Any object will generate a gravitational field in the surrounding space. The expression for the universal gravitation between two objects is Newton’s law of universal gravitation, which is similar in formula to Coulomb’s law. The expression for the gravitational field is like (3), so there should also be a similar mass-energy calculation law in the gravitational field.

For celestial bodies with complex shapes, mass density distributions, and motion states, it is difficult to calculate their mass-energy. However, for celestial bodies with simple shapes, mass distributions, and motion states, such as the Earth and the Sun, it is easy to calculate their mass-energy. For example, to measure the mass of the sun, $M_0$, using the centripetal force of a planet orbiting to measure the gravitational force, expressed as

$\frac{m{v}^2}{r}=G\frac{M_{0}m}{r^2}$ (6)

In Equation (6), $m$ is the planet’s mass, $v$ is its orbital speed, and $r$ represents its distance to the Sun.

Assuming a planet has an orbital radius of $r$, an orbital period of $T$, and an orbital speed $v=2\pi r/T$, formula (6) can be used to derive the mass of the central celestial body, the Sun.

$M_0=\frac{{ℰ}_0}{c^2}=\frac{4{\pi }^2}{G}\frac{r^3}{T^2}$ (7)

According to Equation (6), the speed of the planetary revolution around the Sun is

$v=\sqrt{\frac{G{M}_0}{r}}$ (8)

From the perspective of the materiality of the gravitational field, if using the radius $R$ and period $T$ of the Earth’s orbit around the Sun to calculate the mass of the Sun, there is no doubt about the correctness of using this classical formula to calculate the mass of the Sun. However, the result value should not only include the mass of ordinary matter (baryons) inside the Sun but also the mass of the gravitational field between the Sun’s interior and the Earth’s orbit. That is to say, this mass includes the mass of ordinary matter in the internal region of the Sun, as well as the mass of the gravitational field in a sphere with the radius of the Earth’s orbit, but it does not distinguish between the mass of ordinary matter and the mass of gravitational field matter.

If the mass $M_0$ inside a sphere with a radius $R$ is measured using the above method, then according to Newton’s law of universal gravitation, the strength of the gravitational field at the outer radius $r$ of the sphere is

$g_0=-G\frac{M_0}{r^2}=-G\frac{{ℰ}_0}{c^2{r}^2}$ (9)

According to formulas (1), (3), and (4) for calculating the mass-energy density around a spherical charge, corresponding to the gravitational field strength $g$, the mass-energy density formula for the gravitational field matter should be

$ϵ=\frac{1}{2}\frac{g^2}{4\pi G}$ (10)

Substituting Equation (9) into Equation (10), the mass-energy density formula at radius $r$ is

${ϵ}_0=\frac{g_0^2}{8\pi G}=\frac{1}{8\pi G{c}^4}{\left(-\frac{G{ℰ}_0}{r^2}\right)}^2=\frac{G}{8\pi c^4}\frac{{ℰ}_0^2}{r^4}$ (11)

Considering the mass-energy of the gravitational field, that inside a sphere with radius $r$ must be modified. It should contain $M_0$ within the sphere with radius $R$ and add the gravitational field matter within the sphere layer from radius $R$ to radius $r$, as shown in Figure 1 .

The first modification value of mass-energy is

$\Delta {ℰ}_1=\underset{R}{\overset{r}{{\displaystyle \int }}}{ϵ}_0\cdot 4\pi r^2\text{d}r=\underset{R}{\overset{r}{{\displaystyle \int }}}\frac{G}{8\pi c^4}\frac{{ℰ}_0^2}{r^4}\cdot 4\pi r^2\text{d}r$ (12-1)

$=\frac{G}{2c^4}\underset{r=R}{\overset{r}{{\displaystyle \int }}}\frac{{ℰ}_0^2}{r^2}\text{d}r$ (12-2)

So, after the first modification, the total mass-energy inside the sphere with radius $r$ is

${ℰ}_1=M_0{c}^2+\Delta {ℰ}_1$ (13-1)

$={ℰ}_0+\frac{G}{2c^4}\underset{r=R}{\overset{r}{{\displaystyle \int }}}\frac{{ℰ}_0^2}{r^2}\text{d}r$ (13-2)

A critical physical question arises: Will the added mass-energy in ${ℰ}_0=M_0{c}^2$ generate new gravity? Scholar Charles T. Sebens believed [13] that gravitational mass-energy does not produce gravity, but this viewpoint lacks sufficient justification. Since it is mass-energy, there should be no essential difference between it and the mass-energy of ordinary matter. That is to say, $\Delta {ℰ}_1$ should also further generate a gravitational field.

After the first mass-energy modification, it is necessary to modify the gravitational field. So, the strength of the gravitational field is modified as follows

$g_1=-\frac{G{ℰ}_1}{c^2{r}^2}$ (14)

After the first modification, the mass-energy density at radius $r$ should be

${ϵ}_1=\frac{g_1^2}{8\pi G}=\frac{G{ℰ}_1^2}{8\pi c^4{r}^4}$ (15)

The mass-energy within the sphere should be modified for the second time, with a modification value as follows

$\Delta {ℰ}_2=\underset{r=R}{\overset{r}{{\displaystyle \int }}}{ϵ}_1\cdot 4\pi r^2\text{d}r$ (16-1)

$=\underset{r=R}{\overset{r}{{\displaystyle \int }}}\frac{G{ℰ}_1^2}{8\pi c^4{r}^4}\cdot 4\pi r^2\text{d}r$ (16-2)

$=\frac{G}{2c^4}\underset{r=R}{\overset{r}{{\displaystyle \int }}}\frac{{ℰ}_1^2}{r^2}\text{d}r$ (16-3)

So, after the second modification, the mass-energy in the entire sphere with radius $r$ is

${ℰ}_2={ℰ}_0+\Delta {ℰ}_2$ (17-1)

$={ℰ}_0+\frac{G}{2c^4}\underset{r=R}{\overset{r}{{\displaystyle \int }}}\frac{{ℰ}_1^2}{r^2}\text{d}r$ (17-2)

