Point gravitational lenses

14.1.1

Problem 14.1.1: Optical lens vs. gravitational lens

Compare the dependence of refraction angle $\hat{\alpha}$ on impact parameter $p$ for optical and gravitational lenses.

solution

When light is deflected in optical lens, its deflection angle is proportional to impact parameter. In contrast, gravitational field of gravitational lens decreases with distance, and so does deflection angle.
Considering opaque lenses and small angles, the light ray deflection angle is inversely proportional to impact parameter.
Two main conclucions can be drawn from these differences in deflection angle dependencies for optical and gravitational lenses. First, unlike optical lens, gravitational lens has no focal point. Indeed, rays, which passed through the gravitational lens, can not intersect in one point, since rays with larger impact parameters would cross the axis of the system at a greater distance from lens, than rays with smaller impact parameters $p$.
Instead, gravitational lens has focal semiaxis. Moreover, opacity of the lens leads to the existence of shadow area behind the lens, which location is determined by the rays, passed close to the lenses edge.
Second, in contrast to optical lens, the intensity of registered radiation after passing the gravitational lens is infinite not in focal point, but on entire focal semiaxis. However, this singularity exists only in small angles approximation and dissapears if we account for finite size of the source.
It's important to understand the notion of small angles in the context of gravitational lensing. Rays near the axis of optical lens have small deflection angles, so that their impact parameter is small. Small deflection angle of light ray in gravitational lens means that impact parameter is large (much larger, than gravitational radius of the lens, to be more precise).

14.1.2

Problem 14.1.2: Deflection angle: Newronian approach

Obtain the formula $\hat{\alpha}=r_g/p$ for refraction of light ray using the Newtonian theory.

solution

Let's consider small deflections of the ray. This condition is satisfied if kinetic energy of the particle is much larger than its maximum potential energy in the gravitational field: $$ {GmM\over \xi} \ll{mv_0^2\over 2}, $$ or $$ \xi \gg r_g = {2GM\over c^2}. $$ Most of the gravitating bodies have sizes much larger than their gravitational radius, so that this condition is always true. In the first order approximation: $$ \hat{\alpha} \simeq - {\Delta v_z\over c}. $$ To obtain $\Delta v_z$ one needs to consider equations of motion. Equation of motion along $z$ axis reads $$ m\frac{dv_z}{dt} = - f\cos \varphi = - \frac{f\xi}{r}, $$ and thus $$ \Delta v_z = - \xi GM\int_0^\infty {\frac{dt}{r^3 (t)}}. $$ Using $dt = dx/c$ and integrating over $\varphi$, we obtain: $$ \Delta v_z = - \frac{2GM}{\xi c}, $$ i.e. $$ \hat{\alpha} = {2GM\over \xi c^2} ={r_g\over \xi}. $$ Note, that obtained expression is half of the correct expression, derived using general relativity.

14.1.3

Problem 14.1.3: Deflection of light near the Sun

Calculate the angle of refraction of light in the gravitational field of the Sun.

solution

Let's calculate the deflection angle of rays, rassing near the Sun's surface, i.e. for $\xi = R_\odot$. Sun's radius is $R_\odot = 7 \cdot 10^5~km$, its gravitational radius is $r_{g \odot } = 2.96~km$. Thus, $$ \hat{\alpha} = {2 r_g\over \xi} = 8.4 \cdot 10^{-6} = 1.74'' $$

14.1.4

Problem 14.1.4: Refractive index

Propagation of light in gravitational field could be considered as propagation in a medium. Calculate the effective refractive index for such a medium.

solution

Photons travel along the trajectory with zero interval $ds^2 = 0$. Using Schwarzschild metric $$ \displaystyle ds^2 = \left(1 - {r_g\over r}\right) c^2 dt^2 - {dr^2\over 1 - {r_g\over r}} - r^2 \left(d\theta^2 + \sin^2\theta d\varphi^2\right), $$ we can obtain speed of light at different distances from gravitating body: $$ c_g = c \left(1 - {r_g\over r}\right). $$ Then, index of refraction is $$ n_g \equiv {c\over c_g} = \left(1 - {r_g\over r}\right)^{-1} \simeq 1 + {r_g\over r} $$ In general: $$ n_g = \left(1 + {2\Phi\over c^2}\right)^{-1} \simeq 1 - {2\Phi\over c^2}. $$ Note, that this index of refraction is effective and has no direct physical meaning. Obtained expressions are valid when $$ {\left|\Phi \right|\over c^2} \ll 1. $$ Using effective refraction index, one can write eikonal equation and derive Snell's law for spherically layered media: $$ n_g(r)r\sin\nu(r) = const, $$ where $\nu(r)$ is an angle between position vector and tangent to the ray.

