Is the Schwarzschild Metric a Vacuum Solution of the Einstein Equation?
aa r X i v : . [ g r- q c ] A p r Is the Schwarzschild Metric a Vacuum Solution ofthe Einstein Equation?
Horace Crater ∗ The University of Tennessee Space InstituteJuly 14, 2018
Abstract
This paper examines the inhomogeneous Einstein equation for a staticspherically symmetric metric with a source term corresponding to a per-fect fluid with p = − ρ . By a careful treatment of the equation near theorigin we find an analytic solution for the metric, dependent on a smallparameter ε, which can be made arbitrarily close to the Schwarzschildsolution as ε → This paper examines solutions of the inhomogeneous Einstein equation G µν = R µν − Rg µν = − πGT µν ≡ − κT µν , (1)for static spherically symmetric metrics corresponding to a perfect fluid at restwith p = − ρ . Our aim is to develop an analytic solution for the metric thatcan be made arbitrarily close to the Schwarzschild solution but yet retains anonvanishing contribution to the source term, unlike the Schwarzschild solu-tion. At the same time we find that the second solution with the same sourceterm is arbitrarily close to the de Sitter solution. The Einstein equations forthe metric are second order and highly nonlinear. This implies that if one hastwo independent solutions, then their linear combination will not be a solution.Nevertheless, as we emphasize in the first section, under those special circum-stances for the source term in the Einstein equation, the metric can be writtenin terms of a potential φ which satisfies a very simple linear homogeneous sec-ond order differential equation, with two linearly independent solutions. Oneof the potential-like functions ( ∼ /r ) is the Newtonian gravitational potential ∗ [email protected] ∼ r ) is apotential associated with the de Sitter solution (together they compose whatare known as the de Sitter Schwarzschild- solution [1], [2],[3]). The contributionof the latter to the source term is a constant while that of the Schwarzschildsolution has a vanishing contribution. In order to accomplish our aim, we recastthe inhomogeneous Einstein equation as the limiting case of a closely relatedinhomogeneous equation dependent on a small parameter ε . A careful treat-ment of this problem near the origin leads to source terms with two separatenon-vanishing contributions p = − ρ. In this section we first review the static spherically symmetric standard solutionof the vacuum Einstein equation G µν = 0. The equation is second order in theabove sense and has two solutions with one being the Schwarzschild solutionwhile the other a constant. We then remind the reader how if one adds a sourceterm corresponding to a perfect fluid at rest with p = − ρ a second nonconstantsolution emerges in addition to the Schwarzschild one. The pressure and densityare found to be constants and the second solution is one found originally by deSitter for the Einstein equation with a cosmological constant[6],[7] and is a staticform of the time dependent one used in models of inflation and dark energy inmodern cosmology.For a spherically symmetric solution one chooses the coordinates x = t,x = r,x = θ,x = φ. (2)In a vacuum with static conditions as well as spherical symmetry, we use Dirac’sexponential parametrization of the metric,[4], dτ = e ν ( r ) dt − e λ ( r ) dr − r ( dθ + sin θdφ ) ,g = − e ν ( r ) = 1 /g g = e λ ( r ) = 1 /g ,g = r = 1 /g ,g = r sin θ = 1 /g .g µν = 0 , µ = ν. (3)2ith Γ κµν = g κσ g νσ,µ + g µσ,ν − g µν,σ ) = Γ κνµ , (4)the only nonzero Γ ′ s are [4]Γ = ν ′ e ν − λ , Γ = ν ′ , Γ = λ ′ , Γ = Γ = r − , Γ = − re − λ , Γ = cot θ, Γ = − r sin θe − λ , Γ = − sin θ cos θ. (5)With R νσ = Γ λνλ , σ − Γ λνσ , λ +Γ κνσ Γ λκλ − Γ κνκ Γ λσλ , (6)the diagonal elements of the Ricci tensor are R = (cid:18) − ν ′′ + λ ′ ν ′ − ν ′ − ν ′ r (cid:19) e ν − λ ,R = ν ′′ − λ ′ ν ′ + ν ′ − λ ′ r ,R = (1 + rν ′ − rλ ′ ) e − λ − ,R = R sin θ. (7)From this we have that the scalar curvature R = g µν R µν = − (cid:18) − ν ′′ + λ ′ ν ′ − ν ′ − ν ′ r (cid:19) e − λ + (cid:18) ν ′′ − λ ′ ν ′ + ν ′ − λ ′ r (cid:19) e − λ + 2(1 + rν ′ − rλ ′ ) e − λ − r = (cid:18) ν ′′ − λ ′ ν ′ + 2 ν ′ − λ ′ − ν ′ r + 2 r (cid:19) e − λ − r (8)For our model for T µν we take that of a perfect fluid with pressure p , density ρ , and four velocity u µ [5] T µν = pg µν + ( p + ρ ) u µ u ν ,g µν u µ u ν = − . (9)We work in the frame in which the fluid is at rest, u = , and so g u = − ,u = g u = g ( − g ) − / ,u = − g . (10)Thus, the only nonzero elements of T µν are T = pg − g ( p + ρ ) = − g ρ,T = pg ,T = pg , (11) T = pg . G µν = − κT µν (12)now become G = R − g R = κg ρG = (cid:18) − ν ′′ + λ ′ ν ′ − ν ′ − ν ′ r (cid:19) e ν − λ + (cid:18) ν ′′ − λ ′ ν ′ + ν ′ − λ ′ − ν ′ r + 1 r (cid:19) e ν − λ − e ν r = (cid:18) − λ ′ r + 1 r (cid:19) e ν − λ − e ν r = − κe ν ρ, − κρ = e − λ (cid:18) − λ ′ r + 1 r (cid:19) − r , (13)and G = R − g R = − κpg , − κpe λ = ν ′′ − λ ′ ν ′ + ν ′ − λ ′ r − e λ ( (cid:18) ν ′′ − λ ′ ν ′ + 2 ν ′ − λ ′ − ν ′ r + 2 r (cid:19) e − λ − r )= − e λ ( (cid:18) + 4 ν ′ r + 2 r (cid:19) e − λ − r ) , − κp = e − λ (cid:18) − ν ′ r − r (cid:19) + 1 r , (14)and G = R − g R = − κpg , − κpr = (1 + rν ′ − rλ ′ ) e − λ − − r ( (cid:18) ν ′′ − λ ′ ν ′ + 2 ν ′ − λ ′ − ν ′ r + 2 r (cid:19) e − λ − r ) − κp = ( ν ′ r − λ ′ r ) e − λ − ( (cid:18) ν ′′ − λ ′ ν ′ + ν ′ − λ ′ − ν ′ r (cid:19) e − λ )= − (cid:18) ν ′′ − λ ′ ν ′ + ν ′ − λ ′ − ν ′ r (cid:19) e − λ , (15)and the fourth equation, the one for G , gives nothing new beyond that for G . Hence, the above three simultaneous equations become − κρ = e − λ (cid:18) − λ ′ r + 1 r (cid:19) − r , − κp = e − λ (cid:18) − ν ′ r − r (cid:19) + 1 r , − κp = − (cid:18) ν ′′ − λ ′ ν ′ + ν ′ − λ ′ − ν ′ r (cid:19) e − λ . (16)4hese are three nonlinear inhomogeneous equations for two unknown functions( λ and ν ) of r .For empty space ( ρ = p = 0) these equations become those originally solvedby Schwarzschild, that is. e − λ (cid:18) − λ ′ r + 1 r (cid:19) − r = 0 e − λ (cid:18) − ν ′ r − r (cid:19) + 1 r = 0 , − (cid:18) ν ′′ − λ ′ ν ′ + ν ′ − λ ′ − ν ′ r (cid:19) e − λ = 0 . (17)Combining the first two equations implies that λ ′ = − ν ′ ,λ = − ν + λ ( t ) . (18)The third equation then yields ν ′′ + 2 ν ′ + 2 ν ′ r = 0 . (19)We parametrize the exponential metric function ν by introducing the potential-like function φ , ν = 12 ln(1 + 2 φ ) ,ν ′ = φ ′
11 + 2 φν ′′ = φ ′′
11 + 2 φ − φ ′ φ ) = φ ′′
