International Journal of Theoretical and Mathematical Physics 2020, 10(6): 120-129 DOI: 10.5923/j.ijtmp.20201006.02

About the Existence of Black Holes

Doron Kwiat

Israel

Abstract The existence of black holes is based on singularity in the Schwarzschild metric at the Schwarzschild radius. Another singularity is at r = 0. These singularities of a spherically symmetric non-rotating, uncharged mass of radius R, are considered here. Considering Newton's shell theorem, the gravitational potential falls off linearly with r for r < . The point r = 0 is an infinitesimally small location in space. No mass can be indefinitely condensed to this point. Thus, when investigating the concept of a mass, one has to consider its finite radius. By including the shell theorem for𝑅𝑅 r < , the singularity at r 0 is removed. It is also shown, that a situation where rs < R but r > is possible. Therefore, the existence of a photon sphere light ring does not necessarily indicate a black hole. It is shown, that the condition𝑅𝑅 for a → 3 𝑝𝑝ℎ 𝑅𝑅 gravitational collapse is R < r , and not R < r . Further in this work, the question of maximal density is considered and 2 compared to the quantum limit of𝑠𝑠 mass density put𝑠𝑠 by Planck's units as dictated from dimensional analysis. There are two main claims here: 1. Black holes (if exist) will result only if the Schwarzschild radius is larger than 2/3R. 2. General relativity leads to quantization of gravity. Keywords Newton's Shell theorem, Schwarzschild singularities, Photon sphere, Planck's units, Quantization of Gravity

whether this is a true singularity one must look at quantities 1. Introduction that are independent of the choice of coordinates. One such important quantity is the Kretschmann invariant [19], which The Schwarzschild solution appears to have singularities 2 is given by Rα R . For a Schwarzschild black hole of at r = 0 and r = r , with r = which cause the metric to s s 2 mass M and radius𝛼𝛼𝛼𝛼𝛼𝛼𝛼𝛼 R, the Kretschmann invariant is 𝐺𝐺𝐺𝐺 2 2 𝛽𝛽𝛽𝛽𝛽𝛽 2 diverge at these radii. There is no problem as long as R > rs. 48 842 𝑐𝑐 = 4 6 2 4 . Obviously it is independent of the The singularity at r = rs divides the Schwarzschild 𝐺𝐺 𝑀𝑀 𝐺𝐺 radius R, and is always greater than 0 (provided we accept coordinates in two disconnected regions. The exterior 𝐾𝐾𝑟𝑟 𝑐𝑐 𝑅𝑅 ≈ 𝜌𝜌 𝑐𝑐 Schwarzschild solution with r > rs is the one that is related the physical requirement that the density is never infinite). to the gravitational fields of stars and planets. The interior At r = 0 the curvature can become infinite, if and only if, 𝜌𝜌 Schwarzschild solution with 0 ≤ r < rs , which contains it represents a point particle of zero mass. One cannot the singularity at r = 0, is completely separated from compress a finite mass into an infinitesimal point (of radius the outer region by the so-called singularity at r = rs . R = 0). Any finite mass should have a finite radius R > 0. The Schwarzschild coordinates therefore give no physical Therefore, when looking into the point r = 0, one needs to connection between these two regions. consider the effects of the outer shells of matter which The singularity at r = rs is an illusion however; it is an surround the r = 0 point. instance of what is called a coordinate singularity. As the General relativity predicts that any object collapsing name implies, the singularity arises from a bad choice of beyond a certain point (for stars this is the Schwarzschild coordinates or coordinate conditions. When changing to a radius) would form a black hole, inside which a singularity different coordinate system, the metric becomes regular at r (covered by an event horizon) would be formed. = rs and can extend the external region to values of r smaller As will be shown in the following, no such singularities than . Using a different coordinate transformation one can exist. then relate the extended external region to the inner region. 𝑠𝑠 As 𝑟𝑟will be shown later on, this so-called singularity is completely removed even for the Schwarzschild coordinates. 2. Spherically Symmetric, Non-rotating, The case r = 0 is different, however. If one asks that the solution be valid for all r, one runs into a true physical Uncharged Spherical Body singularity, or gravitational singularity, at the origin. To see Consider a spherically symmetric, non-rotating, uncharged mass (a stellar object or an elementary particle). * Corresponding author: [email protected] (Doron Kwiat) Under spherical symmetry, and at a remote distance in Received: Nov. 19, 2020; Accepted: Dec. 7, 2020; Published: Dec. 15, 2020 empty space outside the object, the Schwarzschild metric Published online at http://journal.sapub.org/ijtmp [1,2,3] is given by:

International Journal of Theoretical and Mathematical Physics 2020, 10(6): 120-129 121

2 In other words 1 2 𝐺𝐺𝐺𝐺 1 3 𝑟𝑟𝑐𝑐 2 ( ) = = � − � 1 (1) 3 − 2 4 𝐺𝐺𝐺𝐺 𝑀𝑀 ⎛ 2 ⎞ And for any finite mass M 𝜇𝜇𝜇𝜇 � − 𝑟𝑟𝑐𝑐 � 𝜌𝜌 𝑅𝑅 𝑔𝑔 ⎜ 2 2 ⎟ 𝜋𝜋𝑅𝑅 lim ( ) This metric describes very well the− gravitational𝑟𝑟 effects of 0 ⎝ ⎠ that mass on gravitational field curvature, −and𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠dynamics𝜃𝜃 of However, as will be𝑅𝑅⟶ shown𝜌𝜌 𝑅𝑅 later,⟶ the∞ density of any object moving particles, in the empty regions outside the object. must have an upper limit and it cannot become infinitely However, it suffers of two problems: large. To resolve this contradiction, one must accept, that no 2 mass can be condensed to a singular point r = 0. A mass will = 1. At the Schwarzschild radius 2 , the metric has always have some finite density and a finite radius (volume). 𝐺𝐺𝐺𝐺 a singularity. Therefore, when considering gravitation at r = 0, one must 𝑟𝑟𝑠𝑠 𝑐𝑐 2. At r  0, the metric diverges. consider the effects of surrounding mass. Condensed as it The first singularity is not a real one, as it can be removed might be, but still of finite density and surrounding r = 0. by appropriate coordinate transformations [5-13]. In classical mechanics, the shell theorem, proven already The second singularity cannot be removed by coordinate by Isaac Newton [14] states that: transformation. It is an inherent singularity at r 0 in any A spherically symmetric body affects external objects inverse square law central force problem of gravitational gravitationally as though all of its mass were concentrated at field). One so far have not been able to relate→ to the a point at its center. divergency problem of force which is inversely proportional If the body is a spherically symmetric shell (i.e., a hollow to the square of the distance r from center of the field source. ball), no net gravitational force is exerted by the shell on any To solve this problem, we must investigate the behavior of object inside, regardless of the object's location within the the field inside the mass. shell. If any mass source is finite in size, the solution to the A corollary is that inside a solid sphere of constant density, potential as a function of distance must be modified, so that it the gravitational force within the object varies linearly with includes the region where r < R, the radius of the mass. distance from the center, becoming zero by symmetry at the In addition, the Einstein field equation should be related to center of mass. in non-empty space. But to a very good approximation, the Due to the shell theorem, the gravitational potential of a Einstein equation in the inside of even the heaviest stellar spherically symmetric object of mass M and radius R, as a objects can be considered same as empty space. function of distance r, from the object's center is given by: 1 This assumption will only be valid for a perfect fluid (3 2 2) 2 solution, where of the Tμν tensor, only energy density is ( ) = considered, while momentum density, shear stress, 3 𝑅𝑅 − 𝑟𝑟 𝑓𝑓𝑓𝑓𝑓𝑓 𝑟𝑟 ≤ 𝑅𝑅 (2) 𝐺𝐺𝐺𝐺 ⎧ 3 momentum flux and pressure are either null or irrelevant. ⎪ > 𝛷𝛷 𝑟𝑟 − 𝑅𝑅 ⎨ 𝑅𝑅 The result for r < R ⎪is𝑟𝑟 obtained by𝑓𝑓𝑓𝑓 summation𝑓𝑓 𝑟𝑟 𝑅𝑅 of the 3. Shell Theorem potential at some point p⎩ inside the mass from its spherical shell between R and r, and the remaining mass inside sphere Let R be the radius of an uncharged non-rotating of radius r. spherically symmetric mass, of given density ( ), with The potential at point p due to the shell is given by r being the radial distance from center (at r = 0). 2 2 4 3 𝜌𝜌 𝑟𝑟 ( ) = 2 ( ) (3) Assuming this object has a volume = and radius 3 3𝜋𝜋𝑟𝑟 The potential at 𝑠𝑠sameℎ𝑒𝑒𝑒𝑒𝑒𝑒 point p due to the remaining inner R, its average density will be = and since 𝛷𝛷 𝑝𝑝 − 𝜋𝜋𝜋𝜋𝜋𝜋 𝑅𝑅 − 𝑟𝑟 𝑣𝑣 4 3 𝑀𝑀 sphere is given by 2 = 4 ( ) it is straightforward to see that 2 0 〈𝜌𝜌〉 𝜋𝜋𝑅𝑅 ( ) = 4 (4) 𝑅𝑅 4 3 The total potential at point p inside the sphere is a 𝑀𝑀 ∫ 𝜋𝜋𝑟𝑟 𝜌𝜌 𝑟𝑟 𝑑𝑑 𝑑𝑑= lim (0) 𝛷𝛷𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑝𝑝 − 𝜋𝜋𝜋𝜋𝜋𝜋𝑟𝑟 3 0 superposition of both ( ), and ( ). Thus, (for 𝜋𝜋 (0) r < R): This means that𝑀𝑀 unless the𝑅𝑅 ⟶density𝑅𝑅 𝜌𝜌 at the center is 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 infinite, the mass must vanish (M = 0). Φ 𝑝𝑝 Φ 𝑝𝑝2 2 ( ) = ( ) + ( ) = 4 (5) 𝜌𝜌(0) 2 6 The only case of M > 0 is possible if and only if is 𝑅𝑅 𝑟𝑟 3 1 infinite. And if we assume physical continuity, then we may 𝛷𝛷𝑝𝑝 𝑟𝑟 𝛷𝛷𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟 ( 𝛷𝛷) 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖= 𝑟𝑟 −2 𝜋𝜋𝜋𝜋𝜋𝜋 2� − � (6) 2 3 2 write 𝜌𝜌 𝐺𝐺𝐺𝐺 3 𝛷𝛷𝑝𝑝 𝑟𝑟 − 𝑅𝑅 � 𝑅𝑅 − 𝑟𝑟 � lim 3 ( ) = 0 4 𝑀𝑀 𝑅𝑅⟶ 𝑅𝑅 𝜌𝜌 𝑅𝑅 𝜋𝜋