After the second modification, the strength of the gravitational field generated by mass-energy ${ℰ}_2$ is

$g_2=-\frac{G{ℰ}_2}{c^2{r}^2}$ (18)

The mass-energy density of the gravitational field is modified to be

${ϵ}_2=\frac{g_2^2}{8\pi G}=\frac{G{ℰ}_2^2}{8\pi c^4{r}^4}$ (19)

The third modification of mass-energy from radius $R$ to $r$ within the sphere layer is

$\Delta {ℰ}_3=\underset{r=R}{\overset{r}{{\displaystyle \int }}}{ϵ}_2\cdot 4\pi r^2\text{d}r$ (20-1)

$=\underset{r=R}{\overset{r}{{\displaystyle \int }}}\frac{G{ℰ}_2^2}{8\pi c^4{r}^4}4\pi r^2\text{d}r$ (20-2)

$=\frac{G}{2c^4}\underset{r=R}{\overset{r}{{\displaystyle \int }}}\frac{{ℰ}_2^2}{r^2}\text{d}r$ (20-3)

In this way, after the third modification, the mass-energy of the entire sphere with radius $r$ is

${ℰ}_3={ℰ}_0+\Delta {ℰ}_3$ (21-1)

$={ℰ}_0+\frac{G}{2c^4}\underset{r=R}{\overset{r}{{\displaystyle \int }}}\frac{{ℰ}_2^2}{r^2}\text{d}r$ (21-2)

By continuously modifying to obtain

${ℰ}_{n+1}={ℰ}_0+\frac{G}{2c^4}{\displaystyle \underset{r=R}{\overset{r}{\int }}\frac{{ℰ}_n^2}{r^2}\text{d}r}$ (22)

Modify infinitely multiple times to obtain the final mass-energy. It’s obvious

$\underset{n\to \infty }{\lim }{ℰ}_{n+1}=\underset{n\to \infty }{\lim }{ℰ}_n=ℰ$ (23)

So, (22) ultimately becomes

$ℰ={ℰ}_0+\frac{G}{2c^4}\underset{r=R}{\overset{r}{{\displaystyle \int }}}\frac{{ℰ}^2}{r^2}\text{d}r$ (24)

It is an integral equation. It is not easy to directly find $ℰ$.

For solving Equation (24), it is necessary to convert it into a differential equation, which requires taking the first-order derivative on both sides of the equation and considering that ${ℰ}_0=M_0{c}^2$ is a constant, the above equation becomes

$\frac{\text{d}ℰ}{\text{d}r}=\frac{G}{2c^4}\frac{{ℰ}^2}{r^2}$ (25)

Solving this differential equation yields

$ℰ=\frac{1}{C+\frac{G}{2c^4}\frac{1}{r}}$ (26)

In the Equation (26), $C$ is the integral constant. $ℰ={ℰ}_0=M_0{c}^2$ when $r=R$, obtained from Equation (26)

$C=\frac{1}{{ℰ}_0}-\frac{G}{2c^{4}R}$ (27)

Substitute (27) into Equation (26) to obtain

$ℰ=\frac{{ℰ}_0}{1-\frac{G{ℰ}_0}{2c^4}\left(\frac{1}{R}-\frac{1}{r}\right)}$ (28)

In other words, the mass-energy of the matter inside a sphere with the radius $r$ is

$ℰ=\frac{M_0{c}^2}{1-\frac{G{M}_0}{2c^2}\left(\frac{1}{R}-\frac{1}{r}\right)}$ (26)

The mass inside a sphere with radius $r$ is

$M=\frac{M_0}{1-\frac{G{M}_0}{2c^2}\left(\frac{1}{R}-\frac{1}{r}\right)}$ (30)

Substitute it into Newton’s law of universal gravitation to obtain

$F=-G\frac{Mm}{r^2}=-\frac{G{M}_{0}m}{r^2\left[1-\frac{G{M}_0}{2c^2}\left(\frac{1}{R}-\frac{1}{r}\right)\right]}$ (31)

According to the above equation, obtain the expression for the strength of the gravitational field.

$g=\frac{F}{m}=-G\frac{M}{r^2}=-\frac{G{M}_0}{r^2\left[1-\frac{G{M}_0}{2c^2}\left(\frac{1}{R}-\frac{1}{r}\right)\right]}$ (32)

The mass-energy density of matter in the gravitational field at the radius is

$ϵ=\frac{g^2}{8\pi G}=\frac{G}{8\pi c^4}\frac{{ℰ}_0^2}{r^4{\left[1-\frac{G{ℰ}_0}{2c^4}\left(\frac{1}{R}-\frac{1}{r}\right)\right]}^2}$ (33)

According to Equation (32), the speed of the object’s motion around the central celestial body at the orbital radius $r$ is

$v=\sqrt{\frac{G{M}_0}{r\left[1-\frac{G{M}_0}{2c^2}\left(\frac{1}{R}-\frac{1}{r}\right)\right]}}=\sqrt{\frac{G{M}_0}{\left(1-\frac{G{M}_0}{2c^{2}R}\right)r+\frac{G{M}_0}{2c^2}}}$ (34)

Formula (34) has an additional factor $1/\sqrt{1-\frac{G{M}_0}{2c^2}\left(\frac{1}{R}-\frac{1}{r}\right)}$ compared to formula (8). Obviously, under the condition of $r\ge R$, this factor is greater than or equal to 1, resulting in the speed of the surrounding celestial bodies

$v\ge \sqrt{\frac{G{M}_0}{r}}$ (35)

Moreover, as the factor $\frac{G{M}_0}{2c^2}\left(\frac{1}{R}-\frac{1}{r}\right)\to 1-0$, the deviation of the speed of surrounding celestial bodies from (8) becomes apparent. Therefore, the formula (8) is not applicable, but it is necessary to use (34).

The shape and internal structure of the celestial bodies are often complex, and their mass distribution density and galaxy contours are not easy to express using simple functions. Even if expressed using simple functions, solving the equation may be difficult.

