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Space Theory

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S. Chen  *

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05 November 2024

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06 November 2024

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Abstract
The author introduces a novel theoretical framework that suggests matter and energy are both converted from curved space. In Space Theory, gravitation is generated by the flow of space, instead of being transmitted by the graviton as in String Theory. This theory also suggests that Newton's gravitational constant, denoted as G, may not be truly constant but could vary over time. The equivalent equation of space is S=Ec2=mc4, and the gravitation of hollow sphere space is Sμν=4πGm=(4/3)π((r+a)3-r3). The Space Theory also predicts that the surface gravitational acceleration of the neutron star Crab Pulsar (PSR B0531+21) is approximately 8.21924883×106 m/s2.
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Subject: Physical Sciences  -   Astronomy and Astrophysics

Introduction

Newton’s law of universal gravitation [1] provides the equation for the gravitational force, which states that:
F = G M m r 2 ,
When a star dies, it undergoes a contraction and blows off its outer envelope, forming a planetary nebula. [3,4] If the star collapses to within its Schwarzschild radius, it forms a black hole. [5]
Figure 1. A star forms a black hole.
Figure 1. A star forms a black hole.
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Before the star collapses, the gravitational force between the planet and the star is represented by F 1 . After the collapse, the gravitational force between the planet and the black hole is represented by F 2 . The gravitational force of a black hole is extremely strong and nothing, not even light, can escape it. [6] Therefore, F 2 is greater than F 1 . In this equation, M 1 represents the mass of the star, M 2 represents the mass of the planet, m represents the mass of the black hole, r represents the distance between their centers of mass, and G is the gravitational constant.
F 1 < F 2 ,
G M 1 m r 2 < G M 2 m r 2 .
As the star collapses into a black hole, it blows off its outer envelope and loses mass. [7] Assuming the star loses 0.2 % of its mass during this process, the mass of the black hole can be represented as 99.8 % of m 1 . This can be rewritten as:
G M 1 m r 2 < G 0.998 M 1 m r 2 .
When the common parameters are removed, the equation can be simplified to:
1 < 0.998 .
How is it even possible that 1 is less than 0.998 ? As the mass decreases from m 1 to m 2 and the distance r between the objects remains unchanged, it suggests that the gravitational constant G has increased. In Einstein’s theory of relativity [2], matter curves spacetime, and the Einstein field equations can be expressed in the following form:
G μ ν = 8 π G c 4 T μ ν .
In this equation, G μ ν is the Einstein tensor and G is the gravitational constant. However, if the gravitational constant G is constantly changing, it raises the question of whether Einstein’s theory is still accurate.

