A Relationship between the Subatomic Particles’ Mass and the Square of the Magnetic Flux Quantum Provides a Novel Significance of the Mass and the Wave Function

1. Relationships Between the Electron Mass and the Square of the Magnetic Flux Quantum and Between the Bohr Radius and the Vacuum Permittivity

The magnetic flux quantum Φ₀ [1] is defined as following (the equation is presented in this way because of the motion of both the proton and the electron around the center of mass in the hydrogen atom)

$$e=h/2{\Phi }_0\to e^2=h^2/4{\Phi }_0^2$$

(1)

where $e$ is the elementary charge of an electron and $h$ is Planck’s constant. The electrostatic force acting on the electron at the Bohr level is

$$e^2/4\pi a_0^2{\epsilon }_0=m_e{v}_e^2/a_0$$

$$e^2=m_e{v}_e^{2}4\pi a_0{\epsilon }_0$$

(2)

where $a_0$ is the Bohr radius, ${\epsilon }_0$ is the vacuum permittivity, $m_e$ is the electron mass, and $v_e$ is the electron velocity at the Bohr radius. The electron’s angular momentum at the Bohr radius is

$${\displaystyle \frac{h}{2\pi }}=m_e{v}_e{a}_0 \to h=2\pi m_e{v}_e{a}_0$$

(3)

By substituting Eq. (3) in Eq. (1) (the squared term), we obtain

$$e^2=h^2/4{\Phi }_0^2$$

$$e^2=m_e{v}_{e}2\pi a_0\times m_e{v}_{e}2\pi a_0/4{\Phi }_0^2$$

(4)

We can then rewrite Eq. (4) as follows:

$$e^2=m_e{v}_e^{2}4\pi a_0\times (\pi a_0{m}_e/4{\Phi }_0^2)$$

(5)

By multiplying both the numerator and denominator of Eq. (5) by ${\epsilon }_0$, we have

$$e^2=m_e{v}_e^{2}4\pi a_0{\epsilon }_0(\pi a_0{m}_e/4{\epsilon }_0{\Phi }_0^2)$$

(6)

According to Eq. (2), the expression in parentheses in Eq. (6) should equal unity; thus, we obtain

$$(\pi a_0{m}_e/4{\epsilon }_0{\Phi }_0^2)=1$$

(7)

We can then substitute the following values published by the National Institute of Standards and Technology Committee on Data for Science and Technology in 2018 (NIST CODATA 2018) [2] in Eq. (7) (SI units):

$$a_0=5.291772109\times 10^{-11} \mathrm{m}$$

$${\epsilon }_0=8.8541\Pi 878\Pi 128\times 10^{-12}{ \mathrm{A}}^2{\mathrm{s}}^4{\mathrm{kg}}^{-1}{\mathrm{m}}^{-3}$$

$${\Phi }_0=2.067833848\times 10^{-15}{ \mathrm{kg} \mathrm{m}}^2{\mathrm{s}}^{-2}{\mathrm{A}}^{-1}$$

$${\Phi }_0^2=4.275936823\times 10^{-30}\Pi \mathrm{k}{\mathrm{g}}^2\Pi {\mathrm{m}}^4{\mathrm{s}}^{-4}{\mathrm{A}}^{-2}$$

$$m_e=9.1093837015\times 10^{-31}\mathrm{kg}$$

These substitutions give us the following relationship:

$${\displaystyle \frac{\pi }{4}}\times {\displaystyle \frac{5.291\Pi 772\Pi 109\Pi \times 10^{-11} \mathrm{m}}{8.854\Pi 187\Pi 812\Pi 8\times 10^{-12}{ \mathrm{A}}^2{\mathrm{s}}^4{\mathrm{kg}}^{-1}{\mathrm{m}}^{-3}}}\times {\displaystyle \frac{9.109\Pi 383\Pi 701\Pi 5\times 10^{-31} \mathrm{kg}}{4.275\Pi 936\Pi 823\times 10^{-30}\Pi \mathrm{k}{\mathrm{g}}^2{\mathrm{m}}^4{\mathrm{s}}^{-4}{\mathrm{A}}^{-2}}}=1$$

(8)

Finally, we can consider the multiplication of the vacuum permittivity and the square of the magnetic flux in the denominator of Eq. (8). We can multiply their units:

$${(\mathrm{A}}^2{\mathrm{s}}^4{\mathrm{kg}}^{-1}{\mathrm{m}}^{-3})(\mathrm{k}{\mathrm{g}}^2{\mathrm{m}}^4{\mathrm{s}}^{-4}{\mathrm{A}}^{-2})=\mathrm{k}\mathrm{g}\Pi \mathrm{m}$$

(9)

We consider this reduction further in the next section.

2. Analysis of Equation #8

• After reducing the units in the denominator of Eq. (8), we obtained units corresponding to those in the numerator in Eq. (8). The result in Eq. (9) implies two options: Either ${\Phi }_0^2$ has units of mass $[\mathrm{k}\mathrm{g}]$ or units of length $[\mathrm{m}]$ or, vice versa, ${\epsilon }_0$ has units of mass $[\mathrm{k}\mathrm{g}]$ or has units of length $[\mathrm{m}]$.The orders of magnitude in Eq. (8) indicate the scale of magnitude of these parameters. The Bohr radius scale is $10^{-11}\mathrm{m}$, and the vacuum permittivity scale is $10^{-12}{\mathrm{F} \mathrm{m}}^{-1}$. and the scale of the square of the magnetic flux quantum is $10^{-30}{[\mathrm{Wb}]}^2$, and the scale of the electron mass is $10^{-31}kg$.
• From this consideration and the two options presented in (a), it is clearly nonsensical for ${\epsilon }_0$ to have units of mass or ${\Phi }_0^2$ to have units of length. No particles with a mass on the scale of $10^{-12}kg$ or a length on the scale of $10^{-30}\mathrm{m}$ exist in the atom domain.

