Gravitational Lensing in the Strong Field Limit for Kar’s Metric

Abstract

In this paper we calculate the strong field limit deflection angle for a light ray passing near a scalar charged spherically symmetric object, described by a metric which comes from the low-energy limit of heterotic string theory. Then, we compare the expansion parameters of our results with those obtained in the Einstein’s canonical frame, obtained by a conformal transformation, and we show that, at least at first order, the results do not agree.

Appendix A: Finding \(\hat {\alpha }(\theta )\) for Kar’s Metric

In order to calculate the deflection angle in the strong field limit, we used Kar’s metric in the form proposed in (17), i.e.,

$$ ds_{str}^{2}=\underbrace{\left( 1-\frac{1}{x}\right)^{\zeta+\sqrt{1-\zeta^{2}}}}_{A(x)}dt^{2}- \underbrace{\left( 1-\frac{1}{x}\right)^{\zeta-\sqrt{1-\zeta^{2}}}}_{B(x)}dx^{2}- \underbrace{\left( 1-\frac{1}{x}\right)^{1+\zeta-\sqrt{1-\zeta^{2}}} x^{2}}_{C(x)}d{\Omega}^{2}. $$

(40)

The deflection angle is calculated using the strong field expansion [8, 19]

$$ \widehat{\alpha}(\theta)=\overline{a}\ln\left( \frac{\theta D_{OL}}{u_{m}}-1\right)+\overline{b}, $$

(41)

where

$$\begin{array}{@{}rcl@{}} \bar{a} & = & \frac{R(0,x_{m})}{2\sqrt{\beta_{m}}}, \\ \bar{b} & = & b_{R}+\bar{a}\ln\left[\frac{2\beta_{m}}{y_{m}}\right]-\pi \end{array} $$

(42)

and

$$ u_{m}=\sqrt{\frac{C(x_{m})}{A(x_{m})}}. $$

(43)

1.1 Calculation of $\bar {a}$:

Using (5), (6), (9), (21) and (40) we obtain that

$$ R(z,x_{m})=\frac{2\sqrt{1-\zeta^{2}}+1}{\sqrt{1-\zeta^{2}}+\zeta}\left\{\frac{\left( \frac{2\sqrt{1-\zeta^{2}}-1}{2\sqrt{ 1-\zeta^{2}}+1}\right)^{\frac{1+\zeta-\sqrt{1-\zeta^{2}}}{2}}-\left( \frac{2\sqrt{1-\zeta^{2}}-1}{2\sqrt{1-\zeta^{2}}+1} \right)^{\frac{3\zeta-\sqrt{1-\zeta^{2}}+1}{2}}}{[(1-y_{m})z+y_{m}]^{\frac{\zeta}{\sqrt{1-\zeta^{2}}+\zeta}}}\right\}. $$

(44)

Evaluating the last expression at z=0 we obtain

$$ R(0,x_{m})=\frac{2\sqrt{1-\zeta^{2}}+1}{\sqrt{1-\zeta^{2}}+\zeta}\left\{\frac{\left( \frac{2\sqrt{1-\zeta^{2}}-1}{2\sqrt{ 1-\zeta^{2}}+1}\right)^{\frac{1+\zeta-\sqrt{1-\zeta^{2}}}{2}}-\left( \frac{2\sqrt{1-\zeta^{2}}-1}{2\sqrt{1-\zeta^{2}}+1} \right)^{\frac{3\zeta-\sqrt{1-\zeta^{2}}+1}{2}}}{y_{m}^{\frac{\zeta}{\sqrt{1-\zeta^{2}}+\zeta}}}\right\} $$

(45)

where

$$ y_{m}=\left( \frac{2\sqrt{1-\zeta}-1}{2\sqrt{1-\zeta^{2}}+1}\right)^{\sqrt{1-\zeta^{2}}+\zeta}. $$

(46)