To understand the concept of gravitational field matter, assume that the mass distribution of ordinary matter inside celestial bodies is uniform, the mass density is ${\rho }_0$, the outline of the bodies is a sphere with a radius of $R_0$, the total mass of ordinary matter is $M_R$, and the total mass-energy is ${ℰ}_R$. Obviously, there is a relation as follows

${ℰ}_R=M_R{c}^2=\frac{4\pi }{3}{\rho }_0{c}^2{R}_0^3$ (36)

The mass-energy of the ordinary matter within a sphere with a radius of $r$ is

${ℰ}_0=M_0{c}^2=\frac{4\pi }{3}{\rho }_0{c}^2{r}^3$ (37)

According to Newton’s law of universal gravitation, the strength of the gravitational field on a sphere with a radius of $r$ is

$g_0=-G\frac{{ℰ}_0}{c^2{r}^2}$ (38)

The corresponding mass-energy density is

${ϵ}_0=\frac{g_0^2}{8\pi G}=\frac{1}{8\pi G}{\left(-G\frac{{ℰ}_0}{c^2{r}^2}\right)}^2=\frac{G}{8\pi c^4}{\left(\frac{{ℰ}_0}{r^2}\right)}^2$ (39)

The first modification value of the mass-energy inside this sphere is

$\Delta {ℰ}_1={\displaystyle \underset{0}{\overset{r}{\int }}{ϵ}_{0}4\pi r^2\text{d}r}=\frac{G}{2c^4}{\displaystyle \underset{0}{\overset{r}{\int }}\frac{{ℰ}_0^2}{r^2}\text{d}r}$ (40)

The mass-energy inside the sphere after the first modification is

${ℰ}_1=M_0{c}^2+\Delta {ℰ}_1$ (41-1)

$=M_0{c}^2+\frac{G}{2c^4}{\displaystyle \underset{0}{\overset{r}{\int }}\frac{{ℰ}_0^2}{r^2}\text{d}r}$ (41-2)

Similar to the modification process in 2.1, the mass-energy value after the ( $n+1$) times modification is

${ℰ}_{n+1}={ℰ}_0+\frac{G}{2c^4}{\displaystyle \underset{0}{\overset{r}{\int }}\frac{{ℰ}_n^2}{r^2}\text{d}r}$ (42)

${ℰ}_{n+1}\to {ℰ}_n\to ℰ$ when $n\to \infty$, so the above formula changes to

$ℰ={ℰ}_0+\frac{G}{2c^4}{\displaystyle \underset{0}{\overset{r}{\int }}\frac{{ℰ}^2}{r^2}\text{d}r}$ (43)

Both $ℰ$ and ${ℰ}_0$ are functions on $r$. This equation is an integral one whose solution cannot be obtained directly through integration.

It is necessary to transform the equation into a differential one to solve $ℰ$. Differentiating both sides of the equation with respect to $r$ yields

$\frac{\text{d}ℰ}{\text{d}r}=\frac{\text{d}{ℰ}_0}{\text{d}r}+\frac{G}{2c^4}\frac{{ℰ}^2}{r^2}$ (44)

Using Equation (37), Equation (44) changes to

$\frac{\text{d}ℰ}{\text{d}r}=\frac{G}{2c^4}\frac{{ℰ}^2}{r^2}+4\pi {\rho }_0{c}^2{r}^2$ (45)

It is a nonlinear ordinary differential equation, and it is impossible to directly find $ℰ$, which requires further changes to a new equation. It needs a Riccati substitution. Suppose there is a new function $u\left(r\right)$ whose first-order derivative to $r$ is $u^{\prime }=\text{d}u/\text{d}r$ and its second-order derivative is $u^{″}={\text{d}}^{2}u/\text{d}{r}^2$, as a substitution.

$ℰ\left(r\right)=-\frac{2c^4{r}^2}{G}\frac{u^{\prime }}{u}$ (46)

Then Equation (45) becomes

$r{u}^{″}+2u^{\prime }+\frac{2\pi G{\rho }_{0}r}{c^2}u=0$ (47)

It is a second-order homogeneous ordinary differential equation with variable coefficients. Set a constant

$\alpha =\sqrt{\frac{2\pi G{\rho }_0}{c^2}}$ (48)

Then Equation (47) becomes

$r{u}^{″}+2u^{\prime }+{\alpha }^{2}ru=0$ (49)

Expanding the function into a Taylor series

$u\left(r\right)=\underset{n=0}{\overset{\infty }{{\displaystyle \sum }}}{a}_n{r}^n$ (50)

Substituting (50) to (49) gives

$\underset{n=0}{\overset{\infty }{{\displaystyle \sum }}}\left(n+2\right)\left(n+1\right)a_{n+2}{r}^{n+1}+2\underset{n=0}{\overset{\infty }{{\displaystyle \sum }}}\left(n+1\right)a_{n+1}{r}^n+{\alpha }^2\underset{n=0}{\overset{\infty }{{\displaystyle \sum }}}{a}_n{r}^{n+1}=0$ (51)

Derive the relationship among coefficients from this

$a_1=0$ (52)

$a_2=-\frac{1}{3!}{\alpha }^2{a}_0$ (53)

$a_3=0$ (54)

$a_4=-\frac{1}{20}{\alpha }^2{a}_2=\frac{1}{5!}{\alpha }^4{a}_0$ (55)

$a_5=0$ (56)

$a_6=-\frac{1}{42}{\alpha }^2{a}_4=-\frac{1}{7!}{\alpha }^6{a}_0$ (57)

$a_7=0$ (58)

$⋮$

$a_{n=2k}=\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{\alpha }^{2k}{a}_0\left(k\ge 0\right)$ (59)

$a_{n=2k+1}=0\left(k\ge 0\right)$ (60)

Summarizing the above relationship can conclude that

$u\left(r\right)=a_0\underset{k=0}{\overset{\infty }{{\displaystyle \sum }}}\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{\alpha }^{2k}{r}^{2k}$ (61-1)

