A Class of Exact Interior Solutions of the Einstein-Maxwell EquationsAuthor(s): J. N. IslamReviewed work(s):Source: Proceedings of the Royal Society of London. Series A, Mathematical and PhysicalSciences, Vol. 353, No. 1675 (Apr. 21, 1977), pp. 523-531Published by: The Royal SocietyStable URL: http://www.jstor.org/stable/79211 .Accessed: 19/04/2012 10:14

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Proc. R. Soc. Lond. A. 353, 523-531 (1977)

Printed in Great Britain

A class of exact interior solutions of the Einstein-Maxwell equations

By J. N. ISLAM

Department of Applied Mathematics and Astronomy, University College, Cardiff, U.K.

(Communicated by S. W Hawking, F.R.S. - Received 28 May 1976)

A class of exact interior stationary solutions of the Einstein-Maxwell equations is found in terms of an arbitrary solution of the flat-space Laplace equation. These solutions represent pressure-free charged matter rotating with constant angular velocity about an axis of symmetry. Some properties of the solution are discussed.

1. INTRODUCTION

Van Stockum (I937) found a class of interior solutions of the Einstein equations representing pressure-free dust rotating with constant angular velocity about an axis of symmetry. These are expressible in terms of a harmonic function, i.e. solu- tion of the axisymmetric flat-space Laplace equation. This solution has recently been extended to the case of differential rotation by Winicour (I 97 5) . In the present paper we extend Van Stockum's solution to the Einstein-Maxwell equations, i.e. we find a class of interior solutions representing zero-pressure charged matter rotating with constant angular velocity about an axis of symmetry. This new solution also depends on a single harmonic function.

In ? 2 the field equations are written down and some aspects discussed. The solu- tion is given in ? 3 and some properties of the solution are described in ? 4. An outline of the derivation of the solution is given in the appendix.

2. FIELD EQUATIONS

The Einstein-Maxwell equations are as follows:

Rlz = 8ic{tT v + Tlv - 2g1tv (T' +T`)},( = 1 1I(1)

F; v + Fviy;j +i FJ; v =0, (2)

F1v; v = 4cJJ/t, (3)

where a semicolon denotes covariant differentiation, R,U and F,l, are respectively the Ricci and the electromagnetic field tensor and JIs the current. T,U and T,p are respectively the energy-momentum tensor of the matter and the electromagnetic

[ 523 ]

524 J. N. Islam

field, with T' = T'Y, T' = T"'. We have set the gravitational constant and the velocity of light equal to unity. T'"# is given as follows:

T -I" = I ( -Yv + kg"v' F,Fafl). (4)

Note that T7'? 0. We consider the matter as consisting of non-interacting charged particles of mass m and charge q. Let n be the number density of the particles. T'itv is then given by the following:

T/MPv-= n,U/lul, (5)

where u/1 is the four-velocity of the particles. Equation (5) reflects the fact that there is no pressure, and that the mass-energy density consists of the rest masses of the particles, since there is no internal energy due to random motion.

We proceed to verify the conservation of the total energy-momentum given by the equation (T'klv + T"'); p = O. (6)

With the use of (2) and (3) one can show, through a standard procedure, that

T "P, -v ;, v == - F/, PJv. (7)

Further, T'fV ; v - m(nuv); u/ i+?mnuvulA; (8)

The first term on the right hand side of (8) vanishes because of conservation of matter. In the second term

U,; VUP = {uiuj/xP + rV uo}uP = dul/ds + rIV uou' = Duj/d8. (9)

With the use of the following equation of motion for a charged particle in the field

Du'A m d =qu F"l, (10)

it is easily seen that the total energy-momentum tensor satisfies the conservation equation (6) provided J' = qnu/t. The conservation of current JP 0 is satisfied by virtue of the conservation of particles.

