Here, we take the hydrogen atom as an example to derive the particle radius. In quantum mechanics, approximation methods are frequently employed to solve problems. Based on the degenerate mass of particles, we construct the following approximate equation:

$m_e\sqrt{1-\frac{a^2}{n^2}}\approx m-V,m=\frac{m_e}{\sqrt{1-v^2}},V=\frac{a{m}_{e}v}{L\sqrt{1-v^2}},L=\sqrt{l\left(l+1\right)},c=\hslash =1$ (1)

Derive the expression for the orbital velocity of the electron from Equation (1):

$v\approx \frac{a}{L}\left(1\pm \sqrt{1-\frac{L^{\text{2}}}{n^2}}\right),L\approx n$ (2)

This corresponds to the velocities at the two extremal points of an elliptical orbit. Assuming the closest point is point-A, and the farthest point is point-B. Therefore, the average radius of the orbit should be:

$\begin{array}{l}\langle r_1\rangle =\frac{r_A+r_B}{2}=\frac{L}{2\left(1+b\right)}\left(\frac{\hslash }{m_1{v}_A}+\frac{\hslash }{m_1{v}_B}\right),L\approx \sqrt{l\left(l+1\right)}\\ v_A\approx \frac{a}{L}\left(1-\sqrt{1-\frac{L^{\text{2}}}{n^2}}\right),v_B\approx \frac{a}{L}\left(1+\sqrt{1-\frac{L^{\text{2}}}{n^2}}\right)\end{array}\}\langle r_1\rangle =\frac{n^2\hslash \sqrt{1-v_1^2}}{m_{10}a}$ (3)

Here, r_A is the distance from point A to the focus, and r_B is the distance from point B to the focus. However, since we need to generalize the equation to two-body particle systems while maintaining applicability to high-speed spinning particles, we have made the following modifications:

$\begin{array}{l}{m}_1{v}_1{r}_1+m_2{v}_2{r}_2=\left(1+b\right)m_1{v}_1{r}_1=L\hslash ,b=\frac{m_1}{m_2}\\ m_1=\frac{m_{10}}{\sqrt{1-v_1^2}},v_1=\frac{a}{\left(1+b\right)n},L=\sqrt{l\left(l+1\right)}\ne 0\end{array}\}\to r_1=\frac{nL\hslash \sqrt{1-v_1^2}}{m_{10}a}$ (4)

Here, m₁₀ is the rest mass of a particle 1, and m₁ is the relativistic mass of the lighter particle, a is the coefficient of attraction between two particles, and r₁ is the orbital radius of particle 1. On the other hand, according to quantum mechanics, the following averaging equation is introduced:

$\begin{array}{l}\langle \frac{1}{r_1^2}\rangle =\frac{1}{n^3\left(l+1/2\right)r_0^2}\\ d_0=\left(1+b\right)r_0,b=\frac{m_1}{m_2},c=1\end{array})\to r_1\approx n{r}_0\sqrt{n\left(l+\frac{1}{2}\right)}\approx n\left(l+\frac{1}{2}\right)\frac{\hslash \sqrt{1-v_1^2}}{m_{10}a}$ (5)

Here, the symbol <1/r²> represents the average value of 1/r². Generally, when we refer to the spin radius and spin rate of a particle system, we are discussing the average orbital radius and orbital velocity of the lighter particle. Therefore, m₁ represents the mass of the lighter particle, while m₂ denotes the mass of the heavier particle. The mass ratio b cannot exceed 1. In conclusion, our improvement mainly lies in adding the relativistic factor to the Equation, which can explain why the radius of electrons is so small.

According to Figure 1 , the interaction force between moving charges at low velocities is given by:

$\begin{array}{l}F=\frac{q_1{q}_2{a}_e}{d^2\sqrt{1-v_{1\pm 2}^2}}\approx \frac{q_1{q}_2{a}_e}{d^2}\left(1+\frac{v^2}{2}\right)\approx q_2{E}_1+q_{2}v\times B_1,a_e\approx \frac{1}{137},\\ c=1,v\approx v_{1\pm 2}=\frac{v_1\pm v_2}{1\pm v_1{v}_2},E_1=\frac{q_1{a}_e}{d^2},B_1=\frac{q_{1}v{a}_e}{2d^2}=\frac{I_1{a}_e}{2d^2},I_1=q_{1}v\end{array}$ (6)

At high velocities, due to the periodic positional alternation of two charges in circular motion—where additive and subtractive superposition occur once per revolution—the expression for the average magnetic force between them is:

$\langle F_B\rangle =q_{2}v\times B_1=\frac{q_1{q}_2{a}_e}{d^2}\left({\beta }_{1\pm 2}-1\right),{\beta }_{1\pm 2}=\frac{1}{\sqrt{1-v_{1\pm 2}}}$ (7)

We often encounter situations where particles have different radii, such as protons and electrons forming hydrogen atoms. If the electron’s radius is very small, its spin rate will be very high. First, assuming the electron and proton spin antiparallel (with S = 0). Since the electron’s spin period is much shorter than the proton’s, when the proton’s valence charge is closest to the electron, the electron’s valence charge is also precisely closest to the proton, as shown in Figure 2(A) . Meanwhile, the situation in Figure 2(B) may also occur. When the configuration is as shown in Figure 2(A) , we first neglect the influence of other charges and only consider the magnitude of electrostatic attraction between the two closest charges, giving the following formula:

$F_{1e}\approx \frac{a_e}{d^2}\frac{1}{{\left(1-\Sigma \right)}^2},a_e\approx \frac{1}{137},\Sigma =\frac{r_p+r_e}{d},\Delta =\frac{r_p-r_e}{d}$ (8)

When considering only magnetic forces (excluding electrostatic forces), the resultant magnetic force exerted by charge 3 on charges 4, 5 and 6 is given by (see Figure 2(A) ):

$\begin{array}{l}{\langle F_{3B}\rangle }_{3*\left(4+5+6\right)}=a_e\left[-\frac{q_3{q}_4\left({\beta }_{e+p}{}_{-4}-1\right)}{{\left(d-2r-r_4\right)}^2}+\frac{q_3{q}_5\left({\beta }_{e+p+}{}_5-1\right)}{{\left(d-2r+r_5\right)}^2}-\frac{q_3{q}_6\left({\beta }_{e-p}-1\right)}{d^2}\right],\\ {\beta }_{e+p-4}=\frac{1}{\sqrt{1-v_{e+p-4}}},v_{e+p-4}=\frac{v_{e+p}-v_4}{1-v_{e+p}{v}_4},v_{e+p}{}_{+5}=\frac{v_{e+p}+v_5}{1+v_{e+p}{v}_5},v_{e+p}=\frac{v_e+v_p}{1+v_e{v}_p}\end{array}$ (9)

Here, positive charges denoted by “+”, negative charges by “−”; and “+” in results indicates repulsion, “−” indicates attraction. Due to the differing spin periods of protons and electrons, Charge 3 in Figure 2(A) may exhibit either subtractive superposition (50% probability) or additive superposition (50% probability) with Charge 6.