$=\frac{a_0}{\alpha r}\underset{k=0}{\overset{\infty }{{\displaystyle \sum }}}\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{\left(\alpha r\right)}^{2k+1}$ (61-2)

$=a_0\frac{\sin \left(\alpha r\right)}{\alpha r}$ (61-3)

Substituting it into (46) obtains

$ℰ\left(r\right)=-\frac{2c^4{r}^2}{G}\frac{u^{\prime }}{u}$ (62-1)

$=\frac{2c^{4}r}{G}\left[1-\left(\alpha r\right)\cdot \text{ctg}\left(\alpha r\right)\right]$ (62-2)

Set

$\chi =\alpha r$ (63)

Then (62) becomes

$ℰ\left(r\right)=\frac{2c^4}{G\alpha }\chi \left(1-\chi \cdot \text{ctg}\chi \right)$ (64)

When $r$ reaches the radius of the sphere, $R_0$, i.e., $r=R_0$, then $\chi =\alpha r$ becomes

${\chi }_0=\sqrt{\frac{3G{ℰ}_{0R}}{2c^4{R}_0}}=\sqrt{\frac{3G{M}_{0R}}{2c^2{R}_0}}$ (65)

The mass-energy of ordinary matter inside the sphere is ${ℰ}_{0R}=M_{0R}{c}^2$, and the total mass-energy of matter in a gravitational field is

$ℰ\left(R_0\right)=\frac{2c^4{R}_0}{G}\left(1-{\chi }_0\cdot \text{ctg}{\chi }_0\right)$ (66)

The total mass-energy $ℰ\left(R_0\right)$ inside the sphere is not equal to that of ordinary matter ${ℰ}_{0R}$.

If $r=R=R_0$, the value ${ℰ}_0$ in Equation (28) equals $ℰ\left(R_0\right)$ in Equation (66), thus establishing a relationship

$ℰ=\frac{M_0{c}^2}{1-\frac{G{M}_0}{2c^2{R}_0}\left(1-\frac{R_0}{r}\right)}$ (67-1)

$=\frac{2c^4{R}_0}{G}\frac{1-{\chi }_0\text{ctg}{\chi }_0}{1-\left(1-{\chi }_0\text{ctg}{\chi }_0\right)\left(1-\frac{R_0}{r}\right)}$ (67-2)

This formula expresses the relationship between the gravitational matter and the ordinary matter of the sphere with radius r due to the presence of gravitational field matter inside and outside the sphere.

If $r\to \infty$, the above expression tends towards

${ℰ}_{\infty }=\frac{2c^4{R}_0}{G}\left(\frac{\text{tg}{\chi }_0}{{\chi }_0}-1\right)$ (68)

The graph of this function resembles the one shown in Figure 2 .

$y=\frac{\text{tg}x}{x}-1$ (69)

In this function, the independent variables $x=n\pi +\pi /2$ are singularities (where $n=0,1,2,3,\cdots$). At these singularities, ${ℰ}_{\infty }=\pm \infty$. The graphs of functions (68) or (69) show that within some ranges, the gravitational matter of celestial bodies takes on negative values, indicating repulsive forces between the internal matter of the body, which is inconceivable.

The radius $R_0$ approaches its minimum values where ${\chi }_0\to n\pi +\pi /2-0$.

$R_{0n}=\frac{6}{{\left(2n+1\right)}^2{\pi }^2}\frac{G{M}_{0R}}{c^2}+0\left(n=0,1,2,3,\cdots \right)$ (70)

The curve with potential physical significance is the first one corresponding to $n=0$. Due to ${\chi }_0\to \pi /2-0$, obtained from Equation (65)

$R_0\ge \frac{6}{{\pi }^2}\frac{G{M}_{0R}}{c^2}$ (71)

The minimum value of the celestial body’s radius can only be

$R_{0\min }=\frac{6}{{\pi }^2}\frac{G{M}_{0R}}{c^2}$ (72)

However, this minimum radius is the largest value in formula (70).

Since gravitational field matter is related to ordinary matter and the structure factor $G{M}_0/2c^2{R}_0$ of celestial bodies, this is particularly significant for dense objects. Black holes are usually such dense objects with strong gravitational fields, making the mass of gravitational field matter non-negligible.

The measured mass of black holes using surrounding celestial bodies always yields only part of their gravitational mass rather than the entire gravitational mass of the objects or their actual mass.

The structure of black holes is discussed below from two different perspectives.

1) According to the theory of special relativity

To find the relationship between the mass and structural radius of ordinary matter, suppose there is a black hole with the mass of ordinary matter being $M_{0R}$ and its radius being $R_0$. According to formula (67-2), when measured on the surface of the black hole, the mass of the black hole is

$M_0=\frac{2c^2{R}_0}{G}\left(1-{\chi }_0\cdot \text{ctg}{\chi }_0\right)$ (73)

If an object with a rest mass of $m_0$ is launched radially from the surface of a black hole with an initial speed of $v_0$, and its speed at infinity distance is zero, with gravitational potential energy also being zero, an equation based on the law of conservation of mechanical energy is given by

$\frac{m_0{c}^2}{\sqrt{1-\frac{v_0^2}{c^2}}}-\frac{G{M}_0{m}_0}{R_0\sqrt{1-\frac{v_0^2}{c^2}}}=m_0{c}^2$ (74)

This equation can turn into

$1-\frac{G{M}_0}{c^2{R}_0}=\sqrt{1-\frac{v_0^2}{c^2}}$ (75)

Since celestial bodies are black holes, even if an object’s initial speed reaches that of light, it cannot escape the gravitational pull of the black hole. Therefore, the condition must meet with $v_0=c$. The equation can turn into

$1-\frac{G{M}_0}{c^2{R}_0}=0$ (76)

2) According to the photon redshift perspective

One can also imagine that a photon with energy $h\nu$ is emitted radially from the surface of a black hole. Due to gravitational redshift, its energy becomes zero at infinity distance, which sets up an equation as the following

$h\nu -\frac{G{M}_0}{R_0}\frac{h\nu }{c^2}=0$ (77)

The solution to this equation is also Equation (76). It yields the same solution from two different starting points.