Model

In Einstein’s theory of relativity [2], matter curves spacetime. What if the situation were reversed, and both matter and energy were converted from curved space?
Matter can release energy through annihilation, fission, and fusion [2], as matter and energy are different forms of the same thing. Therefore, matter curves spacetime because it is converted from curved space, and the flow of space released by matter creates gravitational force. The greater the amount of space released, the stronger the gravitational force generated. If one object releases much more space than another object, the flow of space will narrow the distance between the two objects. It is the space that moves, not the objects.
Figure 2. Gravitation.
Figure 2. Gravitation.
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When two galaxies are very far apart, the space they release accumulates in the middle, and this expands their distance. Therefore, the expansion of the universe and the phenomenon of redshift [8] are caused by the increase in space.
Figure 3. Expansion of the universe.
Figure 3. Expansion of the universe.
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Thus, space curves in one dimension, converting matter and energy.
Figure 4. Curved space.
Figure 4. Curved space.
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Since an electron is matter and there are only two directions in one dimension, matter curves in two different directions, creating two types of electric charges: positive and negative. Like charges repel each other because they occupy the same position in one dimension, while unlike charges attract each other because they occupy opposite positions in one dimension. Since they are all matter, they can only attract or repel each other, not annihilate.
Figure 5. Electric charges.
Figure 5. Electric charges.
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Quarks have three color charges [9]: red, green, and blue, which are converted by matter curving in three different dimensions. When a gluon is transferred between quarks, a color change occurs in both. For example, if a red quark emits a red–antigreen gluon, it becomes green; if a green quark absorbs a red–antigreen gluon, it becomes red. Quantum chromodynamics [10] proves that these three values of quark color can be converted into each other: they are different forms of the same thing.
Figure 6. Quark color.
Figure 6. Quark color.
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Since there are only three values of quark color, space has only three dimensions. Therefore, space curves in six different directions of three dimensions, creating six types of matter: ordinary matter, antimatter, two types of dark matter, and two types of anti-dark matter.
Matter and antimatter in opposite directions attract and annihilate each other, while matter and dark matter perpendicular to each other in another dimension have no electromagnetic force. Both matter, antimatter, dark matter, and anti-dark matter are affected by gravitation.
During the Big Bang, matter and antimatter moved in opposite directions, so antimatter ended up on the other side of the universe. Currently, there is only dark matter and anti-dark matter in the observable universe. Particle collision experiments [11] prove that ordinary matter can be converted into antimatter: they are different forms of the same thing.
Figure 7. Six types of matter.
Figure 7. Six types of matter.
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Matter is converted from curved space, and it also curves time.
In the double-slit experiment, an interference pattern emerges as the particles build up one by one. [12] This occurs because matter curves time, causing the particle from the past to interfere with the particle from the future.
In which-way experiment, if particle detectors are positioned at the slits, showing through which slit a photon goes, the interference pattern will disappear. [13] The mass of the observer is much greater than the particle, and when the observation occurs, the time of the observer engulfs the time of the particle, similar to how a black hole swallows a star, resulting in wave function collapse.
The Wheeler’s delayed choice experiment demonstrates that extracting “which path” information after a particle passes through the slits can appear to retroactively alter its previous behavior at the slits. [14] Because matter curves time, the particle from the future can interfere with the particle from the past, meaning that the present behavior can have an impact on the past.
The quantum eraser experiment further shows that wave behavior can be restored by erasing or making permanently unavailable the “which path” information. [15] Since the time of the particle has no connection with the time of the observer, no wave function collapse occurs.
Figure 8. Matter curves time.
Figure 8. Matter curves time.
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Every elementary particle is matter, and they are all converted from curved space, which gives them rest mass. However, the current commonly accepted physical theories imply or assume that the photon is strictly massless [16]. The Lorentz factor γ is defined as [17]:
γ = 1 1 v / c 2 .
In Newton’s law of universal gravitation [1], the gravitational acceleration is:
r ¨ = G m 0 / r 2 1 v / c 2 .
In Einstein’s theory of relativity [2], the stress-energy tensor T α β for a non-interacting particle with rest mass m 0 and trajectory x p t is given by:
T α β x , t = m 0 v α t v β t 1 v / c 2 δ x x p t .
If the photon has no rest mass, it would not be subject to gravitation and could escape from a black hole, contradicting observations. [6]
Of course, there is no doubt that the Standard Model is 100 % accurate, so the blame is on the photon. The mistake, therefore, lies with the photon, which should behave precisely as predicted by scientists. The photon is both a wave and a particle, both matter and antimatter, and both has and does not have rest mass.
According to the standard model in particle physics [18], neutrino has zero rest mass and is a spin-half particle. Unfortunately, experimental observations by the Super-Kamiokande Observatory and the Sudbury Neutrino Observatories have shown that the neutrino actually has a non-zero rest mass, revealing the limitations of the Standard Model. [19,20,21]
Furthermore, as is widely known, dark matter possesses mass. Is this mass conferred by the Higgs field? If the answer is yes, does the Higgs boson composed of both matter, antimatter, and dark matter? If the answer is no, where does the mass of dark matter originate from?