11 + 2 φ − ν ′ . (20)Thus, using e ν = (1 + 2 φ ) , Eq. (19) becomes φ ′′ + 2 φ ′ r = 0 . (21)This linear second order homogeneous equation is an equidimensional one andhas the general solution of φ = k r + k . (22)Our metric is thus e ν = 1 + 2 k r + 2 k = − g e λ = e − ν +2 λ = e +2 λ k r + 2 k . (23)5n order for the metric to become Minkowskian at r → ∞ we must have k = 0 = λ . (24)Matching g to − − φ for large r where φ is the Newtonian potential − M G/r gives k = − M G, (25)the Schwarzschild radius, and hence the usual Schwarzschild solution of g = − − φ = − M Gr ,g = 1 / (1 − M G/r ) , (26)with the remaining components the same as for free space.The other set of exact solutions and that which is the focus of this paper isfound by assuming that p = − ρ = 0 , (27)so that Eq. (11) gives us T µν = − ρg µν , (28)and the Einstein equation becomes G µν = − κT µν = κρg µν . (29)Just as with the Schwarzschild solution with ρ = p = 0 , combining the first twoequations of (16) implies that ν ′ = − λ ′ ,ν = − λ + ν ( t ) . (30)In this case we absorb the factor ν ( t ) into a redefinition of the time scale usedin the metric. Thus we have ν = − λ and the last two equations of (16) become − κp = − e ν (cid:18) ν ′ r + 1 r (cid:19) + 1 r − κp = − (cid:18) ν ′′ + 2 ν ′ + 2 ν ′ r (cid:19) e ν . (31)Note that these two equations determine both the metric function ν ( r ) and thepressure (and thus the density) so that one does not have a freedom of choicefor the pressure and density beyond Eq. (27).We parametrize the exponential metric function ν by introducing a potential-like function φ just as in Eq. (20) so that the last of the two crucial Einsteinequations in (31) become − κp = − (cid:18) ν ′′ + 2 ν ′ + 2 ν ′ r (cid:19) e ν = φ ′′ + 2 φ ′ r . (32) In [10] a source term for a perfect fluid but with no pressure term is considered. Theydemonstrate that the point charge is a completely stable object, without any ad hoc pressureterms required, and its mass is completely determined by its field interactions. φ ′′ + 2 φ ′ r = − κp = − (1 − φ ) (cid:18) − φ ′ (1 − φ ) r + 1 r (cid:19) + 1 r = 2 φ ′ r + 2 φr (33)This leads to the second order linear homogeneous equation φ ′′ = 2 φr . (34)Note the difference between this equidimensional equation and (21) for the ho-mogeneous case. Thus one has the general solution of a linear combination ofa harmonic oscillator (for positive k ) with the Newtonian potential, φ ( r ) = 12 kr − M Gr . (35)Our metric is thus g = − e ν = − kr − M G/r = 1 /g g = e λ = e − ν = 1 / (1 + kr − M G/r ) = 1 /g ,g = r = 1 /g ,g = r sin θ = 1 /g .g µν = 0 , µ = ν. (36)This corresponds to what is called the Schwarzschild de Sitter space [3]. Withoutthe Newtonian term it corresponds to the solution obtained by de Sitter for theEinstein equation with a cosmological constant[6],[7] if ρ is a constant..As it turns out, the Einstein equations for a perfect fluid plus the equationof state p = − ρ with no assumption about their space-time dependence requires them to be constants. The value of the constant is determined by Eq. (32) − κp = ( φ ′′ + 2 φ ′ r ) = 3 k,ρ = − p = − kκ . (37)The contributions to the pressure and density from the Newtonian part of thepotential vanishes. Note because of this determination that ρ and p from theEinstein equation 12 be constant, the use of Eqs. (29) and (11) implies thatthis is equivalent to starting with the Einstein equation with a cosmological constant Λ and no source term G µν + Λ g µν = 0 , (38)7here Λ = − κρ = 3 k, so that a positive cosmological constant corresponding toa negative density .This vanishing of