The discussion so far is quite general, in the sense that it is independent of any space-time symmetries of the metric. To specialize to the case we are interested in we start with the most general axisymmetric stationary metric

ds2 = fdt2 - 2kdOdt - 1d02 - elt(dp2 + dz2), (1 1)

wheref, k, I and u are all functions of p and z only. Setting (Xo, X1, X2, X3) = (t,p, z, 0), the components of the four-velocity of the rotating dust are as follows

0dt 21- U= 2 3dO = - = (f-2Q1c-Q2k)- , u1 = 0 u2 = 0 u3 = - = Qu0 (12) ds ds

where Q is the constant angular velocity. It can be verified, since all the functions are independent of t and 0, that the equation of continuity (nul");/, = 0 is satisfied identically. FAIV is given in terms of a four-potential A, as follows

F, = -A ^, (13)

Interior solutions of Einstein-Maxwell equations 525

where a comma denotes partial differentiation, in which case (2) is satisfied identi- cally. In the present case the components of A4# can be taken as follows

(Ao A1, A2, A3) = (v, ?, 0? X) (14) where 0 and X are related to the electric and magnetic potentials respectively. The The t- and 0-components of (10) vanish identically, while the p- and z-components give the following equations

lm(fp -2Qkp -Q21p) (f-2Qk-Q2j)-i = q($ + QX ), (15a)

m (fz-2Qkz - Q21z) (f- 2Qk - Q21)- = q((bz + QXy) (15b)

withfp =_ f/8p, etc. If Q is assumed to be constant, (15 a, b) imply

f-2Qk-Q21 = (- (V+QX)+B} (16)

where B is an arbitrary constant. An important combination of the field equations is the following

e/tID(RO+R3)=Dp +Dzz = 0, (17)

vhere D2 = fl + k2. By a standard procedure due to Weyl (I 9I7), the application of wrhich to the stationary vacuum Einstein equations was first noted by Lewis (I 932),

one can introduce a coordinate system, without destroying the form (11) of the metric, such that D = p.

For our purpose it is more convenient to write down the rest of the field equations in terms of the functions L-, K-k + kQl, F = f-2Qk-Q21 and k = 0 + XQ%. This amounts to transforming to a coordinate system rotating with angular velocity Q.

Note that FL+K2=fl+k2 = p2 (18)

Our sign convention for the Ricci tensor is as follows

-P~~ FX~~~, (1 9) Ritv ( ,, )(jAv/g), - (Ig -g),# v v-/J, FAV,( 9

where g is the determinant of the metric. With the use of (18), the rest of the field

equations can be written as follows

2el-pR0 - (1 (KLp - LKP) + ((KLz - LKz)}

=16 nnezp K- 4

- Lp + Kyp) Xp + (-L + Kx) X} (20a)

2e/'p{(kI + Q + Q(f- Qk) RO}(K - QL)' = 1 (FKp - KFp)} + ( (FKz - KF)

4 K (2f2 + Wf2) -4 -(Rfp Xp + kfz Xz)) p p zp

(20b)

- eXll =-2(PP + z) + 2p-P + (FpLp + K

=-2{L( 2p+ 2) + 2K(Xp 3>p-X,z%) + F(X2-X2)}+ 4CnmnelI (20 c)

526 J. N. Islam

-eR2 =-2 +#)--P + p2(Fz Lz + Kz

=-L {L(p - f-2K (X2Kp ip - Xz Vfz) + F(Xy2- x_)} + 4mn1eL, (20d)

R12 =2p Iz + 42(Fp LZ + Fz lp + 2Kp Kz)

2 = p-2 {L#P 34 + K(#P Xz + z XP) + FyP X:} (20 e)

p2

Note that in the new variables F, L, K and f the equations (20 a-e) are independent of Q. The functions F, L and K are related by (18). The p- and z-components of the Maxwell equation (3) vanish identically, while the t- and 0-components are as follows

(-LVfi p + KXP),p + (- LVrz + KXz)z - - (- LVr + Ks,) = 4np2e/J0 =- 4np2e#qnF-A, (21a)

(KV,p +FX,P)P + (KVrz+Fxz)2-- (Kfp +FxFp) = 47p2(J3-QJO) 0. (21b)

Equations (20 a-e) and (21 a, b) are to be solved for the four unknowns,a, n and two of F, L, and K.

3. TwES O TLAUTION

Let 6 be a function satisfying the equation

1 6 - 6PP + zz - 1= 0. (22)

p

This function can be expressed in terms of a harmonic function as follows.

zz + p 0. (23)

The solution, the derivation of which is described in the appendix, is given by the following.

