The field equation, black holes and vacuum energy

In a previous article the transformation of space and time from an inertial system to another inertial system were obtained without using rigid measuring-rods and clocks as primitive entities. In this work the field equation is derived for the new transformation as well as the necessary mass for an object inside the black hole and in the neighborhood of the event horizon to reach the center (core) of a black hole. Using the new transformation the value of the vacuum energy density is found to be equal to 1.016124 × 10J m−3 and the contribution of the vacuum mass for the field equation is found to be 1.58 × 10−26[kg m−3] for the Hubble parameter equal to 67.6 [km s−1Mpc−1].


Introduction
In Ferreira [1] transformation of space and time from an inertial system to another inertial system were obtained without using rigid measuring-rods and clocks as primitive entities. Section 2 shows the transformation obtained.
In sections 3 to 4 the energy-momentum tensor T and the field equation are derived with the transformation obtained in [1]. The invariance of the speed of a particle in the transformation is used to obtain the energy-momentum tensor.
Sections 5 and 6 are on black holes. It is assumed that black holes have no singularities, instead they have cores. The transformation obtained in [1] is used in a black hole with zero angular moment to determine the time for a particle to go from the event horizon to the core of the black hole. It is assumed that a core has a radius equal to Planck's length in the center of the black hole. The necessary mass for an object inside the black hole and in the neighborhood of the event horizon to reach the core is given in section 6.
Weinberg [2] states that anything that contributes to the energy density of the vacuum acts just like a consmological constant. In this work energy density of the vacuum and the cosmological constant are distinct things. I section III of [2] we found for a class of cosmological models described by Friedmann. Section 7 determines the value of the vacuum energy density, which is found to be equal to 1.016124 × 10 113 J m −3 . Sections 8 and 9 show that the value of the contribution of the vacuum mass density (denoted by ∆ρ v ) for the field equation is not constant and calculates ∆ρ v which is found to be 1.

The Transformations
Let S be a reference system moving with respect to the reference system S with constant speed u in the direction x+. The axis-y and the axis-z are parallel to axis-y and axis-z, respectively. Let us suppose that the zero of t coincides with the zero of t and the origin of x , y , z coincides with x, y, z when t = 0. The Lorentz transformation is and The transformation found in Ferreira [1] is and t = 1 − u 2 c 2 1/2 t (10)

Energy-momentum tensor
In the special relativity the interval s ab between two events in system S is defined by and s ab 2 has the same value when calculated in the system S . In the differential form (11) becomes where x 0 = ct, x 1 = x, x 2 = y and x 3 = z Let the relation between time t in the system S and time t in the system S be presented in [1] in the differential form the equation (13) can be rewritten as for a particle traveling with constant velocity u in S (here we assume that S is moving in relation to S with velocity u and u 1 = u). Substituting u by its components we have Since and using (13) we have or Let be the cases in (12) where the differential interval components dx 2 and dx 3 are zero. Comparing (12) with (19) we see that with the transformation given in (13) other inertial observers will not find the same value for ds 2 . Let us consider the vector dx = (dx 0 , dx 1 , dx 2 , dx 3 ), the time increment dt and define the four-velocity vector This vector is invariant in the proposed transformation in [1]. The vector U can be rewritten as where The mass m in the system S becomes γm in the system S . Multiplying by γmc both members of the module of (21) we obtain γmcu = (γmc 2 , γmcv) (23) Now consider the general case In the system S the volume contraction and the mass increase change the mass density where ρ 0 is the mass density in the system S. Since u 0 = c and for T µµ with µ = 0 we have where g will be given in section 4, for the time being g = 1. For a perfect fluid (a special case) there is no heat conduction (T 0µ = T µ0 = 0 for µ = 0) and the fluid has no viscosity (T µν = 0 for µ = ν).

