Proc. NatL Acad. Sci. USA Vol. 79, pp. 6404-6408, October 1982 Physics
Gravitational field of a charged mass point (exactly linear equations in general relativity/finite electromagnetic mass/square root of Poisson equation/vanishing Schwarzschild radius) C. L. PEKERIS Department ofApplied Mathematics, The Weizmann Institute, Rehovot, Israel Contributed by C. L. Pekeris, June 14, 1982 ABSTRACT Adopting, with Schwarzschild, the Einstein gauge One may consider, in the abstract, the metric of a mass-less (lg,,l = -1), a solution ofEinstein's field equations for a charged point charge, similar to Schwarzschild's treatment of a charge- mass point ofmass M and charge Q is derived, which differs from less mass point. The gravitational field is then due entirely to the Reissner-Nordstr0m solution only in that the variable r is re- the mass equivalent of the electric energy distribution, which placed by R = (r3 + a3)'3, where a is a constant. The Newtonian is everywhere as positive as is the mass M in the Schwarzschild gravitational potential 4' (2/c2)(l - goo) obeys exactly the Pois- solution. By putting M equal to zero in the Reissner-Nordstr0m son equation (in the R variable), with the mass density equal to solution, we are led to the conclusion that in the Newtonian (E2/4w2), E denoting the electric field. qk also obeys a second approximation, the gravitational acceleration is everywhere di- linear equation in which the operator on 4, is the square root of rected radially outwards! the Laplacian operator. The electrostatic potential 4 (= Q/R), be shown in the sequel that the solution to this paradox A, and all the components of the curvature tensor remain finite It will at the origin of coordinates. The electromagnetic energy of the is the fact that for a given charge it is physically impossible to point charge is finite and equal to (Q2/a). The charge Q defines make the mass M in the Reissner-Nordstr0m solution vanish, a pivotal mass M* = (Q/G"/2). If M < M*, then the whole mass because the charge, by itself, generates an electromagnetic is electromagnetic. If M > M*, the electromagnetic part of the mass which is part ofM or constitutes the whole ofthe mass M. mass M,,, equals [M - (M2 - M*2)1/21, whereas the material part The electromagnetic mass vanishes only when the charge of the mass M,,,, equals (M2 - M*2)1/2. When M > M*, the con- vanishes. stant a is determined, following Schwarzschild, by shrinking the In the following a solution of Einstein's field equations for "Schwarzschild radius" to zero. When M < M*, a is determined a charged mass point is derived, in which the gravitational field so as to make the gravitational acceleration vanish at the origin. stemming from the mass equivalent of the electric energy dis- tribution is everywhere attractive, as it should be. The solution is based on the method used by Schwarzschild in deriving his 1. Introduction solution for the metric of a neutral mass point (3). The line element in the case ofspherical symmetry and time Writing the metric in the form independence is d&2 = e (dx0)2 - eA(dxl)2 - e"(dG2 + sin2Gd42), [5] ds2 = e c2dt2 - eAdr2 - r2(d62 + sin2Od4)2), [1] where V, A, and ,u are functions of r. Schwarzschild adopts the where v and A are functions of r, the Reissner-Nordstr0m so- coordinates lution of Einstein's field equations for a charged mass point is given by x0 = ct, xi = (r3/3)x= x2 =-cosO, x3 = O [6] 2GM GQe2 whereby Eq. 5 becomes e =-2 + 42r2 =e-e, [2] crcr dS2 = evc2dt2 - eAdx2 - eA(d02 + sin20d42). [7] where M denotes the mass and Q the charge ofthe particle (1, In the case of spherical symmetry and time independence, 2). The + sign in front ofthe last term in Eq. 2 is puzzling. We only the diagonal components of the Einstein tensor G. are know that the solution of Einstein's field equations for a geo- different from zero. Moreover, because G2 = G3, Einstein s desic in the case ofcircular orbits yields a centrifugal