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Modification of Gravitational Field Equation due to Invariance of Light Speed and New System of Universe Evolution
Modification of Gravitational Field Equation due to Invariance of Light Speed and New System of...
Yang, Jian Liang
Hindawi Advances in Astronomy Volume 2021, Article ID 5579060, 14 pages https://doi.org/10.1155/2021/5579060 Research Article Modification of Gravitational Field Equation due to Invariance of Light Speed and New System of Universe Evolution Jian Liang Yang College of Physics, Zhengzhou University, Zhengzhou 450001, China Correspondence should be addressed to Jian Liang Yang; firstname.lastname@example.org Received 16 February 2021; Revised 4 March 2021; Accepted 9 March 2021; Published 20 March 2021 Academic Editor: Ghulam Abbas Copyright © 2021 Jian Liang Yang. ,is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We make a systematic examination of the basic theory of general relativity and reemphasize the meaning of coordinates. Firstly, we prove that Einsteinʼs gravitational ﬁeld equation has the light speed invariant solution and black holes are not an inevitable prediction of general relativity. Second, we show that the coupling coeﬃcient of the gravitational ﬁeld equation is not unique and can be modiﬁed as 4πG to replace the previous − 8πG, distinguish gravitational mass from the inertial mass, and prove that dark matter and dark energy are not certain existence and the expansion and contraction of the universe are proven cyclic, and a new distance-redshift relation which is more practical is derived. After that, we show that galaxies and celestial bodies are formed by gradual growth rather than by the accumulation of existing matter and prove that new matter is generating gradually in the interior of celestial bodies. For example, the radius of the Earth increases by 0.5mm every year, and its mass increases by 1.2 trillion tons. A more reasonable derivation of the precession of planetary orbits is given, and the evolution equation of planetary orbits in the expanding space-time is also given. In a word, an alive universe unfolds in front of readers and the current cosmological diﬃculties are given new interpretations. tides, the moon still has an unexplained retreat, and the 1. Introduction increase of the day length is also inconsistent with the Although general relativity has made some remarkable prediction of the tide theory. Recently, Melissa Ness and her achievements, some basic problems have not been well colleagues have observed that there is a ﬁne X-shaped box solved, such as the physical meaning of the coordinates of structure in vortex galaxies similar to the Milky Way . Schwarzschild metric, whether general relativity is the Melissa Ness said that this structure implies that large curved theory of space-time or the theory of gravity in ﬂat galaxies are not formed by the merger of small galaxies, space-time, whether the constant speed of light is also because once the merger occurs, the structure will inevitably tenable in the gravitational ﬁeld, the singularity problem of be destroyed, and we must abandon the existing theory of the ﬁeld equation, and whether the existence of black holes is galaxy formation and establish a new logic system. ,e true. However, only these basic problems have plagued the observations of Martinez-Lombila and others  show that the radius of disk galaxies similar to the Milky Way galaxy is development of general relativity but also led to some confusion in practice; for example, on the one hand, the expanding at a speed of 500 m/s; such a high speed cannot be radial coordinates of Schwarzschild metric are not inter- the speed at which matter accumulates at the edge. If matter preted as the normal radius, while, on the other hand, the accumulates at this speed at the edge, it should be the same radial coordinates on the solar surface are treated as the everywhere on the disk. Obviously, the current theoretical radius of the sun in calculating the curvature of light on the framework cannot explain