14.1.5

Problem 14.1.5: Shadow area

Determine the dependence of a ray's shifting from the axis of symmetry after the refraction on a nontransparent lens. Find the region of shadow and estimate its size, considering the Sun as a lens.

solution

Let's consider the case, when source is located at infinite distance from the lens, so that input beam is parallel. Deflection can be characterized by deflection angle vector: $$ \vec{\hat{\alpha}} = - {2r_g \vec{\xi}\over \xi^2}, $$ which is constructed in a way to account for deflection of ray towards the lens. Furthermore, if considering only small angle, the ray can be represented as two asymptotes. The deflection of the ray, thus, is $$ \vec \rho(\vec{\xi},x) = \vec{\xi} + x\vec{\hat{\alpha}} = \vec{\xi}\left(1 - {2r_g D_d\over \xi^2}\right). $$ Note the differences of this dependence from the corresponding dependence for optical lens: $$ \vec \rho(\vec{\xi},x) = \vec{\xi} \left(1 - {D_d \over F}\right). $$ All rays, which are parallel to the axis of optical lens, intersect in one point $D_d = F$, independently on impact parameter. This is not true for gravitational lens, since from $\rho = 0$ we have $$ D_d = {\xi^2\over 2r_g}, $$ so that location of intersection point depends on impact parameter. Since $\xi$ is in range from $R$ (radius of the lens) to $\infty$, intersection points fill semiaxis $D_d \ge x_{\min} = R^2 /2r_g$. Segment $D_d \le x_{\min }$ is, thus, in shadow. Let's estimate $x_{\min }$ for the Sun. Using obtained expression for deflection angle $$ \hat{\alpha} = {2 r_g\over \xi} $$ we obtain $$ x_{\min\odot} = {R_\odot\over \hat{\alpha}_{\odot}} \sim 10^{11}~\mbox{km}. $$

14.1.6

Problem 14.1.6: Scales

What scales of angles and distances determine the position of the images of the light source after the passage through the gravitational lens? Consider two cases: 1) the source and the lens are at cosmological distances from the observer; 2) the distance from the observer to the lens is much smaller than the distance to the source.

solution

Introducing the new quantity $$ \alpha_0^2 = 2r_g {D_{ds}\over D_s D_d}. $$ the lens equation can be rewritten as $$ \vartheta^2 - \beta\vartheta - \alpha_0^2 = 0. $$ It can be shown, that $\alpha_0$ determines the distance between the images. The characteristic distance scale is $$ \xi_0 = \alpha_0 D_d, $$ which is just characteristic value of impact parameter. Let's consider the first case. Given the distances are of order of Hubble length $$ {D_d D_{ds}\over D_s} = \eta{c\over H_0}, $$ $$ {D_{s}D_d\over D_{ds}} = \eta'{c\over H_0}, $$ one can substitute it and rewrite in dimensionless form: $$ \alpha_0 \approx 1.67 \cdot 10^{-6}\sqrt{M\over M_\odot}\sqrt{h\over\eta'}~arcsec, $$ $$ \xi_0 \approx 0.022\sqrt{M\over M_\odot}\sqrt{\eta\over h}~\mbox{pc}. $$ When $D_d \ll D_{ds} \approx D_s$ we obtain $$ \alpha_0 \approx 3 \cdot 10^{ - 3} \sqrt {\frac{M}{M_ \odot }} \sqrt {\frac{1\mbox{kpc}}{x}}~arcsec, $$ $$ \xi_0 \approx 4 \cdot 10^{13} \sqrt {M\over M_\odot} \sqrt {{x\over 1\mbox{kpc}}}~\mbox{sm}. $$

14.1.7

Problem 14.1.7: Multiple images

Show that when the gravitational lens is placed between source and observer in the general case the two images of the source would be observed. How are the images placed relative to the lens and observer?

14.1.8

Problem 14.1.8: Einstein ring

How should source, gravitational lens and observer be placed relative to each other in order to observe the Einstein ring? Calculate the radius of the ring.

solution

Setting $\beta = 0$ in lens equation (i.e., considering the case when source, lens and observer lie on the same line), we obtain: $$ \vartheta_E^2 = 2r_g{D_{ds}\over D_s D_d}. $$ In this case two observed images degenerate into ring, which is called the Einstein ring. Its radius is $$ l_E = \vartheta_E D_d = \sqrt{2r_g {D_d D_{ds}\over D_s}}.$$ The lensing equation can be rewritten in a more convenient form using angle $\vartheta_E$: $$ \beta = \vartheta - {\vartheta_E^2 \over \vartheta}, $$ while location of images in this notation is determined by the equation $$ \vartheta_{\pm} = {1\over 2}\left(\beta \pm \sqrt{\beta^2+4\vartheta_E^2}\right). $$

