The five types of strong interaction have covered all strong interaction forms known in the world. This paper aims at comprehensively studying the internal symmetrical structures of hadrons and the force action from the aspect of the representative particles of the five types of strong interaction. The basic research thought is as follows:

1) Propose relevant hypothesis;

2) Infer and demonstrate according to the hypothesis;

3) Compare theoretical value and experimental value;

4) Obtain the conclusion.

The detailed analysis process includes nine steps:

1) Hypothesis for basic strong interaction;

2) Analysis of strong interaction of nucleon (including p and n);

3) Analysis of strong interaction of Σ particle family;

4) Analysis of strong interaction of π meson;

5) Analysis of strong interaction of K meson;

6) Analysis of strong interaction of atomic nucleus (e.g., deuteron and triton);

7) Analysis of strong interaction of Δ (1232);

8) Analysis of strong interaction of η particle and η ′ ( 958 ) particle;

9) Conclusion.

1) Charge type quantity of strong force: quark quantity of neutron or proton is 3, so the charge type quantity of the corresponding strong force is 3; one neutron and one proton can form the simplest atomic nuclear structure, so the charge type quantity of the corresponding strong force is 2. The corresponding charge type of the two types of the most basic strong interaction is 3 or 2.

{ ∑ 1 ≤ i ≤ n n = 3 , 2 e i = 0 e i ⋅ e i = 0 (1-1)

where e i is charge type; hadrons are usually neutral and can present force action only under the distance small enough, so the sum of charge types is 0.

It can be proved that the formula (1-1) is unsolved in the range of real number. We must extend the domain from the real to complex number, and adjust the formula to have a real root.

{ ∑ 1 ≤ i ≤ n n = 3 , 2 e i ( ∗ ) = 0 e i ⋅ e i ∗ = 0 (1-2)

e i and e i ∗ are conjugate complex number each other.

The following formula is set:

{ e j = a j + b j i e j ∗ = a j − b j i (1-3)

a and b are real numbers.

Formula (1-3) are put into the formula (1-2) to obtain:

{ ∑ 1 ≤ i ≤ n n = 3 , 2 a j + b j i = 0 ( a j + b j i ) ( a j − b j i ) = 1 (1-4)

When n = 2,

{ a 1 , 2 = ± 1 b = 0 (2-1)

A group of solutions of charge type of π meson type strong force

e = ± 1 (2-2)

When n = 3, considering that neutrons and protons can combine to form stable atomic nuclei, that is to say, there must be a set of solutions (2-1),

Formulae (1-4), (2-1) are combined to solve.

{ a 1 = ± 1 , b 1 = 0 a 2 = ∓ 1 2 , b 2 = ∓ 3 2 i a 2 = ∓ 1 2 , b 3 = ∓ 3 2 i (2-3)

A group of solutions of charge type of nucleon type strong force,

{ e 1 = 1 + 3 i 2 e 2 = 1 − 3 i 2 e 3 = − 1 ,         { e 1 ∗ = − 1 − 3 i 2 e 2 ∗ = − 1 + 3 i 2 e 3 ∗ = 1 (2-4)

when n = 3, the strong interaction between three forced particles can be decomposed into any two forced particles. Then we deduce the ratio of n = 3 to n = 2’s Strong force intensity by formula (1-1).

∑ j ≠ m n = 3 e j e m ∗ 2 e + e − = ( ∑ j n = 3 e j ) ( ∑ j n = 3 e j ∗ ) − ∑ j n = 3 e j e j ∗ ( e + + e − ) 2 − e + 2 − e − 2 = 3 2 (3)

2) Present two types of symmetrical structures by deuteron as shown in Figure 2.

1) Strong force potential shall have universal attributes as the same as gravitational potential and electrostatic potential, namely:

V ^ s ( r ) ∝ − a r (4)

2) Special attributes of strong force potential: the most significant difference between hadrons and atoms is that hadrons only have ground state, but atoms have multiple states. In order to ensure the stability of hadrons and avoid external energy excitation, a condition for limiting state transition shall be added on the basis of Formula (1), namely:

V ^ s ( r ) ∝ − a r F ( φ ) (5)

where φ is the ground state function of hadrons, and F ( φ ) is the variable for limiting state transition.