such a rapid expansion of the surface of the sun, resulting in conceptual confusion. In radius of the disk. ,ere is also the problem of dark matter addition, there are some new observations that are not and dark energy; the reason why we need them is that the accommodated by the current gravity theory. As Lorio  observed phenomena do not conform to the prediction of pointed out, there is an unexplained increase in the distance the theory, but, no one has seen them really. ,en, whether between the Sun and the Earth, and after considering the they are real or the theory itself needs to be modiﬁed is also 2 Advances in Astronomy an unavoidable problem. ,e latest observation data of In this paper, we use natural units, the speed of light of Nielsen and others  show that the universe is expanding at ﬂat space-time c � 1, and it is agreed that ﬂat space-time a constant speed rather than accelerating, so whether the linear element is universe accelerates or decelerates or expands at constant 2 μ ] 2 2 2 2 2 2 ds � g dx dx � dt − dr − r dθ + sin θdφ . (1) μ] speed still needs to be reconsidered. Besides, some new studies of frontier disciplines [5, 6] have shown that 1 billion According to general relativity, in a spherically symmetric years ago, the brightness of the sun was less than half of what gravitational ﬁeld, in the coordinate system (t, r, θ, φ), the it is today, the Earth is an ice ball, and the mountain is not as general form of space-time line elements is [7–11] high as it is today, and 2.7 billion years ago, the air pressure 2 μ ] 2 on the Earth was only half of todayʼs. ,ese seem to be purely ds � g dx dx � B(r, t)dt − Q(r, t)dtdr μ] (2) geophysical problems, which can only be reasonably 2 2 2 2 − A(r, t)dr − D(r, t)dθ + sin θdφ . explained from the perspective of cosmology because the evolution of the Earth is an epitome of the evolution of the ,e condition of this formula is only spherical symmetry, universe and the Earth must be reﬂected by cosmological which is applicable to the gravitational ﬁeld of both static and events. On the contrary, the phenomena on the Earth can be oscillating gravitational sources. In this paper, we will just deal used to test the cosmological theory more accurately and with the static gravitational ﬁeld, which is what Newtonian people do not have to go far to test the theory of cosmology. gravity describes. For the static case, g no longer contains μ] In a word, we are faced with some new problems that cannot time. Besides, the static case requires time version to be be avoided. We will see that when the speed limit of light, that symmetric, so g � Q(r) � 0. ,erefore, for the static case of is, the speed of light always 1 (in natural units), is still satisﬁed spherical symmetry, the space-time line element is in the gravitational ﬁeld, the above problems can be solved in a 2 μ ] 2 2 2 2 2 package. ,e author thinks that it is a great mistake of general ds � g dx dx � B(r)dt − A(r)dr − D(r)dθ + sin θdφ . μ] relativity that the invariance of the speed of light in the (3) gravitational ﬁeld is not emphasized in the past, and it is this fault that leads to a series of misconceptions and absurd results; We just have to solve for three functions B(r), A(r), and for example, it is necessary to admit singularity as physical D(r). In order to ensure that the meaning of coordinates is reality, which will never be allowed in other parts of physics. In always clear and unchanged, this paper will not continue to a word, it is shameless to tie the correctness of general relativity simplify (3) into the so-called standard form through co- with some wrong conclusions such as big bang and black holes, ordinate transformation but directly solve with the gravi- and it is shameless to praise mistakes as successes. Leading to tational ﬁeld equation. Firstly, determine the external the big bang, black holes and all kinds of other singularities are solution that satisﬁes the vacuum ﬁeld equation R � 0, and μ] not the success of general relativity, but its failure. ,e reason is then the source internal