the source term for the Newtonian portion of the metricis contrary to what is expected based on what occurs in the Poisson equationfor a point mass density ∇ Φ = 4 πGM δ ( r ) =4 πG̺, Φ = − GMr . (39)Is there a point mass at the origin in the case of the Einstein equation? Thiswould seem to be implied by the above Newtonian-Poisson connection. Onemay be tempted to replace φ ′′ + φ ′ r with ∇ φ and proclaim that − κρ ≡ − κ̺/ κp = −∇ φ = − πM Gδ ( r ) − k but for regions that do not exclude the origin(in contrast see Eq. ( ?? ) below) this is not consistent with the other expressionfor the pressure of − κρ ≡ − κ̺/ κp = − φ ′ /r − φ/r = − k . This calls for a more careful treatment of the Einstein equation near theorigin. In order to treat the problem at the origin more carefully and achieve the aimof this paper, we view the Einstein equations in their reduced forms given in(31) as the limit for small ε of the following pair − κp = − e ν (cid:18) ν ′ ¯ r + 1¯ r (cid:19) + 1¯ r , − κp = − (cid:18) ν ′′ + 2 ν ′ + 2 ν ′ ¯ r (cid:19) e ν . (40)where ¯ r ≡ (cid:0) r + ε (cid:1) / . (41)We shall solve these equations instead of the reduced forms (31) of the Einsteinequations and view the proper solutions of the Einstein equation as the limit of Normally one starts with a cosmological constant and then shows the equivalence toa solution of the ordinary Einstein equation in the presence of a perfect fluid with ρ = − p = − Λ κ [11] . In this paper, we start with ρ = − p with no assumption about their spacetime dependence and show that the Einstein equations then force them to be space-timeindependent. This is a subtle difference not emphasized in most text books. Unlike startingwith a cosmological constant where the first of Eqs. ( 31) would be (as in [3]), − r = − e ν (2 rν ′ + 1) which can be readily solved, in the approach given here that first equationcannot be solved since p is an unknown. Instead one must use the second of Eqs. (31) toeliminate p , solve directly for V and then determine p from that solution. The reason for the introduction of ̺ ≡ ρ is that in the limit, − g = 1 − φ → − − κρ = − πGρ = −∇ Φ which disagrees with the ordinary source termin the Poisson equation by a factor of 2 . Thus, we take ρ ≡ ̺/ . ε of the solutions of the modified equations. Thus, as before, using Eq.(20) the two crucial Einstein equations in (31) become − κp = κρ = ( φ ′′ + 2 φ ′ ¯ r ) = − e ν (cid:18) − φ ′ (1 − φ ) ¯ r + 1¯ r (cid:19) + 1¯ r = 2 φ ′ ¯ r + 2 φ ¯ r , (42)and this leads to φ ′′ = 2 φ ¯ r = 2 φ ( r + ε ) . (43)Clearly, one solution is φ ( r, ε ) = κ r + ε ) . (44)Using the connection φ = κ φ Z drφ , (45)between the first and second solution of a homogeneous second order differentialequation, the second solution is φ ( r, ε ) = 2 κ ( r + ε ) κ Z r dr ′ ( r ′ + ε ) , (46)in which both κ and κ are constants. Redefine κ /κ as κ and choose thelower limit to be r = ∞ so that, performing the integration, φ ( r, ε ) = κ ( r + ε ) ε [ 1 ε (cid:16) arctan rε − π (cid:17) + rε ( r + ε ) ] (47)= κ ( r + ε ) ε [ − ε (cid:16) arctan εr (cid:17) + rε ( r + ε ) ]If we let ε → , we should get (to match with the Newtonian solution forlarge r φ ( r,
