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IOAA 2007, Теория — Задача 4. GRAVITATIONAL LENSING

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Theoretical Competition — 1st IOAA — December 5, 2007 · 5 декември 2007 г. · 10 т.

Условие

Hubble Space Telescope photograph of the galaxy cluster Abell 2218 showing many galaxies and thin arcs of gravitationally lensed background galaxies; the credit strip inside the image reads “Gravitational Lens in Abell 2218”, “HST · WFPC2”, “PF95-14 · ST ScI OPO · April 5, 1995 · W. Couch (UNSW), NASA”.

The deflection of light by a gravitational field was first predicted by Einstein in 1912 a few years before the publication of the General Relativity in 1916.

A massive object that causes a light deflection behaves like a classical lens.

This prediction was confirmed by Sir Arthur Stanley Eddington in 1919.

Lensing geometry: observer O at the bottom, lens L on the vertical dotted line above O at distance D_L, source plane at distance D_S with the source S and the images S1 (right) and S2 (left); a light ray from S is deflected by angle φ at impact parameter ξ from L and reaches O; angles β (to S) and θ (to S1) measured from the vertical at O. — Figure 1

Consider a spherically symmetric lens, with a mass $M$ with an impact parameter $\xi$ from the centre. The deflection equation in this case is given by:

\phi = \frac{4GM}{\xi c^{2}} \quad \text{, a very small angle}

In figure 1, the massive object which behaves like a lens is at L. Light rays emitted from the source S being deflected by the lens are observed by observer O as images $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ . Here, $\phi, \beta$ , and $\theta$ are very very small angles.

a) For a special case in which the source is perfectly aligned with the lens such that $\beta = 0$ , show that a ring-like image will occur with the angular radius, called Einstein radius $\theta_{E}$ , given by:

\theta_{E} = \sqrt{\left(\frac{4GM}{c^{2}}\right)\left(\frac{D_{S} - D_{L}}{D_{L}D_{S}}\right)}

[2 т.]

b) The distance (from Earth) to a source star is about 50 kpc. A solar-mass lens is about 10 kpc from the star. Calculate the angular radius of the Einstein ring formed by this solar-mass lens with the perfect alignment. [1 т.]

c). What is the resolution of the Hubble space telescope with 2.4 m diameter mirror? Could the Hubble telescope resolve the Einstein ring in b)? [2 т.]

d). In figure 1, for an isolated point source S, there will be two images $\left(\mathrm{S}_{1}\text{ and }\mathrm{S}_{2}\right)$ formed by the gravitational lens. Find the positions ( $\theta_{1}$ and $\theta_{2}$ )of the two images. Answer in terms of $\beta$ and $\theta_{E}$ . [2 т.]

e). Find the ratio $\dfrac{\theta_{1,2}}{\beta}$ ( $\dfrac{\theta_{1}}{\beta}$ or $\dfrac{\theta_{2}}{\beta}$ ) in terms of $\eta$ . Here $\theta_{1,2}$ represents each of the image positions in d.) and $\eta$ stands for the ratio $\dfrac{\beta}{\theta_{E}}$ . [2 т.]

f). Find also the values of magnifications $\dfrac{\Delta\theta}{\Delta\beta}$ in terms of $\eta$ for $\theta = \theta_{1,2}$ ( $\theta = \theta_{1}$ or $\theta = \theta_{2}$ ), when $\Delta\beta \ll \beta$ , and $\Delta\theta \ll \theta$ . [1 т.]