122 Doron Kwiat: About the Existence of Black Holes

4. The Identification of ( ) as the 6. Modified Schwarzschild Metric Gravitational Potential Inserting the expression for ( ): 2 2 𝚽𝚽 𝐫𝐫 ( ) = 4 (for (r R) In the Schwarzschild metric, the 00 term is a function of 2 6 𝛷𝛷 𝑟𝑟 𝑅𝑅 𝑟𝑟 the distance r and one may write: 3 𝑔𝑔 𝛷𝛷(𝑟𝑟) = −4𝜋𝜋𝜋𝜋𝜋𝜋 � − (for� (r > R)≤ 2 ( ) 3 00( ) = 1 + 2 (7) 𝑅𝑅 𝛷𝛷 𝑟𝑟 𝛷𝛷The𝑟𝑟 Schwarzschild− 𝜋𝜋𝜋𝜋𝜋𝜋 � 𝑟𝑟 �metric inside the sphere (r ≤ R) First argument – dimensions𝑔𝑔 𝑟𝑟 � 𝑐𝑐 � becomes ( ) is a function of distance r, and must have the 3 1 2 dimensions of 2 . It must obey the constraint of 2 = 1 1 2 2 3 𝑟𝑟𝑠𝑠 𝑟𝑟 lim𝛷𝛷 𝑟𝑟∞ ( ) 0 as should be the case at infinite 1 2 𝑅𝑅 2 2 𝑅𝑅 2 2 2 distance in empty𝑐𝑐 flat space. 𝑑𝑑𝑑𝑑 ( ( �) − � − � � �� 𝑑𝑑𝑑𝑑 ) (9) 𝑟𝑟⟶ 𝛷𝛷 𝑟𝑟 ⟶ ( ) 2 Its units should be [G][Kg]/[m] in order to make 2 𝛷𝛷 𝑟𝑟 while outside− 𝑐𝑐 theℎ sphere𝑟𝑟 𝑟𝑟 𝑑𝑑 𝑑𝑑(r− > 𝑟𝑟R)𝑑𝑑 it𝜃𝜃 becomes− 𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃𝜃𝜃𝜑𝜑 dimensionless. 𝑐𝑐 We therefore have a good reason to assume that ( ) is 2 = 1 2 𝑠𝑠 the gravitational potential due to a spherically symmetric 1 𝑟𝑟 object, independent of the size and of the density 𝛷𝛷of𝑟𝑟 that ( ( )𝑑𝑑𝑑𝑑2 � −2 2� 𝑑𝑑𝑑𝑑 2 2 2) (10) 2 𝑟𝑟 object. Exactly as described by eq. 2. There are− now𝑐𝑐 ℎ three𝑟𝑟 𝑟𝑟 different𝑑𝑑𝑑𝑑 − 𝑟𝑟 𝑑𝑑 possibilities,𝜃𝜃 − 𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃depicted𝜃𝜃𝜑𝜑 in the Second argument- Einstein equation under spherical following: symmetry A a situation where both and are inside the objects For a curved spacetime, based on the Riemann tensor, radius R. < < R. This𝑠𝑠 is a standard𝑝𝑝ℎ object. it can be shown that ( ) must have the form of a 𝑟𝑟 𝑟𝑟 B - a situation𝑠𝑠 𝑝𝑝ℎ where is inside the object's radius R gravitational potential (see Appendix 1). (a standard𝑟𝑟 object)𝑟𝑟 while is outside the object. < R 𝛷𝛷 𝑟𝑟 𝑠𝑠 Following these arguments, it is justified to identify ( ), < . 𝑟𝑟 as the gravitational potential of the field created by a 𝑝𝑝ℎ 𝑠𝑠 C - a situation of a black𝑟𝑟 hole, where both and 𝑟𝑟 are spherically symmetric non-rotating uncharged mass. 𝛷𝛷 𝑟𝑟 𝑝𝑝ℎ 𝑟𝑟 < < outside R. . Therefore, B represents𝑠𝑠 a possible𝑝𝑝ℎ situation where a photon sphere light ring exists𝑟𝑟 around𝑟𝑟 a 𝑠𝑠 𝑝𝑝ℎ 5. Inside the Sphere (r < R) standard (not𝑅𝑅 a black𝑟𝑟 hole)𝑟𝑟 body. 3 For said cases, the photon sphere radius r = r may In the case where r < R (inside the mass), ≠ 0 where ph 2 s 16 G fall inside or outside R. = 2 π 11 27𝕂𝕂 G=6.6743 × x10 and so ≈ 37 x10 , so unless A Standard object R rph >R the order of 6 x1017 Kg/m3. For elementary particles,𝕂𝕂𝜌𝜌 the 𝑟𝑟𝑠𝑠 neutron for instance, has density of approximately 3 x1017 Case A is a standard object, with both , and falling 3 inside R. Kg/m . So, for all practical calculations one may assume 𝑠𝑠 𝑝𝑝ℎ = 0 inside any mass (that is, for r < R). Case B is a white object, with falling𝑟𝑟 inside𝑟𝑟 R but falling outside R. Light is trapped in the photon sphere, For the internal pressure p, as long as the internal pressure 𝑠𝑠 𝑝𝑝ℎ 𝕂𝕂is much less than 1024 , p = 0 . For the Sun for causing the object to have a white𝑟𝑟 glowing ring around𝑟𝑟 it. instance, the internal core pressure is p1011 . There is no gravitational collapse. Finding the solution of𝑃𝑃𝑃𝑃 the𝕂𝕂 homogeneous differential Case C is a true black body, with both and falling equation with = 0, p = 0 , will 𝑃𝑃𝑃𝑃lead to the outside R. 𝑠𝑠 𝑝𝑝ℎ non-homogeneous solution with = constant. But we will Case A, both and are inside R𝑟𝑟 (holds𝑟𝑟 for most concentrate on the𝕂𝕂 homogeneousρ 𝕂𝕂 solution, since 0. stellar objects – but not for black holes). One obtains: 𝑠𝑠 𝑝𝑝ℎ In case of a constant , the non𝕂𝕂 -homogeneous solution 𝑟𝑟 𝑟𝑟 3 1 2 can be found. 𝕂𝕂 ≈ 2 = 1 1 2 2 3 This is the case for a perfect𝕂𝕂 fluid. 𝑟𝑟𝑠𝑠 𝑟𝑟 Under the assumption of 0, for the interior of the 𝑑𝑑𝑑𝑑 1 � −2 𝑅𝑅 � 2− 2�𝑅𝑅� �2� 𝑑𝑑𝑑𝑑2 2 2 ( ( ) ) (11) sphere, the time separation inside the sphere is given by 2 d = g dx dx : 𝕂𝕂 ≈ Picking the− 𝑐𝑐coordinateℎ 𝑟𝑟 𝑟𝑟 𝑑𝑑 system𝑑𝑑 − 𝑟𝑟 𝑑𝑑so𝜃𝜃 that− 𝑟𝑟the𝑠𝑠𝑠𝑠𝑠𝑠 radius𝜃𝜃𝜃𝜃𝜑𝜑 r is along μ ν the x axis, the and terms are zero μν 2 2 ( ) 2 τ = 1 + 2 In other words, for r > R 𝛷𝛷 𝑟𝑟 𝜃𝜃 𝜑𝜑 1 2 2 2 2 2 2 2 2 2 ( ( )𝑑𝑑𝑑𝑑 � 𝑐𝑐 � 𝑑𝑑𝑑𝑑 ) (8) = 1 (12) 𝑟𝑟𝑠𝑠