3) a value: for the strong nuclear force as the most basic force for maintaining the internal stability of hadrons, the intensity thereof is approximate to the product of the reduced Planck constant and light velocity, and this indicates that strong nuclear force is certainly related to fluctuation. The following hypothesis is made according to such relation:

a) Strong force intensity of π meson

a = 2 × ℏ c 2 = ℏ c (6)

b) Strong force intensity of nucleon

a = 3 × ℏ c 2 = 3 2 ℏ c (7)

4) Ground state function of hadrons: the state function of hadrons is unique and cannot be influenced by strong force potential, approximate to or as the same as that of protium atom (Schrodinger’s equation is adopted to calculate the ground state function of protium atom for the most accurate comparison).

φ ( r , a ) = exp ± r / a (8-1)

where r and a are the variables related to distance.

The state function is normalized:

φ ( r , a ) = exp ± ( r / a − 1 ) (8-2)

5) F ( φ ) formula

F ( φ ) = φ 2 = exp ± 2 ( r / a − 1 ) (9)

6) Solution of strong force potential: two solutions of strong force potential can be obtained according to Formulae (4)-(9).

a) Strong force potential formula of π meson

V ^ S ( r ) = − ℏ c r exp − 2 ( r / a − 1 ) (10)

or

V ^ S ( r ) = − ℏ c r exp + 2 ( r / a − 1 ) (11)

b) Strong force potential formula of nucleon

V ^ S ( r ) = − 3 / 2 ℏ c r exp + 2 ( r / a − 1 ) (12)

V ^ S ( r ) = − 3 / 2 ℏ c r exp − 2 ( r / a − 1 ) (13)

Five issues shall be explained for 1) and 2).

1) In actual conditions, π meson only has strong force potential formula (10);

2) The representative particle carrying (11) includes positive and negative K mesons;

3) In actual conditions, nucleon only has strong force potential formula (12);

4) The representative particle carrying (12) includes Σ particles family;

5) Strong force potential has four types, represented by Formulae (10), (11), (12) and (13).

Dirac equation [4] can be adopted to describe the internal movement of hadrons

( i γ ∂ ∂ x μ − m k ) u k ( x ) = 0 ( k = n , p , e , v ) (14)

γ μ is Dirac matrix.

Dirac eigenfunction of nucleon, established on three-quark model, is combined with positive/negative charge type Dirac function of electrons under the action of Coulomb field of protons. Three-quark model is three-body symmetrical structure and positive/negative charge is two-body symmetrical structure, so a 2/3 conversion factor shall be applied between them when the three-body problem is converted into two-body problem for relevant research. The factors which represent the attributes of nucleons, such as mass and carrying force, shall be multiplied with this conversion factor, namely:

H ^ i = c α ⋅ p + 2 3 ( m i c 2 β − V ^ e ( r ) − V ^ s ( r ) ) (15)

where i = p , n respectively represent proton and neutron, V e ( r ) and V s ( r ) respectively represent Coulomb field and strong interaction in nucleons.

In this way, we can establish a Hamilton operator for describing the internal state of the moving nucleons by simple mass and potential energy. When 〈 H 〉 = 0 is true, namely: nucleon energy “is developed from nothing”, we can obtain the intrinsic equation of nucleons.

Internal Odd Symmetrical Structures of Neutrons and Protons are as shown in Figure 3.

Firstly, we analyze protons [5] : protons carrying momentum move under strong force field and coulomb field in nucleons; then, we solve the common eigenstate function of bound states ( 0 < E < m p c 2   or   m n c 2 ) and ( H ^ , K ^ , J 2 , J z ) .

Firstly, Hamilton operator of proton is as follows:

H ^ = c α ⋅ p + 2 3 ( m p c 2 β − 3 × ( z 1 z 2 k e 2 r 1 + z 2 z 3 k e 2 r 2 + z 1 z 3 k e 2 r 3 + 3 / 2 ℏ c r e 2 ( r / r 0 − 1 ) ) ) (16)

where

p = − i ℏ ∇ ,   α = ( 0 α α 0 ) ,   β = ( I 0 0 − I ) (17)

α and I are respectively second-order pauli matrix and unit matrix, as shown below:

α r = α ⋅ r ⇀ r = α ⋅ r ⇀ r ( 0 1 1 0 ) = α r ( 0 I I 0 )

ℏ K ^ = β ( Σ ⋅ l + ℏ ) = ( ℏ + ∂ l ) ( 1 0 0 − 1 )

where 1 is the angular momentum operator of the track, and ℏ ∑ / 2 is spin angular momentum operator, namely:

∑ = α ( I 0 0 I ) (18)

z 1 z 2 r 1 + z 2 z 3 r 2 + z 1 z 3 r 3 is the sum of potential energy of every two quarks of the

three quarks, and coefficient 3 is the sum of the observed values of three different quarks.