From (9) we obtain that

$$ \beta_{m}=\frac{1}{4}\frac{\left[\left( 2\sqrt{1-\zeta^{2}}+1\right)^{\zeta+\sqrt{1-\zeta^{2}}}-\left( 2\sqrt{1-\zeta^{2}} -1\right)^{\zeta+\sqrt{1-\zeta^{2}}}\right]^{2}}{(\sqrt{1-\zeta^{2}}+\zeta)^{2}(3-4\zeta^{2})^{\sqrt{1-\zeta^{2}} +\zeta-1}} $$

(47)

and taking into account (21) it is possible to express R(0,x _m ) and β _m and terms of the photon sphere as,

$$ R(0,x_{m})=\frac{2x_{m}}{k}\left( \left( 1-\frac{1}{x_{m}}\right)^{\frac{1-k}{2}}-\left( 1-\frac{1}{x_{m}}\right)^{\frac{k+1}{2}} \right). $$

(48)

$$ \beta_{m}=\frac{({x^{k}_{m}}-(x_{m}-1)^{k})^{2}}{k^{2}(x_{m}-1)^{k-1}x^{k-1}_{m}} $$

(49)

then, from (12)

$$ \bar{a}=\frac{x_{m}[x^{k-1}_{m}-(x_{m}-1)^{k}x^{-1}_{m}]}{{x^{k}_{m}}-(x_{m}-1)^{k}}=1. $$

(50)

1.2 Calculation of u _m :

From (13) the impact parameter is calculated as,

$$ u_{m}=\sqrt{\frac{C(x_{m})}{A(x_{m})}}, $$

(51)

then, using (40), (21) and taking into account that A(x _m ) = y _m

$$ u_{m}=\frac{(x_{m}-1)^{\frac{1}{2}+\sqrt{1-\zeta^{2}}}}{x_{m}^{-\frac{1}{2}-\sqrt{1-\zeta^{2}}}}=\frac{1}{2}\frac{(2\sqrt{1-\zeta^ {2}}-1)^{\frac{1}{2}-\sqrt{1-\zeta^{2}}}}{(2\sqrt{1-\zeta^{2}}+1)^{-\frac{1}{2}-\sqrt{1-\zeta^{2}}}}. $$

(52)

1.3 Calculation of b _R :

The term b _R is calculated by (14), which corresponds to the regular part of the integral (10), i. e.,

$$ b_{R}=I_{R}(x_{m})={{\int}^{1}_{0}}g(z,x_{m})dz={{\int}^{1}_{0}}[R(z,x_{m})f(z,x_{m})-R(0,x_{m})f_{0}(z,x_{m})]dz. $$

(53)

However, as is pointed out in Ref. [8], b _R can not be calculated analytically but by an expansion of I _R (x _m ) in powers of some metric’s parameter. In this sense, to calculate the regular term for Kar’s metric (17), we made and expansion of I _R (x _m ) in powers of ζ around ζ=0³ and used the first order expansion as our b _R . Mathematically this idea is

$$ b_{R}=\sum\limits_{n=0}^{\infty}\frac{1}{n!}\frac{d^{(n)}I_{R}(x_{m})}{d\zeta^{n}}(\zeta-0)^{n} $$

(54)

and taking terms up to first order in ζ we have that⁴

$$ I_{R}(x_{m})=I_{R}(x_{m})_{\zeta=0}+\left[\frac{dI_{R}(x_{m})}{d\zeta}\right]_{\zeta=0}\zeta. $$

(55)

In this sense, in order to calculate the regular term b _R , it is necessary to find the functional form of R(z,x _m ), f(z,x _m ), R(0,x _m ) and f ₀(z,x _m ). The form of R(z,x _m ) and R(0,x _m ) are already shown in (44) and (45). From (6), (7) and (22) we have that

$$ f(z,x_{m})=\frac{1}{\sqrt{y_{m}-[(1-y_{m})z+y_{m}]\frac{C_{m}}{C}}}\\ $$

(56)

and

$$ f_{0}(z,x_{m})=\frac{1}{\sqrt{\beta_{m}}|z|}=\frac{2(\sqrt{1-\zeta^{2}}+\zeta)(3-4\zeta^{2})^{\frac{\sqrt{1-\zeta^{2}}+\zeta-1}{2}}}{ (2\sqrt{1-\zeta^{2}}+1)^{\sqrt{1-\zeta}+\zeta}-(2\sqrt{1-\zeta^{2}}-1)^{\sqrt{1-\zeta}+\zeta}}\frac{1}{|z|}, $$