− 𝑐𝑐 ℎ 𝑟𝑟 𝑟𝑟 𝑑𝑑𝑑𝑑 − 𝑟𝑟 𝑑𝑑𝜃𝜃 − 𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃𝜃𝜃𝜑𝜑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 � − 𝑟𝑟 � International Journal of Theoretical and Mathematical Physics 2020, 10(6): 120-129 123

while for r R: In the case of infinitesimally small r, we see that the time component becomes independent of the distance r from the 3 1 2 ≤ 2 = 2 1 1 (13) origin. In any case, the divergence at r 0 is removed. 2 3 𝑟𝑟𝑠𝑠 𝑟𝑟 The proper time , must remain well defined. Therefore, 2 𝑅𝑅 rs 𝑅𝑅 → Obviously, 𝑑𝑑𝑑𝑑for r = R,𝑑𝑑𝑑𝑑 dτ� =− 1 � −, fo�r both.� �� an upper limit on the density must exist for a given radius R, otherwise the expression𝑑𝑑𝑑𝑑 for becomes undefined. Investigating the condition � − 𝑅𝑅 � Notice that this assertion leads to the constraint 2 rph 1 𝑑𝑑𝑑𝑑 2 1 1 > 0 (16) 3 4 2 𝑟𝑟 𝑐𝑐 we see that in −Case B,� for− all� � �[0, ] and as long as In other words, there is an upper𝜋𝜋𝜋𝜋 𝑅𝑅limit on the density , of 2 𝑅𝑅 𝑅𝑅 𝜌𝜌 ≤ there is no singularity in the Schwarzschild metric. any object with radius R. When 0, the density may 3 𝜌𝜌 2 𝑟𝑟2∈ 𝑅𝑅 increase indefinitely, but as will be shown next, there is an 𝑠𝑠 At = (and = ). dτ becomes zero, and the 𝑅𝑅 → 𝑟𝑟 ≤ 𝑅𝑅 3 upper limit on the density. 2 object evaporates𝑠𝑠 into photons𝑝𝑝ℎ (dτ = 0). Thus, > is 𝑟𝑟 𝑅𝑅 𝑟𝑟 𝑅𝑅 3 the condition for a gravitational collapse. As𝑠𝑠 long as 2 𝑟𝑟 𝑅𝑅 < the object is a standard object and does not 7. Density must have an Upper Limit 3 undergo𝑠𝑠 any gravitational collapse. This, in contrast to the Simple dimensional arguments show that the physical current𝑟𝑟 𝑅𝑅assertion (for r > R), is based on Eq. 37, which states phenomena where quantum gravitational effects become that > , is the condition for a gravitational collapse. relevant are those characterized by the Planck length We see that as long as the photon sphere is inside the 𝑠𝑠 35 = 3 = 1.616 10 m. Here is the Planck constant object𝑟𝑟 ( rph𝑅𝑅 [0, ]) , the Schwarzschild radius ℏ𝐺𝐺 − 2 that𝑝𝑝 governs the scale of the quantum effects, G is the [0, ]. In other words, as long as the photon sphere does𝑠𝑠 ℓ �𝑐𝑐 𝑥𝑥 ℏ 3 ∈ 𝑅𝑅 𝑟𝑟 ∈ Newton constant that governs the strength of the not show, the object is stable and does not suffer gravitational force, and c is the speed of light, that governs gravitational𝑅𝑅 collapse. Once the photon sphere shows up the scale of the relativistic effects. The Planck length is many (rph > ), the object undergoes gravitational collapse. 2 times smaller than what current technology is capable of Anywhere in the region where [ , ] the object observing. physical effects at scales that are so small. 𝑅𝑅 3 undergoes gravitational collapse, and since r = 3/2 , Because of this, we have no direct experimental guidance for 𝑟𝑟𝑠𝑠 ∈ 𝑅𝑅 𝑅𝑅ph the photon sphere is outside the object, creating a bright light building a quantum theory of gravity [16]. ring outside R, at r = . 𝑅𝑅 Suppose there exists a quantum minimum for distance. r 0 We call it the Planck length and denote it by . Near the sphere's center,𝑝𝑝ℎ where and r < R, one has 𝑟𝑟 It is given by 𝑝𝑝 2 = 2 1 (14) ℓ → 35 𝑟𝑟𝑝𝑝ℎ = = 1.616 10 m. And, as long as𝑑𝑑𝑑𝑑, rph <𝑑𝑑𝑑𝑑 , � −> 0.𝑅𝑅 It� may be re-written as: ℏ𝐺𝐺 − Defineℓ𝑝𝑝 � in𝑐𝑐 addition the𝑥𝑥 Planck mass 4 𝑅𝑅 𝑑𝑑𝑑𝑑 2 8 = 1 2 (15) Planck's mass = = 2.176 10 Kg. 𝜋𝜋𝜋𝜋𝜋𝜋 ℏ𝑐𝑐 − 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 �� − 𝑐𝑐 𝑅𝑅 � If this assumption𝑚𝑚𝑝𝑝 is� true,𝐺𝐺 then the𝑥𝑥 minimal spherical volume possible is 4 3 = 3 𝜋𝜋ℓ𝑝𝑝 Let denote the mass𝑉𝑉 of this volume, so its density will be given by 𝑚𝑚′ 3 = 4 ′ 3 𝑚𝑚 𝜌𝜌𝑃𝑃 Since by assumption is the𝜋𝜋ℓ minimal𝑝𝑝 length possible in nature, then for any mass 𝑝𝑝m' the density is the maximum possible. ℓ 𝑃𝑃 It can be showed, that for a classical spherically𝜌𝜌 symmetric Figure 1. For an object of radius R=3(*). The lines represent the behavior object of mass m' and radius R, the general relativistic limit of dτ, for the cases where < R (A-STNDRD) and > 2/3R (B-WH). As gives can be seen, in case B, the inner part of the object is below the horizon 𝑠𝑠 𝑠𝑠 𝑟𝑟 𝑟𝑟 4 (dτ = 0), but some of the the external part is not. This case allows for the = 1 2 photon sphere to be visible outside the body and yet the object is not a black 2 (17) 𝜋𝜋𝜋𝜋𝜋𝜋 hole yet. In this figure, dτ < 0 is set to dτ = 0. *(The dimensions are relative only) Since the expression𝑑𝑑𝑑𝑑 in brackets𝑑𝑑𝑑𝑑 �� − must𝑐𝑐 be𝑅𝑅 real,� we arrive at