From these considerations, we can conclude with a high degree of certainty that the Bohr radius $a_0$and the vacuum permittivity ${\epsilon }_0$ are the same entity:

$${\epsilon }_0\equiv a_0$$

(10)

Moreover, the electron mass $m_e$ and the square of the magnetic flux quantum ${\Phi }_0^2$ are the same entity:

$$m_e\equiv {\Phi }_0^2$$

(11)

c.

The last conclusion is not as strange as it may seem, as the square of the magnetic flux quantum appears in the context of magnetic energy in a current loop and according to Albert Einstein’s special theory of relativity, energy is equivalent to mass.

* See also section #8 further on for the meaning of the vacuum permittivity ${\epsilon }_0$ from a different aspect!

3. Using the Conclusions from Sections 2(c) and 2(d) for the Electron Mass and Vacuum Permittivity Constant

By rearranging Eq. (7) for the electron mass calculation and substituting the values of ${\epsilon }_0,a_0,{\Phi }_0^2$ from NIST CODATA 2018, we obtain

$$m_e={\displaystyle \frac{4}{\pi }}\times {\displaystyle \frac{{\epsilon }_0}{a_0}}\times {\Phi }_0^2={\displaystyle \frac{4}{\pi }}\times {\displaystyle \frac{8.8541878128\times 10^{-12}\mathrm{m}}{5.291772109\times 10^{-11}\mathrm{m}}}\times 4.275936823\times 10^{-30}\mathrm{kg}=9.109383697\times 10^{-31}\mathrm{kg}$$

(12)

This value compares well with the NIST CODATA value of $m_e=9.1093837015\times 10^{-31}\mathrm{kg}$We can obtain another expression for $m_e$ from Eq. (3) with the Compton wavelength ${\lambda }_e$with $(v_e=c\alpha )$

$${\lambda }_e=h/m_{e}c=\alpha 2\pi a_0$$

(13)

where $c$ is the speed of light in vacuum and $\alpha$ is the fine structure constant. Here, we can rearrange Eq. (12) as $a_0=4{\epsilon }_0{\Phi }_0^2{\pi }^{-1}{m}_e^{-1}$ and substitute this term in Eq. (13) for another expression of $m_e$:

$$m_e=\left({\displaystyle \frac{ 8 \alpha {\epsilon }_0}{{\lambda }_e}}\right){\Phi }_0^2$$

(14)

To calculate the value of the vacuum permittivity ${\epsilon }_0$, we rewrite Eq. (7) and substitute the values of ${\Phi }_0^2,m_e,a_0$ from NIST CODATA 2018, which yields

$${\epsilon }_0={\displaystyle \frac{\pi }{4}}\times {\displaystyle \frac{m_e}{{\Phi }_0^2}}\times a_0={\displaystyle \frac{\pi }{4}}\times {\displaystyle \frac{9.1093837015\times 10^{-31}\mathrm{kg}}{4.275936823\times 10^{-30}\mathrm{kg}}}\times 5.291772109\times 10^{-11}\mathrm{m}=8.854187816\times 10^{-12}\mathrm{m}$$

(15)

This numerical value compares well with the NIST value of ${\epsilon }_0=8.8541878128\times 10^{-12}{\mathrm{A}}^2{\mathrm{s}}^4{\mathrm{kg}}^{-1}{\mathrm{m}}^{-3}$.

4. Using the Conclusions from Sections 2(c) and 2(d) for Elementary Charge $e$ and the Planck Constant h

In this subsection, we derive a new expression for the elementary charge $e$ with $m_e=4{\epsilon }_0{\Phi }_0^2{\pi }^{-1}{a}_0^{-1}$ from Eq. (12) and with the fine structure constant $\alpha$ and the speed of light in vacuum $c$.

Here, we substitute $m_e$ in Eq. (2) with $v_e^2=c^2{\alpha }^2$ to obtain an expression of the elementary charge $e$:

$$e^2=m_e{c}^2{\alpha }^{2}4\pi {\epsilon }_0{a}_0=(4{\epsilon }_0{\Phi }_0^2{\pi }^{-1}{a}_0^{-1})(c^2{\alpha }^{2}4\pi {\epsilon }_0{a}_0)$$

$$e=4c\alpha {\epsilon }_0{\Phi }_0$$

(16)

In Eq. (16), we substitute the values of ${\epsilon }_0,{\Phi }_0$, the speed of light in vacuum $c$, and the fine structure constant $\alpha$ from NIST CODATA 2018: $c=2.99792458\times 10^8 \mathrm{m}{\mathrm{s}}^{-1}$ and $\alpha =7.2973525684\times 10^{-3}$, It yields

$$e=1.602176633\times 10^{-19}\mathrm{C}$$

(17)

This value compares well with the NIST CODATA value of $e=1.602176634\times 10^{-19}\mathrm{C}$.

We can substitute $c^2{\epsilon }_0={\mu }_0^{-1}$ (where ${\mu }_0$ is the vacuum permeability) in Eq. (16) to obtain another expression for the elementary charge $e$:

$$e=4\alpha {\Phi }_0{({\epsilon }_0/{\mu }_0)}^{1/2}$$

(18)

By substituting the expression of $e$ from Eq. (16) in Eq. (1), we obtain a new expression for $h$:

$$h=8c\alpha {\epsilon }_0{\Phi }_0^2$$

(19)

By applying the values of ${\epsilon }_0,{\Phi }_0,c,\alpha$ from NIST CODATA 2018 in Eq. (19), we find

$$h=6.626070146\times 10^{-34}k{\mathrm{g} \mathrm{m}}^2{\mathrm{s}}^{-1}$$

This value compares well with the NIST value of $h=6.626070146\times 10^{-34}k{\mathrm{g} \mathrm{m}}^2{\mathrm{s}}^{-1}$. Moreover, we obtain another expression for the Planck constant $h$ by utilizing $c^2{\epsilon }_0={\mu }_0^{-1}$:

$$h=8\alpha {({\epsilon }_0/{\mu }_0)}^{1/2}{\Phi }_0^2$$

(20)