Gravitation

Since there are only three values for the color charge of quarks, so the space has only three dimensions. Therefore, the space released by matter takes the form of a three-dimensional sphere. As this space flows outward, it forms a hollow sphere, and its volume can be written in the following form:
V 3 = 4 3 π r 3 ,
S μ ν = 4 3 π r + r ¨ 3 r 3 .
Where S μ ν is the space released by the matter, r is the radius of the sphere, and r ¨ is the degree of space curvature. The greater the degree of space curvature, the greater the gravitational acceleration of the object. When the outward flow of space occurs, its volume remains constant, allowing for the direct calculation of the gravitational acceleration r ¨ 2 at a different distance using the gravitational acceleration r ¨ 1 at a known distance. The equation can be rewritten to:
S μ ν = 4 3 π r 1 + r ¨ 1 3 r 1 3 = 4 3 π r 2 + r ¨ 2 3 r 2 3 ,
r ¨ 2 = r 1 + r ¨ 1 3 r 1 3 + r 2 3 3 r 2 .
The mass of the Earth is approximately 5.97237 × 10 24   k g , and the average distance from its center to its surface is about 6.37123 × 10 6   m . [22,23] According to Newton’s law of universal gravitation, the value of the gravitational constant is approximately 6.67408 × 10 11   m 3 / k g / s 2 . [24] The gravitational acceleration r ¨ 1 at the Earth’s surface can be calculated using this law and is given by:
r ¨ 1 = G m r 1 2 9.81954911   m / s 2 .
If the distance up to 10 7   m , the gravitational acceleration r ¨ 2 is:
r ¨ 2 = G m r 2 2 3.98600751   m / s 2 .
When the gravitational acceleration r ¨ 1 is equal to the Newtonian gravitational acceleration r ¨ 1 at r 1 , the gravitational acceleration r ¨ 2 at r 2 is:
r ¨ 2 = r 1 + r ¨ 1 3 r 1 3 + r 2 3 3 r 2 3.98601207   m / s 2 .
As you can see, the value of the gravitational acceleration r ¨ 2 calculated using the Space Theory is extremely close to the value calculated using Newton’s law of universal gravitation. This confirms that the Space Theory can be used to accurately calculate the gravitational acceleration. The gravitational acceleration of the Earth at different distances in both models is shown in the following table:
Table 1. Gravitational acceleration of Earth.
Table 1. Gravitational acceleration of Earth.
Distance of Earth 0   m 1.15 × 10 3   m 10 4   m 10 5   m 10 6   m 10 7   m
Newton’s law 3 × 10 8   m / s 2 3986007   m / s 2 39860   m / s 2 398.600   m / s 2 3.986007   m / s 2
Space Theory 106141   m / s 2 104991   m / s 2 96171   m / s 2 29976   m / s 2 398.442   m / s 2 3.986012   m / s 2
The mass of the Moon is approximately 7.342 × 10 22   k g , its mean radius is about 1.737 × 10 6   m , and the time-averaged distance between the centers of the Earth and Moon is about 3.844 × 10 8   m . [25,26,27] When considering different distances, the gravitational acceleration of the Moon in two different models is shown in the following table:
Table 2. Gravitational acceleration of Moon.
Table 2. Gravitational acceleration of Moon.
Distance of Moon 0   m 1.27 × 10 2   m 10 5   m 10 6   m 10 7   m 3.844 × 10 8   m
Newton’s law 3 × 10 8   m / s 2 490.01   m / s 2 4.900109   m / s 2 0.04900   m / s 2 0.00003316   m / s 2
Space Theory 24496   m / s 2 24369   m / s 2 487.60   m / s 2 4.899863   m / s 2 0.04899   m / s 2 0.00003316   m / s 2
The mass of the Sun is approximately 1.9885 × 10 30   k g , with a mean radius of about 6.96342 × 10 8   m , and the mean distance between the centers of the Earth and the Sun is about 1.496 × 10 11   m . [28,29] The table below shows the gravitational acceleration of the Sun in two different models at varying distances:
Table 3. Gravitational acceleration of Sun.
Table 3. Gravitational acceleration of Sun.
Distance of Sun 0   m 6.65 × 10 5   m 10 7   m 10 9   m 10 10   m 1.496 × 10 11   m
Newton’s law 3 × 10 8   m / s 2 1327140   m / s 2 132.71   m / s 2 1.3271   m / s 2 0.005929   m / s 2
Space Theory 7356638   m / s 2 6693449   m / s 2 1181938   m / s 2 132.71   m / s 2 1.3271   m / s 2 0.005929   m / s 2