(57)

where

$$ C_{m}=\left[\frac{2\sqrt{1-\zeta^{2}}+1}{2}\right]^{2}\left[\frac{2\sqrt{1-\zeta^{2}}-1}{2\sqrt{1-\zeta^{2}}+1}\right]^{ 1+\zeta-\sqrt{1-\zeta^{2}}}, $$

(58)

$$ C=\frac{[(1-y_{m})z+y_{m}]^{\frac{1+\zeta-\sqrt{1-\zeta^{2}}}{\zeta+\sqrt{1-\zeta^{2}}}}}{[1-[(1-y_{m})z+y_{m}]^{\frac{1}{\zeta+\sqrt{ 1-\zeta^{2}}}}]^{2}}. $$

(59)

As (55) shows, the value of I _R (x _m ) reduces to that of Schwarzschild for ζ=0. Therefore, for 0≤z≤1 (|z| = z) we obtain

$$ I_{R}(x_{m})_{\zeta=0}=2{{\int}^{1}_{0}}\left[\frac{1}{|z|\sqrt{1-\frac{3}{2}z}}-\frac{1}{|z|}\right]dz=2\ln{6(2-\sqrt{3})}=0.9496. $$

(60)

On the other hand,

$$\begin{array}{@{}rcl@{}} \left[\frac{dI_{R}(x_{m})}{d\zeta}\right]_{\zeta=0}&=&{{\int}^{1}_{0}}\left[\frac{d}{d\zeta}(R(z,x_{m})f(z,x_{m}))-\frac{d}{d\zeta}(R(0, x_{m})f_{0}(z,x_{m}))\right]_{\zeta=0}dz\\ &=&{{\int}^{1}_{0}}[f(z,x_{m})\frac{d}{d\zeta}R(z,x_{m}) +R(z,x_{m})\frac{d}{d\zeta}f(z,x_{m})\\ &&-f_{0}(z,x_{m})\frac{d}{d\zeta}R(0,x_{m})-R(0,x_{m})\frac{d}{d\zeta}f_{0}(z,x_{m})]_{\zeta=0}dz\\ &=&{{\int}^{1}_{0}}[f_{S}(z,x_{m})\left[\frac{d}{d\zeta}R(z,x_{m})\right]_{\zeta=0} +2\left[\frac{d}{d\zeta}f(z,x_{m})\right]_{\zeta=0}\\ &&-f_{0S}(z,x_{m})\frac{d}{d\zeta}R(0,x_{m})-2\left[\frac{d}{d\zeta}f_{0}(z,x_{m})\right]]dz, \end{array} $$

where f _S (z,x _m ), f _0S (z,x _m ) are those of Schwarzschild (Cfr. [8]). Therefore,

$$\begin{array}{@{}rcl@{}} \left[\frac{dI_{R}(x_{m})}{d\zeta}\right]_{\zeta=0}&=&{{\int}^{1}_{0}}[\frac{\left[\frac{d}{d\zeta}R(z,x_{m})\right]_{\zeta =0}}{z\sqrt{1-\frac{2}{3}z}}-\frac{\left[\frac{d}{d\zeta}R(0,x_{m})\right]_{\zeta=0}}{z}\\ &&+2\left[\frac{d}{d\zeta}f(z,x_{m})\right]_{\zeta=0}-2\frac{\left[\frac{d}{d\zeta}f_{0}(z,x_{m})\right]_{\zeta=0}}{z}]dz, \end{array} $$

where all derivatives, evaluated at ζ=0, are:

$$\begin{array}{@{}rcl@{}} \left[\frac{d}{d\zeta}R(z,x_{m})\right]_{\zeta=0}&=&-2-2\ln\left( \frac{2}{3}z+\frac{1}{3}\right)\\ \left[\frac{d}{d\zeta}R(0,x_{m})\right]_{\zeta=0}&=&-2+2\ln(3)\\ \left[\frac{d}{d\zeta}f_{0}(z,x_{m})\right]_{\zeta=0}&=&\frac{\ln(3)-1}{|z|}\\ \left[\frac{d}{d\zeta}f(z,x_{m})\right]_{\zeta=0}&=&-\frac{1}{2}\frac{\ln{(3)}\left[\frac{7}{3}z^{3}-2z^{2}\right] +z(2z+1)(1-z)\ln(2z+1)}{z^{3}(1-\frac{2}{3}z)^{\frac{3}{2}}}. \end{array} $$

(61)

Thus,

$$\begin{array}{@{}rcl@{}} \left[\frac{dI_{R}(x_{m})}{d\zeta}\right]_{\zeta=0}&=&{{\int}^{1}_{0}}\left[\frac{2-2\ln(3)}{z}-\frac{2+2\ln\left( \frac{2}{3}z+\frac{ 1}{3}\right)}{z\sqrt{1-\frac{2}{3}z}}\right]dz\\ &&+{{\int}^{1}_{0}}\left[\frac{\ln{(3)}\left[\frac{7}{3}z^{3}-2z^{2}\right]+z(2z+1)(1-z)\ln(2z+1)}{z^{3}(1-\frac{2}{3}z)^{\frac{3}{2}}} +\frac{2\ln(3)-2}{z}\right]dz\\ &=&{{\int}^{1}_{0}}\left[\frac{2-2\ln(3)}{z}-\frac{2+2\ln\left( \frac{2}{3}z+\frac{1}{3}\right)}{z\sqrt{1-\frac{2}{3}z}}\right] dz\\ &&+\underbrace{{{\int}^{1}_{0}}\left[\frac{-2\ln(3)z^{2}+z(2z+1)(1-z)\ln(2z+1)}{z^{3}(1-\frac{2}{3}z)^{\frac{3}{2}}}+\frac{2\ln(3)-2}{z }\right]dz}_{i}\\ &&+\frac{7\ln(3)}{3}{{\int}^{1}_{0}}\frac{dz}{\left( 1-\frac{2}{3}z\right)^{\frac{3}{2}}}. \end{array} $$

(62)

The latter integrals were calculated numerically as

$$\begin{array}{@{}rcl@{}} {{\int}^{1}_{0}}\left[\frac{-2\ln(3)z^{2}+z(2z+1)(1-z)\ln(2z+1)}{z^{3}(1-\frac{2}{3}z)^{\frac{3}{2}}}+\frac{2\ln(3)-2}{z}\right] dz&=&-2.3980,\\ {{\int}^{1}_{0}}\left[\frac{2-2\ln(3)}{z}-\frac{2+2\ln\left( \frac{2}{3}z+\frac{1}{3}\right)}{z\sqrt{1-\frac{2}{3}z}}\right] dz&=&-3.457723875,\\ \frac{7\ln(3)}{3}{{\int}^{1}_{0}}\frac{dz}{\left( 1-\frac{2}{3}z\right)^{\frac{3}{2}}}&=&\frac{14\sqrt{3}\ln(3)}{(3+\sqrt{3})}. \end{array} $$

(63)

Therefore, b _R for Kar’s metric is

$$ b_{R}=2\ln(6(2-\sqrt{3}))-0.226\zeta. $$

(64)

Finally, the deflection angle for Kar’s metric is

$$ \hat{\alpha}=-\ln\left[\frac{\theta D_{OL}}{u_{m}}-1\right]+2\ln(6(2-\sqrt{3}))-0.226\zeta-\pi+\ln\left[\frac{2\beta_{m}}{y_{m}}\right], $$

(65)

where u _m , β _m and y _m are given by (52), (22) and (46) respectively.