5. Radius of the Proton (Hydrogen Nucleus)

For further development, it is necessary to find the proton’s radius. According to the Physicist Arthur Beiser on the nucleus motion in his book ‘Concepts of Modern Physics’ [3], the nucleus of a hydrogen atom (proton) revolves around the center of mass shared with the electron. The rotation of both the electron and nucleus arises from considerations of momentum conservation in an isolated system and is taken into account by a computational correction called the reduced mass of the electron. The notion of reduced mass played an important part in the discovery of the deuterium and also corrects a small but definite discrepancy between the predicted wavelengths of the spectral lines of hydrogen and the measured ones. The center of mass is very close to the axis of the nucleus because of its larger mass; thus, we can assume that the trajectory depicted by the nucleus while revolving around the center of mass lies at a distance almost equivalent to the nucleus radius. We will denote this radius as the proton radius, validated in the final result. As a side note, this radius is not equivalent to the proton’s charge radius; however, there is a connection between these two parameters, which will be clarified in Section 5b. To find the proton’s radius, we will use known formulas generated for the Natural units of the Stoney [4] and Planck [5] scales. We will start with the Stoney scale, from which we will move to the Planck scale.

The Stoney length $l_{\mathrm{s}}$ in natural units is

$$l_{\mathrm{s}}=\sqrt{{\displaystyle \frac{e^{2}G}{4\pi {\epsilon }_0{c}^4}}}$$

(21)

The Stoney mass $m_{\mathrm{s}}$ from the natural units is

$$m_{\mathrm{s}}=\sqrt{{\displaystyle \frac{e^2}{4\pi {\epsilon }_{0}G}}}$$

(22)

where $G$ is the gravitational constant. Rewrite Eq. (22) for the gravitational constant $G$:

$$G={\displaystyle \frac{e^2}{4\pi {\epsilon }_0{m}_{\mathrm{s}}^2}}$$

(23)

By substituting the relation $e^2=2hc\alpha {\epsilon }_0$ introduced by the physicist Arnold Sommerfeld $\alpha =e^2/2hc{\epsilon }_0$[6], in Eq. (23), we obtain

$$G={\displaystyle \frac{2hc\alpha {\epsilon }_0}{4\pi {\epsilon }_0{m}_{\mathrm{s}}^2}}$$

(24)

The orbital angular momentum of the proton at the trajectory around the center of mass should be expressed by the reduced Planck constant. The proton’s velocity at this trajectory is denoted here as $v_p$. An initial estimation of this velocity yield approximately one fifth of the speed of light in vacuum. Hence, it is necessary to add a relativistic element ${(1-v_p^2/c^2)}^{1/2}$ with $v_p=c\beta$:

$$h=m_{p}c\beta 2\pi r_p\sqrt{1-{\beta }^2}$$

(25)

where $m_p$ is the proton mass, $\beta$ is the ratio of $v_p$ to $c$, and $r_p$ is the proton radius. By substituting the expression of $h$ from Eq. (25) in Eq. (24) and reducing the expression, we obtain

$$G={\displaystyle \frac{m_p{c}^2\alpha \beta r_p\sqrt{1-{\beta }^2}}{m_{\mathrm{s}}^2}}$$

(26)

The $\beta$ is similar to the fine structure constant $\alpha$, also known as the electromagnetic coupling constant, and it appears in the electron’s velocity expression at the Bohr radius as $v_e=c\alpha$.

We can divide Eq. (21) by Eq. (22) ($4\pi {\epsilon }_0$ is reduced, and the elementary charge $e$ is partially reduced):

$${\displaystyle \frac{l_{\mathrm{s}}}{m_{\mathrm{s}}}}=\sqrt{{\displaystyle \frac{e^2}{e^2}}\times {\displaystyle \frac{G^2}{c^4}}}={\displaystyle \frac{e}{e}}\times {\displaystyle \frac{G}{c^2}}$$

(27)

Then rearrange Eq. (27) to obtain an expression for $G$:

$$G={\displaystyle \frac{e}{e}}\times {\displaystyle \frac{l_{\mathrm{s}}}{m_{\mathrm{s}}}}{c}^2$$

(28)

By setting the expressions in Eq. (28) and Eq. (26) equal to each other, we have

$${\displaystyle \frac{e}{e}}\times {\displaystyle \frac{l_{\mathrm{s}}}{m_{\mathrm{s}}}}{c}^2={\displaystyle \frac{m_p{c}^2\alpha \beta r_p\sqrt{1-{\beta }^2}}{m_{\mathrm{s}}^2}}$$

(29)

We then divide both sides of Eq. (29) by $\left(e/e\right)$, multiply both sides by $m_{\mathrm{s}}^2$, reduce, and rearrange:

$$m_{\mathrm{s}}\times l_{\mathrm{s}}\times c^2=\left[\left({\displaystyle \frac{1}{e}}{m}_p\beta \right)\left({\displaystyle \frac{\alpha r_p\sqrt{1-{\beta }^2}}{\frac{1}{e}}}\right)\right]\times c^2$$

(30)

Eq. (30) presents a similarity between the right and left flanks (mass component and length component). The expression is split into two parts on the right-hand side of the equation because it contains the solutions corresponding to actual experimental results in the final analysis. The following new expressions are proposed solutions for the Stoney units.

New expression of Stoney mass $m_{\mathrm{s}}$:

$$m_{\mathrm{s}}={\displaystyle \frac{1}{e}}{m}_p\beta$$

New expression of Stoney length $l_{\mathrm{s}}$:

$$l_{\mathrm{s}}=\left({\displaystyle \frac{\alpha r_p\sqrt{1-{\beta }^2}}{\frac{1}{e}}}\right)$$

Note that the $\left(1/e\right)$ expression in Eq. (30) represents a dimensionless number, for instance, the number of charged particles in one Coulomb $\left[C\right]$ unit.

$${\displaystyle \frac{1}{e}}={\displaystyle \frac{1\cancel{C}}{1.602176634\times {10}^{-19}\cancel{C}}}=6.2415090744\times {10}^{18}$$

(31)

This number, as a multiplier, creates a quantity of charged particles (in our case, the number of protons contained within the Stoney mass, which corresponds to a quintillion protons) or, as a divisor, creates the smallest length (in our case, a contracted radius of the proton within the Stoney mass under internal attraction forces, which corresponds to a quintillionth of the proton radius that represents the Stoney length). It is displayed in the following equations as in Eq. (31), indicating that the value is dimensionless.