solution is determined. simple: there is no singularity in real nature. No matter how In order to reﬂect the invariance of light speed, we re- much you boast big bang and black holes, they cannot be true. quire A(r) � B(r). From the following solving process, we ,e author thinks that if these absurd things are not stripped can see that such a solution not only exists but also is unique. away from general relativity, there will be no real progress in Equation (3) provides general relativity, the ﬁeld of astrophysics will be dominated by g � B(r), all kinds of idealism, and more and more young students will be misled into the wrong way. In order to deal with these g � − A(r), problems systematically, to get to the bottom and bring order out from chaos, this paper begins with the most basic problem, g � 0(μ≠ ]), μ] that is, solving the metric of the spherically symmetric grav- itational ﬁeld represented by coordinates in the usual sense. g � − D(r), 2. Spherically Symmetric Static Metric g � − D(r)sin θ, Represented in Usual Coordinates g � − , We just have to solve for the metric form in the usual (4) D(r) spherical coordinates; the form in other coordinates can be obtained by coordinate transformation. Indices g � − , μ 0 1 μ, ], λ, α, β � 0, 1, 2, 3. Space-time coordinates x � (x , x , D(r)sin θ 2 3 0 1 2 3 x , x ) � (t, r, θ, φ) and x � t, x � r, x � θ, x � φ repre- sent the usual time, radius, and pole angles, respectively. g � , ,ey have the same meaning as in quantum mechanics or B(r) electrodynamics. In the language of the observational theory of general relativity, t is the time recorded by a stationary g � − , A(r) observer at inﬁnite distance, r is the distance the observer measures from the origin to another point, and θ, φ are the μ] g � 0(μ≠ ]). polar angles measured by the observer. Advances in Astronomy 3 According to the deﬁnition of connection, inﬁnity, there must be C � 1/4, namely, AB � D /4 D. λ λα ] μ α Γ � (1/2)g (zg /zx + tzg /zx n − qzg /zx ), the And inserting A � D /4B D into equation (8) gets μ] αμ α] μ] repeating indices up and down means summing from 0 to 3, ′ ′ D D and it is not hard to ﬁgure out all of its nonzero connections ′ (10) B + B − � 0, 2 D 2 D as follows [2–7]: which is a diﬀerential equation with respect to B. Writing 1 zD 1 2 Γ � − sin θ, D � l , the general solution of (10) is given by B � 1 + C /l. 2A zr C is an integral constant. Because we must return to Newton gravitation in the distance, we have C � − 2GM. G cos θ 3 2 Γ � , Newtonʼs gravitational constant M is the mass of the source. sin θ It is important to insert B � 1 − 2GM/l and 1 zA ′ ′ 1 A � D /4B D � l /(1 − 2GM/l) into any one of (6)–(8). Γ � , You can obtain an identity with respect to l � l(r); namely, 2 zr no matter what l � l(r) is, this identity can be tenable, so we 1 zB can pick an appropriate l(r) so that A � B. And letting Γ � , 2B zr (5) A � B, namely, 1 − 2GM/l � dl/dr, we obtain 1 zB r � C + l + 2GM ln(l − 2GM), (11) Γ � , 2A zr where C is the integral constant and can be decided by the 2 3 continuity of l(r) on the surface of the source. In Section 4 of Γ � Γ 12 13 this paper, C will be calculated. So far, we obtain the ex- terior line element: 1 zD Γ � − , 2A zr 2GM 2GM 2 2 2 2 2 2 2 ds � 1 − dt − 1 − dr − l dθ + sin θdφ , l l Γ � − sin θ cos θ. (12) According to the deﬁnition of curvature tensor, where l � l(r) can be inversely solved from (11). It can be λ ] λ λ α λ α λ R � zΓ /zx − zΓ /zx + Γ Γ − Γ Γ ; for μ≠ ], we μ] μλ μ] μλ α] μ] αλ seen from (11) that when l � 2GM, r becomes negative, have R � 0, which means that the vacuum ﬁeld equation is μ] which means l − 2GM is always greater than 0 and so there is automatically satisﬁed. It is not hard to ﬁgure out all of the no horizon and no black hole. And considering nonzero components of R . Write A � dA/dr, μ] lim (ln x/x) � 0, we have r � l for l ⟶ ∞. 