
wave way, but rather in the usual way, and the wave phenomena become
invisible.
If we assume that the length of the standing wave is equal to 2R
λ
′= ,
where
R
is the radius of the proton, then from the equality of the wave energy
and the rest energy of the proton we obtain:
2
cc
R
ν
λ
′==
′, 2
2
p
hc
Mc h
R
ν
′
==, 16
6.6 10
2p
h
RMc
−
==⋅
m,
here
ν
′ is the oscillation frequency, p
M
is the mass of the proton.
Another way to estimate the radius of the proton assumes that the difference
between the rest energy of the neutron and the proton is due to the electrical
energy of the proton charge. In this case, it should be:
2
2
0
()
4
np
ke
MMc
R
πε
−= , (12)
where n
M
is the mass of the neutron, e is the elementary charge, 0
ε
is the
vacuum permittivity.
In (12) for the case of the uniform distribution of the charge in the volume
of the proton 0.6k=, as a result the estimation of the proton radius gives the
value of 16
6.68 10R−
=⋅ m.
In [6] and [7], the radius of the proton was found from the condition that the
limiting angular momentum of the strong gravitation field inside the proton is
equal in magnitude to the spin of the proton. This leads to the following
formula:
16
2
56.7 10
21
p
ΓM
Rc
−
==⋅ m. (13)
8
In (13) the strong gravitational constant
Γ
is used. According to [4], this
constant is determined from the equation of electric force and the force from
the strong gravitation field, acting in the hydrogen atom on the electron with
the mass e
M
, which is located in the ground state on the Bohr radius B
R
:
2
22
0
4
pe
BB
Γ
MM
e
RR
πε
=,
2
29
0
1.514 10
4pe
e
ΓMM
πε
==⋅
m3·kg–1·s–2, (14)
In addition to the attractive forces from gravitation and the charges of the
nucleus and the electron, in the hydrogen atom the electron substance in the
form of the rotating disc is influenced by the repulsive forces acting away from
the nucleus. One of these forces is the electric force of repulsion of the charged
substance of the electron cloud from itself. In the rotating non-inertial reference
frame in which an arbitrary part of the electron substance is at rest, there is also
the force of inertia in the form of the centrifugal force, which depends on the
velocity of rotation of this substance around the nucleus. In the first
approximation, these forces are equal in magnitude, which leads to (14).
We shall remind that the idea of strong gravitation was introduced into
science in the works of Abdus Salam and his colleagues [8], [9] as the
alternative explanation of the strong interaction of the particles. Assuming that
hadrons can be represented as Kerr-Newman black holes, they estimated the
strong gravitational constant as 27
6.7 10⋅ m3·kg–1·s–2.
With the help of the strong gravitation constant (14) we can express the fine
structure constant:
1
137.035999
pe
ΓMM
c
α
==
=.
Another estimate of the radius of the proton follows from the equality of the
rest energy and the absolute value of the total energy, which, taking into
account the virial theorem, is approximately equal to the half of the absolute
value of the strong gravitation energy associated with the proton [4]:
9
2
2
2
p
p
k
Γ
M
Mc
R
=. (15)
If we take 0.6k= for the case of the uniform mass distribution, then from
(15) it follows that 16
8.4 10R−
=⋅ m.
All of the above estimates are based on the classical approach to the proton
as to the material object of small size in the form of the ball with the radius
R
.
It is assumed that the strong gravitation acts at the level of elementary particles
in the same way as ordinary gravitation at the level of planets and stars.
In the Standard model of elementary particles and in quantum
chromodynamics it is assumed that the nucleons and other hadrons consist of
quarks, and baryons have three quarks, while mesons have two quarks. Instead
of the strong gravitation, the action of gluon fields is assumed to hold the
quarks in hadrons. Quarks are considered to be charged elementary particles,
therefore as the radius of the proton the charge and magnetic root mean square
radii are considered. These radii are determined by the electric and magnetic
interactions of the proton and can differ from each other.
The estimate of the proton charge radius can be made with the help of the
experiments on the scattering of charged particles on the proton target [10]. In
such experiments the total cross sections of interaction of the particles
σ
are
found. For the case of the protons scattering on nucleons with energies more
than 10 GeV we can assume that 2
R
σπ
=, and 30
3.8 10
σ
−
=⋅ m
2. Hence we
obtain 16
7.8 10R−
=⋅ m.
3 The self-consistent model
Our aim will be to find a more exact value of the radius of the proton by
using classical methods. In the calculations we shall use only the tabular data
on the mass, charge and magnetic moment of the proton. The proton will be
considered from the standpoint of the theory of infinite nesting of matter [11],
in which the analogue of the proton at the level of stars is a magnetar or a
charged neutron star with a very large magnetic and gravitational field.
Similarly to the magnetar, the substance of the proton must be magnetized and
held by a strong gravitation field.
10
To take into account the non-uniformity of the substance density inside the
proton we shall use the simple formula in which the substance density changes
linearly increasing to the center:
(1 )
cAr
ρρ
=−, (16)
where c
ρ
is the central density,
r
is the current radius, R
A1
0<< is the
coefficient which should be determined.
Formula (16) should be considered as a first approximation to the actual
distribution of the density of matter inside the proton. Approximate linear
dependence of the density of matter in neutron stars has been shown in [12],
and we assume that this is also true for the proton as an analogue of the neutron
stars.
To estimate the values
A
and the radius
R
we shall consider the integral
for the proton mass in the spherical coordinates:
3
243
(1 ) sin 1
34
c
pc
R
A
R
MArrdrdd
πρ
ρθθϕ
⎛⎞
=− = −
⎜⎟
⎝⎠
∫. (17)
For accurate calculation of state of neutron stars, and thus protons as their
analogues we should consider the curvature of spacetime in a strong
gravitational field, as well as the contribution of the energy of the gravitational
field to the total mass-energy. We shall assume that in (16), in dependency of
matter density on the radius all relativistic effects are taken into account, and
the mass of the proton (17) is the gravitational mass from the point of view of a
distant observer.
