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Deflection of Light Ray Due to a Charged Body Using Material Medium Approach

  • Saswati Roy EMAIL logo and A.K. Sen

Abstract

The gravitational deflection of light ray is an important prediction of general theory of relativity. In this paper we have developed an analytical expression of the deflection of light ray without any weak field approximation due to a charged gravitating body represented by Reissner-Nordström (RN) and Janis-Newman-Winicour (JNW) space-time geometry, using material medium approach. It is concluded that although both the geometries represent the charged, non-rotating, spherically symmetric gravitating body, the effect of charge on the gravitational deflection is just opposite to each other. The gravitational deflection decreases with charge in the RN geometry and increases with charge in the JNW geometry. The calculations obtained here are compared with other methods done by different authors. The formalism is applied to an arbitrarily selected gravitating body, as a test case and compared with the standard Schwarzschild geometry for comparison purposes.

Acknowledgments

SR acknowledges Trishna Bordaloi and Samujwal Das, MSc. Student of Assam University, Silchar who had done MSc. project under this topic. SR deeply acknowledges S Chakraborty, Assam University, Silchar also for useful discussions. Finally we acknowledge grants from the UGC-SAP under which the work was done.

Appendix A

Bending Angle Due to RN Geometry

We rewrite (22) as

(A1)Δψ=2Iπ

where

(A2)I=n(v,q)vvdxx(n(x,q)x)2(n(v,q)v)2=Dvdxx(n(x,q)x)2D2

with D=n(v, q)v.

To evaluate the above integral, we follow a procedure similar to what was done by Sen [38] and Roy and Sen [39].

Thus with the value of refractive index from (12) and D=Dr [where Dr=n(v, q)v, n(v, q) is the refractive index due to RN geometry at the limit of impact parameter] we have

(A3)I=Drvdxx{n0(x)(1+Cx)1x}2Dr2=Drvdxxn02(x)x2D02+n02(x)x2(1+Cx)2n02(x)x2+D02Dr2=Drvdxxn02(x)x2D02[1+n02(x)x2{(1+Cx)21}+D02Dr2n02(x)x2D02]12=Drvdxxn02(x)x2D02[1+K(x)]12

where D0=n0(vv (corresponding to Schwarzschild deflection). And we have also denoted

(A4)K(x)=n02(x)x2{(1+Cx)21}+D02Dr2n02(x)x2D02

Here, we can show that K(x)≪1. To evaluate the value of K(x) of (A4), the value of ((1+Cx)−2−1) is as follows:

(A5)(1+Cx)21=1(1+Cx)21=1(1+q2x(x1))21=2q2x(x1)+q4x2(x1)2(x(x1)+q2)2x2(x1)2=2q2x(x1)+q4(x(x1)+q2)2

Substituting the value of ((1+Cx)−2−1) from (A5) and n0(x)=x/(x−1) we can write the value of K(x) as

(A6)K(x)=n02(x)x2{(1+Cx)21}+D02Dr2n02(x)x2D02=n02(x)x2{(1+Cx)21}+D02D02(1+Cv)2n02(x)x2D02=x4(x1)2{2q2x(x1)+q4(x(x1)+q2)2}+D02{2q2v(v1)+q4(v(v1)+q2)2}x4(x1)2D02=x4{2q2x(x1)+q4(x(x1)+q2)2}+D02(x1)2{2q2v(v1)+q4(v(v1)+q2)2}x4D02(x1)2

At this stage we can show that K(x)≪1. As K(x) is discontinuous at x=v, we can remove its discontinuity and evaluate its value by applying L’Hospital’s rule.

Therefore, from (A1) and (A3) one can write:

(A7)Δψ=2Drvdxxn02(x)x2D02[112K(x)+38K2(x)516K3(x)+35128K4(x)63254K5(x)+]π=2[I0+I1+I2+I3+]π

where, we have introduced some other notations:

(A8a)I0=Drvdxxn02(x)x2D02
(A8b)I1=Drvdxxn02(x)x2D02(1/2K(x))
(A8c)I2=Drvdxxn02(x)x2D02(3/8K2(x))

and so on.

