Blog #11, Worksheet #4.1, Problem 3, Angular Quantities of Microlensing

3) When speaking about microlensing, it is often easier to refer to angular quantities in units of \(\theta_E\). Let’s define \(u = \frac{β}{θ_E}\) and \(y = \frac{\theta}{\theta_E}\)

(a) Show that the lens equation can be written as: \[u = y - y^{-1}\]We know that the lens equation takes the form \[\beta = \theta - \frac{4GM_L \cdot \pi_{rel}}{\theta c^2} \]We also showed in question 2 that \[{\theta_E}^2 = \frac{4GM_L \cdot \pi_{rel}}{c^2} \]Using these facts and rearranging the given quantities \[\beta = u \theta_E \qquad and \qquad \theta = y \theta_E \]We substitute in to find \[u \theta_E = y \theta_E - \frac {{\theta_E}^2}{y \theta_E}\]Canceling all of our \(\theta_E\) we get the desired answer of \[u = y - y^{-1}\]

(b) Solve for the roots of y(u) in terms of u. These equations prescribe the angular position of the images as a function of the (mis)alignment between the source and lens. For the situation given in Question 2(f) and a lens-source angular separation of 100 µas (micro-arcseconds), indicate the positions of the images in a drawing.

We begin by solving for y in terms of u. We start with our equation from part a) \[u = y - y^{-1}\] Rearranging for y we get \[y^2 - uy - 1 = 0 \]which solves quadratically as \[y = \frac {u \pm (u^2 + 4)^{\frac{1}{2}}}{2}\]We begin by solving for u which we know is \(u = \frac{\beta}{\theta_E}\) where \[{\theta_E}^2 = \frac{4GM}{c^2} \cdot \frac{D_s - D_l}{D_s D_l}\]Where \[D_s = 8kpc \qquad D_l = 4kpc \qquad M = .3M_\odot \]When we solved for this in question 2 we found that \(\theta_E \thickapprox 1x10^{-6} \: arcseconds\) and we know that \(\beta = 100 \mu as \) Which means that \[u = \frac{\beta}{\theta_E} = \frac{100x10^{-6}}{1x10^{6}} = 100\]We then solve for y and get \[y_+ = \frac{100 + ((100)^2 + 4)^\frac{1}{2}}{2} \thickapprox 100\]and \[y_- = \frac{100 - ((100)^2 + 4)^\frac{1}{2}}{2} \thickapprox\ -1x10^{-2}\]We know that \[y = \frac{\theta}{\theta_E} = \frac{\theta}{1x10^{-6}}\]So finally solving for theta we get \[\theta_+ = 10^{-4} \: arcseconds\] and \[\theta_- = 10^{-8} \: arcseconds\]