0) = − κ r Z ∞ r dr ′ r ′ = − κ r r = − κ r = − M Gr , (48)so we take κ = 3 M G . (49)Let us check that with this choice our integrated result (47) has this same limit3 M G r + ε ) ε [ − ε (cid:16) arctan εr (cid:17) + rε ( r + ε ) ] → M G r + ε ) ε [ − ε (cid:18) εr − (cid:16) εr (cid:17) (cid:19) + 1 εr (1 + ε /r ) ] → M G r ) ε [ − ε (cid:18) εr − (cid:16) εr (cid:17) (cid:19) + 1 εr − εr ]= 3 M G r ) ε (cid:20) − ε r (cid:21) = − M Gr φ ( r, ε ) = . κ r + ε ) + 3 M G r + ε ) ε [ εr ( r + ε ) − arctan ε/r ] (50)with the corresponding metric given by g = − e ν = − − φ ( r, ε ) = 1 /g g = e λ = e − ν = 11 + 2 φ ( r, ε ) = 1 /g ,g = r = 1 /g ,g = r sin θ = 1 /g ,g µν = 0 , µ = ν. (51)Now, let us determine what the density and pressure are for the limit ofsmall ε . This will provide us with insight into the nature of the source term.The simplest way is to evaluate κρ ≡ κ̺/ φ ′ / ¯ r + 2 φ/ ¯ r . We obtain κ̺ ( r, ε ) / κ [ ̺ ( r, ε ) / ̺ ( r, ε ) / κ ( 2 r ( r + ε ) / + 1)+ 6 M Gε ( r + ε ) / [1 − rε arctan ε/r ] + 6 M G ε [ εr ( r + ε ) − arctan ε/r ] , (52)where ̺ is the density that arises from the oscillator-like part of the solutionwhile ̺ is the density that arises from the Newtonian-like part of the solution.For r = 0 and ε → κ̺ / → M Gε r [ ε r ]+ 6 M G ε [ εr − ε r − εr + ε r ] = 6 M Gε r [ ε r ]+ 6 M G ε [ − ε r ] = 0 , (53)while for r = 0 and ε → κ̺ / → M Gε (1 − π R and using thedivergence theorem, that we obtain the appropriate constant (independent of ε ). From (42) and (43) we find that 10 Z d r̺ ( r, ε ) /
2= 6
M G Z d r [ 1 ε ( r + ε ) / [1 − rε arctan ε/r ] + 12 ε [ εr ( r + ε ) − arctan ε/r ]]= Z d r [ 2 φ ′ ¯ r + 2 φ ¯ r ] = Z d r ( φ ′′ + 2 φ ′ ¯ r ) . (55)For small ε ( ε/R →
0) this integral becomes → Z d r ( φ ′′ + 2 φ ′ ¯ r ) = Z d r ∇ φ = lim ε/R → R πφ ′ ( R )= 4 π lim ε/R → R M G (cid:20) ε − Rε arctan ε/R (cid:21) = 12 πM G lim ε/R →∞ R ε ( εR − εR + 13 (cid:16) εR (cid:17) )= 4 πGM = 4 πG Z d r̺ ( r, ε ) (56)Such would not be the case for the homogeneous equation which would givezero for the integrated density (see below Eq. (37)). Thus, in the limit ε → ε − modified Einstein equations become the actual Einstein equation,the integral of the density over an arbitrarily small volume remains a constantindependent of ε . So, rearranging the terms in ρ , we define δ ( r ,ε ) ≡ κ̺ ( r, ε ) / πGM = 32 πε [ ε/r (1 + ( ε/r ) ) / (1 + 12 1(1 + ( ε/r ) ) / ) − arctan ε/r ( 1(1 + ( ε/r ) ) / + 12 )] , (57)with the property that Z d rδ ( r ,ε ) = 1 . (58)Our δ ( r ,ε ) therefore has the requisite properties for a distribution that in thelimit represents a Dirac delta function. Its value for r = tends to zero as ε → κ = 0 , thesource term is non-zero and has the property of a sharply confined distribution.What makes this source distinct from others [8], which begin with the matchingsolution of Schwarzschild [9] to an incompressible fluid confined within a finitespherical surface, is that within this sharply confined region, p = − ρ = ̺/ In [10] a different approach also leads to a delta function like source term for the Einsteinequation for static, spherically symmetric circumstances. They assume a dust with no pressureterm present, confined in a radius ε composed of charged particles. They demonstrate thatonly in the ε → κρ = κ̺/ φ ′′ + 2 φ ′ ¯ r = ∇ φ ( r, ε )= 4 πG̺ = 4 πGM δ ( r ,ε ) ,φ ( r, ε ) = . κ r + ε ) + 3 M G r + ε ) ε [ εr ( r + ε ) − arctan ε/r ] → κ r − GMr (59)