2 x⟶∞ 2 2 2 ″ ′ A � d A/dr , A � (dA/dr) , and note that R � R sin θ; we are left with the following three equa- 33 22 3. Link with the Mechanics of Special Relativity tions about B(r), A(r), and D(r): ,e invariance of the speed of light described by (11) is easy ′ ′ ′ ′ ′ ′ B ′ B B A D B to see. Suppose that the photon moves in the radial direction, (6) R � − + − + + � 0, 2A 2AB 2A 2A D 2B dφ � dθ � 0, ds � 0, and from (11), we can get 2 2 dr /dt � B(r)/A(r) � 1, namely, dr/dt � ± 1, which shows 2 2 that radial speed of light is constant. And now we look at the ′ ′ ′ ′ ′ ′ ′ ′ D B D B A D B ⎝ ⎠ ⎛ ⎞ R � + + + − + � 0, 2 2 light moving tangentially and set θ � π/2, dr � 0, ds � 0; D 2B 2A D 2B 2D 4B the tangential speed of light is given from (12) √��������� (7) byrdφ/dt � (r/l) 1 − 2GM/l. And (11) tells us that the √��������� smaller 1 − 2GM/l than 1, the larger r/l than 1, so the √��������� ″ ′ ′ ′ D D A B deviation of (r/l) 1 − 2GM/l from 1 is actually very small, R � + − + − 1 � 0. (8) and you can still think of it as 1. It is in this sense that we say 2A 4A A B that (11) describes the invariant speed of light, not strictly From R × (1/B) + R × (1/A) � 0 we obtain constant. ,is slight change of the tangential speed of light 00 11 causes light to bend near a celestial body; otherwise, it travels ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ AB + A B D D D (AB) D D in straight lines. ,e previous result is − + + � − + √��������� 2AB D D 2AB D D 2D rdφ/dt � 1 − 2GM/r which can be obtained from (28), and it is not hard to ﬁnd that when r � 2GM, the tangential D speed of light is zero and the deviation from 1 is severe; + � 0. although it means also that light can bend, it does not reﬂect 2D the invariance of the speed of light. (9) ,e correctness of (12) is not only in the invariance of the Equation (9) is a diﬀerential equation with respect toAB, speed of light but also in the natural connection with rel- its general solution is AB � C D /D, and C is the integral ativistic mechanics under weak ﬁeld approximation. 1 1 ′ ″ constant. Since A � B � 1, D � r , D � 2r, D � 2 at Equation (12) provides 4 Advances in Astronomy dv Γ � 0, 2 € _ _ _ € � r − rφ e + (2rφ + rφ)e , r φ dt GM dl GM 2 1 0 dv Γ � Γ � � , 2 2 11 01 2 2 (13) � 2r _r € + 2r φ _ φ € + 2rr_φ _ , dr (21) (1 − 2GM/l)l l dt GM d(mv) dv dm dv dv Γ � . � m + v � m + mv . 2 2 dt dt dt dt l 2 1 − v dt ,e dynamic equation describing the motion of free Using (20), we have particles is the geodesic equation, the proper time must be 2 2 2 2 2 eliminated when solving for acceleration, and the geodesic 1 − r φ _ r r _φ _ φ € φ _ rr _ GM (22) € _ r + + − rφ � − , 2 2 2 2 equation after the elimination of proper time is 1 − v 1 − v 1 − v r 2 μ ] λ ] λ μ d x dx dx dx dx dx μ 2 2 3 (14) + Γ · − Γ · · � 0, r − rr_ rr _φ _ r € r φ _ r _ ]λ ]λ (23) dt dt dt dt dt φ € + + + 2r _φ _ � 0, dt 2 2 2 1 − v 1 − v 1 − v which is derived in detail in post-Newtonian mechanics . eliminating φ € in the use (22) ×(1 − r_ )− (23) ×rφ _ r _ gets Let the particle move on the plan θ � π/2, set μ � 1, 3, and write dr/dt � v , dφ/dt � v /r; we have r φ GM GM (24) r − + − v � 0, 2 2 2 2 2 lv r r d r GM GMv 1 1 2 0 2 1 2 � − Γ − Γ r _ + 2Γ r_ + Γ φ _ � − + + , 2 00 11 01 33 2 2 2 dt l l r which is exactly equation (17), and inserting it into (23), we (15) get 2 GM € (25) ′ φ + v v − v v � 0, d φ 2l 1 2GM r φ r φ 3 0 2 3 � − 2Γ r _φ _ + 2Γ r_φ _ � − v v + v v . r r r φ r φ 2 13 01 2 rl 1 − 2GM/l dt l r which is (19) omitting the higher-order small GMv v /r . So r φ (16) far, it can be seen that (12) in the weak ﬁeld approximation In the weak ﬁeld or in the distance, 2GM/l≪ 1, l ⟶ r, can link well with the special relativistic mechanics. 2 2 l ⟶ r , l ⟶ 1, (15) and (16) become, respectively, And again, when the particle moves along the radial direction, if v � 1, the acceleration of the particle is equal to 2 2 d r GM GMv zero, which means that the invariance of light speed and the (17) + − − � 0, 2 2 2 dt r r light speed limit are uniﬁed. d φ 2 2GM (18) + v v − v v � 0. 