International Centre for Theoretical Physics

IC/8T/22T INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS OH THE REDUCTION OF SYMMETRY FOR STATIC FLAT SPACE-TIME IN SOME GENERAL CYLINDRICAL-LIKE COORDI1IATES Ashfaque H. Bokhari and Housheen A. Bokhari INTERNATIONAL ATOMIC ENERGY AGENCY UNITED NATIONS EDUCATIONAL, SCIENTIFIC AND CULTURAL ORGANIZATION 1987 MIRAMARE-TR1ESTE IC/8T/227 1. INTRODUCTION While studying the reduction of symmetry it turned out that the maximal symmetry corresponds to anti-de Sitter, de Sitter and the Minkowski International Atomic Energy Agency metrics and the minimal to the Schwarzechild or the metrics of the form and obtained earlier [l]. The symmetry of any space-time in relativity is United Nations Educational Scientific and Cultural Organization expressed In terms of the number of independent Killing vector fields possessed by the metric of that space-time [2],[3] , [h]. Thus in these terms INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS the maximal and the minimal symmetry obtained earlier gave either 10 or h independent Killing vector fields and nothing in between. The problem taken in the above case [l], was, in fact, aimed at finding all possible ON THE REDUCTION OF SYMMETRY FOTC STATIC FLAT SPACE-TIME metrics which had independent Killing vector fields filling all the gaps in between numbers 10 and It. In group-theoretic terms this problem is IN SOME GENERAL CYLINDPICAL-LIKE COORDINATES * translated as finding all the subgroups of the maximal group having 10 parameters down to the subgroups having 9, 8, ..,, to U parameters respectively. This attempt, however, was unsuccessful but still (juite Ashfaque 2. Bokhari ** interesting in the sense that it lead to an important result, i.e. any International Centre for Theoretical Priysics, Trieste, Italy reduction from a general static, spherically symmetric space-time metric could yield only the Minkowski or the Echwarzachild type symmetry with no and Intermediate symmetry. This result was published in 1987 in the form of a HoUGheen A. Bokhari theorem [l]. F.G. Margala College for Women, F-TA Islamabad, Pakistan. In this paper the same problem is addressed. This problem is different from the earlier problem in the sense that the space-time metric has general cylindrical symmetry containing the maximal symmetry in itself, ABSTRACT whereas in the earlier problem the metric was spherically symmetric and had the minimal symmetry in itself yielding the maximal symmetry afterwards for Hat static metric in terms of general cylindrical-like coordinates different cases. Since we are interested in the classification of static is considered and symmetry in reduced r,tep by step. It turns out that the metrics only, we require that the metric considered be static. maximal and the minimal symmetry remains the same a^ that of the Minkowskl or the SchwarzHchild type. As soon as the dimensions of the metric are reduced, the symmetry turns out to be 6 or 3 In terms of independent 2. METRIC AND ITS STEPWISE SYMMETRY REDUCTION Killing vector fields, respectively, not yet filling all the gaps between 10 to 8 to k or from 10 to 8 to 6 to 1+ to 3 to 1. Consider the metric of the form of a general cylinder in polar type coordinates, given by MIRAMAKE - TRIESTE September 1987 where v is a function of r only and prime represents partial differentiation with respect to the coordinate r. The symmetry of this * To be submitted for publication. space-time is worked out in terms of the independent Killing vector fields ** Permanent address: Mathematics Department, Quaid-i-Azam University, Islamabad, Pakistan. -2- T ' possessed by this metric. The Killing vector K is the one relative Eqs.(ll*) with Eq.(l3) then yield to which the Lie derivative of the metric tensor £ is zero, i.e. The vector K satisfying Eq.(2) is called a Killing vector. To find (15) all the possible Killing vectors ve solve the ten Killing equations Koif using the values of K° and K1 from Eq.(l5) into Eq.(lii) (M and that of K and K3 into Eq.(6) it turns out that (5) (6) (16) (7) Thus the set of equations given by Eq.(15) reduces to (8) +• *C = o (9) (17) (10) Nqw Eq.s(^) together- with Eq.(17) give (11) (12) (16) where 0,1,2,3 correspond to t, r, 6 and W coordinates respectively. Differentiating Eq.(8) and (10) with respect to B and r coordinate, Bow Eq.(T) can be easily solved to Rive respectively,and subtracting which yields two cases namely; I : 0 + 0 and II : 0 = 0. We first consider (19) Case I From the Killing equations it la easily Been that and SO. (£0) K, CU) -It- -3- Using Eqs.