force (v2/ field equations provide three equations for the determination r), which is given exactly (and not only in the Newtonian ap- of the three functions v, A, and A. However, because of the proximation) by Bianchi identities, these three equations are not independent. v2 c2 dev Hence, one additional relation is required. It is the additional relation r -=-2 dr ~~~~~~~~~~[3] customary (4) to take as en=r2, [8] From Eq. 2 we have by arguing that we are at liberty to choose as the r-coordinate v2 GM GQ2 any function of r-namely, e">. However, this choice is not r =r2 2c2r32 ~~~~~~~~~[4]mandatory. Schwarzschild adopts instead ofEq. 8 the "Einstein determinantal equation" (5) The second term in Eq. 4 would imply that the electric charge creates a repulsive gravitational force, which is strange. IgMvI = g = -1, [9] or by Eqs. 5 and 6, The publication costs ofthis article were defrayed in part by page charge I + A + 2, = 0. [10] payment. This article must therefore be hereby marked "advertise- ment" in accordance with 18 U. S. C. §1734 solely to indicate this fact. It can be shown from his analysis that for the case ofempty space 6404 Downloaded by guest on October 1, 2021 Physics: Pekeris Proc. Natl. Acad. Sci. USA 79 (1982) 6405 and under the gauge of Eq. 9, the function tk obeys the differ- be shown to reduce in our case to the diagonal form ential equation = -F"aF_ - e2 d2(e3/2M) 4rc2Tg V ((,i2 [21] = [11] ~~~2 dx2 from which it follows that enAe =R°2, [12] I TI = -T2 = -T3 = T = e2F(')2 - - 2 -2 [22] where 1 2 3 ~~8,iTC2 87Tc2 R = (r3 + a3)1"3, [13] 3. Einstein's field equations for a charged mass point and a is an arbitrary constant. Note that Eq. 8 is a special case of Eq. J2 valid for a = 0. Einstein's field equations Schwarzschild's solution for a neutral mass point is (equation G.,> = - = - [23] 14 of ref. 3) RAP. 2 RgAV KTAV where the gravitational coupling constant K iS given by = (- )c2dt2 - (1- >dR2 K = (87G/C2), [24]
can be taken from the paper by Dingle (8), who worked out the - R2(d62 + sin26d42), [14] components ofthe Einstein tensor GM3, for the case of a diagonal metric. Denoting by dashes differentiation with respect to x, where R is given in Eq. 13, and m = (GM/c2). He sets a = 2m, we obtain so as to shrink the "Schwarzschild radius" to zero.t P2 mlIKI =-;eA -A A + A [25] 2. The electromagnetic energy-momentum tensor (e4 2 )
The Metric tensor for the line element [7] and the Schwarzschild -KT2 = 3KT3 = e1 + 1ii coordinates [6] is
gi, = -eA, g22 = -eM(sin6)-2, g33 = -eysin20, goo = e' [15] -+ 4' - A - A'v' + i2 + 'V')I, [26] -e-A g22 = -eAsin2o, g1= p ek 4. 3k A4 g33 >P = _e-M(sin)-2, g00 = e [16] [27] In the electromagnetic field equation On subtracting Eq. 27 from Eq. 25, and using Eq. 10, we [17] get V ax ax., 3 2 d2(e32M) only the electrostatic potential 00 (-F) is different from zero. K(T - To) = e-A(Ac \2 [28] Here (D is a function of r. It follows from Eq. 17 that the only 2 3 dx2 surviving components of the electromagnetic field tensor are It follows that when (T' - T) vanishes, as in Schwarzschild's axF case of a neutral mass point or in our case of a charged mass = = [18] point, when the energy-momentum tensor is given by Eq. 22, F0 -F10 -x the expression d2(e3/2L)/dx2 vanishes, and consequently e" is If R'A" denotes the electromagnetic tensor density, then the given by Eqs. 12 and 13. Eq. 22 now reads condition of the vanishing of (aa"/dax3) in empty space re- 2~~~~~E duces, in our case ofspherical symmetry, to the single equation TI = -T2= -T3= To= 81Tc2R4' [29] (7) whereas Eq. 20 becomes aR= (googFol g) d(F 1 d(F E ax, ax [30] d dl d(D\ dx R2 dR R4' = - d(e-A Fol)= - e2at 10. [19] dx dx \~dxl (F=-. [31] The integral of Eq. 19 is R d (F approaches asymptotically (c/r), so that we can identify the go=-Ee- [20] constant E with the electric charge Q. dx We note that at the origin of coordinates (r = 0), where R E being a constant. is equal to