In (17) there are three unknown quantities, to obtain which two more
equations are required. We shall assume the virial theorem to be valid and
equate the rest energy of the proton to the half of the absolute value of the
energy of the static field of strong gravitation:
22
00
11
216
p
M
cdVGdV
Γ
επ
∞∞
=− =
∫∫
, (18)
11
where
2
8
G
Γ
επ
=− is the energy density of the strong gravitation field
according to [4], G is the gravitational acceleration or strength of gravitational
field.
In (18), the integration of the energy density of the field should be done
both inside and outside of the proton. The value G inside the proton can be
conveniently found by integrating the equation for the strong gravitation field
ρ
π
Γ4−=⋅∇ G, which is part of the equations of the Lorentz-invariant theory
of gravitation [13]. After integrating over the spherical volume with the radius
R
r
≤, and then using the Gauss theorem, that is making transition to
integrating over the area of the indicated sphere inside this proton, in view of
(17) we obtain:
2
44
ii i
dV dS r G
Γ
dV
ππρ
∇⋅ = ⋅ = =−
∫∫ ∫
v
GGn ,
⎟
⎠
⎞
⎜
⎝
⎛−−= 4
3
1
3
4rA
Γcr
Gi
ρπ
. (19)
Outside the proton the gravitational acceleration is equal to:
3
r
MΓpr
Go−= . (20)
Substituting (19) and (20) in (18), we obtain the relation:
R
MΓ
RARA
RΓcM p
cp 41123645
1
4
2
22
5222 +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛+−=
ρπ
. (21)
In (21) we can eliminate the value c
ρ
using (17), which give the
dependence of A on R in the form of the quadratic equation:
12
0
9
4
15
2
3
2
36
7
414
222
22 =−+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛+−+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−cR
MΓ
cR
MΓ
RA
cR
MΓ
RA ppp .
The analysis of this equation shows that it has the following solution:
27
45360
13
9453
2
36
7
2
222
2
cR
MΓ
MΓcRMΓ
cR
MΓ
RA
p
ppp
−
−+−
=, (22)
on condition that when
213
0.3 0.371
35
p
Rc
ΓM
<<≈
, then accordingly
01AR<<
.
We shall now turn to the magnetic moment of the proton. As in [4], we
assume that the magnetic moment of the proton is equal to the magnetic
moment, which is formed due to the maximum rapid rotation of the charged
substance of the proton. In spherical coordinates, the magnetic moment can be
approximately calculated as the sum of the elementary magnetic moments of
the separate rings with their radius
θ
sinr, which have the magnetic moment
due to the current di flowing in them from the rotation of the charge:
22 22
5
22 2
sin sin
45
sin (1 ) sin 1 .
15 6
mm
Lqc
qc
dq
PdP r di r dt
R
dAR
rArrdrd
dt
πθπθ
πωρ
ϕ
πθρ θθ
== = =
⎛⎞
=− =−
⎜⎟
⎝⎠
∫∫ ∫
∫
(23)
The angular velocity L
d
dt
ϕ
ω
= of the maximum rotation of the proton can
be found from the condition of limiting rotation, with the equality of the
centripetal force and the gravitation force at the equator: R
R
MГ
L
p2
2
ω
=. Further
13
we believe that for the charge density and the substance density the equation
pc
qc
M
e
=
ρ
ρ
holds, and we use (17). This gives the following:
()
()
RA
RARMГe
Pp
m3430
564
−
−
=. (24)
4 Conclusions
The relation (24) together with (22) allow us to find the radius of the proton
16
8.73 10R−
=⋅ m, as well as the value 0.48
AR
=. From (17) we obtain then the
central substance density 17
9.4 10
c
ρ
=⋅ kg/m3, which exceeds the average
density of the proton 1.57 times. The maximum angular velocity of rotation of
the proton in view of (23) is equal to 23
6.17 10
L
ω
=⋅
rad/s. At the same time, if
the spin of the proton in the approximation of the uniform density of substance
would be equal to the standard value for the spin of the fermion
2
0.4 2
p
LMR
ω
==
=, then the angular velocity of rotation 23
1.03 10
ω
=⋅
rad/s
would correspond to this spin.
For comparison with the experimental data we shall point to the results of
calculations of electron scattering from [14], where the charge radius
16
8.7 10
E
R−
=⋅ m is obtained taking into account only the scattering on
protons, 16
8.71 10
E
R−
=⋅ m – taking into account the data on the pion
scattering, and 16
8.8 10
E
R−
=⋅ m – taking into account the data on the neutron
scattering. In [3] the charge radius 16
8.4184 10
E
R−
=⋅
m was found in the study
of the coupled system of the proton and the negative muon. The study of the
scattering cross section of polarized photons by protons [15] gives the charge
radius 16
8.75 10
E
R−
=⋅ m and the magnetic radius 16
8.67 10
M
R−
=⋅ m. The
charge radius 16
8.77 10
E
R−
=⋅ m and the magnetic radius 16
7.77 10
M
R−
=⋅ m
of the proton are listed on the site of Particle data group [16]. In the database
CODATA [17] the proton charge radius is equal to 16
8.775 10
E
R−
=⋅
m.
14
The value 16
8.73 10R−
=⋅ m obtained in the framework of the self-consistent
model is close to the experimental values of the radius of the proton, which
confirms the possibility of applying the idea of strong gravitation to describe
the strong interaction of elementary particles.
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