The gravitational constant $G$ in Stoney units from Eq. (28) with the proposed new expressions is:

$$G={\displaystyle \frac{\left(\frac{\alpha r_p\sqrt{1-{\beta }^2}}{\frac{1}{e}}\right)}{\left(\frac{1}{e}{m}_p\beta \right)}}\times c^2$$

(32)

We can then set Eq. (23) and Eq. (32) equal to each other and substitute the square of the Stoney mass term $m_{\mathrm{s}}^2={\left({\displaystyle \frac{1}{e}}{m}_p\beta \right)}^2$ in the denominator of Eq. (23) as:

$${\displaystyle \frac{e^2}{4\pi {\epsilon }_0{\left(\frac{1}{e}{m}_p\beta \right)}^2}}={\displaystyle \frac{\alpha r_p\sqrt{1-{\beta }^2}}{\frac{1}{e}\left(\frac{1}{e}{m}_p\beta \right)}}\times c^2$$

(33)

* See also section #9 further on for the new expressions proposed for the Stoney mass and length, developed from a different aspect!.

Multiplying both sides of Eq. (33) by $4\pi {\epsilon }_0{m}_p\beta$, reducing, and rearranging yields

$$e^2=2\left[m_{p}c\beta 2\pi r_p\sqrt{1-{\beta }^2}\right]c\alpha {\epsilon }_0$$

(34)

The expression of Eq. (34) shows the equivalence of $e^2=2hc\alpha {\epsilon }_0$, where the right-hand side (in brackets) contains the expression of the Planck constant $h$ with the proton parameters introduced in Eq. (25). This result confirms the choice of the proposed solutions for the Stoney units of mass and length from Eq. (30). Although this option was based on a logical consideration, there are additional combinations that could be chosen that yield incorrect results. We multiply the numerator and denominator of Eq. (32) by ${\alpha }^{-1/2}$ to obtain the gravitational constant $G$ at the Planck scale

$$G={\displaystyle \frac{\left(\frac{{\alpha }^{1/2}{r}_p\sqrt{1-{\beta }^2}}{\frac{1}{e}}\right)}{\frac{1}{{\alpha }^{1/2}}\left(\frac{1}{e}{m}_p\beta \right)}}\times c^2$$

(35)

Note: The difference between the Stoney and Planck units arises from the need to multiply Planck units by the square root of the fine structure constant ${\alpha }^{1/2}$. We obtain the following expressions.

New expression of Planck mass $m_{\mathrm{pl}}$:

$$m_{\mathrm{pl}}={\displaystyle \frac{1}{{\alpha }^{1/2}}}\left({\displaystyle \frac{1}{e}}{m}_p\beta \right)$$

New expression of Planck length $l_{\mathrm{pl}}$:

$$l_{\mathrm{pl}}=\left({\displaystyle \frac{{\alpha }^{1/2}{r}_p\sqrt{1-{\beta }^2}}{\frac{1}{e}}}\right)$$

By using the Planck mass in natural units and the new expression of Planck mass, we can derive the expression and value of $\beta$. The Planck mass defined by natural units is

$$m_{\mathrm{pl}}=\sqrt{{\displaystyle \frac{hc}{2\pi G}}}$$

(36)

$$\begin{array}{cc}\text{The new expression for the Planck mass from Eq. (35) is}& m_{\mathrm{pl}}={\displaystyle \frac{1}{{\alpha }^{1/2}}}\left({\displaystyle \frac{1}{e}}{m}_p\beta \right)\end{array}$$

(37)

Setting Eq. (36) and Eq. (37) as equal:

$$\sqrt{{\displaystyle \frac{hc}{2\pi G}}}={\displaystyle \frac{1}{{\alpha }^{1/2}}}\left({\displaystyle \frac{1}{e}}{m}_p\beta \right)$$

(38)

Then rearrange Eq. (38) to obtain an expression for $\beta$:

$$\beta =\sqrt{{\displaystyle \frac{hc\alpha }{2\pi G}}}\times {\displaystyle \frac{1}{1/e\times m_p}}$$

(39)

In Eq. (39), we substitute the values of $h,c,\alpha ,m_p$ and the following values from NIST CODATA 2018:

$1/e=6.2415090744\times {10}^{18}$ and $G=6.6743\times {10}^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$

We obtain $\beta =0.178090537$The relationship of $\beta$ to the nuclear research is through the strong coupling constant in QCD as follow

$${\alpha }_s=2/3\beta \to {\alpha }_s=2/3\times 0.178090537=0.118727$$

This value compares well with the value obtained experimentally [7,8], ${\alpha }_s\approx 0.1187$.

Using the Planck length from natural units and the new expression for the Planck length, we can derive the expression and value of the proton radius $r_p$.

$$l_{\mathrm{pl}}=\sqrt{{\displaystyle \frac{hG}{2\pi c^3}}}$$

(40)

The new expression for the Planck length from Eq. (35) is

$$l_{\mathrm{pl}}=\left({\displaystyle \frac{{\alpha }^{1/2}{r}_p\sqrt{1-{\beta }^2}}{\frac{1}{e}}}\right)$$

(41)

By setting the expressions in Eq. (40) and Eq. (41) as equal, we obtain

$$\sqrt{{\displaystyle \frac{hG}{2\pi c^3}}}=\left({\displaystyle \frac{{\alpha }^{1/2}{r}_p\sqrt{1-{\beta }^2}}{\frac{1}{e}}}\right)$$

(42)