4. The Light Speed Invariant Solution within a r φ r φ 2 2 3 dt r r Spherically Symmetric Static Gravity Source Ignoring the higher-order small quantity 2GM/r , (17) Now, we solve the interior A(r), B(r), and D(r). In order to becomes keep the constant speed of light, we still require A(r) � B(r) d φ 2 inside the source. From the following solution’s process, we (19) + v v � 0. r φ 2 2 know such a solution not only exists but also is unique. Note dt r that the constant speed of light means that the speed of light It is not diﬃcult to prove that (17) and (19) are just the passing through the cavity in the source is 1, but that passing relativistic Newton equations of gravity: through the medium is 1. ,e equation of gravitational ﬁeld in the source is d(mv) GMm � − e , (20) dt (26) R � cT − Tg , μ] μ] μ] √����� where m � m / 1 − v is the motion mass of the particle, 2 2 2 _ _ v � r_e + rφe , v � r _ + r φ , and e , e are the base vectors. where c is the coupling constant, and that we do not write its r φ r φ Prove as follows: from theoretical mechanics, we know speciﬁc value here is to lay a hint for the following Advances in Astronomy 5 modiﬁcation of the constant. And T � (ρ + p)u u − pg the identity with respect to l � l(r); that is, this is true re- μ] μ ] μ] is the energy-momentum tensor of the source. Note that gardless of the function of l(r), regardless of the value of c. μ μ ] μ μ] μ] u � dx /ds, u � g u , u u � 1, g g � 4, T � g T � So, we can pick a function l(r) by the following equation (41) μ μ] μ μ] μ] ρ − 3p. to make A(r) � B(r): i i And for the static source, u � dx /ds � 0, i � 1, 2, 3. ] 2 0 2 l l ,en, u � g u � 0, 1 � B(dt/ds) � B(u ) , 2GM i i] − 2 2 √� � 1 − exp − cA pl + l l ρdldl 0 2 u � g u � B, T � pA, T � p D, T � p D sin θ, 0 00 11 22 33 l l 0 e e T � ρB; from (26), we have (35) − 1 ′ ′ ′ ′ ′ ′ − 1 2 B ′ B B A D B c � 1 + l c ρl dl l . R � − + − + + � (ρ + 3p)B, 2A 2AB 2A 2A D 2B 2 (27) On the other hand, we know T � 0, for the static μ;] source; it is [2–7] 2 2 ′ ′ ′ ′ ′ ′ ′ ′ D B D B A D B ⎛ ⎝ ⎞ ⎠ √� � zp z R � + + + − + 2 2 (36) D 2B 2A D 2B � − (ρ + p) ln B. 2D 4B μ μ (28) zx zx � (ρ − p)A, 5. Modification of the Coupling Constant and ″ ′ ′ ′ D D A B c the Internal D(r) R � + − + − 1 � (ρ − p)D. (29) 2A 4A A B 2 When the pressure at the surface of the gravitational source In the use of (R /2B) + (R /2A) + (R /D) � ρc, we is set as zero, it can be seen from (39) that once the geodesic 00 11 22 get equation is required to return to Newtonian gravity under the weak ﬁeld approximation, the coupling constant c must ″ ′ 1 2D D 1 2 D + − − − 2cρ � 0, (30) be − 8πG, which is the previous result. But this result should ′ ′ ′ A D 2 D A D D be considered as a mistake because it leads to a lot of sin- gularities that should not occur. For example, when the ratio which can be looked as a diﬀerential equation with respect to of the mass to the radius of an object is 2GM/R> 8/9, the 1/A. And writing D � l , in the requirement of ensuring D � pressure inside the object becomes inﬁnite [7–9], which is 0 and A being limited at origin, the solution of (31) is given obviously absurd [10–16]. ,e root of all kinds of singularity by is the improper selection of − 8πG. As we will see, when the − 1 c pressure is taken negative, the coupling constant is identiﬁed ′ (31) A � 1 + ρl dl l , as 4πG, which not only avoid singularity of Schwarzschild l 0 metric but also remove in a package cosmological problems. 2 l 2 − 1 − 1 2 where l � (dl/dr) . Writing A � (1 + l c ρl dl) , we I need to say a few words about the negative pressure. have Einstein did not refuse the negative pressure. In the book %e Meaning of Relativity, Princeton University Press Published, ′ ′ ′ ′ ″ A � A l + 2A l l , 1 1 1922, (page 117), for T � ρu u , Einstein said, we “shall μ] μ ] add a pressure term that may be physically established as ′′ ′ ″ l � 2l + 2ll , follows. Matter consists of electrically charged particles. On (32) the basis of Maxwellʼs theory these cannot be conceived of as l dB electromagnetic ﬁelds free from singularities. In order to be B � . consistent with the facts, it is necessary to introduce energy dl terms, not contained in Maxwellʼs theory, so that the single Insert these into (30), we obtain electric particles may hold together in spite