ARS 2008

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Subject: search.filters.subject.deflection angle, cosmological constant, Schwarzschild de Sitter metric, light ray trajectory

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    Deflection of light with cosmological constant
    (University of Kelaniya, 2008) Jayakody, J.A.N.K.; de Silva, L.N.K.
    A beam of light is deflected when it passes near a massive object such as the Sun or a black hole, due to the gravitational influence of the massive body. Without the cosmological constant, the total deflection angle of light is 28 = 4m (11, where r0 is the closest approach of ro the light ray from the centre of the massive body and m = GAf . Here G is the gravitational c constant and M is the mass of the central object. Taking into account the Schwarzschild de Sitter geometry, some authors [21 have found recently, that the cosmological constant contributes to the deflection of light. In this paper, we study the effect of the cosmological constant on the deflection of light when it passes near a massive object. The null-geodesic equation in Schwarzschild-de-sitter metric where can be written as 2m Ar2 3 f(r)=1----[ -l r 3 ' where l and h are constants, which can be written in another form of u" + u = 3mu2 (11. The zeroth order solution and the first order solution of the light ray trajectory are respectively given below. 1 u0 = -cos [IJ ro 1 E 2 2E u=-cos"'---cos "'+--['1 where E =3m.Now, assume a second order solution for the light ray trajectory in the following form. 1 £ 2 2£ 2 u = -cos th ---cos th + --+ £ w 'I' 2 'I' 0 ' r0 3r0 3r0 􀁑 where w = w(r0,). We find that the solution up to the second order for the light ray trajectory can be written as, 1 E 2 2£ r0 sin [ 5£ 2 E 2 • ( 5£ 2 A J ] u = -cos --- 2 cos +-- 2 + --4 ---4 sm2 + --4 +- cot . r0 3r0 31() 2 6r0 12r0 9r0 3 n n Taking the limits as, r 􀀋 00' u 􀀋 O, <1> 􀀋 2 + 8 '<1> 􀀋 -2-8 we find the two asymptotes of the trajectory, which give the total deflection angle as, [􀀉 _ 8£ 2 + 2A102 ] 28 = 3ro 9ro 2 3 (1-􀁒+ Ar0 33r0 2£ J A is shown to be a function of E 2 , and substituting A= E 2 k in the above equation, and neglecting the terms ofe , of order higher than four , we find, 28 = 4£ - 3 2£ 3 - 16£ 4 k + 64£ 4 3r0 27r􀁓3 27 81r0 4 • Writing the above equation in terms of A, we obtain, Substituting E = 3m, 28 =4m _ 16m2 A_ 32m3 + 64m4 ro 3 ro 3 ro 4 , which is the angle of deflection in the presence of the cosmological constant, neglecting fifth and higher order terms of £ .