We then rearrange Eq. (42) for the proton radius $r_p$:

$$r_p=\sqrt{{\displaystyle \frac{hG}{2\pi c^3\alpha }}}\times {\displaystyle \frac{1}{e}}\times {\displaystyle \frac{1}{\sqrt{1-{\beta }^2}}}$$

(43)

Substitutingin Eq. (43) $\beta =0.178090537$ and the values $h,G,c,\alpha$ and $\left(1/e\right)$ from NIST CODATA, yields $r_p=1.200094665\times 10^{-15}\mathrm{m}$

This value compares well with the value obtained experimentally [9], explained below:

• The proton radius obtained in Eq. (43) complies with the experimental formulation that assumes a spherical nucleus.The Fermi equation for the nuclear radius $R_{\mathrm{n}}$: $R_{\mathrm{n}}=R_0{A}^{1/3}$, Where$A$is the atomic number and $R_0$ is from experimental results:$R_0=1.2\times 10^{-15}\mathrm{m}$. For the Hydrogen,$\mathrm{A}=1$ and $R=1.2\times 10^{-15}\mathrm{m}$.
• The proton charge radius $r_{pcr}$ represents the maximum distance from the proton axis that the electron or the muon reaches in their penetration to the proton. This radius is:

  $$r_{pcr}=4\beta \times r_p{(1-{\beta }^2)}^{1/2}$$

Substituting the following in the expression for $r_{pcr}$; $r_p=1.200094665\times 10^{-15}\mathrm{m}$ and $\beta =0.178090537$, we obtain $r_{pcr}=8.4123564\times 10^{-16}\mathrm{m}$. This value is similar to the NIST value of $r_{pcr}=8.414\times 10^{-16}\mathrm{m}$.The proton’s Compton wavelength from Eq. (25) is

$${\lambda }_p={\displaystyle \frac{h}{m_{p}c}}=2\pi \beta r_p\sqrt{1-{\beta }^2}$$

(44)

Substituting the values of $\beta$ and $r_p$ in Eq. (44) yields

$${\lambda }_p=2\pi \times 0.178090537\times 1.200094665\times 10^{-15}\mathrm{m} \times 0.984014106=1.3214098539\times 10^{-15}\mathrm{m}$$

This value compares well with the NIST CODATA value of

$${\lambda }_p=1.32140985539\times 10^{-15}\mathrm{m}$$

The last result shows that $\beta$ combined with the proton radius $r_p$ obtained from Eq. (43) and used in Eq. (44) is entirely consistent with the value of the proton’s Compton wavelength ${\lambda }_p$ from NIST CODATA 2018, confirming the validity of our approach.

To obtain the gravitational constant $G$, we utilize Eq. (32) and substitute the $m_p,c,\alpha$ from NIST CODATA 2018 and also $1/e=6.2415090744\times {10}^{18}$,$\beta$ and $r_p$:

$$G={\displaystyle \frac{\left(\frac{ r_p\sqrt{1-{\beta }^2}}{6.2415090744\times {10}^{18}}\right)\alpha }{\left(6.2415090744\times {10}^{18}\times m_p\right)\beta }}\times c^2$$

(45)

It yields $G=6.6743\times {10}^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$

Which compares well with the NIST CODATA value $G=6.6743\times {10}^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$

6. The Radius of the Neutron

The ratio between the proton mass and neutron mass is the same as the ratio between the neutron Compton wavelength and proton Compton wavelength. The values of $m_p,m_n,{\lambda }_n,{\lambda }_p$ are substituted from the NIST CODATA 2018 in the following ratios:

$$\begin{gathered} {\displaystyle \frac{m_p}{m_n}}={\displaystyle \frac{1.67262192369\times 10^{-27}K\mathrm{g}}{1.67492749804\times 10^{-27}K\mathrm{g}}}=0.998623478 \\ {\displaystyle \frac{{\lambda }_n}{{\lambda }_p}}={\displaystyle \frac{1.31959090581\times 10^{-15}\mathrm{m}}{1.32140985539\times 10^{-15}\mathrm{m}}}=0.998623478 \end{gathered}$$

This result indicates that the ratio is also appropriate for the ratio between the neutron and proton radii.

$${\displaystyle \frac{m_p}{m_n}}={\displaystyle \frac{r_n}{r_p}} \to r_n={\displaystyle \frac{m_p}{m_n}}\times r_p$$

(46)

By substituting the values of $m_p,m_n$ from NIST CODATA 2018 and the radius $r_p$ from Eq. 43 in Eq. (46) for the neutron radius, we obtain:

$$r_n={\displaystyle \frac{1.67262192369\times 10^{-27}k\mathrm{g}}{1.67492749804\times 10^{-27}k\mathrm{g}}}\times 1.200094667\times 10^{-15}\mathrm{m}=1.19844271\times 10^{-15}\mathrm{m}$$

The proton and neutron are almost identical in size, and the $\beta$ is found to be related to both radii. Consequently, $v_n=v_p=c\beta$, Validated by the neutron Compton wavelength ${\lambda }_n$ with $\beta$ and $r_n$, as follows

$${\lambda }_n={\displaystyle \frac{h}{m_{n}c}}=2\pi \beta r_n\sqrt{1-{\beta }^2}$$

(47)

Substituting the values of $\beta$ and $r_n$ in Eq. (47) for ${\lambda }_n$, we obtain

$${\lambda }_n=2\pi \times 0.178090537\times 1.19844271\times 10^{-15}\mathrm{m}\times 0.984014105=1.31959090471\times 10^{-15}\mathrm{m}$$

This value matches well with the NIST CODATA value of ${\lambda }_n=1.31959090581\times 10^{-15}\mathrm{m}$The result obtained by combining $\beta$ with the neutron radius $r_n$ given by Eq. (46) and substituting in Eq. (47) is entirely consistent with the NIST CODATA 2018 value of the neutron Compton wavelength ${\lambda }_n$, confirming the validity of our approach.