When the distance is zero, Newton’s gravitational acceleration becomes infinite, which is obviously incorrect. In contrast, the gravitation of the Space Theory is more accurate and does not require the use of Newton’s constant of gravitation. The Schwarzschild radius [5] is a physical parameter that appears in the Schwarzschild solution to Einstein’s field equations. It corresponds to the radius defining the event horizon of a black hole and can be expressed as:
r s = 2 G m c 2 .
The Schwarzschild radius of Earth is approximately 0.00887   m . However, when the distance is 1150   m , the Newton’s gravitational acceleration exceeds approximately 3 × 10 8   m / s 2 , it is greater than the speed of light in vacuum, approximately 2.9979 × 10 8   m / s , and this leads to the formation of a black hole, which is obviously wrong.
Expanding the formula, the gravitational acceleration of the Space Theory can be derived as:
S μ ν = 4 π r 2 r ¨ 1 + r ¨ r + r ¨ 2 3 r 2 .
Then introduce a new variable β to represent 1 + r ¨ / r + r ¨ 2 / 3 r 2 . The equation can be simplified to:
S μ ν = 4 π r 2 r ¨ β .
The expression implies that β can approach 1 as closely as desired by increasing the distance r to infinity.
lim r β = 1 .
At the surface of the Earth, the ratio of the gravitational acceleration r ¨ to the distance r is approximately 0.00000153 , which is negligible and can be omitted. The formula can be rewritten as:
S μ ν = 4 π r 2 r ¨ .
Since the accelerations of the two formulas are equal at long distances, the formula can be simplified to:
r ¨ = S μ ν 4 π r 2 = G m r 2 .
After removing the same parameters, the gravitation of the Space Theory S μ ν thus takes the form:
S μ ν = 4 π G m .
The result demonstrates that the space released by matter per kilogram is precisely equal to 4 π G , providing evidence that matter is converted from curved space, indicating that they are different forms of the same thing. The further the distance between the objects, the closer the values of the formulas.
Figure 9. Gravitational acceleration of Earth.
Figure 9. Gravitational acceleration of Earth.
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In Einstein’s theory of relativity [2], matter curves spacetime, and the Einstein field equations can be expressed in the following form:
G μ ν = 8 π G c 4 T μ ν ,
κ = 8 π G c 4 .
Where G μ ν is the Einstein tensor, T μ ν is the stress-energy tensor, c is the speed of light in vacuum, and κ is the Einstein constant of gravitation.
In the geometrized unit system, the value of 4 π G is set equal to unity. Therefore, it is possible to remove the Newton’s constant of gravitation from the equations to avoid mistakes, especially since the Newton’s law of universal gravitation does not apply to black holes. The equation can be rewritten as:
4 π G = 1 ,
κ = 2 c 4 .
Since the theory is that matter and energy are both converted from curved space, the form should also be reversed. As space S is proportional to mass m and the fourth power of the speed of light in vacuum c 4 , the equivalent equation for space S can be expressed as:
S = E c 2 = m c 4 ,
S μ ν = 4 π G c 4 S .
Of course, under ordinary circumstances, if the Newton’s “constant” of gravitation doesn’t change too much, it can still be used to calculate the gravitational acceleration, the form can be rewritten to:
S μ ν = 4 π G m = 4 3 π r + r ¨ 3 r 3 ,
r ¨ = 3 G m + r 3 3 r .
On the surface of Earth, the gravitational acceleration is:
r ¨ = G m r 2 9.81954911   m / s 2 .
r ¨ = 3 G m + r 3 3 r 9.81953396   m / s 2 .
And the relative error is:
η 0.00000154284 .
As you can see, this value is still extremely close to the original. What’s more, this formula does not introduce any new variables.
In 1665, Newton extended the binomial theorem [1] to include real exponents, expanding the finite sum into an infinite series. To achieve this, he needed to give binomial coefficients a definition with an arbitrary upper index, which could not be accomplished through the traditional factorial formula. Nonetheless, for any given number n , it is possible to define the coefficients as:
n k = n n 1 n 2 n k + 1 k ! = n k k ! .