$M_0=\frac{c^2{R}_0}{G}$ (78)

Substituting Equation (78) into Equation (73) obtains

$\frac{\text{tg}{\chi }_0}{{\chi }_0}=2$ (79)

This equation has infinitely many solutions. To solve this equation, let the function

$y=\frac{\text{tg}x}{x}-2$ (80)

The intersection points of this function with the X-axis are the solutions to Equation (79): $x_1,x_2,x_3,\cdots$. Its graph resembles the one shown in Figure 3 .

Use computer mathematical function graphing tools to obtain these solutions.

${\chi }_{01}=1.165561185207$ (81-1)

${\chi }_{02}=4.604216777200$ (81-2)

${\chi }_{03}=7.789883751144$ (81-3)

$⋮$

According to formula (63), each solution corresponds to a radius. However, since each solution is isolated by an infinitely deep potential well and an infinitely high potential barrier, the state of the black hole cannot transition between these radii. Therefore, solutions except for ${\chi }_{01}$ are not to be considered. Perhaps it poses a more profound physical significance hidden in this suspense that needs further exploration.

According to Equation (65), the critical radius of the black hole corresponding to ${\chi }_{01}$ is

$R_C=\frac{3}{2{\chi }_{01}^2}\frac{G{M}_R}{c^2}=1.104132278276\times \frac{G{M}_R}{c^2}=\kappa \frac{G{M}_R}{c^2}$ (82)

where

$\kappa =\frac{3}{2{\chi }_{01}^2}=1.104132278276$ (83)

Among all the solutions, ${\chi }_{01}$ is the smallest, corresponding to the maximum radius of the black hole.

If the distance $R_0$ in formula (67) exactly equals the critical structural radius of the black hole, $R_C$, then $M_0=M_{0R}$, yields

$M=\frac{M_0}{1-\frac{G{M}_0}{2c^2}\left(\frac{1}{R_0}-\frac{1}{r}\right)}=\frac{M_0}{1-\frac{G{M}_0}{2c^2{R}_0}+\frac{G{M}_0}{2c^{2}r}}$ (84-1)

$=\frac{2c^2}{G}\frac{R_{0}r}{r+R_0}=\frac{2c^2}{G}\frac{R_{C}r}{r+R_C}$ (84-2)

$=\frac{2\kappa M_{0R}r}{r+\kappa \frac{G{M}_{0R}}{c^2}}$ (84-3)

That is, the minimum mass of a black hole is

$M_{\min }=\frac{2\kappa M_{0R}r}{r+\kappa \frac{G{M}_{0R}}{c^2}}\left(r\ge R_{BH}\right)$ (85)

At the surface of a critical black hole, the relationship between gravitational mass $M$ and ordinary mass $M_{0R}$ is such that the gravitational mass $M$ is $\kappa$ times the ordinary mass $M_{0R}$, that is $\kappa M_{0R}$. If observing the critical black hole from an infinite distance, the gravitational mass is $2\kappa M_{0R}$. It proves that half of a critical black hole’s gravitational mass is inside the black hole while the other half is outside.

This is somewhat similar to the mass-energy distribution characteristics of a single electron. So, a single electron is an electric field black hole.

On the other hand, according to Equation (65), setting ${\chi }_0\to \pi /2-0$ yields

$R_0\to \frac{6}{{\pi }^2}\frac{G{M}_{0R}}{c^2}+0$ (86)

Substituting it into (67-2) obtains

$\underset{{\chi }_0\to \pi /2}{\lim }M=\underset{{\chi }_0\to \pi /2}{\lim }\frac{2c^2{R}_0}{G}\frac{1-{\chi }_0\text{ctg}{\chi }_0}{1-\left(1-{\chi }_0\text{ctg}{\chi }_0\right)\left(1-\frac{R_0}{r}\right)}=\frac{2c^{2}r}{G}\left(r\ge R_C\right)$ (87)

It is the limit value of black hole mass measured at a distancer from the black hole’s center that is impossible to reach, of course, so there always exists a relationship as

$M<\frac{2c^{2}r}{G}\left(r>R_C\right)$ (88)

So, if a black hole’s radius reaches the value $6G{M}_{0R}/{\pi }^2{c}^2$, no object can escape its gravitational pull, even if it travels at the speed of light, regardless of how far away it is.

If the ordinary mass of a black hole is $M_{0R}$, then its gravitational mass, $M$, as measured from outside the black hole entity, would be

$\frac{2\kappa M_{0R}r}{r+\kappa \frac{G{M}_{0R}}{c^2}}\le M<\frac{2c^{2}r}{G}\left(r>R_C\right)$ (89)

As shown in formula (89), the gravitational mass of a black hole increases with the increase in radius $r$. In contrast, the gravitational mass of celestial bodies calculated according to Newton’s conventional law of universal gravitation does not increase with an increase in measurement radius $r$.

Equation (89) indicates that at least half of the black hole’s gravitational mass is distributed outside the black hole, or in other words, most of the gravitational mass spreads out in space outside the black hole.

Combining (82) and (86) obtain the range of structural radii for the black hole

$\frac{6}{{\pi }^2}\frac{G{M}_{0R}}{c^2}<R_{BH}\le \kappa \frac{G{M}_{0R}}{c^2}$ (90)

It is just a range for the radius of the black hole structure, but its specific expression remains unknown.

The Earth’s ordinary mass is $M_e=5.965\times {10}^{24}\text{kg}$, with an Earth radius of $R_e=6.37\times {10}^6\text{m}$. This results in

${\eta }_e=\frac{G{M}_e}{2c^2{R}_e}=1.04\times {10}^{-9}\ll 1$ (91)

According to formula (65), this factor is negligible unless precise calculation is necessary. Therefore, the Earth’s mass-energy can be expressed as $ℰ=M_e{c}^2$.