7. Additional Expressions for the Proton and Neutron Masses and Radii

We divide Eq. (3) (with $(v_e=c\alpha )$ by Eq. (25) as follows:

$${\displaystyle \frac{m_{e}c\alpha 2\pi a_0}{m_{p}c\beta 2\pi r_p\sqrt{1-{\beta }^2}}}=1$$

(48)

By rearranging Eq. (48) and solving for the proton mass, we obtain

$$m_p={\displaystyle \frac{m_e\alpha a_0}{\beta \times r_p\sqrt{1-{\beta }^2}}}$$

(49)

Substituting $m_e=4{\epsilon }_0{\Phi }_0^2{\pi }^{-1}{a}_0^{-1}$ from Eq. (12) in Eq. (49) yields the following expression for the proton $m_p$

$$m_p={\displaystyle \frac{4}{\pi }}\times {\displaystyle \frac{\alpha }{\beta }}\times {\displaystyle \frac{{\epsilon }_0}{r_p\sqrt{1-{\beta }^2}}}\times {\Phi }_0^2$$

(50)

Substituting ${\lambda }_e=\alpha 2\pi a_0$ and ${\lambda }_p=\beta 2\pi r_p\sqrt{1-{\beta }^2}$ in Eq. (48) yields

$${\displaystyle \frac{m_e}{m_p}}={\displaystyle \frac{{\lambda }_p}{{\lambda }_e}} \mathrm{Or} \to m_p={\displaystyle \frac{m_e{\lambda }_e}{{\lambda }_p}}$$

(51)

We substitute ${\lambda }_e=\alpha 2\pi a_0$ and $m_e$ from Eq. (12) in Eq. (51) to obtain another expression for $m_p$:

$$m_p=\left({\displaystyle \frac{ 8 \alpha {\epsilon }_0}{{\lambda }_p}}\right){\Phi }_0^2$$

(52)

Substituting the values of $\alpha ,{\epsilon }_0,{\lambda }_p, {\Phi }_0^2$ from NIST CODATA 2018 in Eq. (52), it gives the proton mass

(53)

$m_p={\displaystyle \frac{8\times 7.297352569\times 10^{-3}\times 8.8541878128\times 10^{-12}\mathrm{m}}{1.32140985539\times 10^{-15}\mathrm{m}}}\times 4.27593682\times 10^{-30}\mathrm{k}\mathrm{g}=1.672621922\times 10^{-27}\mathrm{k}\mathrm{g}$ This value matches well with the NIST CODATA value of $m_p=1.67262192369\times 10^{-27}\mathrm{k}\mathrm{g}$.

We can rearrange Eq. (47) to obtain the orbital angular momentum of the neutron:

$${\displaystyle \frac{h}{2\pi }}=m_{n}c\beta r_n\sqrt{1-{\beta }^2}Orh=m_{n}c\beta 2\pi r_n\sqrt{1-{\beta }^2}$$

(54)

We then divide Eq. (3) by Eq. (54), as follows:

$${\displaystyle \frac{m_{e}c\alpha 2\pi a_0}{m_{n}c\beta 2\pi r_n\sqrt{1-{\beta }^2}}}=1$$

(55)

Rearranging Eq. (55) for the neutron mass, gives

$$m_n={\displaystyle \frac{m_e\alpha a_0}{\beta r_n\sqrt{1-{\beta }^2}}}$$

(56)

Substituting $m_e$ from Eq. (12) in Eq. (56) yields an expression for the neutron mass $m_n$:

$$m_n={\displaystyle \frac{4}{\pi }}\times {\displaystyle \frac{\alpha }{\beta }}\times {\displaystyle \frac{{\epsilon }_0}{r_n\sqrt{1-{\beta }^2}}}\times {\Phi }_0^2$$

(57)

Substituting ${\lambda }_e=\alpha 2\pi a_0$ from Eq. (13) and ${\lambda }_n=\beta 2\pi r_n\sqrt{1-{\beta }^2}$ from Eq. (47) in Eq. (55) yields

$${\displaystyle \frac{m_e}{m_n}}={\displaystyle \frac{{\lambda }_n}{{\lambda }_e}} \mathrm{Or} \to m_n={\displaystyle \frac{m_e{\lambda }_e}{{\lambda }_n}}$$

(58)

$$\begin{array}{cc} \mathrm{We} \mathrm{substitute} {\lambda }_e=\alpha 2\pi a_0 \mathrm{and} m_e=4{\epsilon }_0{\Phi }_0^2{\pi }^{-1}{a}_0^{-1} \mathrm{from} \mathrm{Eq}. \left(12\right) \mathrm{in} \mathrm{Eq}. \left(58\right)\mathrm{for}\mathrm{another}\mathrm{expression} \mathrm{of}{m}_n& m_n=\left({\displaystyle \frac{ 8 \alpha {\epsilon }_0}{{\lambda }_n}}\right){\Phi }_0^2\end{array}$$

(59)

Substituting the values of $\alpha ,{\epsilon }_0, {\lambda }_n, {\Phi }_0^2$ from NIST CODATA 2018 in Eq. (59) for the neutron mass $m_n$:

$$m_n={\displaystyle \frac{8\times 7.2973525693\times 10^{-3}\times 8.8541878128\times 10^{-12}\mathrm{m}}{1.31959090581\times 10^{-15}\mathrm{m}}}\times 4.275936823\times 10^{-30}\mathrm{k}\mathrm{g}=1.6749274973\times 10^{-27}\mathrm{k}\mathrm{g}$$

(60)

This value compares well with the NIST CODATA 2018 value of $m_n=1.67492749804\times 10^{-27}\mathrm{k}\mathrm{g}$.