The Pochhammer symbol · k is used to represent a falling factorial, which is defined as a polynomial:
n k = n k _ = n n 1 n 2 n k + 1 = k = 1 n n k + 1 .
This formula holds true for the usual definitions when n is a nonnegative integer. For any complex number n and real numbers x and y with x > y , the following equation holds:
x + y n = k = 0 n k x n k y k = x n + n x n 1 y + n n 1 2 ! x n 2 y 2 + n n 1 n 2 3 ! x n 3 y 3 + .
The series for the cube root can be obtained by setting n = 1 / 3 , which gives:
x + y 1 3 = x 1 3 + 1 3 x 2 3 y 1 9 x 5 3 y 2 + 5 81 x 8 3 y 3 + .
At long distances where r > 3 G m 3 , the equation becomes:
r ¨ = r 3 + 3 G m 3 r = r + 1 3 r 2 3 G m 1 9 r 5 3 G m 2 + 5 81 r 8 3 G m 3 + r ,
r ¨ = G m r 2 1 r G m r 2 2 + 5 3 r 2 G m r 2 3 + .
Then introduce a new variable β , the equation can be simplified to:
r ¨ = G m r 2 + β ,
β = 1 r G m r 2 2 + 5 3 r 2 G m r 2 3 + .
At the surface of the Earth, the variable β is approximately 0.00001513417 , which is negligible and can be omitted. This is the reason why it is extremely close to Newton’s value at long distances.
lim r β = 0 .
At short distances where 3 G m 3 > r , the form of the equation is different:
r ¨ = 3 G m + r 3 3 r = 3 G m 1 3 + 1 3 3 G m 2 3 r 3 1 9 3 G m 5 3 r 6 + 5 81 3 G m 8 3 r 9 + r ,
r ¨ = 3 G m 3 r 1 1 3 r 3 G m 3 2 + 1 9 r 3 G m 3 5 5 81 r 3 G m 3 8 + .
In summary, formulas inversely proportional to the distance r raised to the power of n can all be replaced with the hollow sphere model:
x n + n y 1 n = x n 1 n + 1 n x n 1 n 1 n y + = x + y x n 1 + ,
lim x y x n = lim x n + 1 y + x n + 1 n + 1 x .
In Chapter 7 of the book “Why String Theory?”, it is mentioned that there is no direct experimental evidence for the String Theory. [30] To avoid encountering the same issue of unverifiability as String Theory, the author makes a prediction about the surface gravity of neutron star.
The Crab Pulsar (PSR B0531+21) has a mass of 1.4 solar masses (M☉) and a radius of approximately 10500   m . [31] The solar mass (M☉) is a standard unit of mass in astronomy, equal to approximately 1.98847 × 10 30   k g . [28] On the surface of Crab Pulsar, the gravitational acceleration is:
r ¨ = G m r 2 1.68523274 × 10 12   m / s 2 ,
r ¨ = 3 G m + r 3 3 r 8.21924883 × 10 6   m / s 2 .
Newton’s prediction of 1.68523274 × 10 12   m / s 2 , which is significantly greater than the speed of light in vacuum, 2.9979 × 10 8   m / s , and this leads to the formation of a black hole, clearly deviates from what is expected. In contrast, the predicted value from the Space Theory, 8.21924883 × 10 6   m / s 2 , aligns more closely with reality.
In addition, the distance can also be determined by the gravitational acceleration, it can be rewritten as:
r = G m r ¨ ,
r = G m r ¨ 1 12 r ¨ 2 1 2 r ¨ .
And the product of variables, G and m , becomes:
G m = r ¨ r 2 ,
G m = r ¨ r 2 + r ¨ 2 r + 1 3 r ¨ 3 .
As you can see, when the acceleration is very small, the results between the two formulas will be very close. Due to the modification of the Newton’s law, the orbital velocity also requires a corresponding adjustment, and it can be expressed in the following form:
lim r x r = lim r 2 x + r 2 r ,
r ˙ = 2 G m + r 2 r .
The mass of the Earth is approximately 5.97237 × 10 24   k g , and the mass of the Moon is approximately 7.342 × 10 22   k g . [23,25] The time-averaged distance between the centers of the Earth and Moon is about 3.844 × 10 8   m . [27] The orbital velocity is:
r ˙ = G m 1 + m 2 r 1024.54383   m / s ,
r ˙ = 2 G m 1 + m 2 + r 2 r 1023.84606   m / s .
The modified formula resolves the issue of approaching infinity as the distance nears zero. Similarly, the distance of the orbit can also be determined by velocity, it can be rewritten as:
r = G m r ˙ 2 1 2 r ˙ 2 .
When the orbital velocity is 1024   m / s , the distance is:
r = G m 1 + m 2 r ˙ 2   3.84808   × 10 8   m ,
r = G m 1 + m 2 r ˙ 2 1 2 r ˙ 2 3.84284 × 10 8   m .
To ensure the accuracy of calculations, the author recommends using the Decimal Module in Python, and suggests setting precision to 100 digits.