In the space near the Earth’s surface, the mass-energy density of the gravitational field is given by

$ϵ=\frac{g^2}{4\pi G}=1.146\times {10}^{11}\text{J}/{\text{m}}^{\text{3}}$ (92)

or mass density is

${\rho }_e=\frac{ϵ}{c^2}=\frac{1}{4\pi G}\frac{g^2}{c^2}=1.273\times {10}^{-6}\text{kg}/{\text{m}}^{\text{3}}$ (93)

The gravitational field matter is negligible compared to ordinary matter on Earth. However, there is indeed gravitational field matter inside and outside the Earth.

The sun is a typical example of a star. The mass of the sun is $M_{sun}=2.0\times {10}^{30}\text{kg}$, and its radius is $R_{sun}=6.963\times {10}^8\text{m}$.

As calculated in Equation (91)

${\eta }_{sun}=\frac{G{M}_{sun}}{2c^2{R}_{sun}}=1.06\times {10}^{-6}\ll 1$ (94)

The mass of the sun, calculated using formula (67) with the condition $r=\infty$, is $M_{sun}=M_R=2.0\times {10}^{30}\text{kg}$.

When calculating the solar total mass, the mass of the gravitational field matter is negligible. Thus, the solar total mass equals the mass of ordinary matter unless precise calculation is necessary.

A typical neutron star forms from the collapse of a star with a mass of about 1.4 times that of the Sun, that is $M_n=2.8\times {10}^{30}\text{kg}$. The collapse process leads to a supernova explosion, during which the lost mass is roughly 5.0 × 10⁻⁴ times of the star’s pre-collapse mass, which is negligible in rough calculations. The radius of a neutron star is approximately 1.1 × 10⁴ m. Thus

${\eta }_n=\frac{G{M}_n}{2c^2{R}_n}=0.094323$ (95)

Utilizing formula (65), determine

${\chi }_0=\sqrt{\frac{3G{M}_n}{2c^4{R}_n}}=0.53195$ (96)

Substitute this result into formula (68) to calculate the desired value.

${ℰ}_{n\infty }=\frac{2c^4{R}_0}{G}\left(\frac{\text{tg}{\chi }_0}{{\chi }_0}-1\right)=2.842\times {10}^{47}\text{J}$ (97)

The total mass of a neutron star measured from an infinite distance is

$M_{n\infty }=\frac{{ℰ}_{\infty }}{c^2}=3.158\times {10}^{30}\text{kg}=1.1278M_n$ (98)

The gravitational mass of a neutron star is 1.1278 times that of the ordinary matter of the star before it exploded, $M_n$. It means that after a star undergoes a supernova explosion and becomes a neutron star, its total mass increases to 1.1278 times. What has increased is the mass of the gravitational field.

Does this violate the law of conservation of energy? Of course not. In fact, after the collapse of a celestial body, the total mass-energy increases while the gravitational potential energy decreases, and the total energy before and after changes in the system remains conserved.

Galaxies have shapes such as spherical, ellipsoidal, and spiral. Even if galaxies are spiral-shaped, there is also a bulge-dominated spiral (NGC7814), disk-dominated spiral (NGC6503), and gas-dominated dwarf (NGC3741). While it is theoretically possible to derive a series of formulas to calculate the gravitational field matter and other physical quantities based on observed data, the complexity of these calculations poses significant challenges. Nevertheless, the mentioned formulas can still help make approximate estimations.

Take the Milky Way as an example. Currently, there is no precise value for the total mass of visible matter in the Milky Way, but it is estimated to be on the order of magnitude represented by $M_G\approx {10}^{42}\text{kg}$, with a disk radius of $R_G=3.9\times {10}^{20}\text{m}$. At the center of the Milky Way, a supermassive black hole is the point with the highest mass density.

Preliminary estimation indicates

${\eta }_G=\frac{G{M}_G}{2c^2{R}_G}\approx 9.5014245\times {10}^{-7}$ (99)

According to Equation (65) obtains

${\chi }_0=\sqrt{\frac{3G{M}_G}{2c^2{R}_G}}=1.68832\times {10}^{-3}$ (100)

Use formula (68) to perform the calculation as follows

$M_{G\infty }=\frac{2c^2{R}_G}{G}\left(\frac{\text{tg}{\chi }_0}{{\chi }_0}-1\right)$ (101-1)

$=1.00000243\times {10}^{42}\text{kg}\approx 1.00000243M_G$ (101-2)

The calculation results show that the gravitational mass of the galaxy is slightly greater than its ordinary mass. So, the structure factor of the entire galaxy is tiny, the macroscopic gravitational field matter of the visible matter is negligible, and the conventional Newton’s law of universal gravitation is applicable.

A spiral galaxy consists of three parts: the galactic nucleus, the disk, and the spiral arms. The majority of the mass in the Milky Way consists of stars, black holes, interstellar dust, gas, and other matter.

Based on the luminosity pattern [1] and assuming that the mass density of the disk is $\sigma ={\sigma }_0{\text{e}}^{-\beta r}$, the mass within a radius $r$ on the disk’s surface can be expressed as follows

$M={\displaystyle \underset{0}{\overset{R}{\int }}\sigma \text{d}\left(\pi r^2\right)}=2\pi {\sigma }_0{\displaystyle \underset{0}{\overset{r}{\int }}{\text{e}}^{-\beta r}r\text{d}r}=\frac{2\pi {\sigma }_0}{{\beta }^2}\left[1-\left(1+\beta r\right){\text{e}}^{-\beta r}\right]$ (102)

If $r\to \infty$, the formula above yields $M\to M_G$, that is

$M_G=\frac{2\pi {\sigma }_0}{{\beta }^2}$ (103)

Using formulas (8) and (102), obtain

$v=\sqrt{\frac{GM}{r}}={\left\{\frac{2\pi G{\sigma }_0\left[1-\left(1+\beta r\right){\text{e}}^{-\beta r}\right]}{{\beta }^{2}r}\right\}}^{\frac{1}{2}}=v_{0}y$ (104)

Here, $x=\beta r$, $v_0=\sqrt{2\pi G{\sigma }_0/\beta }=\sqrt{\beta G{M}_G}$, y can be expressed by the function as follows

$y={\left[\frac{1-\left(1+x\right){\text{e}}^{-x}}{x}\right]}^{\frac{1}{2}}$ (105)

Its graph resembles as shown in Figure 4 , where the unit for the Y-axis is $v_0=\sqrt{\beta G{M}_0}$.