Above potential energy expression includes six physical quantities, namely z 1 , z 2 , z 3 , r 1 , r 2 , r 3 , which cannot be directly solved through Dirac equation. Therefore, the potential energy expression must be simplified.

Firstly, we shall notice that all charges are integer multiples of e in coulomb field, but the basic definition is forcedly divided into three parts in quark sub- model, namely:

3 ( z 1 , z 2 , z 3 ) = ± 1 , ± 2 , ± 3 (19)

or

z 1 , z 2 , z 3 = ± 1 3 , ± 2 3 , ± 1 (20)

Proton charge is e, namely:

z 1 + z 2 + z 3 = 1 (21)

The following formula can be obtained through one group of values:

z 1 = − 1 3 ,   z 2 = 2 3 ,   z 3 = 2 3 (22)

In proton, three quarks carrying color charges are equivalent to positron- electron, and the only difference is that two symmetrical objects are changed into three objects. We can find the central distance of several quarks according to the symmetry condition in nucleon. r₁, r₂ and r₃ respectively represent the central distances of the three quarks, and r is the radius of nucleon, as shown in Figure 4:

r 1 = r 2 = r 3 = 3 r (23)

Namely:

〈 V e ( r ) 〉 = 〈 ( − 1 / 3 × 2 / 3 × 2 + 2 / 3 × 2 ) 3 r 〉 = 0

Hamilton is changed as follows:

H ^ = c α ⋅ p + 2 3 ( m p c 2 β − 3 / 2 ℏ c r e 2 ( r / r 0 − 1 ) ) (24)

Formula (22) is converted into the following formula through the algebraic relation of ∂ , β , ∑ :

H ^ = c α r ⋅ p r + i ℏ c 1 r ∂ r K ^ β + 2 3 ( m p c 2 β − 3 / 2 ℏ c r e 2 ( r / r 0 − 1 ) ) (25)

where p r is radial momentum operator.

p r = 1 r ( r p − i ℏ ) = i ℏ ( 1 r + ∂ ∂ r ) (26)

J = l + ℏ 2 ∑ (27)

H ^ , K ^ , J 2 , J z can be proven to be mutually commuted, and the eigenvalues thereof are as follows:

J 2 = j ( j + 1 ) ℏ 2 , j = 1 2 , 3 2 , ⋯ J z = m j ℏ , m j = j , j − 1 , ⋯ K ^ = K = ± ( j + 1 2 ) = ± 1 , ± 2 , ⋯

As a result, common eigenfunction of ( H ^ , K ^ , J 2 , J z ) can be divided into two types:

1) K = j + 1 2

φ = ( φ 1 φ 2 φ 3 φ 4 ) = ( ϕ j m j A f ( r ) ϕ j m j B g ( r ) ) = ( ϕ j m j A F ( r ) r i ϕ j m j B G ( r ) r ) (28-1)

2) K = − ( j + 1 2 )

φ = ( φ 1 φ 2 φ 3 φ 4 ) = ( ϕ j m j B f ( r ) ϕ j m j A g ( r ) ) = ( ϕ j m j B F ( r ) r i ϕ j m j A G ( r ) r ) (28-2)

and f ( r ) = F ( r ) r , g ( r ) = G ( r ) r

For convenient calculation, ϕ j m j A and ϕ j m j B g ( r ) are set as two-component ( J 2 , J z ) eigenfunction ( θ , φ function, irrelevant to r).

Formula (28) is put into intrinsic equation of energy.

H φ = E φ

The following formula is obtained

p r F ( r ) r = − i ℏ 1 r d F d r (29)

to establish F ( r ) , G ( r ) simultaneous equations:

{ d F d r − K r F = ( 2 3 m c 2 + E ℏ c + 1 r e − 2 ( r / r 0 − 1 ) ) G d G d r + K r G = ( 2 3 m c 2 − E ℏ c − 1 r e − 2 ( r / r 0 − 1 ) ) F (30)

where α = e 2 / ℏ c is fine structure constant.