(l8) and (19) in Eqs.(lT) we obtain Eq(22)can be used to determine the form of i(r) and w[r). It turns out that faf xe t0V°y (£6) Using Eqs.(5)i(8},(9) and (11) with other equations it is easily seen that p(£), h(8) and D(ttr,<0 are all constants .namely a and a, respectively with a and ag beuigaero and b(6) = (a, cos 8 *• a, 8iti 6). Using all (21) these results in Eq.(25) and rearranging the parameters one obtains Differentiating Eqs.(5) and (10) with respect to 6 and t coordinate!and then adding gives I 3 J (22) To be able to find a solution of Eq.(20) we use Eq.(22) there to get, where a is a separation const ant, 0<>,0,sC<iJ. We first consider the case B > 0 = 1 J / This gives (27) (23) At this stage one can easily integrate Eq.(lO) with respect to the 6 coordinate to obtain which shows that in the case of a generalised cylindrical like symmetric static metric the number of independent Killing vectors turn' out to be ten giving the maximally symmetric group of n(n + l)/2 parameters. In the other two cases i.e. a>0 j* l,.a$e, in Eq.(20)' a1=0=a2 reducing the number of Killing vectors to eight in Eq.(27). In the case a*0 eVV6t2fSr"where S and B are integration constants. Using all these values in the Bet of Killing equations, Eqs. (21), one obtains Case II A look back at Eq.(3) gives case II in which 0=0, This constraint reduces the number of independent Killing vector fields to U in only one step giving $ - O • (28) The other Killing equations vith Eq.(20) become 1 1 fe° fe " le" ? (29) - * A Jfc and the rest of the equations remain the same. Following the sane procedure as above it is easily seen that in this case the independent Killing vectors k = (25) turn out to be four Had have the form -6- -5- Case Ill.b Here also the independent Killing vector fields are six as above with 0 the only difference being that C = 1 in the above case. Case IILd Here the independent Killing vector fields are three having the form: (30) Motice that in this cose the number of parameters of this group are the same as those of the Schwarsschild one. The difference, however, is that here the (35) direction of symmetry has been shifted, in some sense, and rotations have been converted into translations. Notice that in this case, c = 0 reduces another dimension of the (3+1) manifold. Case III The next possibility for obtaining a. subgroup of the maximal group 3. DISCUSSION AND CONCLUSION is obtained through a coordinate transformation given by In the work on reintroduction of concept of force in relativity (3D [5),[6],[Tj it was shown that the reduction of symmetry from maximum to This transformation, in which e becomes a constant, yields minimum yields only 10 or k parameter groups with no intermediate class for a metric of the Schwartschild type. The metric considered there [5] was not (32) so general in the sense that the above statement could be generally true for The form of Eq.(l) due to Eq.{32) becomes any general spherically symmetric, static metric. Thus the most general (in some sense) static and spherically symmetric metric [1] was considered (33) which contained the metric given in[6] as its aubmetric. However, it turned out that the reconsidered metric also gave the same result for quite a number Clearly the metric given by Eq.(l) under the transformation given by of cases as discussed thoroughly [l] and [8J . Eci.(31) becomes unstable lowering one dimension of the (3 + l) dimensional manifold.In this case there ar« three possibilities, namely, III-a c > 0, To be able to achieve the goal of obtaining independent Killing III«b c = 1, and III-d c = 0. In case Ill'a the possible number of vector fields from 10... h ... 3 ... 1, we consider a model whose metric independent Killing vector fields are six given by is given by Eq.(l). For the metric of this model the independent Killing vector fields characterizing tha group have been explicitly worked out. In the first step, it is seen, as claimed, that one gets the maximum symmetry of h(k+l)/2. No sooner one shifts to the second case the symmetry reduces in one step to four being minimal in It-dimensions. This is not exactly like that obtained earlier [1], c The next step here is important, In the sense that it tells that the lower dimensional groups are obtained only when the metric becomes -7- -8- KEFERENCES unstable. Here the important thing to note is that space-time becomes maximally symoetric giving n(n+l)/£ independent Killing vector [l] Ashfaque H.

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