a, the electrostatic potential 4) does not become in- The electromagnetic energy-momentum tensor T"', given finite but has the finite value of (Q/a). by 41rc2TAMV = -9g3FILaF;, + '14 e"FPF (Gaussian units), can It will be convenient henceforth to change from the variable x (=r3/3) to the variable R [= (r3 + a3)1/3]. Eq. 26 then becomes t I am indebted to N. Rosen for bringing to my attention the paper by d2e" 2 de" 2GQ2 Abrams (6), where the vanishing of the Schwarzschild radius in the original Schwarzschild solution is stressed. dR2 R dR c4R4' [32] Downloaded by guest on October 1, 2021 w406& A Physics: Pekeris Proc. Natl. Acad. Sci. USA 79 (1982) while either one of Eqs. 25 or 27 leads to With Eq. 44, Eq. 38 takes on the form de" e" 1 GQ2 eV= R-2(R - a)(R - mem). [45] dR R R c4R3 Since, by Eq. 44, a is greater than me,n, and furthermore, since These are two linear equations for the same function ev. How- the minimum value of R is a, it follows that ev is positive every- ever, the equations are not independent, the operator on e" in where and that it vanishes only at the origin r = 0. Eq. 32 being the square of the operator on ev in Eq. 33: The term gravitational radius is applied to the radius where /d 1\ goo vanishes. We shall call the constant a also the gravitational Eq. 32= d+ - Eq 33, [34] radius, because in our solution the singularity occurs at R = a. It should be noted that neither the material mass M, nor where Eqs. 32 and 33 represent either side of Eqs. 32 and 33, the electromagnetic mass Mem is directly observed but only respectively. Eq. 34 is a manifestation ofthe Bianchi identities. their sum M We take as a solution of Eqs; 32 and 33 M = Mmat + Mem, [46] 2GMe, GQ2 Therefore, we cannot at this stage evaluate the gravitational e1v 1 2GMmatc211 c2Rc [35] radius of the electron, for example, for which Q equals the el- ementary electric charge e. Indeed, it follows from Eq. 44 that where MMt is the material component of the mass, and Mem m-m,,,lt + mem = t + denotes the integrated mass equivalent of the electric energy (m2 K2)12, [47] distribution, showing that Eq. 44 applies only to cases where m 2 K, or Mem = Q2/(c2a), (GQ2/c4) = (GaMem/c2). [36] M M* =Q(G ). [48] The term [-2GMem/(c2R)] in Eq. 35 was chosen so as to nnake In the case of the elementary charge e, M* is equal to 1.86 x the gravitational field stemming from the distribution of celec- 10-6g, which is greater than the mass ofthe electron by afactor trostatic energy everywhere attractive, as will be shown lhater. of 2.04 x 1021. Writing For masses less than M* (m < K), Schwarzschild's criterion for determining the gravitational radius is inapplicable because = = mtt (GM",,.t/c2), mem (GMem/c2) [37] the function ev has no real root then, as becomes clear when we in Eq. 35, we obtain write Eq. 38 in the form 2mmat 2mem amem ell ~~+ =[(- )2 +(K2-m2)] [49] R R R2 2m am~w It is seen from 1, as well as from 56 and 57 below, = 1- Rm + m2 m = (mmat + mem). [38] Fig. Eqs. R R2 that starting with large values of M and going in the direction It follows from Eq. 10 that ofdecreasing M, the material component of the mass M,,,a de- creases, whereas the electromagnetic component increases. At eA =R-4e-P, [39] M = M*, Mmat^, vanishes, whereas Mem = M. Therefore it is rea- and noting that sonable to assume that when M < M*, the entire mass is elec- tromagnetic. With the electrostatic potential equal to (QI dx' = dx = R12dR, [40] R), by Eq. 31, and therefore continuous everywhere, the elec- we obtain from Eq. 5 the line element for a charged mass point trostatic energy density is finite everywhere. Furthermore, because all of the equivalent mass. is located outside the origin ds2 = evc2dt2 - e-vdR2 - R2(dO2 + sin26d42), [41] and is spherically symmetrical, it follows that the gravitational acceleration must vanish at the origin. Now it can be shown that where ev is given by Eq. 38. in the case ofcircular orbits, with v = R(d4/dt), a relation sim- 4. The gravitational radius ilar to Eq. 3 holds exactly: v2 C2de" [50] The solution (Eq. 35) contains an arbitrary integration cons tant R 2dR a, that appears in two places: in the expression for R in Eq1. 