When distance effects are neglected, the average magnetic interaction force between Charge 3 and Charge 6 is given by:

$\begin{array}{l}{\langle F_{3*6}\rangle }_{e-p}=-\frac{q_3{q}_6{a}_e}{d^2}\left[1+\left({\beta }_e{}_{-p}-1\right)\right],{\langle F_{3*6}\rangle }_{e+p}=-\frac{q_3{q}_6{a}_e}{d^2}\left[1+\left({\beta }_e{}_{+p}-1\right)\right],\\ \langle F_{3*6}\rangle =\frac{{\langle F_{3*6}\rangle }_{e-p}+{\langle F_{3*6}\rangle }_{e+p}}{2}=-\frac{q_3{q}_6{a}_e}{d^2}\left[1+\left({\beta }_e{\beta }_p-1\right)\right],\frac{{\beta }_e{}_{-p}+{\beta }_e{}_{+p}}{2}={\beta }_e{\beta }_p\end{array}$ (10)

Thus, at long distances (when proton radius effects are negligible), the total combined interaction force between protons and electrons is given by:

$\begin{array}{l}\left({\beta }_p-{\beta }_2{\beta }_{p2}+{\beta }_1{\beta }_{p2}-1\right)\approx \frac{{\mu }_p}{d}\\ \left(-{\beta }_e+{\beta }_4{\beta }_{e2}-{\beta }_5{\beta }_{e2}+1\right)\approx \frac{{\mu }_e}{d}\end{array}\}\langle F_{ep}\rangle \approx -\frac{a_e}{d^2}\left(q_e{q}_p+\frac{{\mu }_p{\mu }_e}{d^2}\right)$ (11)

Equation (11) shows that when particles are far apart, their magnetic interaction approaches zero. This explains why electrons exhibit purely charge-based forces at long distances, while allowing for annihilation effects at close range.

As analyzed in Figure 2 , particles with negligible radii exhibit near-complete cancellation of gravitational influence regardless of spin orientation variations. The dominant contribution to gravitational modulation stems exclusively from unpaired charges—termed valence charges of protons or electrons. From relativistic principles, we derive the following governing rule:

$\begin{array}{l}{\beta }_{e\pm p}=\left(1\pm v_e{v}_p\right){\beta }_e{\beta }_p,\left({\beta }_{e+p}-{\beta }_{e-p}\right)=2v_e{v}_p{\beta }_e{\beta }_p,\\ \left({\beta }_{e+p}+{\beta }_{e-p}\right)=2{\beta }_e{\beta }_p,{\beta }_{e-p}={\beta }_{e+p}\frac{1-v_e{v}_p}{1+v_e{v}_p}\end{array}$ (12)

When evaluating exclusively the valence-originating magnetogravitic coupling through ensemble averaging over Figure 2 (Only calculate F_3*6). Figure 2(A)-(H) ’s topological configurations, the S = 0 proton-electron system exhibits a mean magnetic attraction of:

$\begin{array}{l}{\langle F_B\rangle }_{ep,s=0}=\frac{1}{8}\langle F_B\rangle \left(A+B+C+D+E+F+G+H\right)\\ \approx -\frac{a_e}{2d^2}\left[\frac{1+{\Sigma }^2}{{\left(1-{\Sigma }^2\right)}^2}\left({\beta }_{e-p}-1\right)+\frac{1+{\Delta }^2}{{\left(1-{\Delta }^2\right)}^2}\left({\beta }_{e+p}-1\right)\right]-F_{0B}\\ \approx -\frac{a_e}{d^2}\left[\frac{1+{\Sigma }^2-2r_p{r}_e\left(1+v_e{v}_p\right)/d^2}{{\left(1-{\Sigma }^2\right)}^2}{\beta }_p{\beta }_e-1\right]-F_{0B},\\ {\beta }_{e-p}\gg 1,{\beta }_{e+p}\gg 1,d\gg r_p+r_e,F_{0B}=F_{B3*\left(4+5\right)}+F_{B\left(4+5+6\right)*\left(1+2\right)}\end{array}$ (13)

In the formula, F_B3*(4+5) represents the resultant magnetic force between charge 3 and charges (4,5). Since gravitational and repulsive forces nearly cancel out, the magnitude F_0B is significantly smaller than the valence charge magnetic force and can generally be neglected in strong interaction calculations.

Applying the same computational method, the average magnetic attraction between protons and electrons at S = 1 is given by:

$\begin{array}{c}{\langle F_B\rangle }_{ep,s=1}\approx -\frac{a_e}{2d^2}\left[\frac{1+{\Sigma }^2}{{\left(1-{\Sigma }^2\right)}^2}\left({\beta }_{e+p}-1\right)+\frac{1+{\Delta }^2}{{\left(1-{\Delta }^2\right)}^2}\left({\beta }_{e-p}-1\right)\right]-F_{0B}\\ \approx -\frac{a_e}{d^2}\left[\frac{1+{\Sigma }^2-2r_e{r}_p\left(1-v_e{v}_p\right)/d^2}{{\left(1-{\Sigma }^2\right)}^2}{\beta }_e{\beta }_p-1\right]-F_{0B}\end{array}$ (14)

The hyperfine splitting energy (ΔE = 5.87 μeV) between the ground-state hydrogen’s singlet (S = 0) and triplet (S = 1) configurations—arising from electron-proton spin-spin magnetic coupling—manifests observationally as the 1420.4 MHz (21 cm) spectral line:

$\Delta V=d\left({\langle F_B\rangle }_{S=0}-{\langle F_B\rangle }_{S=1}\right)\approx \frac{4a_e}{d}\frac{r_e{r}_p{v}_e{v}_p{\beta }_e{\beta }_p}{d^2}$ (15)

Evidently, this energy difference is proportional to the product of the electron and proton spin magnetic moments. A more precise calculation requires quantum mechanical (see Section 2.5). Here, we only estimate the order of magnitude. In the equation, d = a₀ ≈ 0.53 × 10⁻¹¹ m (Bohr radius), a_e/d = 27.2 eV, v_e ≈ 1. The magnetic potential energy difference ΔV ≈ 5.87 × 10⁻⁶ eV. Proton radius: 0.83 fm; Proton spin rate: 0.93c (see Section 2.9). Thus:

$\begin{array}{l}{v}_e{r}_e{\beta }_e\approx 7\times {10}^{-14}\text{m}\\ v_e{\beta }_e=\frac{1}{\sqrt{1/v_e^2-1}}\end{array}\}\to \{\begin{array}{l}{r}_e={10}^{-19}\text{m},v_e\approx 0.999\cdots \left(12\text{nines}\right)\\ r_e={10}^{-20}\text{m},v_e\approx 0.999\cdots \left(14\text{nines}\right)\\ r_e={10}^{-21}\text{m},v_e\approx 0.999\cdots \left(16\text{nines}\right)\end{array}$ (16)

The above calculations indicate that the radius of an electron is very small and its spin rate is very high. In Section 2.7, we will present an alternative method for calculating the electron’s radius and magnetic moment. Finally, we give that the gravitational coupling coefficient between particles of opposite charge is expressed as:

$\begin{array}{l}{\langle a\rangle }_{s=1}\approx a_e+\frac{a_e}{2}\left[\frac{1+{\Sigma }^2}{{\left(1-{\Sigma }^2\right)}^2}\left({\beta }_{1+2}-1\right)+\frac{1+{\Delta }^2}{{\left(1-{\Delta }^2\right)}^2}\left({\beta }_{1-2}-1\right)\right]-d^2{F}_{0B}\\ {\langle a\rangle }_{s=0}\approx a_e+\frac{a_e}{2}\left[\frac{1+{\Sigma }^2}{{\left(1-{\Sigma }^2\right)}^2}\left({\beta }_{1-2}-1\right)+\frac{1+{\Delta }^2}{{\left(1-{\Delta }^2\right)}^2}\left({\beta }_{1+2}-1\right)\right]-d^2{F}_{0B}\end{array}$ (17)

When two particles possess identical spin radii and rotation rates while carrying opposite charges, they form mutually reciprocal particles. Examples include positronium, protonium, or quarkonium systems. The gravitational coupling coefficient between their constituent particles is given by:

$\begin{array}{l}{\langle a_{p\bar{p}}\rangle }_{s=1}\approx a_e+\frac{a_e}{2}\frac{1+4r^2/d^2}{{\left(1-4r^2/d^2\right)}^2}\left({\beta }_{2p}-1\right)-d^2{F}_{0B}\\ {\langle a_{p\bar{p}}\rangle }_{s=0}\approx a_e+\frac{a_e}{2}\left({\beta }_{2p}-1\right)-d^2{F}_{0B},{\beta }_{2p}=\frac{2v_p}{1+v_p^2}\end{array}$ (18)

This explains why the energy state is always lower when positive and negative magnetic moments have parallel spins (this is also known as pairing energy).

Previously, we did not understand which factors were related to particle mass, leading to a series of difficulties. In 2015, our team revisited the statistical laws governing particle mass and discovered that the mass of a particle is strictly equal to the sum of the total energy and total potential energy of its internal orbiting particles. Based on this discovery, we derived the relativistic form of the Hamiltonian and expressed the eigenvalue equation of the particle as [10] :

$\begin{array}{l}M=m+V\to \left(m+V-M\right){\psi }_{x,y,z,t}=0\\ H=M-m_0=m+V-m_0,E\psi =H\psi \end{array}\}i\hslash \frac{\partial }{\partial t}\psi =\left(m+V-m_0\right)\psi$ (19)

Here, M represents the rest mass of the particle system, H stands for Hamiltonian, m is the sum of the energies of all orbital particles within the particle, this refers to the relativistic mass. V is the sum of the potential energies between all orbital particles, with attractive potential energies counted as negative values and repulsive potential energies counted as positive values. When the spin rate of a particle is very low, Eq. 19 becomes the Schrodinger equation:

$\begin{array}{l}m=\frac{m_0}{\sqrt{1-v^2}}\stackrel{v\to 0}{\to }{m}_0\left(1+\frac{v^2}{2}\right),\frac{m_0{v}^2}{2}\approx \frac{p^2}{2m}\\ H=m+V-m_0=\frac{m_0{v}^2}{2}+V,E\psi =H\psi \end{array}\}\to i\hslash \frac{\partial }{\partial t}\psi =\left(\frac{p^2}{2m}+V\right)\psi$ (20)

Referring to the work achievements of Sommerfield and Dirac, we present the following equation [11] .

$\begin{array}{c}M=\frac{m_{10}}{\sqrt{1+\frac{a^2/{\left(1+b\right)}^2{\left(l+1\right)}^2}{{\left[\frac{n}{l+1}-1+\sqrt{1-\frac{a^2}{{\left(1+b\right)}^2{\left(l+1\right)}^2}}\right]}^2}}}\\ +\frac{m_{20}}{\sqrt{1+\frac{a^2{b}^2/{\left(1+b\right)}^2{\left(l+1\right)}^2}{{\left[\frac{n}{l+1}-1+\sqrt{1-\frac{a^2{b}^2}{{\left(1+b\right)}^2{\left(l+1\right)}^2}}\right]}^2}}}\end{array}$ (21)

Here, n is the principal quantum number, l is the orbital angular momentum quantum number, M is the rest mass of the quark particle, and a is the gravitational coefficient of the n-th principal shell, b = m₁/m₂ is the dynamic mass ratio of the orbital particles. Regardless of the nature of the gravitational force that forms bound-state particles, the quantum rules they follow are consistent. Therefore, this equation is applicable to all two-body particles, including binary star systems. Essentially, each particle behaves as a microscopic magnet. The distinct coupling configurations and resultant spin combinations manifest as orientation variations between these magnetic dipoles, reflecting differential charge motion trajectories. According to relativistic principles, such variations induce modifications in the gravitational coupling coefficients. While the equation form remains invariant, the coefficients exhibit subtle quantum-state-dependent variations—this fundamentally accounts for hydrogen’s hyperfine structure. Crucially, incorporating this conceptual advancement into Equation (21) renders them universally applicable across all quantum states of all particles.

In 2023, Liu Zhe proposed an energy splitting equation for quarkonium triplet states to enable precise calculation of the mass of each quantum state in quarkonium systems:

$\begin{array}{l}\Delta V_{LS}=\frac{a_{nl}{\mu }_1{\mu }_2\sigma }{2m_1^2}\frac{l\left(l+1\right)}{f\left(f+1\right)}\langle LS\rangle \langle \frac{1}{d^3}\rangle \\ \langle \frac{1}{d^3}\rangle =\frac{m_0^3{a}_{nl}^3}{{\left(1+b\right)}^3{n}^3{\hslash }^3}\frac{1}{l\left(l+1\right)\left(l+1/2\right)}\\ A=\frac{m_0{a}_{nl}^4{\mu }_1{\mu }_2\sigma {\beta }_{nl}}{{\left(1+b\right)}^3{n}^{3}l\left(l+1\right)\left(2l+1\right)\sqrt{1-v_{nl}^2}}\end{array}\}\Delta V_{LS}=\{\begin{array}{l}+A{l}^2/\left(l+2\right),f=l+S\\ -A,f=l+0\\ -A{\left(l+1\right)}^2/\left(l-1\right),f=l-S\end{array}$ (22)

Here σ is the error coefficient, μ₁ is the spin magnetic moment of particle 1; μ₂ is the spin magnetic moment of particle 2. Theoretically, the error coefficient is related to the coupling method of the particle’s spin magnetic moment. When calculating the interaction coefficient, we have already taken the spin coupling method into account. Even if there are errors, they can be adjusted by selecting different gravitational coefficients. Therefore, we let σμ₁μ₂ = 1. The computational results of this equation demonstrate remarkable accuracy, substantially exceeding our previous studies, though minor discrepancies persist for low-n states. A year later, Liu Ziyi proposed a groundbreaking insight: the results computed from Equation (22) only modified the potential energy, which is not entirely equivalent to mass. This was a remarkably bold proposition—though it contradicted established textbook theories, the calculated results showed excellent agreement with experimental data, especially with improved accuracy for low-n states. According to Liu Ziyi’s method, the gravitational coefficient for each quantum state should be calculated using the following equation:

$\begin{array}{l}\frac{1}{{\beta }_{nl}}=\sqrt{1-\frac{a_{nl}^2}{{\left(1+b\right)}^2{n}^2}},\langle \frac{1}{d_{nl}}\rangle =\frac{m_1{a}_{nl}}{n^2\hslash \left(1+b\right)}\\ \frac{1}{{\beta }_{nlf}}=\sqrt{1-\frac{a_{nlf}^2}{{\left(1+b\right)}^2{n}^2}},\langle \frac{1}{d_{nlf}}\rangle =\frac{m_1{a}_{nlf}}{n^2\hslash \left(1+b\right)}\end{array}\}\{\begin{array}{l}\Delta V_{LS}=\frac{a_{nl}}{d_{nl}}-\frac{a_{nlf}}{d_{nlf}}\to \\ a_{nlf}^2=\left(1-\frac{n^2\Delta V_{LS}}{m_{nl}{a}_{al}^2}\right)\frac{a_{nl}^2{\beta }_{nl}}{{\beta }_{nlf}}\end{array}$ (23)

In the equation a_nlf represents the gravitational coefficient for the n-th principal layer, the l-th sub-layer, and the total spin f. By solving Equation (23), we obtain the analytical expression for the square of a_nlf:

$a_{nlf}^2=\frac{K^2}{2{\left(1+b\right)}^2{n}^2}\left(\sqrt{\frac{4{\left(1+b\right)}^4{n}^4}{K^2}+1}-1\right)\{\begin{array}{l}K=\left[1-n^2\frac{\left(1+b\right)\Delta V_{LS}}{m_{nl}{a}_{al}^2}\right]a_{nl}^2{\beta }_{nl}\\ \frac{1}{{\beta }_{nl}}=\sqrt{1-\frac{a_{nl}^2}{{\left(1+b\right)}^2{n}^2}}\end{array}$ (24)

To obtain the mass in the f-state, the gravitational coefficient derived from the above equation must be substituted into Equation (21). For quarkonium states, when b = 1, the following relation holds:

$\begin{array}{l}M=\frac{2m_{10}}{\sqrt{1+\frac{a^2/4{\left(l+1\right)}^2}{{\left[\frac{n}{l+1}-1+\sqrt{1-\frac{a^2}{4{\left(l+1\right)}^2}}\right]}^2}}}\\ a_{nlf}^2=\frac{K^2}{8n^2}\left(\sqrt{\frac{64n^4}{K^2}+1}-1\right)\end{array}\}K=\{\begin{array}{l}\left[1-\frac{l{a}_{nl}^2}{4n\left(l+1\right)\left(l+2\right)\left(2l+1\right)}\right]a_{nl}^2{\beta }_{nl},f=l+S\\ \left[1+\frac{a_{nl}^2}{4nl\left(l+1\right)\left(2l+1\right)}\right]a_{nl}^2{\beta }_{nl},f=l+0\\ \left[1+\frac{\left(l+1\right)a_{nl}^2}{4nl\left(l-1\right)\left(2l+1\right)}\right]a_{nl}^2{\beta }_{nl},f=l-S\end{array}$ (25)

According to Equation (25), we can calculate the mass of each quantum state in quarkonium systems. For example, substituting a = 7.31, n = 5, and l = 4 into Equation (25), we immediately obtain three distinct gravitational coefficients. These are then substituted and m₁₀ = 2282 into Equation (25) to derive the triplet mass of the quantum state with n = 5 and l = 4:

$a_{nlf}=\frac{K^2}{8n^2}\left(\sqrt{1+\frac{64n^4}{K^2}}-1\right)\to \{\begin{array}{l}{a}_{545}\approx \text{7}\text{.215874,}{M}_{545}\approx 3159.72,f=l+S=5\\ a_{544}\approx \text{7}\text{.344173,}{M}_{544}\approx 3097.58,f=l+0=4\\ a_{543}\approx \text{7}\text{.577588,}{M}_{543}\approx 2978.17,f=l-S=3\end{array}$ (26)

The calculation results show that the mass states 3097 and 2980 surprisingly belong to the same triplet group. Before this, there was significant controversy regarding the classification hierarchy of the 2980 quantum states. Our calculations reveal that accompanying the 3097 state is a 3160 mass datum—representing a previously undiscovered quantum state. We maintain that this provides an ideal opportunity for experimental verification of our theoretical computations’ accuracy. Using the same method, substitute a = 7.291, n = 6, l = 4 and m₁₀ = 2282 into Equation (25):

$a_{nlf}=\frac{K^2}{8n^2}\left(\sqrt{1+\frac{64n^4}{K^2}}-1\right)\to \{\begin{array}{l}{a}_{645}\approx \text{7}\text{.196991,}{M}_{645}\approx 3555.6,f=l+S=5\\ a_{644}\approx \text{7}\text{.325426,}{M}_{644}\approx 3508.9,f=l+0=4\\ a_{643}\approx \text{7}\text{.567511,}{M}_{643}\approx 3415.3,f=l-S=3\end{array}$ (27)

To ascertain the precise mass of the quark c, we conducted extensive calculations and comparisons. This comparative verification work, spanning over a decade, involved extensive computational efforts. Table 1 below presents one representative example from our numerous comparative studies [11] .

The calculations indicate that only when the mass of the quark c is 2282 can it align with the experimental values. Consequently, the mass number, spin number, and charge number of the quark c are identical to those of the ordinary Λc particle. This discovery reveals that certain high-spin particles exhibit strong interactions at close range while behaving like ordinary particles at larger distances. Precisely because these ordinary particles carry a unit charge (q = 1), and our equations are calculated based on this unit charge, the results achieve unprecedented agreement with experimental data. We therefore conclude that quarks are such ordinary particles whose charges must be integer-valued rather than fractional. Research on other quarkonium states corroborates this finding, for instance, studies on the Υ particle suggest that the mass of the bottom quark is 5580, with spin and mass numbers very close to those of the ordinary Λ_b particle. Therefore, when Λ_b transforms into the bottom quark, its mass changes from 5500 to 5580, and its charge number increases by 1, analogous to the change in mass when a neutron transforms into a proton.

According to Equation (25), we get the following Table 2 [11] .

The number 0 in the table indicates that this quantum state does not exist.

In Table 2 , a is the corresponding gravitational coefficient. The underlined data in the table represent the mass numbers that have been experimentally discovered. According to the calculations, the experimental value 3885 corresponds to the triplet state with n = 9 and l = 4, therefore, it is not listed in Table 2 .

In previous studies of the J/ψ particle, we always considered the ground state to be n = 1, not only were the margins of error considerable, but there was also a plethora of experimental data that defied explanation, which the physics community termed as exotic energy levels [12] [13] . The research outcomes of our team have adeptly resolved these conundrums, with all experimental data aligning impeccably with the calculated values. It is evident that the so-called anomalous energy levels previously discussed in the physics community were merely a result of the lack of correct computational methods.