Electromagnetism

The same formula can also be used to calculate the electrostatic force. Interpret the formula as describing the propagation of photons in three-dimensional space, extending as a hollow sphere, with the number of particles remaining constant, and the density decreasing with distance, causing changes in the electric field. Therefore, the electric field can also be represented using the formula:
lim r x r 2 = lim r 3 x + r 3 3 r ,
E = 3 k e q + r 3 3 r .
where E is the electric field of photons, k e is the Coulomb constant, and q is the quantity of each charge. The Coulomb constant is about 8.98755179 × 10 9   N m 2 / C 2 , and the electric charge of a single electron is approximately 1.602176634 × 10 19   C [32]. If the distance of an electron is 1   m , the electric field is:
E = k e q r 2 1.43996426069 × 10 9   N / C ,
E = 3 k e q + r 3 3 r 1.43996425861 × 10 9   N / C .
The further the distance between the objects, the closer the values of the formulas.
Figure 10. The electric field of an electron.
Figure 10. The electric field of an electron.
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In the geometrized unit system, where = e = G = k e = 1 and ε 0 = 1 / 4 π . For the convenience of calculation, here Z represents the atomic number of the atom, and the quantity of charge is set to 1 . The Bohr radius, symbolized a 0 , is approximately 52.9   p m . When the distance is equal to the Bohr radius, the electric field of the hydrogen atom Z = 1 is:
E = 1 r / a 0 2 = 1 ,
E = 3 + r / a 0 3 3 r / a 0 0.58740105 .
The electric field is shown in the following table:
Table 4. The electric field of the Hydrogen atom.
Table 4. The electric field of the Hydrogen atom.
r / a 0 0 0.25 0.5 1 2 4
Coulomb’s law 16 4 1 0.25 0.0625
Space Theory 1.4422 1.1947 0.962 0.5874 0.2239 0.0615
The variation is illustrated in the diagram:
Figure 11. The electric field of a Hydrogen atom.
Figure 11. The electric field of a Hydrogen atom.
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And the orbital velocity is:
r ˙ = 1 r / a 0 = 1 ,
r ˙ = 2 + r / a 0 2 r / a 0 0.85559967 .
The velocity is shown in the following table:
Table 5. The orbital velocity of a Hydrogen atom.
Table 5. The orbital velocity of a Hydrogen atom.
r / a 0 0 0.25 0.5 1 2 4
Newton’s law 2 1.4142 1 0.7071 0.5
Space Theory 1.1892 1.0891 1 0.8555 0.6704 0.4925
The revised formula shows that the orbital velocity of an electron near the atomic nucleus increases by only 39 % compared to when it’s at the Bohr radius, which is completely different with the value in classical physics. After the correction, the electron will no longer fall into the nucleus when the distance is zero.
Figure 12. The orbital velocity of a Hydrogen atom.
Figure 12. The orbital velocity of a Hydrogen atom.
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Carbon, hydrogen, nitrogen, oxygen, phosphorus, and sulfur are the six most important chemical elements whose covalent combinations make up most biological molecules on Earth. [33] The following table shows atomic radius, Van der Waals radius and covalent bond from theoretical models. [34,35,36,37,38,39]
Table 6. Atomic radius from theoretical models.
Table 6. Atomic radius from theoretical models.
Number Symbol Name Radius Bond
Calculated Empirically Van der Waals Single Double Triple
1 H Hydrogen 53 pm 25 pm 120 pm 32 pm
2 He Helium 31 pm 120 pm 140 pm
6 C Carbon 67 pm 70 pm 170 pm 75 pm 67 pm 60 pm
7 N Nitrogen 56 pm 65 pm 155 pm 71 pm 60 pm 54 pm
8 O Oxygen 48 pm 60 pm 152 pm 63 pm 57 pm 53 pm