of the mutual repulsion between their elements, charged with electricity of dB − 2 2 � − cA pl + l l ρdl. (33) one sign. For the sake of consistency with this fact, Poincare Bdl 0 has assumed a pressure to exist inside these particles which balances the electrostatic repulsion. It cannot, however, be To make B(r) continuous on the boundary of source, the asserted that this pressure vanishes outside the particles. We solution of (33) is shall be consistent with this circumstance if, in our phe- l l nomenological presentation, we add a pressure term. ,is 2GM − 2 2 B � 1 − exp − cA pl + l l ρdldl, (34) must not, however, be confused with a hydrodynamical l 0 pressure, as it serves only for the energetic presentation of where l � l(r ) is the value on the boundary and r is the the dynamical relations inside matter.” From this statement, e e e radius of the source. it can be seen that Einstein did not equate pressure as a It is important that, similar to the external solution, source of gravity with the dynamic pressure of a ﬂuid but substituting (32) and (34) into any of (28)–(30), you can get regarded it as a phenomenological representation of all the 6 Advances in Astronomy action within matter, including the electromagnetic force. It 14 l � r − πGρr , e e is not surprising that a negative value is taken. Now solve for D(r) in the source with negative pressure and in the meantime determine the coupling constant. For 3 r � l + πGρl , (42) e e e the convenience of calculation, we use the average density 9 instead of the density of each point that is to take the interior 3 3 5 ρ � const. ,e density itself is a statistical average, such l � r − πGρr . e e e treatment is equivalent to treating the whole celestial body as a statistical volume element, so it is suitable for some not too Inserting (42) into (40), we obtain large celestial bodies as such the sun, and it is also an ap- 2 2 4πρ 80π ρ G proximation for larger celestial bodies. 3 5 (43) M � r − r + · · · . e e When ρ is regarded as a constant, p � − ρ is the solution 3 9 of (36). Since the geodesic under the weak ﬁeld approxi- Writing 4πρr /3 � M , we have e i mation must return to Newtonian gravity, there must be 5G Γ ⟶ GM/r on the surface of the source, and we conclude 2 M � M − M + · · · , that, from (33), the coupling constantc � 4πG. Notice that r under the weak ﬁeld approximation l ⟶ r, A ⟶ 1, (44) 2 3 5G 4π ρl dl ⟶ M, and p � − ρ ⟶ − 3M/4πr . 2 M � M + M + · · · , With p � − ρ, c � 4πG, and l as the ﬁxed value, the r e e integral of both sides of (35) is easy, and we get where M is the inertial mass and M is the gravitational mass. ���������� � dl 1 − 2GM/l 4πGρl ,e reason why M is called inertial mass is that ρ represents � 1 + . (37) the inertial density of matter measured in comoving coordinate dr 3 1 + 4πGρl /3 system, while M is introduced from the perspective of gravity, ′ so it is natural to call it gravitational mass. Ensuring B � 0 for r � 0, which is because the force on the particle at the origin is zero, the solution of (37) is Equation (44) distinguishes gravitational mass from ���� � ����������� ���� � inertial mass, which is the result of an in-depth discussion in 3 4πGρ 1 − 2GM/l this paper, and for high-density celestial bodies, the dif- arctgl � r . (38) 4πGρ 3 1 + 4πGρl /3 ference is obvious. We see l � r − 7GM/6 + · · ·, and so far, the integral e e On the other hand, we require Γ to be also continuous constant C in (11) can be decided according to the con- on the boundary, which is a necessary condition to ensure tinuity on the boundary; that is (11) is applied to the surface the continuity of gravity on the boundary. So, there exists on of the source the boundary C � r − l − 2GM ln l − 2GM � r − l ���������� � 3 e e e e e zB zl 8πGρl 1 − 2GM/l 2GM 1 1 e e (45) Γ � � � Γ � . 