The field equation
The equation describing how matter generates gravity is modeled by Provided that k is a scalar we have that G is a tensor of the same rank of T and symmetric. The tensor T has zero divergence to satisfy the law of conservation of momentum-energy Like T the tensor G must be a divergence-free tensor. When the space and time is flat the tensor G vanishes. Only the Riemann curvature tensor and metric construct G and the tensor G must be linear in the curvature. With these requirements G is unique, it is called the Einstein curvature tensor. The field equation then is We assume that k is a constant real number and that for v << c the equation (33) becomes the Newtonian gravity theory. Poisson's equation for Newtonian gravity is or where Φ is the gravitational potential, G is the Newtonian gravitational constant, G is the gravitational parameter found in [1] and ρ is the mass density. The path of a free-falling particle for v << c satisfies the equation of motion Using (33), (35) and (36) we can determine the value of k and the field equation becomes The Newtonian gravitational constant is Substituting into (37) g by γ 2 and G by G we obtain where and for T µµ with µ = 0 we have

On black holes
Consider (i) a spheric object of mass M with its center in the origin of the system S and (ii) an observer stationary at a distance d from the object such that the gravitational attraction exercised by the object on the observer is negligible. The gravitational attractions on the object and the observer by other masses of the system S are negligible. There are no other masses for a distance from the mass M less than d. In this case the metric in this region can be expressed as The general form of an isotropic metric is Let us consider radially moving particles and we have for a time-independent metrics we obtain Around the object of mass M at a distance d the space is empty and T is zero. Therefore G is zero too. Since G is expressed in terms of the Ricci curvature tensor with Γ µ να = 1 2 g µβ g βα,ν + g βν,α − g νβ,α From (46) to (50) we can determine A(r) and B(r) and For a weak field Therefore and The (56) metric has a singularity at The radius given in (57) is called Schwarzschild radius. Now let us consider the object at the origin of the system S at a time t 0 − ∆t. Let us suppose that the mass of the object considered is M − ∆M at t 0 − ∆T and at the instant t 0 a mass ∆M has been integrated to the mass of the object. Let be t 1 > t 0 . Let us suppose that the radius of the object of mass M at t 1 is equal to the Schwarzschild radius, the mass density of the object is uniform, immediately after t 1 , when its volume is being collapsed. The volume variation is given by The differential dV sph is in contact with the surface of the object. Since where R is the Schwarzschild radius of the object. From (60) and seeing that and The minus sign in (61) is because the differential volume moves inward The equation (64) gives the time for the contraction of the volume of mass M for the observer in the system S. Also, this is the time spent for a particle to go from the event horizon to the distance to the center of the black hole equal to Planck's length. The time differential dt in the volume dV sph is and the time in S for the contraction associated to the volume dV sph is The reason of the integration upper limit not to be zero is given in the next section.
6 The core of the black hole Let γm p be the mass of the particle, ∆r the distance of the particle to the center of the black hole and ∆t = ∆r c Therefore the contraction from R to l P lanck requires an object near to the surface of radius R with mass equal to 1.088255 × 10 −8 kg (or 6.104659 × 10 18 GeV ) to reach the core. The ratio M/n by the volume of the core is 3M 8nπ(l P lanck ) 4 = 6.154000 × 10 95 Kg/m 3 (77) and the energy density is 3M c 2 8nπ(l P lanck ) 4 = 5.530939 × 10 112 J/m 3 This suggests that the black hole masses are integer multiples of (77).

The vacuum energy
The classic Casimir result [4] for the force between perfectly conducting parallel plates is where a is the distance between the plates. The difference of the energy density outside the plates and inside (between) the plates is The energy density of a black hole of mass M minus the vacuum energy density is 8 On the constancy of the vacuum energy Let us consider the Friedman's equation or where ρ is the matter density inside a sphere of radius r which is expanding about its center, m is the mass of a particle on the surface of the sphere, G is the Newtonian gravitational constant, R is the scale factor, H is the Hubble parameter, k is a dimensionless constant related to the curvature of the universe, c is the speed of light and x is where V is the volume of the sphere. The equation (90) along with (92) governs the evolution of the scale factor of the universe. Let us substitute ρ by ρ + ∆ρ v in (90) 9 The value of ∆ρ v Let us consider a control volume, denoted by cv1, with volume V which has energy density equal to ρ v and is moving with velocity v in system S. This control volume is stationary in the system S . In S the volume is contracted to