124 Doron Kwiat: About the Existence of Black Holes

the restriction: obtains: 4 3 2 2 4 0 1 2 0 (18) 4 ( ) = (29) 0 3 𝜋𝜋𝜋𝜋𝜋𝜋 ℓ𝑝𝑝 𝜋𝜋𝜌𝜌 ℓ𝑝𝑝 And therefore 2 − 𝑐𝑐 𝑅𝑅 ≥ By definition, the𝜋𝜋 ∫ average𝑟𝑟 𝜌𝜌 𝑟𝑟classical𝑑𝑑𝑑𝑑 distance is given 2 by the integral over the normalized density: (19) 4 2 〈𝑟𝑟 〉 𝑐𝑐 2 1 2 ( ) 0 0 (30) For an object of any given𝜌𝜌 ≤ mass𝜋𝜋𝜋𝜋𝑅𝑅 m' and radius R we have ℓ𝑝𝑝 3 = = (20) Thus 〈𝑟𝑟 〉 ≝ ℓ𝑝𝑝 ∫ 𝑟𝑟 𝜌𝜌 𝑟𝑟 ⁄𝜌𝜌 𝑑𝑑𝑑𝑑 4 3 𝑚𝑚′ 𝑚𝑚′ 4 3 4 2 = 0 (31) Since for any mass𝜌𝜌 m' of𝑉𝑉 radius𝜋𝜋𝑅𝑅 R one must have, by Eq. 0 3 16 above, 𝜋𝜋𝜌𝜌 ℓ𝑝𝑝 and so 𝑝𝑝 3 2 𝜋𝜋ℓ 𝜌𝜌 〈𝑟𝑟 〉 = (21) 1 4 3 4 2 = 2 = (32) 𝑚𝑚′ 𝑐𝑐 3 The result is that for any𝜋𝜋𝑅𝑅 mass m'𝜋𝜋 𝜋𝜋𝑅𝑅 𝜌𝜌 ≤ Hence, the actual measured classical√ radius𝑝𝑝 R is given by 2 〈𝑟𝑟〉 �〈𝑟𝑟 〉 ℓ 1 ( ) (22) = = (33) 𝑅𝑅𝑅𝑅 3 Obviously, the smaller the radius𝐺𝐺 R, the smaller the 𝑚𝑚′ 𝑅𝑅 ≤ The above result shows𝑅𝑅 〈𝑟𝑟how〉 √theℓ 𝑝𝑝lower limit of the allowed mass m'. classical gravitation theory by Einstein, is related to the For the minimal possible length (according to Planck) Planck length, which is a quantum phenomenon posed by = one obtains: dimensional analysis of the universe constants. 2 𝑅𝑅 ℓ𝑝𝑝 (23) Therefore, classical relativity and the relationship between ℓ𝑝𝑝 𝑐𝑐 the universal constants leads to quantization of space. Recall now, by Planck's𝑚𝑚′ ≤dimensionality𝐺𝐺 analysis) that = and = so ℏ𝐺𝐺 ℏ𝑐𝑐 8. Quantization by Planck's Units and 𝑝𝑝 𝑝𝑝 2 ℓ � 𝑐𝑐 𝑚𝑚 � 𝐺𝐺 = (24) Maximal Density Limit

𝑚𝑚𝑝𝑝 𝑐𝑐 R 0 Therefore ℓ𝑝𝑝 𝐺𝐺 In the limit where , one needs to consider quantum 2 limits and the uncertainty principle. = (25) → Planck's mass = = 2.176 10 8 Kg. ′ ℓ𝑝𝑝 𝑐𝑐 Therefore, for any mass m' (with radius𝑝𝑝 ) ℏ𝑐𝑐 − 𝑚𝑚 ≤ 𝐺𝐺 𝑚𝑚 𝑝𝑝 Planck's length𝑚𝑚 = � 𝐺𝐺 = 1.616𝑥𝑥 10 35 m. 𝑝𝑝 (26) ℓ ℏ𝐺𝐺 − ′ However, dimensional𝑝𝑝 analysis can only determine the And since 𝑝𝑝 ℓ � 𝑐𝑐 𝑥𝑥 𝑚𝑚 ≤ 𝑚𝑚 Planck's units up to a dimensionless multiplicative factor. = = (27) For instance, use h instead of . 𝑚𝑚′ 𝑚𝑚𝑝𝑝 Based on Planck's units one obtains Planck's maximal We have the result that is the 𝑃𝑃maximal possible 𝜌𝜌 𝑉𝑉 ≤ 𝑉𝑉 𝜌𝜌 density as Planck's mass/Planck'sℏ volume (assuming density, namely Planck density. 𝑃𝑃 is the lowest possible physical distance, leads to the maximal In other words, for any given𝜌𝜌 radius R, the mass m'(R) 𝑃𝑃 𝑝𝑝 𝜌𝜌 ℓ becomes smaller and smaller with R, but when one reaches possible physical density assumption). +96 the smallest possible radius , the mass must be smaller = 3 = 1.23074 10 (34) 4 3 𝑚𝑚𝑝𝑝 than the Planck mass , and 𝑝𝑝so, the density will always be ℓ (If one uses 𝜌𝜌the𝑃𝑃 Planck𝜋𝜋ℓ𝑝𝑝 ⁄ constant instead𝑥𝑥 of the reduced smaller than the Planck density𝑝𝑝 . One can reduce the𝑚𝑚 radius R, but the density will never Planck's constant, the Planck density will be modified by a 𝑃𝑃 exceed the Planck density. 𝜌𝜌 factor of 2 ). 3 Since one must assume that Planck's density is the ( ) = (28) maximum 𝜋𝜋possible density allowed, we now ask, what ℓ𝑝𝑝 should the minimum classical radius R be, in order for 𝑝𝑝 Assume next, that𝑙𝑙𝑙𝑙𝑙𝑙 the𝑅𝑅→ ℓsphere𝜌𝜌 𝑅𝑅 has �density𝑅𝑅 � 𝜌𝜌𝑃𝑃 ( ), which . (any classical density cannot exceed the Planck varies with distance r from its center. density). 𝑃𝑃 Assume the sphere is of minimal possible radius𝜌𝜌 𝑟𝑟 . 𝜌𝜌 Therefore:≤ 𝜌𝜌 2 We need to calculate the radius R of a quantized particle𝑝𝑝 3 ℓ = = 3 (35) by its average normalized density so that we obtain a reduced 4 2 4 𝑚𝑚𝑝𝑝 average radius. 𝑐𝑐 Doing the calculation,𝜌𝜌 𝜋𝜋𝜋𝜋 𝑅𝑅we≤ find𝜌𝜌𝑃𝑃 that𝜋𝜋ℓ 𝑝𝑝one must have By comparing the integrated variable density over the 1 R = 0.93325 10 35 m. Planck radius, to a volume, with constant density 0 one 3 − 𝑝𝑝 𝜌𝜌 ≥ √ ℓ 𝑥𝑥 International Journal of Theoretical and Mathematical Physics 2020, 10(6): 120-129 125