In our calculations, we discovered that some particles lack the n = 1 quantum state. Such particle systems have been named “Suspended State Particles” by Liu Yun. This is a significant discovery as it provides an explanation for why many particles (such as the electron) exhibit only a single quantum state. How do we determine whether a quantum state exists for a particle? First, we assume the spin rate and radius of the quark and aim to obtain data consistent with experimental values. The calculations indicate that the spin rate of quark c is about 0.999436c and a spin radius of 3.66 × 10⁻¹⁷ m = 0.0366 fm. Under these conditions, according to Equation (18), we omit the less significant terms in the formula and obtain the equation for the gravitational coefficient between quark c and anti-quark c (with S = 1):

$\begin{array}{l}{\langle a_{c\bar{c}}\rangle }_{S=1}\approx a_e+\frac{a_e}{2}\frac{1+{\left(2r/d\right)}^2}{{\left[1-{\left(2r/d\right)}^2\right]}^2}\left({\beta }_{c+c}-1\right)-d^2{F}_{0B}\\ \approx \frac{a_e}{2}\frac{1+{\left(2r/d\right)}^2}{{\left[1-{\left(2r/d\right)}^2\right]}^2}{\beta }_{c+c}\approx 6.465\frac{1+{\left(2r/d\right)}^2}{{\left[1-{\left(2r/d\right)}^2\right]}^2},{\beta }_{c+c}=\frac{1}{\sqrt{1-v_{c+c}^2}}\end{array}$ (28)

Using the same computational method, we determine that the radius of the bottom quark (quark b) is approximately 4.1 × 10⁻¹⁷ m (0.041 fm), with a spin rate of about 0.999437c. Only these precise values yield results consistent with experimental data. Consequently, based on Equation (18), the gravitational coefficient equation between a quark b and its antiparticle (anti-quark b) is:

${\langle a_{b\bar{b}}\rangle }_{S=1}\approx \frac{a_e}{2}\frac{1+4r^2/d^2}{{\left(1-4r^2/d^2\right)}^2}{\beta }_{b+b}\approx 6.4815\frac{1+4r^2/d^2}{{\left(1-4r^2/d^2\right)}^2},d=\left(1+b\right)r_1$ (29)

Table 3 is the result of our calculation of the gravitational coefficient between quarks using the above Equation (25) and Equation (29) [11] .

The underlined numbers are the mass states that have been discovered.

With the gravitational coefficient Equation (29), we can determine which energy levels of the Υ particle can exist stably. This method is called the Approximation Correction Method, which we developed from the work of the American physicist Feynman. This rule applies to all particles, including hydrogen atoms.

First, having confirmed the genuine existence of the n = 7 quantum state, we substitute m₀ = 5580 and a = 7.478 into Equation (5), calculating the radius of the Υ-particle at principal quantum number n = 7 to be approximately 0.188 fm. Using Equation (21), we determine the mass of this state to be about 9460.

Second, we estimate the radius for the n = 6 principal quantum level based on quantum mechanical principles. The radius r₆ of the n = 6 level is extrapolated from the n = 7 radius r₇ through the relation: r₆ = (36/49)r₇ ≈ 0.145 fm.

Third, we perform radius correction for the n = 6 state: Input the estimated radius into Equation (29) to obtain the strong interaction coefficient a = 8 for the n = 6 quantum state. Substitute a = 8 back into Equation (5) to derive a corrected radius value.

Fourth, we adjust the gravitational coefficient for the n = 6 quantum state by: Inserting the revised radius into Equation (29). Computing an updated gravitational coefficient.

Fifth, we implement an iterative correction procedure. Through multiple cycles of mutual refinement: If errors progressively amplify → This indicates the Υ-particle cannot maintain stable configuration at n = 6 level, confirming its non-existence. If errors consistently diminish and converge to zero → This verifies a stable, physically realizable quantum state.

Our computational results demonstrate that the Υ-particle cannot stably exist in the n = 6 quantum state, confirming the non-existence of this state.

For decades, extraordinarily precise measurements of cosmic 21 cm hydrogen line frequencies have been achieved, reaching remarkable levels of accuracy. However, despite exhaustive computational efforts through various methodologies, persistent minor discrepancies between theoretical predictions and experimental values have remained unexplained. In different particle systems, the form of spin coupling varies due to differences in particles’ spin magnetic moments and masses. However, spin does not alter the fundamental quantum rules—it only modifies the magnitude of interaction coefficients. Based on this theory, we incorporate all particle systems into the quantum mechanical framework governed by Equation (21), including the hyperfine structure splitting rules of hydrogen atoms.

When n = 1, l = 0, j = 1/2, s₂ = 1/2, it corresponds to the ground state of the hydrogen atom. Proton-electron spin coupling—with quantum mechanics prescribing the resultant hyperfine splitting as:

$\begin{array}{l}{V}_{JS}=-\frac{a_{nl}{g}_p{g}_e}{2m_1^2}\frac{l\left(l+1\right)}{j_1\left(j_1+1\right)}\langle \frac{1}{d^3}\rangle \langle IJ\rangle \\ \langle IJ\rangle =\frac{1}{2}\left(F^2-J^2-I^2\right),\hslash =1\\ \langle \frac{1}{d^3}\rangle ={\left[\frac{m_1{a}_{nl}}{\left(1+b\right)n}\right]}^3\frac{1}{l\left(l+1\right)\left(l+1/2\right)}\end{array}\}\to V_{JS}=\{\begin{array}{l}-\frac{m_1{a}_{nl}^4{g}_e{g}_p}{3{\left(1+b\right)}^3{n}^3},f=j_1+s_2=1\\ +\frac{m_1{a}_{nl}^4{g}_e{g}_p}{{\left(1+b\right)}^3{n}^3},f=j_1-s_2=0\end{array}$ (30)

To minimize computational errors and achieve higher precision, it becomes necessary to account for variations in the gravitational coupling coefficient a, thus requiring the reformulation of Equation (30) as follows:

$a_{nlf}^2=\frac{K^2}{2{\left(1+b\right)}^2{n}^2}\left(\sqrt{1+\frac{4{\left(1+b\right)}^4{n}^4}{K^2}}-1\right),K=\{\begin{array}{l}\left[1+\frac{a_{nl}^2{g}_e{g}_p}{3{\left(1+b\right)}^{2}n}\right]a_{nl}^2{\beta }_{nl}\text{,}f=1\\ \left[1-\frac{a_{nl}^2{g}_e{g}_p}{{\left(1+b\right)}^{2}n}\right]a_{nl}^2{\beta }_{nl}\text{,}f=0\end{array}$ (31)

The following parameters were adopted for calculations:

Electron g-factor: g_e = 2 × 1.00115965218076;

Proton g-factor: g_p = 5.58569469/1836.15266 ≈ 1/328.7241358335;

Electron rest mass: m_e = 510,998.928 eV;

Proton-to-electron rest mass ratio: b₀ ≈1/1836.152669971942;

Dynamic mass ratio (n = 1): b ≈ 1/1836.1038351084 = 0.0005446315077;

Gravitational coupling coefficient (Lamb shift corrected): a ≈ 1/137.0379202;

Relativistic correction factor (n = 1, l = 0): β₁₀ = 1.000026597013;

Using the Planck constant h = 6.62606957(11) × 10³ J·s and the elementary charge e = 1.602176565 (35) × 10⁻¹⁹ C, we derive the following conversion relationship:

1 eV = (e/h) Hz ≈ 241,798.9349604731 GHz;

Substituting these parameters into Equation (31), the calculated gravitational coupling coefficients a for respective quantum states are as follows: ‬

$a_{nlf}=\{\begin{array}{l}n=1,l=0,f=1\text{,}{a}_{101}=1/137.037927599954025206123\\ n=1,l=0,f=0,a_{100}=1/137.037898000399784881604\end{array}$ (32)

These represent the gravitational coupling coefficients a_nlf between protons and electrons for the two spin configurations. While the difference is minute, it fundamentally revises our previous assumption that the coupling coefficients were identical across quantum states. Substituting these two calculated a-values into Equation (21), we obtain the mass difference for different f-state hydrogen atoms at the n = 1 principal quantum level:

$\begin{array}{l}{E}_{101}\approx -13.5980845804222370525\text{eV}\\ E_{100}\approx -13.5980904547481579742\text{eV}\end{array}\}\Delta E\approx 5.874325920922\times {10}^{-6}\text{eV}$ (33)

Since 1 eV≈ 241,798.934 960 473 1 GHz, the conversion result is: ΔE ≈ ‬1.420405751289589716 GHz.