Отговори

Решение

QUESTION 4 GRAVITATIONAL LENSING

Solution

The lensing geometry of Figure 1 redrawn with the point P on the source plane above L: S2, P, S, S1 on the source line; lens L at distance D_L from observer O and D_S - D_L from the source plane; total distance D_S; deflection angle φ, impact parameter ξ, angles β and θ at O. — Figure 1

a). From figure 1, for small angles, $\tan\theta \approx \theta$ , hence

\mathrm{PS}_{1} = \mathrm{PS} + \mathrm{SS}_{1}

\theta D_{S} = \beta D_{S} + \left(D_{S} - D_{L}\right)\phi

\theta - \beta = \frac{4GM}{\xi c^{2}}\frac{\left(D_{S} - D_{L}\right)}{D_{S}} \qquad \text{………………….(1)}

( 1 point)

Note that $\theta = \dfrac{\xi}{D_{L}}$ also, hence,

\theta^{2} - \beta\theta = \left(\frac{4GM}{c^{2}}\right)\left(\frac{D_{S} - D_{L}}{D_{L}D_{S}}\right) \qquad \text{………………….(2)}

For a perfect alignment in which $\beta = 0$ , we have $\theta = \pm\theta_{E}$ , where

\theta_{E} = \sqrt{\left(\frac{4GM}{c^{2}}\right)\left(\frac{D_{S} - D_{L}}{D_{L}D_{S}}\right)} \qquad \text{………………….(3)}

( 1 point)

b). From equation (3), for a solar-mass lens with $D_{S} = 50\ \mathrm{kpc}$ ,

$D_{L} = 50 - 10 = 40\ \mathrm{kpc}$

\theta_{E} = \sqrt{\left(\frac{4GM}{c^{2}}\right)\left(\frac{D_{S} - D_{L}}{D_{L}D_{S}}\right)} = 0.956 \times 10^{-9}\ \text{radian} = 1.97 \times 10^{-4}\ \text{arc second}

( 1 point)

c.) The resolution of the Hubble space telescope having diameter of 2.4 m is,

\theta_{Hubble} = 1.22\frac{\lambda}{D} = 1.22 \times \frac{5 \times 10^{-7}\ \mathrm{m}}{2.4\ \mathrm{m}} = 2.54 \times 10^{-7}\ \text{radian}

for light of wavelength 500 nm. ( 1 point)

Hence the Hubble telescope could not resolve this Einstein ring. ( 1 point)

d). The quadratic equation (2) has two distinct roots, namely,

\theta_{1} = \frac{\beta}{2} + \sqrt{\left(\frac{\beta}{2}\right)^{2} + \theta_{E}^{2}}

( 1 point)

\theta_{2} = \frac{\beta}{2} - \sqrt{\left(\frac{\beta}{2}\right)^{2} + \theta_{E}^{2}}

( 1 point)

where

\theta_{E} = \sqrt{\left(\frac{4GM}{c^{2}}\right)\left(\frac{D_{S} - D_{L}}{D_{L}D_{S}}\right)}

This implies that there are two images for a single isolated source.

e).

\frac{\theta_{1,2}}{\beta} = \frac{\frac{\beta}{2} \pm \sqrt{\left(\frac{\beta}{2}\right)^{2} + \theta_{E}^{2}}}{\beta} = \frac{1}{2} \pm \sqrt{\left(\frac{1}{2}\right)^{2} + \left(\frac{1}{\eta}\right)^{2}} = \frac{1}{2}\left[1 \pm \frac{\sqrt{\eta^{2} + 4}}{\eta}\right]

(1 point)

f). From equation (2) $\theta^{2} - \beta\theta - \theta_{E}^{2} = 0$

\left(\theta + \Delta\theta\right)^{2} - \left(\beta + \Delta\beta\right)\left(\theta + \Delta\theta\right) - \theta_{E}^{2} = 0

\frac{\Delta\theta}{\Delta\beta} = \frac{\theta}{2\theta - \beta}

(1 point)

\left(\frac{\Delta\theta}{\Delta\beta}\right)_{\theta=\theta_{1,2}} = \frac{\theta_{1,2}}{2\theta_{1,2} - \beta} = \frac{\frac{1}{2}\eta \pm \sqrt{1 + \frac{\eta^{2}}{4}}}{\pm 2\sqrt{1 + \frac{\eta^{2}}{4}}} = \frac{1}{2}\left[1 \pm \frac{\eta}{\sqrt{\eta^{2} + 4}}\right]

(1 point)

Оригинал в Архива: exam_theory.pdf · официални решения: exam_theory_solution.pdf