Compared to Planck's length = 1.61625 10 35 m, This gedanken experiment can reveal whether the the classically derived radius R is smaller than the allowed− approach to will last forever in the eyes of the external ℓ𝑝𝑝 𝑥𝑥 observer. If it does, then the particle will never be able to Planck's length , by an unexplained factor of 0.5774. 𝑠𝑠 These calculations were based on the assumptions of come out through𝑟𝑟 the opposite hole. ℓ16𝑝𝑝 G negligible = 2 , and R 0 (r < R). It seems now, thatπ the classical solution to the metric of a 𝕂𝕂 𝑐𝑐 𝜌𝜌 → 10. Dynamics Near Center spherically symmetric object, puts a lower limit on the radius R of an object. It looks like there is a sort of quantization It is interesting to see what are the equations of motion limit to small scale objects in spacetime. near center (where r<< R). Instead of having , we find 0.5774 . Assuming spherical symmetry [21,22], it is allowed to The reason for this 0.5774 factor is due to the fact that we investigate it along the radial distance r, and assume it is the 𝑝𝑝 𝑝𝑝 have assumed a constant𝑅𝑅 ≥ densityℓ inside𝑅𝑅 the≥ sphere. ℓIf instead x axis. we assume variation in density with radius (highest density 2 = (37) at r 0and then it falls off to zero at ). We need to 2𝜇𝜇 𝛼𝛼 𝛽𝛽 𝑑𝑑 𝑥𝑥 𝜇𝜇 𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥 calculate the radius R of a quantized particle by its average where 𝛼𝛼𝛼𝛼 → ℓ𝑝𝑝 𝑑𝑑𝜏𝜏 − 𝛤𝛤 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 density so that we obtain a reduced average radius. 1 = + (38) By comparing the integrated variable density over the 2 Planck radius, to the same volume, but with a constant 𝜆𝜆 𝜆𝜆𝜆𝜆 The total 𝛤𝛤time𝛾𝛾𝛾𝛾 of flight𝑔𝑔 �𝜕𝜕 in𝜇𝜇 𝑔𝑔 the𝑟𝑟𝑟𝑟 classical𝜕𝜕𝛾𝛾 𝑔𝑔𝑟𝑟𝑟𝑟 −case𝜕𝜕𝑟𝑟 𝑔𝑔 will𝛾𝛾𝛾𝛾 � be given average density 0 one obtains (see paragraph VI on density by (see Appendix 2): upper limit)): 2 Q 3 𝜌𝜌 1 2 = 1 2 1 0 = = (36) Q0 3 − 𝑅𝑅 1 𝑝𝑝 Whereas in the general� relativistic case it𝐺𝐺𝐺𝐺 will be: Since 0.5774 = , it is now √clear where the reduction 𝑇𝑇 𝑐𝑐𝑐𝑐𝑐𝑐 � � − �� 3 𝑅𝑅 〈𝑟𝑟〉 ℓ 2 3 factor came from. 2 = 1 2 1 √ The above result shows how the lower limit of the − 𝑄𝑄𝑄𝑄 classical gravitation theory by Einstein, (Schwarzschild Comparing the𝑇𝑇 two results,�𝑄𝑄 𝑐𝑐𝑐𝑐𝑐𝑐 � . � . −> 𝐺𝐺𝐺𝐺 �� . As an example, for the Sun, with R = 6.963x108m and radius), and the dimensional constraints between the 𝐺𝐺𝐺𝐺𝐺𝐺 𝑅𝑅𝑅𝑅𝑅𝑅 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 3 𝑇𝑇 3 𝑇𝑇 constants of the universe, result in a quantized material Sun density ρ☉ = 150 x10 Kg/m the predicted general space. relativistic cross time will be 6.6 mSec longer than the classical calculated cross time. This result is based on average Q over r from 0 to R. 9. A Tunneled Sphere "Thought For an average Neutron star of radius R=11,000 m and Experiment" density ρ(r) = 4x1017 Kg/m3 , the passage time under general relativity prediction, will last 0.13 mSec longer than Consider a sphere with a thin penetrating tunnel through under classical calculation. center of sphere between opposite surfaces. The sphere of radius R and of mass M will exert a gravitational attractive force on a remote particle. The force 11. Conclusions will be directed towards the center of the sphere, as if all of its mass is concentrated at its center (where r = 0). Using Newton's classical shell theorem, the The attracted particle will be accelerated until it reaches Schwarzschild metric was modified. This removed the the surface, but then will continue its path through the hole. It singularity at r = 0 for a standard object (not a black hole). will continue with no acceleration inside the shell and then For all practical matters, r < R can be treated as an empty come out through the opposite hole. space even for the densest known stellar objects (neutron On its journey, the particle will pass twice the stars) and also for elementary particles (neutrons for instance). Schwarzschild radius = 2 on its way towards the center 𝐺𝐺𝐺𝐺 It was demonstrated how general relativity evidently leads and on its way out. 𝑠𝑠 𝑐𝑐 The metric of this shell𝑟𝑟 will be similar to that of a spherical to quantization of space-time. Both classical and quantum mass. Only that this time, even though is inside the body, mechanical limits on density give the same result. one can reach the Schwarzschild radius independently of 𝑠𝑠 whether it is a black hole or a standard mass.𝑟𝑟 One way of doing the experiment would be to send a Appendix 1 photon through the hole in the mass and compare its time of For a curved spacetime we define the Riemann tensor: arrival with a simultaneously emitted photon traveling outside the mass. One would then be able to compare = + [1] differences in the time of arrival for the two photons. 𝜇𝜇 𝜇𝜇 𝜇𝜇 𝜇𝜇 𝜎𝜎 𝜇𝜇 𝜎𝜎 𝑅𝑅𝛼𝛼𝛼𝛼𝛼𝛼 𝜕𝜕𝛽𝛽 𝛤𝛤𝛼𝛼𝛼𝛼 − 𝜕𝜕𝛾𝛾 𝛤𝛤𝛼𝛼𝛼𝛼 𝛤𝛤𝜎𝜎𝜎𝜎 𝛤𝛤𝛼𝛼𝛼𝛼 − 𝛤𝛤𝜎𝜎𝜎𝜎 𝛤𝛤𝛼𝛼𝛼𝛼

126 Doron Kwiat: About the Existence of Black Holes

Where = are the Christoffel symbols: And so 𝜇𝜇 𝜇𝜇𝜇𝜇1 ( ) ( ) 𝛼𝛼𝛼𝛼 = 𝜎𝜎(𝜎𝜎𝜎𝜎 + ) (39) 1 Γ 𝑔𝑔 2Γ = 1 22 ( ) 2 ( ) (′ 𝑟𝑟) (′ 𝑟𝑟) The Ricci tensor𝑖𝑖𝑖𝑖𝑖𝑖 is𝑙𝑙 defined𝑖𝑖𝑖𝑖 𝑘𝑘 by:𝑖𝑖𝑖𝑖 𝑖𝑖 𝑘𝑘𝑘𝑘 𝑟𝑟 ℎ 𝑔𝑔 𝛤𝛤 𝜕𝜕 𝑔𝑔 𝜕𝜕 𝑔𝑔 − 𝜕𝜕 𝑔𝑔 𝑅𝑅 1 − − � ( )− � = 1 (51) 𝛼𝛼𝛼𝛼 = (40) ℎ( 𝑟𝑟) 2 (ℎ) 𝑟𝑟 ℎ 𝑟𝑟( ) 𝑔𝑔 𝑟𝑟 𝑅𝑅 𝑟𝑟 ℎ 𝑟𝑟 𝜎𝜎𝜎𝜎 𝑟𝑟 𝛼𝛼𝛼𝛼 𝛼𝛼𝛼𝛼𝛼𝛼𝛼𝛼 We have then ℎtwo𝑟𝑟 − equations:− ℎ 𝑟𝑟 𝜕𝜕 𝑙𝑙𝑙𝑙 �𝑔𝑔 𝑟𝑟 � And the curvature scalar𝑅𝑅 R 𝑔𝑔is defined𝑅𝑅 by: ( ) = = ( ) ( ) = (41) ( ) (52) 𝛼𝛼𝛼𝛼 𝛼𝛼𝛼𝛼 𝜎𝜎𝜎𝜎 𝑟𝑟ℎ 𝑟𝑟 1 ( ) For the Einstein equation𝛼𝛼 𝛼𝛼one has: 𝛼𝛼𝛼𝛼𝛼𝛼𝛼𝛼 𝑟𝑟 𝑔𝑔 𝑟𝑟 𝓡𝓡 𝑔𝑔 𝑅𝑅 𝑔𝑔 𝑔𝑔 𝑅𝑅 22𝜕𝜕 𝑙𝑙𝑙𝑙=� ℎ 𝑟𝑟 𝑔𝑔 1𝑟𝑟 � − 𝕂𝕂 (53) 1 8 ( ) 2 ( ) ( ) = (42) 𝑟𝑟 ℎ 𝑟𝑟 2 4 𝑟𝑟 𝜋𝜋𝜋𝜋 Outside the 𝑅𝑅mass (rℎ>R)𝑟𝑟 − − ℎ 𝑟𝑟 𝜕𝜕 𝑙𝑙𝑙𝑙 �𝑔𝑔 𝑟𝑟 � In empty space𝑅𝑅𝜇𝜇𝜇𝜇 − 𝑔𝑔𝜇𝜇𝜇𝜇= 0𝓡𝓡 everywhere𝑐𝑐 𝑇𝑇𝜇𝜇𝜇𝜇 and therefore = 0 and so: = 0 everywhere. ( ) ( ) = (some constant). 𝑇𝑇𝜇𝜇𝜇𝜇 Due to spherical symmetry, the metric must be of the form 𝕂𝕂Substitute in eq. 2 to obtain: 𝑅𝑅𝜇𝜇𝜇𝜇 ( ) ℎ 𝑟𝑟 𝑔𝑔 𝑟𝑟 𝛼𝛼 ( ) ( ) = 1 = 0 (54) ( ) 22 2 2( ) = (43) 𝑔𝑔 𝑟𝑟 𝑟𝑟𝑟𝑟 𝑟𝑟 𝛼𝛼 𝑔𝑔 𝑟𝑟 2 𝑟𝑟 And after𝑅𝑅 few steps:𝛼𝛼 − − 𝛼𝛼 𝜕𝜕 𝑙𝑙𝑙𝑙 �𝑔𝑔 𝑟𝑟 � 𝜇𝜇𝜇𝜇 ℎ 𝑟𝑟 2 2 𝑔𝑔 � � = ( ( )) (55) Where g(r) and h(r) are functions𝑟𝑟 of the distance r from the So finally: coordinate center (located at the center 𝑟𝑟of 𝑠𝑠𝑠𝑠𝑠𝑠the mass.𝜃𝜃 𝛼𝛼 𝜕𝜕𝑟𝑟 𝑟𝑟𝑟𝑟 𝑟𝑟 So, the infinitesimal proper time interval between two ( ) = + (56) events along a time-like path is given by ( ) = (57) 𝑑𝑑𝑑𝑑 𝑔𝑔 𝑟𝑟 𝛼𝛼𝛼𝛼+ 𝛽𝛽 2 = 𝛼𝛼 1 𝜇𝜇 𝜈𝜈 The metric becomes: ℎ 𝑟𝑟 𝛼𝛼𝛼𝛼 𝛽𝛽 = ( ) ( ( ) 2 𝜇𝜇𝜇𝜇 2 2 2 2 2) (44) 2 𝑑𝑑𝑑𝑑 𝑔𝑔 𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥 +

With the flat 𝑐𝑐space metric g = (+ ). 𝑔𝑔 𝑟𝑟 𝑑𝑑𝑑𝑑 − ℎ 𝑟𝑟 𝑟𝑟 𝑑𝑑𝑑𝑑 − 𝑟𝑟 𝑑𝑑𝜃𝜃 − 𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃𝜃𝜃𝜑𝜑 + In order to solve Einstein's equation everywhere, one can = 𝛼𝛼𝛼𝛼 𝛽𝛽 𝛼𝛼 (58) μν 2 − −− 𝛼𝛼𝛼𝛼 𝛽𝛽 no longer assume that the curvature is null anywhere. 𝜇𝜇𝜇𝜇 ⎛ 2 2 ⎞ This is true only outside the mass and so = 0 only for 𝑔𝑔 ⎜ ⎟ 𝑟𝑟 r > R. 𝓡𝓡 Comparing this metric with the asymptotic weak-field 𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃 Assuming the mass density of the 𝓡𝓡object being metric (r ):⎝ ⎠ homogeneous and at total freeze of motion, = 0, except 2 2 1 2 for 00 = , with being the mass density. ≫ 𝑅𝑅 𝜇𝜇𝜇𝜇 𝐺𝐺𝐺𝐺 1 1 8 𝑇𝑇 𝑟𝑟𝑐𝑐 2 = (45) = � − � 1 (59) 𝑇𝑇 𝜌𝜌𝑐𝑐 𝜌𝜌 2 4 − 2 𝜋𝜋𝜋𝜋 𝜇𝜇𝜇𝜇 𝜇𝜇𝜇𝜇 2 ⎛ 𝐺𝐺𝐺𝐺 2 ⎞ 𝜇𝜇𝜇𝜇 𝜇𝜇𝜇𝜇 𝜇𝜇𝜇𝜇 𝜇𝜇𝜇𝜇 � − � 𝑔𝑔 �𝑅𝑅 − 𝑔𝑔 𝓡𝓡� 𝑐𝑐 𝑔𝑔 𝑇𝑇 𝑔𝑔 𝑟𝑟𝑐𝑐 ⎜ 2 2 ⎟ T Under perfect fluid conditions, = 𝜌𝜌𝑐𝑐 , −𝑟𝑟 It is suggested that μν 𝑝𝑝 ⎝ −𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 𝜃𝜃 ⎠ � � 2 ( ) where p is the pressure. 𝑝𝑝 + = 1 + 2 (60) 16 G 𝛷𝛷 𝑟𝑟 = 𝑝𝑝 Define 2 , and so, Therefore, 𝑐𝑐 π 𝛼𝛼𝛼𝛼 𝛽𝛽 � � 16 = 1 (61) 𝕂𝕂 𝑐𝑐 00 = 2 = (46) 𝜋𝜋𝜋𝜋 2 ( ) = For all other terms, 11 = 𝑐𝑐 22 = 33: 𝛽𝛽 2 (62) 𝑅𝑅 𝜌𝜌 𝕂𝕂𝜌𝜌 𝛷𝛷 𝑟𝑟 16 = = This shows us that the structure of the Schwarzschild 𝑅𝑅 𝑅𝑅2 𝑅𝑅 (47) 𝛼𝛼 𝑟𝑟𝑐𝑐 𝜋𝜋𝜋𝜋 metric in Eq. 7, is not only a result of dimensional From the definitions𝑅𝑅 𝑖𝑖𝑖𝑖of the𝑐𝑐 Riemann𝑝𝑝 𝕂𝕂 𝑝𝑝tensor and the Ricci considerations discussed above, but also a result of spherical tensor we obtain symmetry considerations of the Einstein Equation, Eq. 17. = + (48) This result was obtained based on symmetry arguments 𝛿𝛿 𝛿𝛿 𝛿𝛿 𝜆𝜆 𝛿𝛿 𝜆𝜆 only, and it is independent of the weak-field approximation. 00𝜇𝜇𝜇𝜇 = 𝛿𝛿 00𝜇𝜇𝜇𝜇 0𝛾𝛾 0𝜇𝜇𝜇𝜇+ 𝛿𝛿𝛿𝛿 00𝛾𝛾𝛾𝛾 0𝛾𝛾𝛾𝛾 0𝛿𝛿𝛿𝛿 (49) 𝑅𝑅 𝜕𝜕 𝛤𝛤 − 𝜕𝜕 𝛤𝛤 𝛤𝛤 𝛤𝛤 − 𝛤𝛤 𝛤𝛤 Inside the mass (r < R) Following some tedious𝛿𝛿 mathematical𝛿𝛿 𝛿𝛿 𝜆𝜆 work, 𝛿𝛿one𝜆𝜆 arrives at 𝑅𝑅 𝜕𝜕𝛿𝛿 𝛤𝛤 − 𝜕𝜕 𝛤𝛤 𝛿𝛿 𝛤𝛤𝛿𝛿𝛿𝛿 𝛤𝛤 − 𝛤𝛤 𝜆𝜆 𝛤𝛤𝛿𝛿 > 0 [4]: : The only approximation made was the assumption on 16 ( ) ( ) ( ) = 𝕂𝕂 2 ( ) (50) empty space solution. For non-empty space, it will be valid 𝜋𝜋𝜋𝜋 𝑟𝑟ℎ 𝑟𝑟

𝜕𝜕𝑟𝑟 𝑙𝑙𝑙𝑙�ℎ 𝑟𝑟 𝑔𝑔 𝑟𝑟 � − 𝑐𝑐 𝜌𝜌 𝑔𝑔 𝑟𝑟 International Journal of Theoretical and Mathematical Physics 2020, 10(6): 120-129 127