F = constant = Fo, #f = constant = bo, K = a6, X = /3, (24)

where a and ,8 are constants. , is given by one of the following equations

1(X2 -4fl2F 1 (- 2 + 4fl2F V2.4i= F0(

2z ,t= (25)

P2p p

the consistency of which is guaranteed by (22). The number density n is as follows,

8tnmn = (X - 222FO) (6+ ) (26)

Interior solutionfs of Einstein-Maxwell equations 527

while the ratio of charge to mass is related to the constants a, fi and Fo by the follow- ing relation q 2xfF,12

Mn (a2-2,82FO) (2)

It can be verified that the solution given by (24)-(27) satisfies all the field equations (20 a-e) and (21 a, b). Note that for positive mass one must have a2> 2/?2F0. The solution reduces to that of van Stockum if ft = 0.

The constant Fo can be set equal to unity without loss of generality, through a scaling of the time variable t. a can also be set equal to unity by absorbing it in the function 6, but it is useful to retain it. Setting Fo = 1, (27) yields a quadratic for

,8 in terms of q/m and a. The appropriate root of this quadratic is one which tends to zero as q tends to zero. Choosing this root, the expression for the constants appearing in (25) and (26) (with Fo = 1) can be written as follows

l= qal{m + (M2 + 2q2)i} ,

a 2?2 -

2mMc2/{M + (M2 + 2q2)1} + (28)

-2 432 = (4m2 - q2)/{q2 + 2m2 + 2m(m2 2q2)1}.

As in the case of the van Stockum solution, the present method cannot be used to derive an exterior solution, because (16) is obtained under the assumption n = 0, and hence is not valid outside matter.

4. DISCUSSION OF THE SOLUTION

Since there is no pressure, for IqI # m the material source cannot be a finite one, but must represent sources extending to infinity along the axis of symmetry. For

q= m this need not necessarily be the case. However, to show that the solution can represent a finite and bounded source it seems necessary to find an asymptotically flat exterior stationary solution of the Einstein-Maxwell equations that will match smoothly onto the interior solution. This is currently under investigation.

A typical particle in the material is acted on by four forces, namely, gravity, electric, magnetic and centrifugal force. It is interesting to see how these forces balance. One way to see this is to consider the Newtonian limit of (16). Inserting the velocity of light c and noting that Vf ( = 0b + QJ) is constant, equation (16) reads as follows 2f -2Qck - Q21 = constant. (29)

In the Newtonian limit we have

1+C2V +O(C-4), k = O(c-3), 1=p2+O(C2), (30)

where V is the Newtonian potential. Equation (29) implies in this limit,

V - p22 = constant, (31)

which is the usual equation for the gravitational and centrifugal potential. Since the electromagnetic potential does not appear in this equation, this suggests that the

528 J. N. Islam

electric and magnetic forces balance each other. That this is the case can be seen again in the Newtonian limit, in which case the electromagnetic field can be taken as in flat space. To transform conveniently to Cartesian components for this section onily we set (x?, xI, X2, X3) = (t, p, 0, z), i.e. we interchange the numbering of 0 and z, and take A, = (, 0, xo 0). The tensor F,'4 can now be written as follows (again setting c= 1) O0 p 0 z

(F4 v)= - 5p - xp ? (32) -

oz ? 0

Transform these to Cartesian components F, p by the transformation x = p cos 0, y p sin 0. The result is

x y 0 x P Y q5 0 Ex Ey Ez

p p x ~~ y

I_f OP -pX,o DX p- Ex 0 -Bz By, =FV , _X(33)

|-op Xp ? p20 xz -Ey Bz 0 -Bx

-z p %z p xz 0 -Ez -By BX 0

where the latter equation expresses the Cartesian components F', in terms of the components of the electric field E = (Ex, Ey, Fz) and the magnetic induction B- (Bx, By, Bz). The electromagnetic force on the particle can be written as

q{E+v A B}, (34)

where v is the velocity in the Newtonian limit, given by

v = Q(-y, x, O), (35)

for a particle rotating with angular velocity Q about the z-axis. When the force (34) is evaluated by inserting the values of E and B given by (33), it is easily seen to vanish by virtue of 0 + QX = const. Thus we have shown that in the Newtonian limit the electric and magnetic forces balance each other. That this is true in the exact solution is indicated by the fact that by virtue of 3f == constant the right hand side of (16) is constant so that the electromagnetic field does not appear explicitly in this equation of motion. That there is nevertheless a non-trivial electromagnetic field present follows from the fact that the electromagnetic potential cannot be made to vanish (or made equal to a constant) in any frame. In this connection it is interesting to consider the frame which is rotating with angular velocity Q with respect to the original frame. The new frame is given by 0 --0- Qt, the other co- ordinates remaining the same. Then A,-- (qS + QX, 0, X, 0). Since 0 + QX = const., in this frame the field is purely magnetic (X =,8c =A const.).