Now I0 can be evaluate by following the same procedure as Sen [38] and Roy and Sen [39]. According to Roy and Sen [39] I0 can be split into two integrals as I01 and I02. Here, Dk is replaced by Dr. Thus the value of I01 and I02 will be :

(A9)I01=DrD0π2

and

(A10)I02=Dr0azdz1D02z2(1z)2

where we change the variable as z=1x, so that the limits of this integration changes from z=1v=a(say) to z=0. The integral (A10) can be evaluated in terms of elliptical function as expressed by (18) of Sen [38] and finally for a given value of a, its numerical value can be obtained as Roy and Sen [39].

Now substituting the value of K(x) from (A6), n0(x)=xx1 and applying the change of variable as z=1x the integral I1 becomes

(A11)I1=12DrvK(x)xn02x2D02dx=12Drv1xx4(x1)2D02[x4{2q2x(x1)+q4(x(x1)+q2)2}+D02(x1)2{2q2v(v1)+q4(v(v1)+q2)2}x4D02(x1)2]dx=12Drv(x1)x6D02x2(x1)2[x4{2q2x(x1)+q4(x(x1)+q2)2}+D02(x1)2{2q2v(v1)+q4(v(v1)+q2)2}x4D02(x1)2]dx=12Dr0a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]dz

Applying the same procedure, the other integrals I2, I3 etc. are as follows:

(A12)I2=38Dr0a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]2dz
(A13)I3=516Dr0a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]3dz
(A14)I4=3564Dr0a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]4dz
(A15)I5=63256Dr0a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]5dz

Therefore, substituting all the values of I0, I1, I2, I3 etc. (A7) becomes

(A16)Δψ=2[DrD0π2+Dr{0azdz1D02z2(1z)2120a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]dz+380a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]2dz5160a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]3dz+35640a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]4dz632560a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]5dz+}]π=(DrD01)π+2Dr{0azdz1D02z2(1z)2120a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]dz+380a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]2dz5160a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]3dz+35640a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]4dz632560a(1z)1D02z2(1z)2[2q2z2(1z)+q4z4((1z)+q2z2)2+D02z2(1z)2{2q2v(v1)+q4(v(v1)+q2)2}1D02z2(1z)2]5dz+}

The above expression represents the light deflection angle due to charged gravitating mass in RN space-time.

Appendix B

Bending Angle Due to JNW Geometry

Here also we will follow the same procedure as Appendix A. Now with the value of refractive index from (18) and D=Dj [where Dj=n(v, q)v, n(v, q) is the refractive index due to JNW geometry at the limit of impact parameter], we have

(B1)Ij=Djvdxx(n(x,q)x)2Dj2=Djvdxx(111x1+4q2)2/1+4q2x2Dj2=Djv(11x1+4q2)1/1+4q2x21Dj2x2(11x1+4q2)2/1+4q2dx=Djv(11x1+4q2)1/1+4q21x(11x1+4q2)11+4q21x21Dj2x2(11x1+4q2)2/1+4q2dx+Djv1x(11x1+4q2)11+4q21x21Dj2x2(11x1+4q2)2/1+4q2dx=Ij1+Ij2

Now, let

y=Djx(11x1+4q2)1/1+4q2

so that

dy=Djx2(11x1+4q2)1/1+4q2dx+Djx3(11x1+4q2)11+4q21dx=Djx2[(11x1+4q2)1/1+4q21x(11x1+4q2)11+4q21]dx

Thus the limit changes to y=0 and y=Djv(11v1+4q2)1/1+4q2=Dj1Dj=1 as x changes to v and ∞. So,

(B2)Ij1=Djv(11x1+4q2)1/1+4q21x(11x1+4q2)11+4q21x21Dj2x2(11x1+4q2)2/1+4q2dx=01dy1y2=π2

Now by applying the change of variable as z=1x like Appendix A we may write the above integral Ij2 as

(B3)Ij2=Djv1x(11x1+4q2)11+4q21x21Dj2x2(11x1+4q2)2/1+4q2dx=Dj0az(1z1+4q2)11+4q211Dj2z2(1z1+4q2)2/1+4q2dz

Thus from expression (A1) and (B1) the light deflection angle due to JNW space-time can be written as

(B4)Δψ=2[π2+Dj0az(1z1+4q2)11+4q211Dj2z2(1z1+4q2)2/1+4q2dz]π=2Dj0az(1z1+4q2)11+4q211Dj2z2(1z1+4q2)2/1+4q2dz

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Received: 2017-6-1
Accepted: 2017-9-25
Published Online: 2017-11-15
Published in Print: 2017-11-27

©2017 Walter de Gruyter GmbH, Berlin/Boston

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