2 2 for a perfect fluid of zero pressure. 1 1 11 1 2 1 1 = (77) 2 2 2 00 2 2 2 However, one may use a model where the pressure p is 𝑑𝑑 𝑟𝑟 00 𝑑𝑑𝑑𝑑 proportional o the density [18], and so p = where And so𝐴𝐴 𝑑𝑑𝑡𝑡 𝑔𝑔 𝜕𝜕𝑟𝑟 𝑔𝑔 �𝐴𝐴 𝑐𝑐 − 𝑔𝑔 𝐴𝐴 �𝑑𝑑𝑑𝑑 � � is a proportionality factor. 2 1 1 1 2 𝜔𝜔𝜔𝜔 𝜔𝜔 = 2 (78) 2 2 00 2 𝑑𝑑 𝑟𝑟 00 00 𝑑𝑑𝑑𝑑 For 𝑑𝑑𝑡𝑡 1 − 𝑔𝑔 𝜕𝜕𝑟𝑟 𝑔𝑔 �𝑐𝑐 − 𝑔𝑔 �𝑑𝑑𝑑𝑑 � � Appendix 2 𝑟𝑟 4 2 𝑅𝑅 Since by coordinates choice and symmetry, only 00 and ≪ 00 = 2 1 (79) 𝜋𝜋𝜋𝜋𝜋𝜋 𝑅𝑅 11 are non-zero (assuming constant gravitationalλ field 2 g = 0 𝑐𝑐 = 0 λg = 0 Γ And so 00 , 𝑔𝑔which means that− 2 . ). 𝑑𝑑 𝑟𝑟 Γ 2 2 2 The acceleration𝑟𝑟 near the center is zero.𝑑𝑑𝑡𝑡 𝑡𝑡 𝑟𝑟𝑟𝑟 1 1 𝜕𝜕 𝜕𝜕 2 = 00 11 (63) However, if one moves away from the center: 𝑑𝑑 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 1 4 2 1 2 1 𝑑𝑑𝜏𝜏 1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 1 1 (80) = −( 𝛤𝛤 �+� − 𝛤𝛤 � � ) 00 2 3 00 2 0 0 0 0 00 𝜋𝜋𝜋𝜋𝜋𝜋 𝑅𝑅 𝑟𝑟 𝑟𝑟 1 11 𝑟𝑟 𝑟𝑟 1 𝑟𝑟 11 And so 𝑔𝑔 𝑐𝑐 � − �𝑅𝑅� � − = 𝛤𝛤( 0 10 + 𝑔𝑔 0 10𝜕𝜕 𝑔𝑔 1 00𝜕𝜕)𝑔𝑔= − 𝜕𝜕 𝑔𝑔 1 00 (64) 2 2 4 2 2 8 = = (81) 00 2 3 2 3 2 and likewise𝑔𝑔 𝜕𝜕 𝑔𝑔 𝜕𝜕 𝑔𝑔 − 𝜕𝜕 𝑔𝑔 − 𝑔𝑔 𝜕𝜕 𝑔𝑔 𝜋𝜋𝜋𝜋𝜋𝜋 𝑅𝑅 𝑟𝑟 𝜋𝜋𝜋𝜋𝜋𝜋 1 1 2 2 1 11 11 𝑟𝑟 1 𝑐𝑐 1 𝑅𝑅 2 1 𝑐𝑐 11 = ( 1 11 + 1 11 1 11) = 1 11(65) 𝜕𝜕 𝑔𝑔 = �− � − 𝑟𝑟 (82) 2 2 2 2 00 2 𝑑𝑑 𝑟𝑟 00 00 𝑑𝑑𝑑𝑑 2 𝛤𝛤The equation𝑔𝑔 𝜕𝜕 of𝑔𝑔 motion𝜕𝜕 in𝑔𝑔 the− x𝜕𝜕 direction𝑔𝑔 becomes𝑔𝑔 𝜕𝜕 𝑔𝑔 𝑑𝑑𝑡𝑡2 1 1𝑔𝑔 8𝑟𝑟 𝑔𝑔 1 𝑑𝑑𝑑𝑑 =− 𝜕𝜕 𝑔𝑔 �𝑐𝑐 −2 + � � � (83) 2 2 2 2 2 2 3 2 4 1 11 2 1 11 𝑑𝑑 𝑟𝑟 𝜋𝜋𝜋𝜋𝜋𝜋 𝑑𝑑𝑑𝑑 2 = 1 00 1 11 (66) 2 2 2 4 1 2 𝑑𝑑 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑡𝑡 = 𝐴𝐴 �− 𝑐𝑐 2 +𝑟𝑟� �𝑐𝑐 𝐴𝐴 � 𝑑𝑑𝑑𝑑 � � (84) 2 3 2 2 4 𝑑𝑑𝜏𝜏Since � the𝑔𝑔 choice𝜕𝜕 𝑔𝑔 � of�𝑑𝑑𝑑𝑑 �coordinates𝑐𝑐 − � 𝑔𝑔 is 𝜕𝜕such𝑔𝑔 � that�𝑑𝑑𝑑𝑑 � x is 𝑑𝑑 𝑟𝑟 𝜋𝜋𝜋𝜋𝜋𝜋 𝑑𝑑𝑑𝑑 2 represented by r, one can write 𝑑𝑑𝑡𝑡2 − 4𝑐𝑐 𝐴𝐴 𝑟𝑟 �𝑐𝑐 𝐴𝐴1 �𝑑𝑑𝑑𝑑 � � 2 = 2 1 + 2 4 (85) 2 1 2 1 2 3 = 11 2 11 (67) 𝑑𝑑 𝑟𝑟 𝜋𝜋𝜋𝜋𝜋𝜋 𝑑𝑑𝑑𝑑 2 2 00 2 11 4 2 𝑑𝑑 𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑡𝑡 − 𝐴𝐴 𝑟𝑟 � 2𝑐𝑐 𝐴𝐴 �𝑑𝑑𝑑𝑑 � � Near center (where r  0), A = 1 2 depends on From𝑑𝑑𝜏𝜏 the proper-𝑟𝑟time result𝑑𝑑𝑑𝑑 (which holds 𝑟𝑟inside the𝑑𝑑𝑑𝑑 sphere, 𝑔𝑔 𝜕𝜕 𝑔𝑔 � � 𝑐𝑐 − 𝑔𝑔 𝜕𝜕 𝑔𝑔 � � r only through the density ( ). 𝜋𝜋𝜋𝜋𝜋𝜋 𝑅𝑅 where r < R) 𝑐𝑐 We may write − 2 2 2 2 4 1 2 2 𝜌𝜌 𝑟𝑟 2 = 1 1 = (68) ( ) 2 3 00 = ( ) 1 + ( ) (86) 𝜋𝜋𝜋𝜋𝜋𝜋 𝑅𝑅 𝑟𝑟 2 𝑑𝑑 𝑟𝑟 𝑣𝑣 𝑟𝑟 𝑐𝑐 𝑅𝑅 𝑑𝑑𝑑𝑑Define 𝑑𝑑𝑑𝑑for convenience� − � − � � �� −𝑔𝑔 𝑑𝑑𝑑𝑑 Here, ( )𝑑𝑑 𝑡𝑡is the𝑄𝑄 approach𝑟𝑟 𝑟𝑟 � velocity.𝑊𝑊 𝑟𝑟 �Q 𝑐𝑐and� W� depend on r 4 2 1 2 and are defined by: = 2( ) = 1 1 (69) 00 2 3 𝑣𝑣 𝑟𝑟 4 ( ) 𝜋𝜋𝜋𝜋𝜋𝜋 𝑅𝑅 𝑟𝑟 ( ) (87) 3 2( ) And so 𝑔𝑔 𝐴𝐴 𝑟𝑟 − 𝑐𝑐 � − �𝑅𝑅� � 𝜋𝜋𝜋𝜋𝜋𝜋 𝑟𝑟 1 𝑄𝑄 𝑟𝑟( )≝ 𝐴𝐴 𝑟𝑟 (88) = (70) 4( ) 𝑑𝑑𝑑𝑑 1 The result shows how𝑊𝑊 on𝑟𝑟 approach≝ 𝐴𝐴 𝑟𝑟 to center (r  0), 𝑑𝑑𝑑𝑑 = = 𝐴𝐴 (71) where ( ) is assumed constant, the acceleration decreases 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 And 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐴𝐴 𝑑𝑑𝑑𝑑 in proportion with r on one hand, but depends also on the 𝜌𝜌 𝑟𝑟 2 1 2 square of the approach velocity. 2 = 2 2 (72) For small approach-to-center velocity ( 1) 𝑑𝑑 𝑟𝑟 𝑑𝑑 𝑟𝑟 2 From the metric definition: 𝑑𝑑𝜏𝜏 𝐴𝐴 𝑑𝑑𝑡𝑡 2 = 𝑣𝑣 ⁄ 𝑐𝑐 ≪ (89) 2 0 0 1 1 𝑑𝑑 𝑟𝑟 = = 00 + 11 (73) One may approximate though𝑑𝑑𝑡𝑡 for−𝑄𝑄 a slow𝑟𝑟 approach velocity 𝜇𝜇 𝜈𝜈 ( 1): and since𝑑𝑑𝑑𝑑 𝑔𝑔𝜇𝜇𝜇𝜇 𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥 𝑔𝑔 𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥 𝑔𝑔 𝑑𝑑𝑥𝑥 𝑑𝑑𝑥𝑥 2 2 00 = (74) 𝑣𝑣⁄𝑐𝑐 ≪ 2 = ( ) (90) 1 1 𝑑𝑑 𝑟𝑟 11 = = 2 (75) 𝑔𝑔 𝐴𝐴 00 where 𝑑𝑑𝑡𝑡 −𝑄𝑄 𝑟𝑟 𝑟𝑟 4 ( ) One obtains: 𝑔𝑔 − 𝑔𝑔 − 𝐴𝐴 ( ) = 4 ( ) 2 1 2 (91) 3 1 1 1 2𝜋𝜋𝜋𝜋𝜋𝜋 𝑟𝑟 3 11 = 2 00 (76) 𝜋𝜋𝜋𝜋𝜋𝜋 𝑟𝑟 𝑅𝑅 𝑟𝑟 00 𝑄𝑄 𝑟𝑟 (Recall that by the assumption� − 𝑐𝑐 on �the− maximal�𝑅𝑅� �� density 𝑟𝑟 𝑟𝑟 Inserting into Eq. [49] one obtains:𝑔𝑔 4πGρ(r)R2 𝜕𝜕 𝑔𝑔 𝜕𝜕 𝑔𝑔 (Eq. 38), < 1, and so, Q(r) 0. Also, note that c2