In following paragraph, F and G actions are researched around singular point r = 0 , ∞ .

When r → ∞ is true, Formula (30) can be approximately valued as:

d F d r ≈ 2 3 m p c 2 + E 2 ℏ c G ,   d G d r ≈ 2 3 m p c 2 − E 2 ℏ c F

Therefore,

d F d r ≈ 4 9 m p c 2 − E 2 ℏ 2 c 2 F ,   d G d r ≈ 4 9 m p c 2 − E 2 ℏ 2 c 2 G

Obviously, physically allowable progressive action of G and F is as follows:

r → ∞ , F , G ~ e − 2 3 m p c 2 − E 2 ℏ c r (31)

When r → 0 is true, F and G shall be at the size of the same level according to Formula (31).

F → C 1 r s , G ~ C 2 r s (32)

Above formula is put into Formula (30) to obtain:

{ ( s − K ) C 1 − e 2 ( r / r 0 − 1 ) C 2 = 0 ( s + K ) C 2 + e 2 ( r / r 0 − 1 ) C 1 = 0 (32)

C₁ and C₂ cannot be set as 0 at the same time, so the following condition must be met.

| ( s − K ) − e 2 ( r / r 0 − 1 ) e 2 ( r / r 0 − 1 ) ( s + K ) | = 0

Accordingly, the following solution is obtained:

s = ± K 2 − ( e 2 ( r / r 0 − 1 ) ) 2

When applying the global approach to find the movement of nucleon in its own field, it is necessary to notice that its static mass is not changed. Namely, when 〈 H 〉 = 0 or s = 0 is true, we can obtain a nucleon only influenced by its internal action, thus avoiding the influence of external momentum.

( e 2 ( r / r 0 − 1 ) ) 2 = K 2 (33)

In above formula, the left part is a continuous value changed along with r, and the right part is discontinuous value which can be only positive and negative integers. Obviously, the two parts are unequal to each other, and the unique feasible method is to set the following equation:

e 2 ( r / r 0 − 1 ) = constant

Namely:

r = r 0 (34)

Proton only has one state, namely its ground state. When K 2 = 1 is true, the following definition is established according to the relation between mass and fluctuation:

r 0 = 3 2 ℏ m p c (35)

The state equation of proton can be solved as follows:

When K = 1 is true,

F = − G = e − r / r 0 (36)

Above formula can be met.

When K = − 1 is true,

F = − G = e r / r 0 (37)

Function convergence requirement cannot be met.

Neutron has two states around decay, namely: neutron state similar to proton state, and proton-electron state after fission, so there are two Hamilton operators.

{ H ^ 1 = [ c ∂ ⋅ p 1 + 2 3 ( m n c 2 β − ( 3 ( z 1 z 2 k e 2 r 1 + z 2 z 3 k e 2 r 2 + z 1 z 3 k e 2 r 3 ) + 3 / 2 ℏ c r e 2 ( r / r 0 − 1 ) ) ) ] p 1 H ^ 2 = ( c ∂ ⋅ p + m e c 2 − e 2 r ) p 2

(38)

p₁ and p₂ respectively represent the probabilities of the two states, so p 1 + p 2 = 1 is true.

Similarly to Formula (24), above formula can be simplified as follows:

{ H ^ 1 = [ c ∂ r ⋅ p r + i ℏ c 1 r α r K ^ β + 2 3 ( m p c 2 β − ( 1 3 k e 2 r + 3 / 2 ℏ c r e 2 ( r / r 0 − 1 ) ) ) ] p 1 H ^ 2 = ( c α ⋅ p + m e c 2 − e 2 r ) p 2 (39)

Three quarks are symmetrical with each other and meet the mutual mass center as shown in Formula (21), the symmetrical diagram of quarks as shown in Figure 5.

Namely:

〈 V e ( r ) 〉 = 〈 3 × ( − 1 / 3 × 2 / 3 × 2 + 1 / 3 × 2 ) 3 r e 2 〉 = − e 2 3 r

We find p₁ and p₂ as follows:

For converting Formula (39) into the fine structure form similarly with Formula (25) for solution, we firstly need to change s.

