Please use this identifier to cite or link to this item: http://repository.kln.ac.lk/handle/123456789/8032
Title: Cosmological constant in gravitational lensing
Authors: Jayakody, J.A.N.K.
de Silva, L.N.K.
Issue Date: 2011
Publisher: University of Kelaniya
Citation: Jayakody, J.A.N.K. and de Silva, L.N.K., 2011. Cosmological constant in gravitational lensing, Proceedings of the Annual Research Symposium 2011, Faculty of Graduate Studies, University of Kelaniya, pp 64-65.
Abstract: Consider the Schwarzschild de Sitter Metric, 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 ( sin ). 3 3 GM r GM r ds c dt dr r d d rc rc                          (1) The constant term 2 2GM c is recognized as the Schwarzschild radius ( s r ), and typically it is replaced by a constant term2m, where 2 1 2 s GM m r c   and then the equation (1) can be written as follows. 1 2 2 2 2 2 2 2 2 2 2 2 2 1 1 ( sin ). 3 3 m r m r ds c dt dr r d d r r                          (2)  is the cosmological constant. The null-geodesic equation in Schwarzschild-de Sitter metric can be written as, 2 2 2 2 2 2 2 3 2 2 0 3 E l l u l u ml u c        , [1] (3) where E is the energy, l is the orbital angular momentum,  is the cosmological constant, 1 u r  and . du u d   Differentiating (3) with respect to , 2 u(uu 3mu )  0. (4) Neglecting the solution,u  0 which implies u = constant, the equation of a light ray trajectory can be written as, 2 uu  3mu . (5) The zeroth order solution and the first order solution of the equation (5) that represent the light ray trajectory are respectively given below. 0 0 1 u cos r   [2], (6) 2 2 2 0 0 0 1 2 cos cos 3 3 u r r r        [2], (7) where   3m. In general, in the literature, it is assumed that (7) is a solution of equation (3) without considering the limitations imposed. In this paper we discuss conditions under which (7) is a solution of equation (3). Now the orbital angular momentum, 0 l  pr where p is the linear momentum. The linear momentum, E p c  . Therefore, 0. E l r c  (8) Substituting (7) and (8) in (3), we have, 2 2 2 2 2 2 2 2 2 2 0 0 0 0 0 3 2 2 2 2 2 0 0 0 1 2 1 2 sin sin cos cos cos 3 3 3 2 1 2 + cos cos 0. 3 3 3 3 E l l c r r r r r l l r r r                                          (9) By simplifying the above equation and since l  0 we obtain the following equation, 3 3 3 3 2 2 2 2 4 6 5 3 6 6 6 6 5 5 5 0 0 0 0 0 0 0 2 4 4 4 4 0 0 0 8 4 2 4 4 cos cos cos cos cos cos 27 9 9 27 3 3 3 2 0 2 2 3 cos cos 3 2 r r r r r r r m r r r                                          2 2 2 2 2 4 6 5 6 6 6 6 5 2 0 0 0 0 0 3 2 4 5 5 4 4 4 0 0 0 0 0 8 4 2 cos cos cos cos 3 3 18 . 4 4 2 2 1 cos cos cos cos 3 2 m m m m m r r r r r m m m r r r r r                                (10) From (10) it is clear that the solution given by (7) of equation (3) is valid only if  is a constant of order m2, and as we neglect terms of order 2 and above we are justified in assuming (7) as a solution of equation (3). However, it turns out that this particular solution is valid only if  is a constant of order 2 or more in m. If  is a non zero constant and of order one in m, the solution (7) is not valid and we have to seek other solutions.
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