14.1.9

Problem 14.1.9: Source of finite size

How would the Einstein ring change if we take into account the finite size of the source? Estimate the space characteristics of the observed image assuming that the radius of the lens is much smaller than the raius of the lens.

solution

Considering the case of the finite size source, we can first investigate the image of the disk source of radius $R$. Angle distance between the images of two uttermost point of the disk is $\beta$. Moreover, since the problem is axisymmetric, the Einstein ring would have finite width. Since its angular width is $\beta$, it's easy to obtain, that width is equal to the radius of the source projected on the lens plane.
In a more precise manner this can be considered as follows. Let the boundary of the source to be described by $\tilde\rho_S(\varphi)$. If its size is so that $\tilde\rho_S(\varphi)\ll\tilde{l}$, we can obtain $\vec p_{1,2} \simeq \vec{\tilde{\rho_S}} \left\{\frac 1 2 \pm {\frac{\tilde{l}}{ \tilde{\rho_S}}}\right\}.$ For disk of the projected radius $\tilde R_S$ Einstein ring has finite width of $$ \Delta p = p_1 - p_2 \simeq \tilde{R_S}, $$ with the mean radius $\tilde l$, as in the case of point source.

14.1.10

Problem 14.1.10: Source of finite size

Qualitatively consider the general situation, when a source of finite size, lens and observer are not on one line. Estimate the angular sizes of the observed images.

solution

When source, lens and observer do not lie on the same line, the image contracts in the direction of source-lens line projected on lens plane. When source shifts at a distance equals to its radius, width of the image becomes zero in this direction. With further displacement from the lens-observer axis image splits on two parts. Linear dimansions of the image in the lens plane can be estimated as follows: since source and two images are viewed at angle $$ \alpha \simeq {\tilde R_S \over \tilde \rho_S}, $$ the sizes of the images are $$ d_{1,2} \simeq 2p_{1,2} \alpha = 2\tilde R_S \left\{ {1\over 2} \pm {\tilde l\over \tilde \rho_S} \right\}. $$

14.1.11

Problem 14.1.11: Angular shift of the Einstein ring

Calculate the angular shift of the Einstein ring from the circle of the gravitational lens. Estimate it, considering the Sun as a lens. Is this value observable?

solution

Angular gap between ring and lens is $$ \delta ={l_E - R\over D_d}. $$ Assuming, that source is located at a large distance $D_s \gg D_d$, we obtain $$ \delta \approx \sqrt{2r_g\over R} - {R\over D_d}. $$ Maximum value of th gap corresponds to $x = 4x_{\min} - 2R^2/r_g$ and is equal to $r_g /2R$. The estimate of this quantity for the Sun as a lens is: $$ \delta_\odot = {1\over 4} \hat{\alpha}_{\odot} = 0.44''. $$ Such values can be relatively easy observed from space, but remember, that this is the maximum value, so that distance from the lens should be large.

14.1.12

Problem 14.1.12: Double Einstein ring

Recently the exceptional phenomenon was observed using the Hubble space telescope: the double Einstein ring, formed by the influence of the gravitational field of the galaxy on the light from two other more distant galaxies. What conditions are necessary for the observation of this phenomenon?

14.1.13

Problem 14.1.13: Brightness amplification

Calculate the energy amplification coefficient for images produced by the gravitational lens. Determine its peculiarities. Compare with an optical lens.

solution

The surface brightness is conserved under the deflection of light rays. However, angular dimensions of images differ from angular dimension of the source. Amplification, thus, is given by $$ \mu = {dS_{image}\over dS_{source}}. $$ For axisymmetric lenses (which point lenses are) $$ \mu = {\vartheta\over \beta }{d\vartheta \over d\beta}. $$ Substituting $\beta$ from lensing equation and introducing the parameter $u=\beta/\vartheta_E$, we obtain the expression $$ \mu_{\pm} = {u^2+2\over 2u\sqrt{u^2+4}}\pm {1\over 2} $$ for amplification of images, while the total energy amplification is $$ \mu=|\mu_{+}|+|\mu_{-}|={u^2+2\over u\sqrt{u^2+4}} $$ These expression are valid when $x>x_{\min}$, while $\mu = 0$ for $x<x_{\min}$.
When observer is located on the source-lens axis, the amplification tends to infinity $\mu \to\infty$. Thus, unlike optical lens, gravitational lens produces image on any distance from it. When moving away from the lens the intenity increases as $\sqrt {D_d}$.