The following formula is defined:

s 2 = s 1 2 + s 2 2 (40)

The following formula can be obtained:

{ s 1 2 = ( K 1 2 − ( 2 3 9 α + e 2 ( r n / r 0 − 1 ) ) 2 ) p 1 2 s 2 2 = ( K 2 2 − α 2 ) p 2 2 (41)

Now, we need to find a probability in decay intermediate process to determine p₁ and p₂.

There are two charges as 2/3 u quarks in neutron, and each charge may decay into one negative particle. Due to the repulsive interaction between the two particles, the following two groups of potential energy can be found.

Potential energy including coulomb field or potential energy including strong force field and coulomb field (d Quark Decays into u Quark and Probably Releases Two Negative Charges Relevant or Irrelevant to Each Other as shown in Figure 6).

{ V ( r ) e = 2 × k e 2 r − 1 2 k e 2 r = 3 2 k e 2 r = 3 2 α ℏ c r V ( r ) n = ℏ c r + 1 2 k e 2 r = ℏ c ( 1 + 1 / 2 α ) r (42)

Probability is directly proportional to the square of potential energy, namely: the following formula can be obtained according to above formula:

{ p 1 = 1 + 1 2 α 1 + 2 α p 2 = 3 2 α 1 + 2 α (43)

Formulae (41)-(43) are combined to obtain:

s 2 = ( K 1 2 − ( 2 3 9 α + e 2 ( r n / r 0 − 1 ) ) 2 ) ( 1 + 1 2 α 1 + 2 α ) 2 + ( K 2 2 − α 2 ) ( 3 2 α 1 + 2 α ) 2 (44)

When 〈 H 〉 = 0 or s = 0 is true, we obtain a neutron only influenced by its internal action, thus avoiding the influence of external kinetic energy.

0 = ( K 1 2 − ( 2 3 9 α + e 2 ( r n / r 0 − 1 ) ) 2 ) ( 1 + 1 2 α 1 + 2 α ) 2 + ( K 2 2 − α 2 ) ( 3 2 α 1 + 2 α ) 2 (45)

Similarly with Formula (35), the following formula can be defined:

r 0 = 3 2 ℏ m n c (46)

Formula (45) is converted into:

0 = ( K 1 2 − ( 2 3 9 α + e 2 ( m p / m n − 1 ) ) 2 ) ( 1 + 1 2 α 1 + 2 α ) 2 + ( K 2 2 − α 2 ) ( 3 2 α 1 + 2 α ) 2 (47-1)

At the moment, neutron is under ground state, namely:

K 1 2 = K 2 2 = 1

Accordingly, proton-neutron mass ratio is obtained as follows:

m p m n = 1 + 1 2 ln ( 1 + 9 4 α 2 ( 1 − α 2 ) ÷ ( 1 + 1 2 α ) 2 − 2 3 9 α ) (47-2)

The influence of proton recoil and weak decay on α 2 is not considered in Formula (47). Therefore, we will continuously discuss relevant problem.

1) Influence of proton recoil on α 2 :

Diagram of Recoil Movement between Proton and Electron is as shown in Figure 7.

Namely:

{ m p v p = m e v e m p v p r e = m e v e r e = ℏ (48)

E = 1 2 × m e c 2 × 9 4 α 2 ( 1 + m e m p ) (49)

Namely, the influence of proton recoil on α 2 is as follows:

9 4 α 2 ( 1 + m e m p ) (50)

2) Influence of weak decay on α 2 :

z-axis is taken as the observation axis; in consideration of the similarity between weak force and spinning interaction, the weak force formula is set as follows:

V ^ ( r ) w = − 8 π 3 e e e v e σ ( r ) { e e = − e m e c S e ⋅ 1 2 ( α x − i α y ) e v e = − e m e c S v e ⋅ 1 2 ( α x + i α y ) (51)

The calculation method for weak potential is as the same as that for the superfine action between proton and electron under s state of hydrogen atom (1 = 0).