The maximum speed of an object’s circular motion within a galaxy depends on the mass $M_G$ of its visible matter and the attenuation rate in mass density, $\beta$. Under certain matter conditions, the higher the value of $\beta$, the more mass is concentrated at the center, leading to a higher maximum speed.

Of course, the distribution of galaxy mass density does not strictly follow an exponential law pattern. However, interpreting $\beta$ as the degree of mass concentration, it is unnecessary to question whether the above models align with actual celestial structures. The speed curve in the image is very similar to the observed situation. However, the estimated speed is numerically smaller than the observed value. Thus, the problem of a large amount of invisible matter still existing in the galaxy arises. So, what exactly are these invisible substances?

Apart from approximately 200 billion stars and interstellar dust, black holes’ mass is a significant component. Although current estimates suggest about 100 million black holes in the Milk Way, their ordinary mass constitutes a small proportion of the total mass. Nevertheless, due to the celestial structure factor of black holes possibly being close to 1, with a very narrow range of variation from critical radius $R_C=\kappa G{M}_{0R}/c^2$ to limitation radius $R_{lim}=6G{M}_{0R}/{\pi }^2{c}^2$, so black holes significantly affect the gravitational mass of celestial bodies.

For example, the mass of the supermassive black hole at the center of the Milky Way, Sagittarius A*, is about 4.3 million times that of the Sun, i.e., $M_R=8.6\times {10}^{36}\text{kg}$, with an observed radius of approximately $R_{BH}=2.2\times {10}^{10}\text{m}$. This data is controversial; even if calculated using the Schwarzschild radius formula $R_S=2GM/c^2$, it should be $R_S=1.186\times {10}^{10}\text{m}$, and according to the critical radius formula for black holes (82), it should be $R_C=\kappa G{M}_{0R}/c^2=6.374\times {10}^9\text{m}$. The value $R_{BH}=2.2\times {10}^{10}\text{m}$ cannot represent a black hole. Of course, black holes don’t have to be spherical or have uniformly distributed mass, nor are they necessarily static. However, based on $R_{BH}$, the structure factor of celestial body is

${\eta }_s=\frac{G{M}_S}{2c^2{R}_S}=\frac{6.67\times {10}^{-11}\times 8.6\times {10}^{36}}{2\times {\left(3\times {10}^8\right)}^2\times 2.2\times {10}^{10}}=0.1448535$ (106)

Thus

${\chi }_0=\sqrt{\frac{3G{M}_S}{2c^2{c}^2{R}_S}}=0.659212$ (107)

Substitute into formula (68) to obtain

$M_{\infty }=\frac{2c^2{R}_S}{G}\left(\frac{\text{tg}{\chi }_0}{{\chi }_0}-1\right)=10.414\times {10}^{36}\text{kg}=1.211M_S$ (108)

It is evident that this gravitational field mass is 1.211 times greater than ordinary mass.

Since a celestial body becomes a black hole, its radius must be less than or equal to the critical radius. Even if the radius equals the critical radius, according to formula (85), the gravitational mass of the black hole is at least $M_C=\kappa M_{0R}$. As the radius approaches $6G{M}_{0R}/{\pi }^2{c}^2$, the gravitational mass of the black hole increases, with no upper limit.

Therefore, the missing mass in galaxies is likely to be the mass of the gravitational field matter of black holes.

Take the Coma Cluster as an example. Its total visible mass is 10¹⁵ times the mass of the Sun, that is, $M_{Coma}=2\times {10}^{45}\text{kg}$, and its radius is approximately 4 times that of the Milky Way [14] , which is $R_{Coma}=1.56\times {10}^{21}\text{m}$.

The structure factor of celestial body is

${\eta }_C=\frac{G{M}_{Coma}}{2c^2{R}_{Coma}}\approx 4.75071225\times {10}^{-4}$ (109)

Use formula (65) to perform the calculation as follows

${\chi }_0=\sqrt{\frac{3G{M}_{Coma}}{2c^2{R}_{Coma}}}=3.7752\times {10}^{-2}$ (110)

Use formula (68) to perform the calculation as follows

$M_{C\infty }=\frac{2c^2{R}_{Coma}}{G}\left(\frac{\text{tg}{\chi }_0}{{\chi }_0}-1\right)=2.00114\times {10}^{45}\text{kg}=1.00057M_C$ (111)

Therefore, the mass of the galaxy’s gravitational field matter has increased slightly, so the speed formula (8) is still valid.

Suppose the mass density distribution function of clusters is

$\rho ={\rho }_0{\text{e}}^{-\beta r}$ (112)

The ordinary mass within a sphere of radius $r$ is

$M={\displaystyle \underset{0}{\overset{r}{\int }}{\rho }_0{\text{e}}^{-\beta r}4\pi r^2\text{d}r}=\frac{8\pi {\rho }_0}{{\beta }^3}\left\{1-\left[1+\beta r+\frac{1}{2}{\left(\beta r\right)}^2\right]{\text{e}}^{-\beta r}\right\}$ (113)

If $r\to \infty$, the formula above yields $M\to M_C$, that is

$M_{Coma}=\frac{8\pi {\rho }_0}{{\beta }^3}$ (114)

The speed of an object moving in circular motion around the center of the galaxy cluster is

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{G{M}_{Coma}\left\{1-\left[1+\beta r+\frac{1}{2}{\left(\beta r\right)}^2\right]{\text{e}}^{-\beta r}\right\}}{r}}$ (115-1)

$=\sqrt{\frac{\beta G{M}_{Coma}\left[1-\left(1+x+\frac{1}{2}{x}^2\right){\text{e}}^{-x}\right]}{x}}$ (115-2)

Here $x=\beta r$, $v_0=\sqrt{\beta G{M}_{Coma}}$, and $v=v_{0}y$, $y$ can be expressed by the function as follows

$y=\sqrt{\frac{1-\left(1+x+\frac{1}{2}{x}^2\right){\text{e}}^{-x}}{x}}$ (116)

The unit on the Y-axis is $v_0=\sqrt{\beta G{M}_{coma}}$. Its graph resembles as shown in Figure 5 .