We consider in this section the potential-like function φ ( r, ε ) given in Eq. (50).We wish to determine in what sense that, if we choose κ = 0 , the secondportion of φ ( r, ε ) for ε > φ s ( r ) = − GM/r for a given range of r. Let us make this statementprecise. We show (with r s = 2 M G ) that for a positive δ > ,εr s ≡ ǫ < δ. (60)then, (cid:18) rr s (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) φ ( r, ε ) − ( − ) GMr (cid:12)(cid:12)(cid:12)(cid:12) < δ . (61)for r in the range ε < r < ∞ . If δ <
1, then the range for r of agreement betweenthe potentials in the above sense would extend down below the Schwarzschildradius with no upper bound.To show this we consider the Taylor series for φ ( r, ε ) − φ s ( r ) in ε about ε = 0 .φ ( r, ε ) − ( − ) GMr = 3
M G r + ε ) ε [ εr ( r + ε ) − arctan ε/r ] + GMr (62)12he series for arctan ε/r converges for r > ε . Expanding we find that φ ( r, ε ) − ( − ) GMr = 3
M G ε [ εr − r (1 + ε r ) ∞ X n =0 ( − ) n n + 1 (cid:16) εr (cid:17) n +1 ] + GMr = 3
M G ε " − ε ∞ X n =0 (cid:18) n + 3) (2 n + 1) (cid:19) ( − ) n (cid:16) εr (cid:17) n +1 + GMr = − M Gε [ (cid:16) εr (cid:17) ( ∞ X n =0 ( 1(2 n + 3) (2 n + 1) )( − ) n (cid:16) εr (cid:17) n ] + GMr = − M G [ (cid:18) r (cid:19) ( 13 + ∞ X n =1 ( 1(2 n + 3) (2 n + 1) )( − ) n (cid:16) εr (cid:17) n ] + GMr = − M Gr [ ∞ X n =1 ( 1(2 n + 3) (2 n + 1) )( − ) n (cid:16) εr (cid:17) n ]= 3 M Gr [ ∞ X n =0 ( 1(2 n + 5) (2 n + 3) )( − ) n (cid:16) εr (cid:17) n +2 ] < M Gε r (63)Thus, with εr s = ǫ < δ, we have since ǫ r s r < (cid:18) rr s (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) φ ( r, ε ) + GMr (cid:12)(cid:12)(cid:12)(cid:12) = 12 (cid:18) rr s (cid:19) (cid:12)(cid:12)(cid:12) rGM φ ( r, ε ) + 1 (cid:12)(cid:12)(cid:12) = 3 r r s [( ∞ X n =0 ( 1(2 n + 5) (2 n + 3) )( − ) n (cid:16) εr (cid:17) n +2 ]= 3 ε r s [( ∞ X n =0 ( 1(2 n + 5) (2 n + 3) )( − ) n (cid:16) εr (cid:17) n ]= 3 ǫ ∞ X n =0 ( 3(2 n + 5) (2 n + 3) )( − ) n (cid:16) ǫ r s r (cid:17) n ]= ǫ ∞ X n =1 ( 152 (2 n + 5) (2 n + 3) )( − ) n (cid:16) ǫ r s r (cid:17) n ! < ǫ < δ . (64)So, we have demonstrated how the Schwarzschild solution can be viewed as avalid approximation for a bona fide nonlinear solution of the full non-homogeneousEinstein equation. We have found that the general solution of the Einstein equation for the specialcase of p = − ρ ≡ ̺/ k ) , two inde-pendent solutions of a linear second order differential equation for the potential-like function φ . If the behavior about the origin is not handled carefully, the13 r contribution gives a constant density and pressure while the Newtoniancontribution gives rise to no point-like (delta function) source. With κ = 0 , this is the usual no source or vacuum solution. Handled more carefully by themethod we present in this paper we obtain density and pressure terms sharplypeaked about the origin, with a unit volume integral.Do our mathematical solutions of the inhomogeneous Einstein equation haveany physical significance? Superimposed on our point-like density is a densityranging between two constants ( κ and 3 κ ; see Eq. (52)), the source of theharmonic oscillator potential-like function φ ( r, ε ) which behaves like κ r / r. The static and spherically symmetric metric we started withis distinct from the standard time-dependent Friedmann-Lematre-Robertson-Walker (FLRW) metric. It is not intended to relate to the universe as a wholebut rather