15 P Phosphorus 98 pm 100 pm 180 pm 111 pm 102 pm 94 pm
16 S Sulfur 88 pm 100 pm 180 pm 103 pm 94 pm 95 pm
Table 7. Typical bond lengths and bond energies.
Table 7. Typical bond lengths and bond energies.
Bond Length Energy Bond Length Energy Bond Length Energy
C O 143 pm 358 kJ/mo N C 147 pm 305 kJ/mo N O 144 pm 201 kJ/mo
C = O 123 pm 745 kJ/mo N = C 127 pm 615 kJ/mo N = O 120 pm 607 kJ/mo
C O 113 pm 1072 kJ/mo N C 115 pm 891 kJ/mo O O 148 pm 146 kJ/mo
C C 154 pm 347 kJ/mo N N 147 pm 160 kJ/mo O = O 121 pm 495 kJ/mo
C = C 133 pm 614 kJ/mo N = N 124 pm 418 kJ/mo O H 96 pm 467 kJ/mo
C C 120 pm 839 kJ/mo N N 110 pm 941 kJ/mo H H 74 pm 436 kJ/mo
C H 109 pm 413 kJ/mo N H 101 pm 391 kJ/mo
The Lennard-Jones potential characterizes intermolecular interactions within pairs. This potential model accounts for both soft repulsive and attractive (van der Waals) forces and is applicable to electronically neutral atoms or molecules. The commonly employed expression for the Lennard-Jones potential is: [40,41,42]
V r = 4 ε σ r 12 σ r 6 ,
where r represents the distance between two interacting particles, ε represents the depth of the potential well, commonly known as the ‘dispersion energy’, and σ represents the distance at which the particle-particle potential energy V r equals zero, often termed the ‘size of the particle’. The formula is rewritten using the hollow sphere model:
lim r x r 6 = lim r 7 x + r 7 7 r ,
V r = 4 ε 7 σ 6 + r 7 7 r 1 2 2 1 4 .
The bond length of the O-H molecule is 96   p m , so the σ is:
σ = r 1 2 6 85.52627   p m ,
σ = 1 7 1 2 + r 7 r 7 6 85.74948   p m .
When r is 96   p m and σ is 85.52   p m , the result is:
V r ε = 4 σ r 6 1 2 2 1 4 0.99999 ,
V r ε = 4 7 σ 6 + r 7 7 r 1 2 2 1 4 0.99975 .
The variation is illustrated in the diagram:
Figure 13. Lennard-Jones potential of a hydrogen bond.
Figure 13. Lennard-Jones potential of a hydrogen bond.
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The formula can also be further modified:
lim r x r 12 = lim r 13 x + r 13 13 r ,
V r = 4 ε 13 σ 12 + r 13 13 7 σ 6 + r 7 7 .
When r is 96   p m and σ is 85.52   p m , the result is:
V r ε = 4 σ r 12 σ r 6 0.99999 ,
V r ε = 4 13 σ 12 + r 13 13 7 σ 6 + r 7 7 0.98473 .
Through correction, the Lennard-Jones potential at short distances has been limited.
Figure 14. Lennard-Jones potential of a hydrogen bond.
Figure 14. Lennard-Jones potential of a hydrogen bond.
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Table 8. Lennard-Jones potential of a hydrogen bond.
Table 8. Lennard-Jones potential of a hydrogen bond.
Radius 0 25   p m 50   p m 100   p m 200   p m 400   p m
Lennard-Jones 10264104 2407 0.9526 0.0243 0.000382
Space Theory 56.6872 56.6111 48.0927 0.9401 0.0242 0.000382
The Schrödinger equation for a hydrogen atom can be solved by separation of variables. [43] In this case, spherical polar coordinates are the most convenient.
Ψ r , θ , φ = R r Y l m θ , φ = R r Θ θ Φ φ .
The Spherical Harmonic functions indicate the electron’s position around the proton, and the radial function shows the electron’s distance from the proton. The solutions to the radial differential equation are products of the associated Laguerre polynomials, an exponential factor, and a normalization factor.
R r = N n , l r a 0 l × L n , l r × e r n a 0 .
The first six radial functions are provided in Table below.
Table 9. Radial wave functions for wavefunctions of the first three shells.
Table 9. Radial wave functions for wavefunctions of the first three shells.
n l Φ φ Θ θ R r
1 0 1 s 1 2 π 1 2 2 Z a 0 3 2 e Z r a 0