S e ⋅ S v e = 1 4 ℏ 2 (52)

Finally, the energy for weak force decay is calculated as follows:

E ( 1 ) = − 1 3 m e c 2 α 4 (53)

In consideration of the influence of proton recoil, the corrected energy is as follows:

E ( 1 ) = − 1 3 m e c 2 α 4 ( 1 + m e m p ) (54)

The influence on α 2 is as follows:

2 × 1 3 α 2 ( 1 + m e m p ) (55)

The correct proton-neutron mass ratio is as follows:

m p m n = 1 + 1 2 ln ( 1 + 9 4 α 2 ( 1 − α 2 ) ÷ ( 1 + 1 2 α ) 2 ( 1 + m e m p ) ( 1 − 2 3 α 2 ) − 2 3 9 α ) (56)

The experimental value of fine structure constant is set as

α = 0 .0072973525664 to obtain the theoretical value of Formula (52) as 0.99862347888. The experimental value of this constant can be checked in NIST

[1] official website as ( m p m n ) exp = 0.99862347844 ( 51 ) . We can find that the

theoretical value has a spectacular accuracy, thus indicating that it is feasible for us to describe the internal movement state of nucleon by Dirac equation. Therefore, this can be taken as the necessary evidence for us to verify the strong nuclear force potential formula.

When neutron is under the ground state s = 0, the state equation of neutron, solved thereby, can meet Formula (36).

Internal Even Symmetrical Structure of π Meson is as shown in Figure 9.

We firstly analyze π_±, and this is simple for us to understand, because we have solved Dirac intrinsic equations of proton and neutron.

Hamilton operator of π_± is as follows:

H ^ = c α ⋅ p + ( m π ± c 2 β − ( − k e 2 r + ℏ c r e 2 ( r / a 0 − 1 ) ) ) (62)

Notably, two quarks in meson have no electrostatic force action, but under decay state, electron and anti-electron neutrino form a special electrostatic interaction of the same charge, which seems as an electrostatic field between two electrons. Anti-electron neutrino like a mirror projects an electron with the attribute as the same as that of primary electron, and then two electrons have electromagnetic interaction (as shown in Figure 10).

Namely:

p = − i ℏ ∇ , α = ( 0 α α 0 ) , β = ( I 0 0 − I )

When 〈 H 〉 = 0 is true, namely: when we obtain a static π_±, a group of necessary conditions can be obtained as follows:

{ K 2 = 1 a 0 = ℏ m π ± c (63)

When K = 1 is true, F = − G = e − r / r π ± can meet the requirement.

When K = − 1 is true, F = − G = e − r / r π ± cannot meet function convergence requirement.

Hamilton operator of π⁰ is as follows:

H ^ = c α ⋅ p + ( m π ± c 2 β − ( 2 × 4 × k e 2 2 r + ℏ c 2 r e − 2 ( r / a 0 − 1 ) ) ) (64)

Namely:

p = − i ℏ ∇ , α = ( 0 α α 0 ) , β = ( I 0 0 − I )

When 〈 H 〉 = 0 is true, namely: when we obtain a static π_±, a group of necessary conditions can be obtained as follows:

{ K 2 = 1 a 0 = ℏ m π ± c (65)

At the moment,

m π ± m π 0 = 1 − 1 2 ln ( 1 − 8 α ) 1 − 1 2 ln ( 1 + α ) (66)

When K = 1 is true, F = − G = e − r / r π ± can meet the requirement.

When K = − 1 is true, F = − G = e − r / r π ± cannot meet function convergence requirement.

The condition shown in Figure 10(c) is not considered in above formula, and the specific value of such condition and common symmetry condition is exactly the specific value of strong force and electrostatic force, namely: 1 : 8 α , so the above formula is corrected as follows:

m π ± m π 0 = 1 − 1 2 ln ( 1 − 8 α − 80 α 2 / 9 ) 1 − 1 2 ln ( 1 + α ) = 1.034087 (67)

The experimental value of the mass ratio of charged π meson and neutral π meson cannot be found in references, so the experimental value of the mass ratio of the two K mesons is obtained through the mass [3] division:

m π ± exp m π 0 exp = 139.57018 ± 0.00035 134.9766 ± 0.0006 = 1.034032 ± ⋯ (68)

Formula (59) is divided by Formula (60) to obtain:

= 1.034087 1.034032 × 100 % − 1 = 5.3 × 10 − 5 (69)

Error rate is 0.0000053, thus indicating the theoretical value is consistent with the experimental value.