The ratio of proton mass to electron mass is obtained from Eq. (49). After rearranging and substituting the NIST CODATA 2018 values for $\alpha$, $a_0$ and $\beta$ and $r_p$, it gives

$$m_p/m_e=\alpha /\beta \times a_0/r_p\sqrt{1-{\beta }^2}=1836.152675$$

(61)

This ratio can also be obtained from Eq. (51). This value matches well with the NIST CODATA ratio of

$$m_p/m_e=1836.1526734$$

The ratio of neutron mass to electron mass is obtained from Eq. (56). After rearranging and substituting NIST CODATA 2018 values for $\alpha$, $a_0$ and $\beta$ and $r_n$, it gives

$$m_n/m_e=\alpha /\beta \times a_0/r_n\sqrt{1-{\beta }^2}=1838.683661$$

(62)

This ratio can also be obtained from Eq. (58). This value matches well with the NIST CODATA ratio of

$$m_n/m_e=1838.683661$$

8. The meaning of the Permittivity of Vacuum Constant ${\epsilon }_0$ from a Different Aspect

The electron in the Hydrogen atom revolves around the center of mass shared with the proton as was described in section #5. Let’s assume now that the virtual photons, responsible for the force acting between them, collide with the virtual shell that represents the supposed locations of the electron at the Bohr level (The absorption and emission process). What if it is possible to describe the virtual photons’ rapid motion within the shell space in terms of molecules in an ideal gas?

The centripetal force $F_c$ that holds the electron at Bohr radius $a_0$ level [10], is

$$F_c={\displaystyle \frac{m_e{v}_e^2}{a_0}}$$

(63)

And the electric force $F_e$ that holds the electron at Bohr radius $a_0$ level, is

$$F_e={\displaystyle \frac{e^2}{4\pi {\epsilon }_0{a}_0^2}}$$

(64)

The condition for a dynamically stable at Bohr radius $a_0$ level, is

$$F_c=F_e$$

$${\displaystyle \frac{m_e{v}_e^2}{a_0}}={\displaystyle \frac{e^2}{4\pi {\epsilon }_0{a}_0^2}}$$

(65)

Multiplying both sides of Eq. (65) by $\left(a_0/2\right)$, it gives

$${\displaystyle \frac{m_e{v}_e^2}{2}}={\displaystyle \frac{e^2}{8\pi {\epsilon }_0{a}_0}}$$

(66)

We obtain at the left side of Eq. (66) the kinetic energy $E_k$ of the electron at $a_0$ level, as follows

$$E_k={\displaystyle \frac{m_e{v}_e^2}{2}}$$

(67)

where $m_e$ is the electron mass, and $v_e$ is the electron velocity at $a_0$ level.

$$\begin{array}{cc}\text{The potential energy of the electron that equals to the kinetic energy }{E}_k\text{ at the right side of Eq. (66), is}& E_p={\displaystyle \frac{e^2}{8\pi {\epsilon }_0{a}_0}}\end{array}$$

(68)

where $e$ is the elementary charge of the electron, and ${\epsilon }_0$ is the vcuum permittivity constant.

The average kinetic energy $E_k$ of the virtual photons exchanged between the electron and the proton that support the movement of the electron in its trajectory at the Bohr level (as an approximation of molecules in motion in Ideal gas from the mechanical point of view), is

$$PV={\displaystyle \frac{2}{3}}{E}_k$$

(69)

where $V$ is the volume contained within the spherical shell, and $P$ is the pressure that the virtual photons impose on the spherical shell. Substituting $E_k$ from Eq. 67 in Eq. 69, as flollows

$$PV={\displaystyle \frac{2}{3}}\left[{\displaystyle \frac{1}{2}}{m}_e{v}_e^2\right]$$

(70)

The pressure $P$ is defined as the force acting on a given surface. If the spherical shell surface is $A=4\pi a_0^2$, and $a_0$ is the Bohr radius, then the pressure, is

$$P={\displaystyle \frac{F}{4\pi a_0^2}}$$

(71)

where $F$ is the total force imposed by the virtual photons on the inner surface of the spherical shell. Substituting Eq. (71) into Eq. (70) and rearrange, it gives

$$FV={\displaystyle \frac{2}{3}}\left({\displaystyle \frac{1}{2}}{m}_e{v}_e^2\right)4\pi a_0^2 \to FV=2\left[{\displaystyle \frac{m_e{v}_e^{2}4\pi a_0^2}{6}}\right]$$

(72)

The same mathematical development can be applied for the potential energy defined as

$$PV={\displaystyle \frac{2}{3}}{E}_p$$

(73)

Substituting Eq. (68) (from $E_k=E_p$ at Eq. (66)) and Eq. (71) in Eq. (73) and rearrange, it gives

$$FV={\displaystyle \frac{2}{3}}\left({\displaystyle \frac{e^2}{8\pi {\epsilon }_0{a}_0}}\right)4\pi a_e^2\to FV=e^2\times 2\left({\displaystyle \frac{a_0}{6{\epsilon }_0}}\right)$$

(74)

Setting Eq. (74) as equal to Eq. (72) with reducing the integer 2 from both sides

$$e^2\left({\displaystyle \frac{a_e}{6{\epsilon }_0}}\right)={\displaystyle \frac{m_e{v}_e^{2}4\pi a_0^2}{6}}$$

(75)

Now setting the value of the expression in parentheses at the left side of Eq. (75) as equal to unity, here

$$\left({\displaystyle \frac{a_e}{6{\epsilon }_0}}\right)=1 \mathrm{or} {\displaystyle \frac{a_0}{6}}={\epsilon }_0$$

(76)

And then substituting the last expression obtained in Eq. (76) for ${\epsilon }_0$ in the right side of Eq. (75), we obtain the electrostatic force acting on the electron at the Bohr level from Eq. (2) at section #1, as

$$e^2=m_e{v}_e^{2}4\pi a_0{\epsilon }_0$$

(77)

Note, that Eq. (76) shows the relation between the Bohr radius $a_0$ and the vcuum permittivity ${\epsilon }_0$ constant that corresponds with the conclussion from the analysis of Equation #8 at section #2.