to the field produced by a single source. There is some superfi-cial similarity between our solution and the so-called dark energy and inflationsolutions of modern cosmology for κ < φ we obtain is not only proportional to 1 /r but also tothe otherwise ubiquitous r potential correspond to the solution of the Einsteinequation under these circumstances. A positive or negative sign of κ wouldgive a negative or positive pressure and an attractive or repulsive force thatwould increase in magnitude with distance (mimicking the effects of dark en-ergy). Note that in nonrelativistic potential theory a constant density wouldgive rise to an attractive Hooke’s law force. However the context is entirelydifference. There, the Hooke’s law form follows from Gauss’ law applied toan inverse square field. The r potential like function discussed in this paperis completely independent of the 1 /r potential. Another factor to point outis that the functional form of the density or pressure correlated with the r potential is fixed by our solution to the Einstein equations themselves, it is notimposed. The only imposition we made on the density and pressure is thatthey be the negative of one another. From a mathematical point of view thereis no distinction between the solutions discussed here for the inhomogeneousEinstein equation and the one we would have obtained by adding a term + ρg µν to the left hand side of the Einstein equation and viewing it as an addition tothe equation, in analogy to the alternative explanation of dark energy. Thedifference here is that ρ being a constant would be an outcome of the modifiedEinstein equations and not an imposed functional form.It was one of Einstein’s early goals, although he never succeeded, to incor-porate Mach’s principle in his general theory of relativity. It has been generallyregarded that general relativity does not embody Mach’s principle, that is thatgeometry can exist independent of matter. It was the Schwarzschild solutionthat seemed to bring this idea its early but reluctant acceptance. That is, ageometry can arise in the absence of a source term, from the vacuum. Ofcourse, in the practical applications of the Schwarzschild solution to the preces-sion problem of Mercury and the bending of light, it was always assumed thatlooming behind the formal sourceless equation was a real sun. Nevertheless, apossible formal interpretation has been that a curved space exists without an14dentifiable source, thus obviating the need for Mach’s principleOur result has been to replace the Schwarzschild solution to the sourcelessspherically symmetric static environment, which then, as now seems to allowthe existence of non-trivial spacetime curvature in absence of any matter, witha solution that does not correspond to a sourceless environment but yet leadsnevertheless to a metric that can approach the Schwarzschild with arbitraryaccuracy in an asymptotic way. In doing so, for this particular case at least,Mach’s principle, the idea that geometry emerges as an interaction between anidentifiable matter term and geometry is preserved[12]. Acknowledgement 1
The author acknowledges helpful correspondences withProfessor M. Sachs and useful suggestions from Professor G. Longhi, Dr. L.Lusanna, Dr. C. Powell, J. Labello, and S. Rubenstein
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