Interior solutions of Einstein-Maxwell equations 529

Consider now a particular member of the class of solutions. One of the simplest choices is to take y as follows

y = - , r2 -p2+z2. (36)

p 2 Then = py =3 (37)

cqp2 /3p2 so that K = 3>X = r3*(38)

From (25) and (26) we have respectively (with Fo = 1)

/6-=(-a2+4f2 ) (r6 8r8J' (39)

and 8nmn= (a-2- 2 -2)4- ) exp (4fl2 -2) (2-8r8)} (40)

The metric is well-behaved everywhere except on the axis p = 0, where the deter- minant of the metric vanishes. However, this may simply be a co-ordinate singularity. To examine this question further one has to evaluate the curvature invariants, which we have not done. The manifold r = 0 o0n the axis displays a somewhat peculiar behaviour, as is seen by examining the number density n given by (40). The latter is well behaved everywhere except on. r = 0, which is not necessarily a point. Let the manifold r = 0 be approached along the curves Sk defined by

0 = const., p = kJzJ, (41)

where k is a positive constant. On this curve

8imn = (X2-2-2)f(+ 4)> !exp (4fi2-a2) (8k /)4 p (42)

Recall that x2 > 2fl2 for positive mass. Let z2 be in the range 2/?2 < a2 < 4/2. Then (42) implies that n tends to infinity or zero according as r = 0 is approached along curves Sk given by values of k satisfying 8/2 > 1 or 8/2 < 1. The reverse is true if a2 > 4/52. To determine whether r = 0 is a point, consider the spatial distance covered in going from the point p = p1 on Sk to the 'point' r = 0. The spatial dis- tance from p = p1 to p= P2 on Sk is given by (see, for example, Landau & Lifshitz 1975, p. 235)

=p e12(dp2 +dz2)1 = (1+k2)1J' exp (4 (2+-/c2)(W - 1) dp. (43)

For 2,82 < oC2 <4, 2, as Pi - 0 this distance tends to infinity or remains finite accord- ing as (8k2- 1) > 0 or (/2- 1) < 0. The reverse is true if a2> 4fl2. This seems to indicate that r = 0 is not a point but a surface. A similar behaviour is displayed by the van Stockum solution (for , = 0). However, the present solution has a distinct behaviour for a2 = 4,82, in which case number density tends to infinity unambig- uously as r -> 0, and in this case r = 0 is indeed a point. Although It in this case is constant, the metric nevertheless is non-flat because K still has the form given by

530 J. N. Islam

(38). The value a2 = 4/52 corresponds to q = ? 2m, as is clear from (27). This is somewhat surprising, as it would have been more natural to have this distinct behaviour for q = ? m.

APPENDIX

Here we describe the derivation of the solution. In van Stockum's solution one has F _ constant. One could start with the assumption that this holds in the present case, leading to #f = constant through (16). However, we shall see how this same conclusion can be arrived at from more general considerations.

Equations (20c, d) imply

I (F Lz + K2 -FL 2) + 2

x {L(-_ /rp + grz2) + 2K(Xp tfp- Xz Rfz) + F(X2 - X2)1 (A 1 )

Elimination of ,t between equations (20 e) and (A 1) leads to the following equation.

-2pF 2F0z + (p2 - K2) [-F-2IAF +FF-3(Fp +EF) FI + 4F-1i0z ?/A + 2F-2Fz (V'p- r) -4F-2F ]

- F-1KKz /F + 2F-2KFz (Kp Fp + Kz Fz)-F-'Fz (Kp + K2)

+ (--KFz + Kz) AK-4K(Xz AVf + fAX) -4FXz AX

+ 4F-1KKz (i2 - ?r2) - 8F-1KK, 7 f + 4Kz (?rp Xp -V'z Xz) 4Kp (Xp Vf z + Xz ?Ip) + 2Fz (Xp-X2)-4FPXz] = 0 (A 2)

where the operator A is defined by (21). Recall that F is related to Vfi( = 0 + DX) by (16). The three functions K, f and X thus satisfy the coupled system of three equa- tions (20b), (21 b) and (A 2). It can be shown, by substituting forFfrom (16) in terms of V/, that the expression in curly brackets in (A 2) vanishes for the special case q2 = m2. We confine ourselves for the present to this case. Since Vf is undefined to within an additive constant, the constant B in (16) can now be absorbed in Vf and one can set F - 2. Equations (A 2), (20b) and (21 b) can then be written respec- tively as follows