00 2 00 2 7GM B zl zr 3 1 + 4πGρl /3 l e e − 2GM ln r − 2GM ≈ − 2GM ln r . e e (39) �������� � Inserting C into (11), we can complete the calculation of 2 2 Solve (39), M � G k + kl − Gk, in which Mercury precession and ray bending, and the calculated 2 2 5 2 k � 16π ρ l /(9 + 12πGρl ). e e result is that the diﬀerence between the new results and the ,e explicit form of M can be solved under the weak ﬁeld original ones is very small and completely consistent with approximation, and expanding the square root in (39) by the observation. ,e concrete calculation is not written here, Taylor, we obtain readers can do it by themselves. In a word, with the new coupling constant 4πG, the 4πρl 8 e 2 2 5 (40) M � − π ρ Gl + · · · . gravitational ﬁeld equation is now modiﬁed as 3 3 (46) Applying (39) on the boundary and taking the ap- R � 4πGT − Tg , μ] μ] μ] proximation M � 4πρl /3 into, we have and correspondingly the pressure p as the source of grav- ���������� � ���� � ���� � itation takes negative. Multiplying the two sides of (46) with 3 4πGρ 1 − 8πGρl /3 μ] g , we have R � − 4πGT, so the equivalent form of (46) is (41) arctgl � r . e e 4πGρ 3 1 + 4πGρl /3 R − (1/2)Rg � 4πGT . μ] μ] μ] Obviously, for 1 − 8πGρl /3 � 0, r ⟶ ∞; that is to say, 6. Further Interpretation of the Physical e e no matter how big the radius of the celestial body, there is Meaning of the Negative Pressure always l < 3/8πGρ; we do not have to worry about whether l e e has a solution. And expanding the two sides of (41) by Taylor Einstein did not interpret the pressure term in a gravitational and taking second-order approximation, we obtain source as the dynamic pressure of a ﬂuid, but as a Advances in Astronomy 7 phenomenological representation of the pressure within a usually. So, we say that dark energy is just the binding energy matter to balance the electromagnetic force and prevent of matter, rather than an independent existence. charged particles from being disintegrated by electrical re- pulsion, which we should accept. Einstein did not point out 7. The Problems of Schwarzschild Metric that pressure is produced by which power. Today, it is easy to infer and prevent the disintegration of the protons and Schwarzschild metric is neutrons are strong; preventing electronics is the disinte- − 1 2GM 2GM 2 2 2 2 2 2 2 gration of the weak force. ,erefore, the pressure term ds � 1 − dt − 1 − dr − r dθ + sin θdφ . r r should be understood as a phenomenological representation (47) of the combined eﬀects of the strong, weak, electromagnetic, gravitational, and all other forms of action within a matter, Because the precession angle of Mercury orbit predicted representing the total binding energy that holds the matter by (47) is consistent with the observation, it is generally together, represented by the potential energy of the system. believed that it is correct. However, there are also some In other words, if you divide the matter inﬁnitely, and you serious problems with the metric; for example, it exposes the move each part to inﬁnity, the work done is the volume incompatibility of electromagnetic theory and gravitational integral of the pressure, which is negative, and the absolute theory: when a charged particle moves in the radial direction value is equal to the mass of the gravitational source, namely, at a speed dr/dt ≈ 1, near the singularity r � 2GM, pdxdydz � − M, which is like adding a physical condition − 1 2 2 2 ds /dt � (1 − 2GM/r) − (1 − 2GM/r) (dr/dt) < 0, to make the solution of the pressure deﬁnite. In fact, the which is ridiculous because there is no reason to think that form of the gravitational source already determines the the electromagnetic equipment in the gravitational ﬁeld interpretation of p. As gravitational source cannot make the speed of a charged particle close to 1. T � (ρ + p)u u − pg , if p is still understood as the μ] μ ] μ] In order to avoid the defect with (47), textbooks do not common dynamic pressure, then the ﬁeld equation can only be interpret r in (47) as the usual radius but instead refer to it as used to solve for the metric of an ideal ﬂuid, which is obviously the radial parameter with fuzzy meaning [7, 8]. But this is not hoped by general relativity, and the dynamic pressure in a unhelpful and leads only to conceptual confusion since it has solid is generally considered to