13 we as to for a and in the expression for the electromagnetic mass Me,,, in Eq. which take the criterion be applied determining < 36. In the case of a neutral mass point, Schwarzschild put a in the case when M M*. By Eq. 38, = 2m, in order to push the imminent "Schwarzschild singi ular- dev 2m 2K2 - --= = ity" to the origin ofcoordinates (condition 4 in ref. 3). We ;shall =W- =0RRO1 R a, [51] follow Schwarzschild and determine a so as to make e" vamish
at the origin r = 0, where R = a. Putting e' = 0 in Eq. 31 8 for K2K Q2 R = a, we obtain a quadratic equation for a, a = = -2'I M < M*, [52] m Mc2 a2 - 2mmta - K2 = 0, [42] 2m am elv= 1- +-W M a a* I E 6I E FIG. 1. The electromagnetic mass I Mem and the material massMnat as func- tions of the total mass M. M* = Q/ (G112). Q denotes the charge and G the gravitational constant. a denotes the gravitational radius; a* = (GM*/c2). M -, M,,,,; -, Mem;.. , (a/a*). case M > M* when, by Eq. 44, being that the electric energy distribution generates an equiv- alent mass density pe(R) which is equal to (E2/4irc2). [55] with the Newtonian gravitational po- a = 2[M+M*2)M/2],(M2- M > M*. However, the analogy tential is only partial, because general relativity requires that qi satisfy, in addition to the Poisson Eq. 59, a second equation, We also have namely, Mmat = (2 - M*2)112, M > M*, [56] doi GQ2 +p [61] Mem = [M - (2 -M*2)/2], M > M*. [57] dR R 2c2R323GRpe(R),= operator on 4i in Eq. 61 is the It follows from Eq. 57 that (dMem/dM) is negative; hence, M* which follows from Eq. 33. The square root the operator on 4i in Eq. 59. A solution of Eqs. mass which of (=Q/G1 2) is the maximum value ofelectromagnetic 59 and 61 is a point charge ofcharge Q can have. The solutions [38] and [53] for the cases M > M* and M < GMmat GMem GaMem __= + [62] M*, respectively, become identical with the Reissner-Nordstr0m R R 2R2 solution (Eq. 2) if we replace in the latter r by R. What our discussion has brought out is that when M > M*, the mass M which, by Eq. 50, gives consists of the sum of a material component M,,., and an elec- V2 _ d M GMem a V dq Glmat + _ tromagnetic component Myl, given by Eqs. 56 and 57, respec- [63] R dR R2 R2 R tively. When M < M*, the entire mass is of electromagnetic origin. The gravitational field due to the electromagnetic mass Mem is therefore everywhere attractive, as it should be. 5. The Newtonian approximation For large values of (r/a), when R -> r, the gravitational ac- celeration in the Newtonian approximation equals (dqi/dr), Let us define a function 4', later to be identified with the New- which, by Eq. 62, is tonian gravitational potential, by do4 r2 [GMmat + GMem 1 a eV= 1- (2I/c2), [58] dr R2 L 12 R2 R without, as yet, restricting the gravitational field to be weak. G(Mmat + Mem) Substitution of Eq. 58 in Eq. 32 gives --3 [64] d2qi 2 d4 + -41rGp(R), [59] The gravitational acceleration due to the charge is thus directed radially inwards. where 6. Curvature Q2 E42 [60] p(R) =Pe(R) =s24 4C By using an orthonormal frame and E denotes the intensity of the electric field. Since for r ct = e1"2"cdt, wd = el'A dr, [65] >> R -* r, 59 shows that the function asymp- a, Eq. 4' obeys we = rd6, ad = r sinOdO, [66] totically the Poisson differential equation which is satisfied by the Newtonian gravitational potential 4', the only modification the nonvanishing components ofthe Riemann curvature tensor Downloaded by guest on October 1, 2021 6408 Physics: Pekeris Proc. Natl. Acad. Sci. USA 79 (1982) for the metric (Eq. 41), with ev given by Eq. 38 and R by Eq. Table 1. Properties of the metric in cases I and II, when M > M* 13, are for M > M*: and M < M*, respectively 1 d2e" 2m I HI 3amE?,, [67] 2 dR2 R3 R4 M>M* M