The process for solving Dirac equation of K meson is similar with the process for solving the same of π meson. But such process is independently described in following paragraph for three purposes: 1) wide applicable scope of the paper: except K meson which also includes s quark, other particles in this paper are composed of two quarks-u and d quarks, so the introduction to such process can further widen the applicable scope of strong interaction verified in this paper; 2) the internal symmetrical type of charged K meson is odd symmetry, and such structure rarely exists among mesons; 3) the strong force potential of charged K meson is as shown in Formula (11), thus proving the existence of such strong force potential (as shown in Figure 11, Figure 12).

Finally, the formula for the mass ratio of charged K meson and neutral K meson is as follows:

m K ± m K 0 = 1 + 0.5 ln ( 1 − 4 ( 1 − 1 / 3 3 ) α ) 1 − 0.5 ln ( 1 + α ) = 0.991678 (70)

The experimental value of the mass ratio of charged K meson and neutral K meson cannot be found in references, so the experimental value of the mass ratio of the two K mesons is obtained through the mass [3] division:

m K ± exp m K 0 exp = 493.677 ± 0.016 497.648 ± 0.024 = 0.992020 ± ⋯ (71)

Formula (70) is divided by Formula (71) to obtain:

= 0.991678 0.992020 × 100 % − 1 = 3.4 × 10 − 4 (72)

Error rate is 0.000034, thus indicating the theoretical value is consistent with the experimental value.

Classical theory is adopted to solve the combined potential of deuteron. Deuteron is an atomic nucleus formed by the combination of one proton and one neutron under strong action, and the internal structure of deuteron can meet angular momentum conservation and stress balance (as shown in Figure 13(a) and Figure 14).

In consideration of continuous movement of π meson between proton and neutron, the balance state in deuteron is namely the release state thereof. Under such state, two protons make circular movement around nucleus, and π meson locates at the center of the central circle. The mechanical formula for such state can be established as follows:

2 m p v r = ℏ (74)

m p v 2 r = ℏ c 4 r 2 ( e − 2 ( r / a − 1 ) + 3 α ) (75)

The following formula is set:

{ k = r a b = m p m π ± ( 1 − 0.5 ln ( 1 + α ) ) (76)

The experimental values of proton, π meson and fine structure constant can be put into the formula to find constant b.

b = 6.747 (77)

Formulae (73), (74) and (75) are combined and simplified as follows:

1 / 4 k [ e − 2 ( k − 1 ) + 3 α ] = 1 4 b (78)

The potential energy formula of deuteron is as follows:

Kinetic energy of each nucleon is as follows:

V ( r ) = − ℏ c 2 r ( e − 2 ( r / a − 1 ) + 3 α ) (79)

Formulae (76) and (77) are put into the following formula to obtain:

Δ E 1 = V ( r ) − 2 E = − 5.152   MeV (80)

Deuteron has a probability of 1/2 respectively for two states―absorption state and release state. Under absorption state, the sum of the charges carried by protons and neutrons therein is zero, so there is no any interaction; protons and neutrons are respectively under free state, and the binding energy is also zero, and the binding energy of deuteron shall be the mean value under the two states.

Δ E = 1 2 × E 1 + 1 2 × 0 = − 2.576   MeV (81)

But the influence of the mass of π meson located at the center is not considered in above formula. Notably, the mass of this π meson is not a constant. The influence of two factors may be considered for the estimation thereof.

1) π ± → e + υ e is true, so the mass of π meson will not be lower than that of electron. Therein, the probability of π meson under release state in deuteron is 2/3 (internal quark composition of proton is (uud), π meson can be only absorbed by 2 u quarks to generate d quark; the probabilities for π meson under absorption state and under release state are respectively 1/2 * 2/3 = 1/3 and 1 − 1/3 = 2/3).

E π ± > 2 3 E e = 0.3407   MeV (82)

2) Influence of internal stress action inside deuteron on π meson

Δ E π ± = E n − E p − 1 2 Δ E = 1.293 − 1.288 = 0.005   MeV (83)

Formulae (82) and (83) are comprehensively considered to obtain the estimated mass of π meson:

E π ± = 2 3 E e + Δ E π ± = 0.3457   MeV (84)

Finally, the binding energy of deuteron is obtained as follows:

Δ E D the = Δ E + m π ± = − 2.230   MeV (85)

The experimental value is put into the formula:

Δ E D exp = E D exp − E n exp − E p exp = − 2.225   MeV (86)

Error is as follows:

Δ = E D exp − E D the = + 0.005   MeV (87)

Error rate is as follows:

η = ( Δ E D the Δ E D exp − 1 ) × 100 % = ( 2.230 2.225 − 1 ) × 100 % = 0.22 % (88)

The theoretical value is basically consistent with the experimental value.