The experimentally measured value is 1.4204057517667(10) GHz. This is the famous 21-centimeter hydrogen line in cosmic radiation. This study employs Liu Ziyi’s method to calculate the 21 cm hydrogen spectral line frequency, yet obtains results that show remarkable agreement with experimental values. The previous theoretical calculation yielded a precise value of 1.4204034(13) GHz (refer: V.W. Hughes. In ATOMIC PHYSICS 10, edi. by H. Narumi and I. Shimamura. Elsevier Sci. Pub. 1987). The latest CODATA 2023 revision has once again updated the electron-to-proton mass ratio, indicating that our calculations are still not entirely precise. Indeed, achieving such precise theoretical calculations necessarily depends on the accuracy of the Lamb shift measurement. Therefore, these calculations also serve as a means to verify the Lamb shift, while further research on the Lamb shift proves highly beneficial and stimulating.

According to the period equation for elliptical orbits:

$m_1\left(\frac{2\pi }{T}\right)r_1\approx \frac{a}{\left(1+b\right)r_1^2}\to T\approx 2\pi \sqrt{\frac{\left(1+b\right)m_1{r}_1^3}{a}},a=\frac{G{m}_1{m}_2}{c\hslash }$ (34)

Here, m₁ is the mass of the Earth and m₂ is the mass of the Sun. The mass of the Earth is about 6 × 10²⁴ kg, and the mass of the Sun is about 2 × 10³⁰ kg. The force coefficient a between Earth and the Sun is derived from the law of universal gravitation, expressed as:

$a=\frac{G{m}_1{m}_2}{c\hslash }\approx \frac{6.67\times {10}^{-11}\times 2\times {10}^{30}\times 6\times {10}^{24}}{3\times {10}^8\times 1.055\times {10}^{-34}}\left(\text{SI}\right)\approx 2.52\times {10}^{70}$ (35)

The “SI” in the equation indicates that all units are in the International System of Units. According to human measurements and estimates, the average speed of the Earth’s orbit around the Sun is 29.8 km/s ≈ 10⁻⁴c. This value can also be used to calculate the gravitational parameter of the Earth-Sun system. Be careful to convert to natural units, v₁ = 2.98 × 10⁴ m/s ≈ 10⁻⁴, c = 1. This is consistent with the results calculated using quantum mechanical formulas.

$\begin{array}{l}n=\left(1+b\right)\frac{m_1{v}_1{r}_1}{\hslash }=2.52\times {10}^{74},a=\left(1+b\right)n{v}_1\\ n\hslash =\left(1+b\right)m_1{v}_1{r}_1,b\approx 0,v_1={10}^{-4},c=1\end{array}\}a\approx 2.52\times {\text{10}}^{\text{70}}$ (36)

Quantization phenomena are the statistical outcomes of a large number of particles (See Figure 3 ).

The smaller the perihelion of an elliptical orbit radius, the larger its aphelion becomes, consequently increasing orbital uncertainty. This demonstrates that the radius of microscopic particles coincides with planetary radii, strictly obeying point-mass mechanics while simultaneously complying with both quantum mechanical and relativistic principles. The perfect unification of relativity and quantum theory represents the correct path forward for the next stage of physics development. According to Equation (36), the principal quantum levels of the nine planets of the Solar System are calculated (see Table 4 ).

The star cluster mentioned in the table refers to a scattered planetary belt composed of numerous small celestial bodies. From Table 4 , the principal quantum number n does not follow an obvious pattern. However, if planetary masses are hypothetically replaced with identical masses, e.g., imagining free electrons occupying the same orbits as planets, the motion of electrons orbiting the Sun would also exhibit quantized orbits. In this scenario, the calculated n-values show a nearly geometric progression pattern. This regularity arises because the quantum number n is extremely large. This observation suggests that during the early formation of the solar system, an explosive event likely generated numerous dust particles or micro-particles with similar masses. These particles gradually coalesced into planets through quantized orbital aggregation. Calculations indicate that most celestial bodies do not possess quantized orbits with n = 1. This is also a consequence of particles aggregating to form larger objects. The Earth’s aphelion distance is 1.52 × 10¹¹ m, perihelion distance is 1.471 × 10¹¹ m, with an average orbital speed of 2.98 × 10⁴ m/s ≈ 10⁻⁴c. Calculate the value of the orbital angular quantum number l for Earth’s orbit using the equation:

$\begin{array}{l}{r}_A\approx \frac{L\hslash }{m{v}_A},r_B\approx \frac{L\hslash }{m{v}_A},v=\frac{a}{n},m=5.97\times {10}^{24}\text{kg}\\ v_A\approx \frac{a}{L}\left(1-\sqrt{1-\frac{L^{\text{2}}}{n^2}}\right),v_B\approx \frac{a}{L}\left(1+\sqrt{1-\frac{L^{\text{2}}}{n^2}}\right)\end{array}\}\Delta r\approx \frac{2\hslash \sqrt{n^2-L^2}}{mv}=5\times {10}^9\text{m}$ (37)

We use the Bohr radius substitution method to calculate the results as follows:

$\begin{array}{l}A:\Delta r=\frac{2\hslash \sqrt{n^2-L^2}}{mv}\approx \frac{2\hslash \sqrt{n^2-L^2}}{6\times {10}^{24}\text{kg}\times {10}^{-4}}=5\times {10}^9\text{m}\\ B:a_0=\frac{\hslash }{m_{\text{e}}{v}_e}\approx \frac{\hslash }{9.1\times {10}^{-31}\text{kg}\times \left(1/137\right)}=5.3\times {10}^{-11}\text{m}\end{array}\}\frac{A}{B}:n^2-L^2\approx 2.04\times {10}^{145}$(38)

Here, a₀ is the Bohr radius. The calculations reveal that the values of L and n are nearly identical:

$\begin{array}{l}{n}^2-L^2=2.04\times {10}^{145}\\ n=2.52\times {10}^{74}\end{array}\}L\approx \sqrt{6.35\times {10}^{148}-2.04\times {10}^{145}}\approx 2.5195\times {10}^{74}$ (39)

Then, a = 2.5 × 10⁷⁰, v = a/n = 10⁻⁴, n ≈ l ≈ 2.52 × 10⁷⁴, b ≈ 0, substituted into Equation (21) to calculate the total rest mass of the Earth and the Sun combined into a two-body particle:

$M\approx m_{10}\left(1-\frac{v_1^2}{2}\right)+m_{\text{20}}$ (40)

Here, m₁₀ is the rest mass of the Earth, and m₂₀ denotes the rest mass of the Sun. This indicates that after Earth was gravitationally captured by the Sun, it would need to release energy equivalent to its own kinetic energy. This rule is identical to the principle of energy release when an electron is captured by a proton in a hydrogen atom.