For comparison to Eq. (76), let’s set Eq. (15), as follows

${\epsilon }_0=\left({\displaystyle \frac{\pi m_e}{4{\Phi }_0^2}}\right)\times a_0=\left({\displaystyle \frac{\pi }{4}}\times {\displaystyle \frac{9.1093837015\times 10^{-31}\mathrm{kg}}{4.275936823\times 10^{-30}\mathrm{kg}}}\right)\times 5.291772109\times 10^{-11}\mathrm{m}=8.854187816\times 10^{-12}\mathrm{m}$ The expression in the parentheses equals to the following

$${\displaystyle \frac{\pi m_e}{4{\Phi }_0^2}}={\displaystyle \frac{\pi }{4}}\times {\displaystyle \frac{9.1093837015\times 10^{-31}\mathrm{kg}}{4.275936823\times 10^{-30}\mathrm{kg}}}={\displaystyle \frac{1}{5.9765754}}$$

It Shows that ${\epsilon }_0=\left({\displaystyle \frac{\pi m_e}{4{\Phi }_0^2}}\right)\times a_0={\displaystyle \frac{1}{5.9765754}}\times a_0\to {\epsilon }_0\approx {\displaystyle \frac{a_0}{6}}$

This leaves the square of the magnetic flux quantum ${\Phi }_0^2$, as appropriate description of the mass!

9. Developing the New Expressions Proposed for the Stoney Mass and Length from a Different Aspect

Using the term of $e^2=h^2/4{\Phi }_0^2$ from Eq. (1) and substitute the terms of$h=2\pi m_e{v}_e{a}_0$ (with $v_e=c\alpha$) from Eq. (3) and $h=m_{p}c\beta 2\pi r_p\sqrt{1-{\beta }^2}$from Eq. (25) in it, as follows

$$e^2={\displaystyle \frac{m_{e}c\alpha 2\pi a_0\times m_{p}c\beta 2\pi r_p\sqrt{1-{\beta }^2}}{4{\Phi }_0^2}}$$

(78)

Substituting $a_0=4{\epsilon }_0{\Phi }_0^2{\pi }^{-1}{m}_e^{-1}$ from Eq. (7) in Eq. (78), and after reducing and rearranging, it gives

$$e^2={\displaystyle \frac{m_{e}c\alpha 2\pi \left(4{\epsilon }_0{\Phi }_0^2{\pi }^{-1}{m}_e^{-1}\right)\times m_{p}c\beta 2\pi r_p\sqrt{1-{\beta }^2}}{4{\Phi }_0^2}} \to e^2=m_p{c}^2\alpha \beta 4\pi {\epsilon }_0{r}_p\sqrt{1-{\beta }^2}$$

(79)

Rearranging Eq. (79), as follows

$${\displaystyle \frac{e^2}{4\pi {\epsilon }_0}}=\left(m_p\beta \right)\alpha r_p\sqrt{1-{\beta }^2}{c}^2$$

(80)

Dividind both sides of Eq. (80) by the square of the Stoney mass term $m_{\mathrm{s}}^2={\left({\displaystyle \frac{1}{e}}{m}_p\beta \right)}^2$(proposed at Eq. (30) and after reducing and rearrenging, it gives

$${\displaystyle \frac{e^2}{4\pi {\epsilon }_0{\left(\frac{1}{e}{m}_p\beta \right)}^2}}={\displaystyle \frac{\alpha r_p\sqrt{1-{\beta }^2}}{\frac{1}{e}\left(\frac{1}{e}{m}_p\beta \right)}}\times c^2=G$$

(81)

We obtain in Eq. (81) the equality of the Gravitational Constant $G$ acording to Eq. (33).

10. The Squared Values of the Magnetic Flux Quantum Used in the Wave Function, Yield Solutions Which Depict the Flow Pattern of the Magnetic Flux Surrounding Electrons at a Given Energy Level

Using the Normalized Wave Function of the ‘Hydrogen-like atom’ equation [11], here

$$R_{nl}(r)=\sqrt{{\left({\displaystyle \frac{2Z}{n{a}_0}}\right)}^3{\displaystyle \frac{(n-1-l)!}{2n{[(n+l)!]}^3}}}{\left({\displaystyle \frac{2{\rm Z}r}{n{a}_0}}\right)}^l{e}^{-{{\rm Z}r/na}_0}{L}_{n+l}^{2l+1}\left({\displaystyle \frac{2{\rm Z}r}{n{a}_0}}\right)$$

(82)

Substituting $a_0=4{\epsilon }_0{\Phi }_0^2{\pi }^{-1}{m}_e^{-1}$ a rewrite of Eq. (7) in Eq. (78) and after rearranging, it gives

$$R_{nl}(r)={\left({\displaystyle \frac{\pi }{2{\epsilon }_0}}\times {\displaystyle \frac{ Z{m}_e}{n{\Phi }_0^2}}\right)}^{3/2}{\left({\displaystyle \frac{(n-1-l)!}{2n{[(n+l)!]}^3}}\right)}^{1/2}\times {\left({\displaystyle \frac{\pi r}{2{\epsilon }_0}}\times {\displaystyle \frac{ Z{m}_e}{n{\Phi }_0^2}}\right)}^l{e}^{-{{\rm Z}r/na}_0}{L}_{n+l}^{2l+1}\left({\displaystyle \frac{\pi r}{2{\epsilon }_0}}\times {\displaystyle \frac{ Z{m}_e}{n{\Phi }_0^2}}\right)$$

(83)

Now writing the third expression in parentheses at Eq. 83, as follows

$${\left({\displaystyle \frac{\pi r}{2{\epsilon }_0}}\times {\displaystyle \frac{ Z{m}_e}{n{\Phi }_0^2}}\right)}^l$$

(84)

Using it as an example that provides the value of the derivative $r$ at the Electron Wave Function of ‘Hydrogen-like Atom’ equation by rearranging it as follows

$$r={\displaystyle \frac{2{\epsilon }_0}{\pi Z{m}_e}}\times n{\Phi }_0^2 \mathrm{When} l=0\to {\left({\displaystyle \frac{\pi r}{2{\epsilon }_0}}\times {\displaystyle \frac{ Z{m}_e}{n{\Phi }_0^2}}\right)}^0$$

(85)