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Gravitational Field of a Charged Mass Point 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.
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