We have long been perplexed by the formation mechanism of electron magnetic moment. For a moving charge observed perpendicular to its direction of motion, applying the relativistic factor to the electric field and current is equivalent to applying the relativistic factor directly to the charge itself. In multi-charge systems, all orbital charges generate magnetic moments, and the total orbital magnetic moment is the superposition of these individual orbital magnetic moments. Therefore, when observed from the particle’s center-of-mass reference frame, the total orbital magnetic moment for a rotating two-body particle system can be expressed as:

${\mu }_l\approx \frac{e_1{v}_1{r}_1{\beta }_1}{2}+\frac{e_2{v}_2{r}_2{\beta }_2}{2}\approx \frac{\left(e_1+b_{0}b{e}_2\right)L}{2m_{10}\left(1+b\right)},b_0=\frac{m_{10}}{m_{20}},b=\frac{m_{10}{\beta }_1}{m_{20}{\beta }_2}$ (41)

When two magnetic moments approach each other, mutual interference occurs, causing both the particle radii and their magnetic moments to increase, though with differing magnitudes. Precise calculation of these variations remains challenging due to dependencies on multiple factors, including the velocity and directional dynamics of internal charges. Here, we present an approximate computational framework to address this phenomenon. In the magnetic field formed by protons, the radius of the circular motion of each positive charge of neutrons will decrease, while that of each negative charge will increase. Other than that, the speed of the charge’s movement will not change. This leads to an increase in the magnetic moment value formed by each negative charge and a decrease in the magnetic moment value formed by each positive charge (see Figure 4 ).

In a deuterium nucleus, the magnetic moment of the proton exerts an electromagnetic force on the π meson within the neutron, causing it to move away from the neutron. This is equivalent to the π meson experiencing a reduced central gravitational force. Considering the relativistic effect, as shown in Figure 4(A) , the average force on the valence charge within the neutron is expressed as:

$\frac{m_{\pi }{v}^2}{{r^{\prime }}_n}\approx \frac{a_n}{d_n^2}-\frac{1}{{\left(d-r_n-r_p\right)}^2}{\beta }_{n+p},v_{n+p}=\frac{v_n+v_p}{1+v_n{v}_p}$ (42)

We provide an estimation method here. According to Equation (14), we estimated the result as:

$\begin{array}{l}{r^{\prime }}_n\approx \frac{r_n}{1-\frac{d_n^2}{{\left(d-r_n-r_p\right)}^2}\frac{a_e}{a_n}{\beta }_{p+n}},a_n\approx \text{2}.5,d_n=1.\text{2}\text{fm}\\ r_n\approx \text{0}\text{.884}\text{fm},r_p\approx 0.83\text{fm},d=4.32\text{fm},v_{n+p}\approx 0.997\end{array}\}{r^{\prime }}_n\approx 1.0087r_n$ (43)

Here, r' is the changed neutron radius, d_n = 1.2fm is the neutron diameter, as shown in Figure 4(A) . The increase in the neutron radius results in a change in the neutron magnetic moment of approximately:

$\Delta {\mu }_n=v_n\left({r^{\prime }}_n-r_n\right)\frac{e_1+b_{0}b{e}_2}{2\sqrt{1-v_n^2}}\approx 0.0087\times 1.913{\mu }_N\approx -0.0166{\mu }_N$ (44)

In Equation (44), we only considered the change of the magnetic moment of motion of the charge 4 within the neutron. When considering the cumulative variations from all internal charged constituents, the net predicted variation in the neutron’s magnetic moment amounts to approximately −0.035μ_N. Similarly, within the neutron’s magnetic field, the proton’s magnetic moment also undergoes modification. While current theoretical frameworks cannot precisely calculate this perturbation, empirical estimates suggest a shift of about +0.012μₙ. The total magnetic moment of the deuteron spin is:

$\begin{array}{l}{\mu }_D\approx \frac{\left(e_1+b_{0}b{e}_2\right)m_s\hslash }{2\left(1+b\right)m_p}+{\mu }_p-{\mu }_n-0.035{\mu }_N+0.012{\mu }_N\\ n=1,m_s=0,e_1=1,e_2=0,{\mu }_p\approx 2.7928{\mu }_N,{\mu }_n\approx 1.913{\mu }_N\end{array}\}{\mu }_D\approx 0.857{\mu }_N$ (45)

The orbital magnetic moment projection along the spin axis is quantized as m_s. Some scholars believe that the anomalous magnetic moment of the deuteron is due to part of the deuteron being in the 3D state [14] [15] . However, our explanation is more credible because the same method can account for the other particles magnetic moment. Moreover, all composite particles exhibit a universal magnetic moment behavior: the net magnetic moment never exactly equals the strict vector sum of constituent particles’ intrinsic moments, but rather demonstrates subtle deviations. For instance, in the npn particle system (tritium nucleus, ³H), the observed variation measures approximately +0.18616 μ_N.

$\begin{array}{l}{\mu }_{npn}\approx \frac{\left(e_1+b_{0}b{e}_2\right)m_s\hslash }{2\left(1+b\right)m_n}+{\mu }_n-{\mu }_n+{\mu }_p+\Delta {\mu }_p\\ n=1,m_s=0,{\mu }_p\approx 2.7928{\mu }_N,\Delta {\mu }_p\approx 0.18616{\mu }_N\end{array}\}{\mu }_{npn}\approx 2.97896{\mu }_N$ (46)

For the pnp particle system (³He nucleus), the magnetic moment variation measures approximately −0.21575 nuclear magnetons (μₙ). The magnetic moment of the helium-3 nucleus is given by:

$\begin{array}{l}{\mu }_{pnp}\approx \frac{\left(e_1+b_{0}b{e}_2\right)m_s\hslash }{2\left(1+b\right)m_p}+{\mu }_n+{\mu }_p-{\mu }_p+\Delta {\mu }_n\\ n=1,m_s=0,{\mu }_n\approx -1.913{\mu }_N,\Delta {\mu }_n\approx -0.21475{\mu }_N\end{array}\}{\mu }_{pnp}\approx -2.12775{\mu }_N$ (47)

Professor Samuel C.C. Ting conducted experiments exploring the electron’s structure, revealing no structural features at 10⁻¹⁸ m. In 1987, Hans Georg Dehmelt and colleagues used electromagnetic trapping experiments to determine that the electron’s radius is less than 10⁻²² m (Refer to Dehmelt’s 1989 Nobel Prize lecture). For decades, people could not understand why the electron’s radius was so small, and many assumed that it was a point particle, which led to an inability to explain how the electron’s spin and magnetic moment were produced.