Similar to disk galaxies, the maximum speed of an object’s circular motion within a galaxy depends on the mass $M_{Coma}$ of its visible matter and the attenuation rate $\beta$ in mass density. Under certain matter conditions, the higher the value of $\beta$, the more mass is concentrated at the center, leading to a higher maximum speed.

Equation (67) illustrates that the magnitude of gravitational matter depends on the mass of ordinary matter and the structure factor $G{M}_R/2c^2{R}_0$. It indicates that the higher the structure factor of a celestial body, the greater its effect on gravitational matter, which is associated with dark matter. In smaller structure factor objects such as the Earth and the Sun, the factor $G{M}_R/2c^2{R}_0$ is negligible, so it seems that there is no dark matter. However, for celestial bodies such as neutron stars and black holes, the factors $G{M}_R/2c^2{R}_0$ are so large that they cannot be ignored.

Using the speed of outer galaxy objects orbiting the center to calculate the galaxy mass and comparing it with luminous mass, there is inevitably a problem which gravitational mass exceeds luminous mass. A black hole not only has massive ordinary matter of its own but also generates gravitational field matter much more than its ordinary matter. Galaxies and galaxy clusters contain a large number of massive black holes, contributing enormous gravitational field matter.

1) The gravitational field matter is generated by ordinary matter and exists inside and outside the ordinary matter. There is almost no gravitational field matter in spaces far from galaxies, galaxy groups, and galaxy clusters, especially in voids.

It can explain why dark matter always accompanies ordinary matter.

2) Because the gravitational field is a vector, it results from the vector superposition of gravitational fields generated by different parts of ordinary matter. Therefore, in space where the gravitational field cancels out to zero, the density of gravitational field matter is also zero. In space where the gravitational field weakens, the density of gravitational field matter decreases. In space where the gravitational field strengthens, the gravitational field matter increases.

It can explain the presence of dense dark matter halos around galaxies or star groups.

3) The strength of the gravitational field is related to the structure factor of the celestial body, $G{M}_R/2c^2{R}_0$.

It can explain why dark matter always exists inside and outside celestial bodies with high structure factors. In contrast, dark matter appears relatively scarce around loosely structured celestial bodies. Within and surrounding tightly structured celestial bodies, however, dark matter is found to be denser. It is particularly evident around neutron stars and black holes, where the concentration of dark matter is significantly higher.

4) The gravitational field matter only interacts through gravity. It does not participate in electromagnetic, strong, and weak interactions. Consequently, non-gravitational methods cannot detect gravitational field matter. In other words, gravitational field matter is transparent to these non-gravitational forces.

It can explain why dark matter is uneasy to detect using non-gravitational methods.

5) The gravitational field matter does not consist of particles.

This view can explain why experimental methods such as particle colliders cannot detect this substance.

6) The gravitational field matter, like electromagnetic field matter, can excite waves and propagate through space. Static ordinary matter generates static gravitational fields, while dynamic ordinary matter produces dynamic gravitational fields that can propagate over long distances.

It can explain gravitational wave phenomena.

7) The mass-energy of the gravitational field matter can exist in a vacuum.

It can explain why dark matter is cold.

There is literature [15] that states that the mass-energy of matter in a gravitational field is negative, with its calculation formula being

${ℰ}_p=\frac{1}{2}{\displaystyle \underset{0}{\overset{V}{\int }}\rho \varphi \text{d}V}$ (117)

where, $\varphi$ represents the gravitational potential, $\rho$ is the mass density of ordinary matter, and $V$ is the volume of the space in which ordinary matter exists.

Since the gravitational potential $\varphi$ is always negative, the gravitational potential energy ${ℰ}_p$ obtained from the above integral is also always negative. Does this mean that the mass-energy of the gravitational field is negative? It is not. The above expression represents the gravitational potential energy of ordinary matter rather than the mass-energy of the gravitational field matter. While gravitational potential energy is negative, it does not imply that the mass-energy of the gravitational field is negative. The calculation of the mass-energy density of the gravitational field must utilize the formula (10) instead of this formula, which results in a positive mass-energy for the gravitational field matter.

Some scientists believe that dark matter must be some unknown fundamental particles, proposing various hypotheses about it, with two main ones: one called Weakly Interacting Massive Particles (WIMPs), which are relatively heavy; the other called axions, which are very light. However, beyond this, they know nothing. Physicists have debated its properties for decades. But so far, attempts to find these particles have been unsuccessful. It has led some researchers to speculate that particle-based dark matter might not exist; instead, it could be a gravitational field matter generated by celestial bodies with high structure factors, such as neutron stars and black holes [16] .

Due to the inability to find dark matter in particle form and the seemingly unnecessary modification of Newtonian dynamics, gravitational field matter provides a possible alternative to dark matter.

According to general relativity, if an object has mass $M_R$, its radius $2G{M}_R/c^2$ becomes the singularity known as the Schwarzschild radius or event horizon, which is the twice orbital radius of a photon in circular motion around a celestial body.

In contrast, the structure of a celestial body within the Newtonian static gravitational field has critical radius $R_C=\kappa G{M}_{0R}/c^2$, and a minimum radius $R_{\min }=6G{M}_{0R}/{\pi }^2{c}^2$. However, these radii do not numerically equal the Schwarzschild radius.

The critical radius $R_C=\kappa G{M}_{0R}/c^2$ of a black hole means that it can become a black hole only if the structural radius of the celestial body is less than or equal to this value. The minimum radius $R_{\min }=6G{M}_{0R}/{\pi }^2{c}^2$ of a celestial body refers to the limit radius that can only infinitely approach but can never reach when a celestial body collapses.