# Galaxies Lecture 04

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### Gravitational Lensing

Today we’ll talk about gravitational lensing as a scattering problem. Suppose we have a particle which comes within impact parameter $b$ of a scattering body, and is deflected by angle $\theta$ . We found last time that

$\tan \Psi _{0}={-bv_{0}^{2} \over GM}\,\!$ where $\Psi _{0}$ is the angle to the “knee” of the path of the scattered particle. We then noticed that $\theta =2\Psi _{0}-\pi$ , so that

$\tan \left({\theta \over 2}\right)={GM \over bv_{0}^{2}}\,\!$ If this particle is a photon, and deflection angles are small (as they usually are), we can say $\tan \left({\theta \over 2}\right)\approx {\theta \over 2}$ and $v_{0}=c$ , so we have:

$\theta ={2GM \over bc^{2}}\,\!$ This is almost correct, and we haven’t even invoked GR. If we do, the only difference we find is that of a factor of 2:

${\theta ={4GM \over bc^{2}}}\,\!$ We can plug in for our sun to find the deflection angle of light by our sun. We use $b=7\cdot 10^{10}$ , and we find that $\theta =8.5\cdot 10^{-6}\approx 1.4$ ". This is, of course, Einstein’s famous prediciton.\

Let’s set up a more general problem. Suppose an emitting source object (S), a lensing object (L), and an observer (O) are all in a line, so that light from the emitter can be bent around to massive object on its way to the observer. We would like to calculate the angle $\theta _{s}$ that light can be emitted at from the source so that it is lensed to hit the observer. We will call $R_{E}$ the distance from L to the knee of the path of the light. $D_{SL}$ is the distance from S to L, $D_{L}$ is the distance from the observer to L ($D_{S}$ is the distance to S), $\theta$ is the scattering angle, and $\theta _{E}$ is the angle at which the photon hits the observer.\

Obviously, $\theta =\theta _{S}+\theta _{E}$ , and for small angles, $\theta _{S}={R_{E} \over D_{SL}}$ , and $\theta _{E}={R_{E} \over D_{L}}$ . We use that $\theta ={4GM \over c^{2}R_{E}}$ and solve our original equation:

${R_{E}^{2}={4GM \over c^{2}}{D_{SL}D_{L} \over D_{L}+D_{SL}}}\,\!$ This is an equation for the radius of an Einstein Ring. It is a ring, of course, because any path that passes at distance $R_{E}$ around L will be deflected to O. It should be noted that a gravitational lens is not the same as a conventional lens which focusses light to a point. Rather, light that passes closer to the center is deflected more. A piece of glass which would lens light in this way would look like the bases of two wine glasses glued together along their bottom side. This is useful for developing an intuition for how light goes through a gravitational lens.\Getting back to our problem, we can solve for $\theta _{E}$ :

$\theta _{E}=\left({4GM \over c^{2}}\right)^{\frac {1}{2}}\left({D_{SL} \over D_{L}(D_{SL}+D_{L})}\right)^{\frac {1}{2}}\,\!$ If L lies directly in the middle of S and O, then we find that:

$\theta _{E}=\left({4GM \over c^{2}}{1 \over D_{S}}\right)^{\frac {1}{2}}\,\!$ which is just twice the Schwarzschild radius divided by the twice the distance to a lensing object. We can estimate the Schwarzschild radius of the sun ($3km$ ) and the distance to a lensing object ($3kpc$ ), and we get that the angle of an Einstein ring on the sky is about 1 marcsec. This is prohibitively small for current optical telescopes, but using VLBI, radio astronomers have been able to observe these. There have only been a couple of these observed because it is very rare to find collinear S, L, and O. \

Let’s examine what happens if they are not collinear. Instead of getting a ring, we will get a caustic, which is the generalized envelope of images produced by a surface (think sun reflecting off the surface of a swimming pool). Let’s consider the case that a source object (S) passes through a collection of potential scattering bodies before it reaches the observer. In order to be deflected, light must pass within an Einstein radius of one of these bodies. We can calculate the optical depth $\tau$ of the scattering bodies to lensing:

$\tau =N\pi \theta _{E}^{2}\,\!$ where $N$ is the number of scattering bodies per steradian. In general $\tau$ is quite small. We’ll estimate $\tau$ for stars in the bulge of the Milky Way.

$N\approx {\#stars \over \Omega _{bulge}}\,\!$ Now $\#stars={M_{tot} \over \left\langle M_{*}\right\rangle }$ , and $\left\langle M_{H}\right\rangle \approx 0.4M_{\odot }$ , giving us $\#stars\approx 2.5\cdot 10^{10}$ . The bulge on the sky is about $20^{\circ }\times 30^{\circ }\sim {\frac {1}{6}}str$ . Plugging these values in, we find

$\tau _{bulge}\sim 4.5\cdot 10^{-6}\,\!$ That is, we need to look at several million stars to find a lensing event. Now that we have computers looking for these events, we find them, but this would be impossible to do by hand. This is why lensing has only recently become a hot topic in research. If we repeated this calculation for the disk, we get $\tau _{disk}\sim 6\cdot 10^{-6}$ , which is about the same.\

Microlensing is the effect of a lensing object increasing the number of pathways for light to hit an observer, making an object appear to brighten and then dim as a lensing object passes by. These events may be distinguished by many other events by the fact that it does not depend on wavelength.\

Let’s set up one more problem: that of the non-collinear S, L, and O. We’ll define $\theta$ to be the angle the incoming light makes with the path from O to L, $\alpha$ is the angle that light is deflected from the straight line it was following out from the source, and $\beta$ is the angle between the lines $OL$ and $OS$ . We’ll also define the point I to be where O “observes” S to be (where the image is).\We know $\alpha ={4GM \over c^{2}b}$ , and $\theta _{E}^{2}={4GM \over c^{2}}{D_{SL} \over D_{S}D_{L}}$ . $b=D_{L}\theta$ , and we’ll define $\eta \equiv D_{S}\beta =D_{S}\theta -D_{SL}\alpha$ . We then find:

${\beta =\theta -{\theta _{E}^{2} \over \theta }}\,\!$ The above equation is written incorrectly in Binney & Tremaine. We may solve for $\theta$ as well, using $\mu \equiv {\beta \over \theta _{E}}$ :

${\theta \over \theta _{E}}={\mu \pm {\sqrt {\mu ^{2}+4}} \over 2}\,\!$ There is also a corresponding light path which goes on the other side of L and reach O. If we cannot resolve this split, then our telescope will just detect the fact that the solid angle for that source will have increased on the sky, and since photons are conserved per solid angle, we will see that as a brightening. We may actually work out the solid angles represented on the sky on each side of L of our source S (called $A_{+}$ and $A_{-}$ for the brighter (close side) and dimmer (far side) of L). Working out the arithmetic, we find:

${{A_{\pm } \over A_{S}}={\frac {1}{2}}\left(1\pm {\mu ^{2}+2 \over \mu (\mu ^{2}+4)^{\frac {1}{2}}}\right)}\,\!$ 