Based on our derived radius and magnetic moment formulas, the internal charged constituents of an electron can only be very light charged particles, with muons (μ) being the sole candidates satisfying these constraints. Further theoretical investigation reveals that when a muon (mass ≈ 106 MeV) combines with a Z boson (mass ≈ 91 GeV) under radius collapse to 2.53 × 10⁻²¹ m in the n = 413 excited state, the resulting composite system precisely reproduces the electron’s measured mass and magnetic moment. To identify which particles, when combined with the Z boson, yield a radius and magnetic moment consistent with experimental values, we present the calculated results for various candidate particles in Table 5 .

The underlined words are basically consistent with the experiment. 1G = 1000 M.

Assuming the electron is synthesized from μ- and the Z⁰ boson, the electron radius is:

$\begin{array}{l}{r}_e\approx \frac{n\sqrt{1-v^2}\hslash }{\left(1+b\right)m_{\mu }v},n=413,b=0.\left(10N\right)824\\ v=0.\left(16N\right)785c,m_{\mu }\approx 105.65837\text{MeV},c=\text{1}\end{array}\}{r}_e\approx -2.53\times {10}^{-21}\text{m}$ (48)

The magnetic moment of the electron is:

$\begin{array}{l}{\mu }_e\approx \frac{\left(e_1+b_{0}b{e}_2\right)m_s\hslash }{2\left(1+b\right)m_{\mu }}+{\mu }_{\mu }+{\mu }_0,{\mu }_0\approx 0,e_1=-1\\ m_s=412,e_2=0,b\approx 1,{\mu }_{\mu }\approx -0.004842371{\mu }_B\end{array}\}{\mu }_e\approx -1.001159652{\mu }_B$ (49)

The quantum mechanical representation formula of the electronic magnetic moment is:

$s=\frac{1}{2},g_s=2\times 1.00115962\to {\mu }_e=-s{g}_s{\mu }_B\approx -1.00115962{\mu }_B$ (50)

The 2010 Committee on Data for Science and Technology (CDST) provided that:

The average mass of the muon as 105.6583715(35); The magnetic moment of the muon as μ_μ = 8.89059697(22)μ_N ≈ 0.004842371μ_B; The magnetic moment of the electron as μ_e = 1.00115965218618076(27)μ_B.

We propose that while a free electron exhibits a characteristic radius of 10⁻²¹ m, its spatial extension undergoes significant contraction when subjected to strong negative magnetic confinement. This phenomenon arises from the circular motion of the electron’s negative charge within the opposing magnetic field, resulting in measurable radius reduction—consistent with Dehmelt’s experimental observations where confined electrons demonstrated radii below 10⁻²² m [16] [17] .

Because such reactions have been observed experimentally [18] :

$\begin{array}{l}A:\gamma +p^{+}\to n+{\pi }^{+}\to p^{+}+{\pi }^0\\ B:{\pi }^{+}+p^{+}\to {\pi }^{+}+{\pi }^{+}+n\end{array}$ (51)

The reaction process described by Equation (51A) is as follows: When a photon collides with a proton, π and π particles are produced. Subsequently, the π particle combines with the proton to form a neutron. Next, the positively charged π strips the negatively charged π from the neutron, synthesizing a neutral π particle. The neutron, now stripped of mesons, reverts to a proton. Similarly, Equation (51B) illustrates that when a π collides with a proton, it generates π and π particles. The π particle then combines with the proton to form a neutron. Our research has demonstrated that as a particle’s quantum state approaches annihilation, only a single state exists at the sublevel where l = n − 1, then, Equation (21) simplifies to:

$M=\frac{\left(1+b\right)m_{10}}{\sqrt{1-v_1^2}}\left(\frac{1}{b}-v_1^2\right),b^2=\frac{1}{B^2-v_1^2\left(B^2-1\right)},B=\frac{1}{b_0}=\frac{m_{20}}{m_{10}}$ (52)

The following table shows the quantum parameters for the formation of neutrons by various mesons and protons to compare which scenario is most likely. Table 6 shows the results calculated using Equation (52).

Table 6. Compare the possible structures of neutrons.

Table 6 presents the resulting radii and orbital velocities when composite particles formed from different initial particles reach a mass equivalent to that of a neutron. Here p938.3 represents a proton with a mass of 938.3. The experimentally measured neutron radius ranges from 0.8 to 0.9 fm. We incorporated this factor when determining the value of n. The underlined data are in agreement with the experiment. From Table 6, we can get the following results: A neutron is synthesized from a π-meson and a proton, the radius of a neutron is calculated to be 0.884 fm, with a spin rate of v_n = 0.921c, mass ratio b = 0.36, existing in the quantum state of n = 2, so m_s = 1. We present the equation for calculating the magnetic moment of the neutron:

$\begin{array}{l}{\mu }_n=\frac{\left(e_1+b_{0}b{e}_2\right)m_s\hslash }{2\left(1+b\right)m_{\pi }}+{\mu }_p+{\mu }_{\pi },{\mu }_p\approx 2.7928{\mu }_N,m_s=1\\ e_1=-1,e_2=1,b\approx 0.36,b_0=\frac{m_{\pi }}{m_p}\approx 0.148752,{\mu }_{\pi }\approx -0.0274{\mu }_N\end{array}\}{\mu }_n\approx -1.913{\mu }_N$(53)

All magnetic moments discussed herein are defined by their spin-axis projections. The orbital magnetic moment projection along the spin axis is quantized as m_s. Since a neutron transforms into a proton by losing a π meson, it is said that a π meson is exchanged when a neutron and a proton transition between each other. This aligns closely with Hideki Yukawa’s theory.

When a proton and a neutron form a deuteron, the binding energy is approximately 2.224 MeV. Therefore, the calculated orbital velocity of the proton is about 0.0489c. According to Equation (36), the gravitational coefficient between the proton and the neutron is: a = (1 + b)nv₁ ≈ 1/10.22. In Section 2.9, we provide a detailed method for calculating this gravitational coefficient and demonstrate that the gravitational force between protons and neutrons still arises from the rotational motion of charges (See Equation (56)). From this, it can be seen that the exchange of gluons between protons and neutrons is not the essence of the strong interaction. Rather, the rotation of particles and their charge distribution constitute the fundamental cause underlying the formation of the strong interaction.

The calculation of three-body particle systems is complex, but still strictly abide by the quantum rules. For instance, when an atomic nucleus captures two electrons, forming a helium-like ion, the binding energy of the second captured electron involves three-body computations. Our proposed formula for this is:

$\begin{array}{l}{W}_1=E_0\left[Z\left(Z-0\right)-0\right]\\ W_2=E_0\left[Z\left(Z-1\right)-\sigma \right]\\ \sigma \approx {\displaystyle \underset{i=1}{\overset{i=k}{\sum }}\frac{i^2-1}{2^{2n}{i}^{2n}}},n=1\\ E_0\approx 13.6\text{eV},k=Z\end{array}\}\{\begin{array}{l}Z=2,W_2=E_0\left(2-\frac{3}{4^2}\right)\approx 24.65\text{eV}\left(24.6\text{eV}\right)\\ Z=3,W_2=E_0\left(6-\frac{3}{4^2}-\frac{8}{36}\right)\approx 76\text{eV}\left(75.5\text{eV}\right)\\ Z=6,W_2=E_0\left(30-\frac{3}{4^2}-\cdots -\frac{35}{{12}^2}\right)\approx 392.7\text{eV}\left(392.2\text{eV}\right)\end{array}$(54)