≥ 128 Doron Kwiat: About the Existence of Black Holes

2 2 3 [ ] 4 0 4 0 Q = 1/ ]). = = = (106) 2 3 3 3 3 One may solve this under certain approximation: 𝑑𝑑 𝑟𝑟 𝐺𝐺𝐺𝐺 𝐺𝐺 𝜋𝜋𝜌𝜌 𝑅𝑅 𝜋𝜋𝜋𝜋𝜌𝜌 ( ) = 𝑠𝑠𝑠𝑠𝑠𝑠0 constant density inside r < R. Classically,𝑑𝑑𝑡𝑡 the− acceleration𝑅𝑅 𝑟𝑟 − will𝑅𝑅 be 𝑟𝑟 − 𝑟𝑟 In this case Q(r) is nearly a constant independent of r, 2 4 = = (107) and𝜌𝜌 the𝑟𝑟 equation𝜌𝜌 becomes 2 3 0 0 𝑑𝑑 𝑟𝑟 𝜋𝜋𝜋𝜋 2 4 G with 0 = 0 𝑑𝑑𝑡𝑡 − 𝜌𝜌 𝑟𝑟 −𝑄𝑄 𝑟𝑟 2 = (92) 3 𝑑𝑑 𝑟𝑟 This, in comparisonπ to the general relativistic solution: 𝑄𝑄 ρ Which general solution is𝑑𝑑 𝑡𝑡 −𝑄𝑄 𝑟𝑟 2 = ( ) ( ) = + ( ) + ( ) (93) 2 (108) 𝑑𝑑 𝑟𝑟 4 G (r)R2 Even though𝑟𝑟 𝑡𝑡 𝐵𝐵(r)𝐶𝐶 is𝐶𝐶𝐶𝐶𝐶𝐶 not 𝑄𝑄constant,𝑡𝑡 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 the term𝑄𝑄 𝑡𝑡 in 𝑑𝑑𝑡𝑡 −𝑄𝑄 𝑟𝑟 𝑟𝑟 � � c2 the denominator of ( ) is of the order of, or πlessρ than, 9.2 10 9. Q(r) may𝑄𝑄 be approximated as: REFERENCES − 𝑄𝑄 𝑟𝑟4 ( ) 4 ( ) ( ) 1 2 (94) [1] Kevin Brown, Reflections on Relativity, ISBN-12: 𝑥𝑥 3 1 9.2 10 9 1 3 𝜋𝜋𝜋𝜋𝜋𝜋 𝑟𝑟 3 𝜋𝜋𝜋𝜋𝜋𝜋 𝑟𝑟 9781257033027 Lulu.com. 2010. − 𝑟𝑟 𝑄𝑄 𝑟𝑟 ≅ � − 𝑥𝑥 � − �𝑅𝑅� �� ≅ ( ) = ( ) + ( ) (95) [2] Landau, L. D.; Lifshitz, E. M.. The Classical Theory of Fields. r(0) = Q D Course of Theoretical Physics. 2 (4th Revised English ed.). Assuming𝑟𝑟̇ 𝑡𝑡 zero− 𝐶𝐶start�𝑄𝑄 𝑠𝑠𝑠𝑠𝑠𝑠velocity,�𝑄𝑄 𝑡𝑡 𝐷𝐷�𝑄𝑄 𝑐𝑐𝑐𝑐𝑐𝑐 �𝑄𝑄 𝑡𝑡one must Pergamon Press. (1950). have D = 0. Hence ̇ −� [3] Buchdahl H. A., Isotropic Coordinates and Schwarzschild Metric, International Journal of Theoretical Physics, Volume ( ) = ( ) (96) 24, Issue 7, pp.731-739, (1985). r(0) = R Also, .𝑟𝑟 𝑡𝑡 𝐵𝐵 − 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 �𝑄𝑄 𝑡𝑡 [4] Davie R, Schwarzschild metric 1-3, Therefore https://www.youtube.com/watch?v=D1mmLjR-szY. = (97) [5] José P. S. Lemos, Diogo L. F. G. Silva. "Maximal extension ( ) = ( ) ( ) (98) of the Schwarzschild metric: From Painlevé-Gullstrand to 𝑅𝑅 𝐵𝐵 − 𝐶𝐶 Kruskal-Szekeres", arXiv:2005.14211. (2020). GM Assume the acceleration r(0) = , 𝑟𝑟 𝑡𝑡 𝐵𝐵 − 𝐵𝐵 − 𝑅𝑅 𝑐𝑐R𝑐𝑐𝑐𝑐2 �𝑄𝑄 𝑡𝑡 [6] P. Painlevé, “La mécanique classique et la théorie de la ( ) = ( ) relativité”, Comptes Rendus de l’Académie des Sciences 173, ̈ − (99) 677 (1921). Using these assumptions, one obtains: 𝑟𝑟̈ 𝑡𝑡 𝐵𝐵 − 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 �𝑄𝑄 𝑡𝑡 [7] A. Gullstrand, “Allgemeine Lösung des statischen

( ) = + 2 ( ) 1 (100) Einkörperproblems in der Einsteinschen Gravitationstheorie”, 𝐺𝐺𝐺𝐺 Arkiv för Matematik, Astronomi och Fysik 16, 1 (1922). Furthermore, 𝑟𝑟if 𝑡𝑡the object𝑅𝑅 𝑄𝑄 arrives𝑄𝑄 �𝑐𝑐𝑐𝑐𝑐𝑐 at� 𝑄𝑄the𝑡𝑡 center− � (r = 0) at some later time T, one has: [8] A. S. Eddington, “A comparison of Whitehead’s and Einstein’s formula”, Nature 113, 192 (1924).

( ) = 0 = + 2 ( ) 1 (101) 𝐺𝐺𝐺𝐺 [9] D. Finkelstein, “Past-future asymmetry of the gravitational field of a point particle”, Phys. Rev. 110, 965 (1958). The time to𝑟𝑟 𝑇𝑇cross the diameter𝑅𝑅 𝑄𝑄𝑄𝑄 �2R𝑐𝑐𝑐𝑐𝑐𝑐 (from�𝑄𝑄 𝑇𝑇surface− � to surface through the center) is thus [10] G. Lemaître, “L’Univers en expansion”, Annales de la 2 3 Société Scientifique de Bruxelles 53A, 51 (1933). 2 = 1 1 (102)

− 𝑄𝑄𝑄𝑄 [11] J. Synge, “The gravitational field of a particle”, Proceedings Where 𝑇𝑇 �𝑄𝑄 𝑐𝑐𝑐𝑐𝑐𝑐 � − 𝐺𝐺𝐺𝐺 � of the Royal Irish Academy, Section A: Mathematical and Physical Sciences 53, 83 (1950). ( ) 4 G (r) Q r = 4 G (r)R2 1 r 2 3 1 1 c2π ρ 3 R [12] M. Kruskal, “Maximal extension of Schwarzschild metric”, π ρ Phys. Rev. 119, 1743 (1959). In this case � − � − � � �� 2 4 [13] G. Szekeres, “On the singularities of a Riemannian manifold”, = ( ) (103) Publicationes Mathematicae Debrecen 7, 285 (1959). 2 3 𝑑𝑑 𝑟𝑟 𝜋𝜋𝜋𝜋 Classically, a falling𝑑𝑑 object𝑡𝑡 − of mass𝑟𝑟 𝜌𝜌 m,𝑟𝑟 starting at r < R, [14] I. Newton, Philosophiae Naturalis Principia Mathematica. will be accelerated towards center according to the potential: London. pp. Theorem XXXI. (1687). ( ) = (3 2 2) (104) [15] R. Arens, "Newton\'s observations about the field of a 2 3 𝐺𝐺𝐺𝐺 uniform thin spherical shell ".Note di Matematica, Vol. x, And the acceleration𝛷𝛷 𝑟𝑟 will− be𝑅𝑅 𝑅𝑅 − 𝑟𝑟 Suppl. n. 1, 39-45. (1990). 2 ( ) = = = 2 (105) [16] Carlo Rovelli, Scholarpedia, 3(5): 7117. 2 2 3 𝑑𝑑 𝑟𝑟 𝜕𝜕𝜕𝜕 𝑟𝑟 𝐺𝐺𝐺𝐺 doi.org/10.4249/scholarpedia.7117 (2008). 𝐹𝐹 𝑚𝑚 𝑑𝑑𝑡𝑡 −𝑚𝑚 𝜕𝜕𝜕𝜕 −𝑚𝑚 𝑅𝑅 𝑟𝑟 International Journal of Theoretical and Mathematical Physics 2020, 10(6): 120-129 129

[17] I. Gkigkitzis, I. Haranas, O. Ragos, "Kretschmann Invariant Kerr-Newman Black Hole". The Astrophysical Journal. The and Relations between Spacetime Singularities, Entropy and American Astronomical Society. 535 (1): 350–353. Information". arXiv preprint arXiv:1406.1581, (2014). [20] A. Zee, "Einstein Gravity in a Nutshell",Princeton University [18] R. Moradi, J. T. Firouzjaee, R. Mansouri. "Cosmological Press, 2013. black holes: the spherical perfect fluid collapse with pressure in a FRW background", Classical and Quantum Gravity. 32, [21] Ray D'Inverno, ""Introducing Einstein's Relativity", Oxford 21 (2015). University Press, 1992. [19] Richard C. Henry (2000). "Kretschmann Scalar for a [22] Misner, K. Thorne and A. Wheeler, "Gravitation", Freeman and Co. 1975.

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