be zero. let alone (36). Can p be been used as the usual radius when calculating the pre- understood as a thermal pressure? No, because the thermal cession angle and the bending angle of light; there should be motion is absorbed by ρ in the form of thermal kinetic energy no other explanation. and cannot be repeated to appear. p is called pressure only Now, we calculate the ordinary pressure given by the because it appears in the equation of motion in the form of Schwarzschild interior metric, from which we can see the pressure; of course, p includes the eﬀect of common pressure. defects of the Schwarzschild metric and the necessity of μ] ,e equation of motion refers to T � 0. ;] modifying the ﬁeld equation. According to the deﬁnite of When the ﬁeld equation with the coupling constant 4πG pressure, it refers to the stress per unit area. It may as well let is applied to the universe, T � (ρ + p)u u − pg repre- μ] μ ] μ] the celestial body be a ﬂuid with ρ � const, P denotes the sents the energy-momentum tensor of space of the universe, common stress, and the common pressure given by the and taking the statistical average of ρ and p, we have p � − ρ, interior solution of Schwarzschild is which is just the equation of state of the dark energy said r r r e e e zB 1 P � ρ Γ dr � ρ dr � ρ dB c 00 A zr A r r r ����������� ����������� ρ 8πGρ 2 2 � 1 − r d 3 1 − 8πGρr /3 − 1 − 8πGρr /3 (48) 4 3 ��������� � 2 3/2 2 2 2 2 3ρ 8πGρr ρ 8πGρr 8πGρr ρ 8πGρr e e � − 1 − + 1 − 1 − − 1 − . 8 3 2 3 3 8 3 In weak ﬁeld approximation, P � ρ(GM/r n − q But, the common pressure given by (39) is c e 2 3 GMr /r ), which is Newton’s result. And at the center, (48) r r r r e e e e zB zB gives ��������� � P � ρ gdr � ρ Γ dr � ρ dr � ρ dl c 00 2 B zr B zl 2 r r r r 3ρ 8πGρr ρ 8πGρr ρ e e (49) P � − 1 − + 1 − − . 8 3 2 3 8 8πGl 1 + 4πGρl /3 � ρ dl � ρ ln , 2 2 Obviously, for 1 − 8πGρr /3 � 0, P < 0, and for e c 1 + 4πGρl /3 1 + 4πGρl /3 1 − 8πGρr /3 � 1/9, P � 0, which are abnormal because the e c (50) pressure should not be zero anyway. 8 Advances in Astronomy − 1 − 1 in which l � l(r) and l � l(r ) satisfy (38). ,ere is no parameter of today H � H(t ) � 70 km · s · Mpc into e e 0 0 singularity in (50). Under weak ﬁeld approximation, l ⟶ r, the above equation, we get t � 1.37 × 10 years, that is, 13.7 2 3 P � ρ(GM/r n − qGMr /r ), which is just the result of billion years, the same as the previous theoretical results. c e e Newton. It can be seen from (54) that the cyclical period of expansion ����� � and contraction of the universe is 2π/ω � 3π/Gρ � 2 × 10 years, namely, 200 billion years, so the universe is currently in 8. The Application of (46) in Cosmology the expansion stage and will begin to contract in 36.3 billion years. Contraction is the reverse course of expansion. In the comove coordinate system (t, l, θ, φ), the metric that Now, we derive the new relation between distance and describes cosmic space is the Robertson–Walker metric: redshift given by (46). Similar to the previous operation, 2 2 2 2 2 2 2 2 2 letting the light given out by distant galaxy at the time t in ds � dt − R (t) dl + l dθ + l sin θdφ , 1 − kl past and reach the Earth at the time t of today, its redshift z � (λ − λ /λ ) � (R(t )/R(t)) − 1. λ is wavelength. We (51) 0 e e 0 may as well put today’s R(t ) � 1. Note that the subscript 0 where R(t) is the cosmic scale factor and k is a constant. l is represents today. And diﬀerentiating 1 + z � 1/R(t), we get the radial coordinate, and in other books it is denoted by r, dR dR dt just to distinguish it from the usual radius, here l instead of r. dz � − � − . (57) Rdt R When the new ﬁeld equation (46) is applied to the universe, R (t) that is, combined with the Robertson–Walker metric, the And the derivative of equation (52) gives following two equations are given: 2 2 € _ 4πGρ/3H � − RR/3H , where H � dR/Rdt, R � dR/dt. dR 4πG Writing today’s 4πGρ/3H � q , H(t ) � H , and ap- 2 0 0 0 0 (52) + k � − ρR , plying (52) to today, we have dt 3 k � − H 1 + q , 0 0 dρ dR (53) R + 3 (p + ρ) � 0. (58) dt dt 4πGρ k 2 2 2 H � − − � H