Triton, He-3 and He-4 have complex internal spatial structure, so the binding energy can be only estimated.

He-3 has two protons and a neutron, when the neutron release a pion, the strong nuclear force will spread between three protons, Strong interaction of He-3 is equivalent to the strong interaction of three pairs of deuteron (as shown in Figure 13(c)), and the binding energy thereof is three times of that of deuteron. Considering the influence of the electrostatic potential difference, the binding energy of he-3 can be obtained.

Δ E He-3the = 2 Δ E D the + Δ E − ( 3 / 2 + 2 ) α 4 k m π ± ( 1 − 0.5 ln ( 1 + α ) ) = − 2.230 × 2 − 2.576 − 0.5136 = − 7.550   MeV (89)

Strong interaction of triton is similar with that of He-3 (as shown in Figure 13(b)), but it is necessary to consider the repulsive interaction between two protons in He-3, and its binding energy is slightly lower than that of He-3.

Δ E He-3the = Δ E Dthe + 2 Δ E − ( 3 / 2 + 2 ) α 2 k m π ± ( 1 − 0.5 ln ( 1 + α ) ) = − 2.230 − 2.576 × 2 − 1.0272 = − 8.411   MeV (90)

The internal spatial structure of He-4 is a normal triangular pyramid (as shown in Figure 13(d)), and each face of the normal triangular pyramid is equivalent to the internal symmetrical structure of a He-3, so its binding energy is four times of that of He-3. Additionally, two faces are lack of π mesons, so the influence of two pairs of proton-proton force must be considered. Specifically, the binding energy is measured and calculated as follows:

Δ E He-4the = 4 Δ E He-4the + 2 ( Δ E He-4the − Δ E T the ) = − 7.550 × 4 + 2 × ( 7.550 − 8.411 ) = 28.48   MeV (91)

The comparison between the theoretical value and thee experimental value of the binding energy of three atomic nucleuses is shown in the following Table 2.

Remarks: 1) The experimental value of He-3 binding energy cannot be found, so it is obtained through deducting the mass [1] of two neutrons and one proton from the mass [1] of He-3. 2) The experimental value of triton cannot be found, so it is obtained through deducting the mass [1] of two neutrons and one proton from the mass [1] of triton.

According to the comparison between the theoretical value and the experimental value of the binding energy of three atomic nucleuses, we find significant error and high error rate, but we cannot deny that such thought can be used for analyzing the internal nucleus action of atomic nucleuses, thus presenting important reference value.

Δ (1232) particles family includes four types of particles, respectively Δ++ (1232), Δ+ (1232), Δ0 (1232) and Δ− (1232). Δ (1232) particles family plays a special role, and forms the complete baryon family of u and d quark structure together with protons and neutrons. According to the mass information analysis, Δ (1232) particles family, different from proton and neutron which are usually directly composed of quarks, is composed of proton or neutron and two π mesons. In other words, the strong interaction thereof belongs to baryon-meson strong interaction.

Δ (1232) particles family has internal odd symmetrical structure, as the same as the internal symmetrical structure of proton and neutron (as shown in Figure 15).

Δ (1232) internal symmetrical structure can be also described by Dirac equation, but the specific process is similar with the process for the mentioned particles, so it will not be repeatedly explained in this chapter.

Theoretical values of the masses of the four types of particles are as follows:

m Δ + + : m Δ + : m Δ 0 : m Δ − = ( 1 + 0.5 ln ( 1 − 2 ( 3 − 3 ) 3 α ) ) − : ( 1 + 0.5 ln ( 1 − 2 3 α ) ) −                         : ( 1 + 0.5 ln ( 1 − 2 3 α ) ) − : ( 1 + 0.5 ln ( 1 − 2 3 α ) ) −                   = 1.00066 : 1 : 1 : 1 (92)

The experimental values of four types of Δ (1232) particles are all 1232 ± 2 MeV, and the specific value thereof is basically as the same as Formula (92). m 0 = m p + 2 m π ± is assumed to estimate the masses of the four types of Δ (1232) particles, and the estimated value is compared with the experimental value (as shown in Table 3).

The theoretical value is basically consistent with the experimental value.