2 0 2 s 1 2 π 1 2 1 2 2 Z a 0 3 2 2 Z r a 0 e Z r 2 a 0
2 1 2 p 1 2 π 6 2 cos θ 1 2 6 Z a 0 3 2 Z r a 0 e Z r 2 a 0
3 0 3 s 1 2 π 1 2 2 81 3 Z a 0 3 2 27 18 Z r a 0 + 2 Z r a 0 2 e Z r 3 a 0
3 1 3 p 1 2 π 6 2 cos θ 4 81 6 Z a 0 3 2 6 Z r a 0 Z r a 0 2 e Z r 3 a 0
3 2 3 d 1 2 π 10 4 3 cos 2 θ 1 4 81 30 Z a 0 3 2 Z r a 0 2 e Z r 3 a 0
The radial probability density function is:
P r = Ψ 2 4 π r 2 .
For the 1 s orbital, the radial wave function is given as:
P r = 1 π Z a 0 3 2 e Z r a 0 2 4 π r 2 = 4 Z a 0 3 r e r Z a 0 2 .
The uv differentiation formula is:
u v = u v + u v = 0 .
Remove the constant terms, take the derivative with respect to r , and set it to zero to obtain the radius of maximum probability.
r = 1 ,
e r Z a 0 = Z a 0 e r Z a 0 .
Which gives:
1 r Z a 0 e r Z a 0 = 0 ,
r = a 0 Z .
Slater’s rules [44] provide a value for the screening constant, denoted by s , which relates the effective and actual nuclear charges as:
Z e f f = Z s .
Stars obtained their energy from nuclear fusion of hydrogen to form helium. [45] In this case, the radius for maximum probability will change from 52.9   p m to 31.1   p m . [34] This means that as the distance between the two atoms changes, the radius for maximum probability will also change. Which gives:
Z e f f = 1 + 1 s α ,
Set the distance between the two atoms as x , and 0 α 1 , the formula can be modified to:
α = 1 + x / a 0 3 3 x / a 0 .
The shielding constant s of a 1 s electron is 0.3 [44], and the bond length in a H2 molecule is 74   p m [39], so the radius for maximum probability of a H2 molecule is:
Z e f f 1.1071 ,
r = a 0 Z e f f 47.7825 .
The variation is illustrated in the diagram:
Figure 15. The radius for maximum probability (nuclear fusion).
Figure 15. The radius for maximum probability (nuclear fusion).
Preprints 138642 g015
And the radial probability density function P r is:
Figure 16. The radial probability density (1s).
Figure 16. The radial probability density (1s).
Preprints 138642 g016
Since the revised orbital formula prevents electrons from falling into the nucleus, the model needs to be recalculated.
Figure 17. Hydrogen and Helium.
Figure 17. Hydrogen and Helium.
Preprints 138642 g017
And the orbital velocity is:
r ˙ = 2 + r / a 0 2 r / a 0 ,
lim r 0 r ˙ = 2 4 1.189207115 .
The sum of E pointing toward the nucleus is:
E = Z E p Z 1 E e ,
E p = 3 + r a 0 3 3 r a 0 ,
E e = 4 π 2 0 π 2 f θ d θ .
where f θ is:
f θ = sin π 2 θ 3 + cos θ 2 r a 0 3 3 cos θ 2 r a 0 .
The author used the quad function from the SciPy library to perform numerical integration, this function is part of the scipy.integrate module and is widely used for calculating definite integrals efficiently. [46] The maximum radial probability in the 1 s orbital of helium occurs at a distance of 31   p m , which gives:
E p 0.88779184 ,
E e 0.26614705 ,
E = 2 E p E e 1.50943663 .
And the effective nuclear charge is:
s = 0.3 ,
Z e f f = Z s = 1.7 .
When considering E p as 1 unit of charge, the result is:
s = E e E p 0.29978542 0.3 ,
Z e f f = E E p 1.70021457 1.7 .

Discussion

This paper is not yet complete. As of this writing, graviton has not been found yet.

Funding

None.

Acknowledgments

I would like to thank the scientists who study String Theory and make outstanding contributions to physics. The thickening speed of physics journals has surpassed the speed of light, but this doesn’t violate the theory of relativity, because those journals aren’t transmitting any information.

Conflict of Interest

None.

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