[- 2g-1KK, AVf - 2r-2KKz (Vf2 + /f22) + 8fr-2K?fz (KP Vrf + K, V'z) + ( 2?r-lKVz + Kz) AK - 2?~1r-fz (Kp + K2) -4K(Xz Ai{ + /fz AX) - 4V-2xZ AX + 4Vr-2KKZ (V2 - V2) -8-2 KK ? fz + 4Kz (Vfp X -b z Xz) -4Kp (XpXpfz+ Xzfrp) + 4ffz (X2-X2)- 8?Vp Xp Xz] = ?, (A 3)

- 2K?0r/\? + Vf2A/K + 2K(Vf 2 + V0f2) + 4Vfr2(Vfrp Xp + ?0fd Xz) = ? ( 4 ~~~~~~~~ +4/2zX +/~~ 0 (A 4)

KA?;f + 3 2AX + Kpp + KZ z + 23b(p Xp + zXz) =0. (A 5)

Note that p appears explicitly in (A 3, 4, 5) only in the operator A. This fact and the form of these equations suggests that one should look for a solution in which K, i- and X are all functions of 6 satisfying equation (22) (i.e. A6 = 0). With this assump-

Interior solutions of Einstein-Maxwell equations 531

tion one gets the following consistency relations from (A 3, 4, 5) respectively (with

K6 OKIOE, etc.).

[- 2 -KKr ifrg + 2~/r-2KK6 ifr2 - 2Vf fr'r K- 22 VKV/tr Kg

+ Kg KX - 4Kyg V/rb - 4(KVf r + f2 xb) x,- 4Vf3/fg X2 - 4Kr V X6] = 0, (A 6)

f 2K6 - 2V K?rK/ + 2KM!6 2+ 4Vf2 X6 1 = 0 (A 7)

KV g + 2f2X66 + Kg V/rg + 2/x f = 0. (A 8)

Equations (A 6) and (A 8) can be integrated as follows

-2V -1Kir #Kg +_Kg2- 4Kxgf -2t2'x2 = u, (A 9)

Kf + V22X? = v, (A 10)

where u and v are arbitrary constants. With the use of (A 10), (A 7) can be integrated as follows

Vf2K6 -2VfKj6 +4vVr = w, (A 11)

where w is an arbitrary constant. Substituting for xy from (A 10) and for Kg from (A I1) into (A 9) one gets the following equation

6v2 8vw W2

~2 2 f3+224=U (A 12)

(A 12) implies first that u = v = w = 0, in which case (A I1) implies K = (const) yf 2. The latter implies that F and K are proportional. For any metric for which this is the case one can find a linear transformation involving t, 0 such that K vanishes, leading to a static metric, which is not of interest. The alternative solution of (A 12) is that

Vf = const. From (A 10) and (A I1) we have Kg = const. # 0, xy = const. =# 0. This then leads to the solution we have given in ? 3. Although the solution is here derived under the assumption q2 = Mi2, it turns out that the solution is valid also for q2 2 in2.

This is because the terms that vanish (in the curly brackets in (A 2)) for q2 - m2 also vanish when Vf = constant. The fact that the solution is valid for q2 2 M2 can be verified directly by substituting in (20a-e) and (21 a, b), as we have done.

I am grateful to B. Schutz and N. C. Wickramasinghe for hospitality at University College, Cardiff, and to the S.R.C. for a grant.

I thank an anonymous referee for useful suggestions on an earlier version of the paper.

REFERENCES

Landau, L. D. & Lifshitz, E. M. 1975 The classical theory of Fields, 4th ed., Oxford: Per- gamon Press.

Lewis, T. 1932 Proc. R. Soc. Lond. A 136, 176. van Stockum, W. 1937 Proc. R. Soc. Edinb. 57, 135. Weyl, H. 1917 Ann. Physik 54, 117. Winicour, J. 1975 J. Math. Phys. 16, 1806.