What does microlensing look like?

We are bounded in a nutshell of Infinite space: Blog Post #11, Worksheet # 4.1, Problem #3: What does microlensing look like?

3. When speaking about microlensing, it is often easier to refer to angular quantities in units of $\theta_E$. Let’s define $u \equiv \beta/\theta_E$  and $y \equiv \theta/\theta_E$.

(a) Show that the lens equation can be written as: $$u = y - y^{-1}$$

(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 $\mu as$ (micro-arcseconds), indicate the positions of the images in a drawing.

(a) From previous problem, the equation for a lens is given as $$\beta = \theta - \alpha^{\prime} ,$$ establishing how the differences in angle between the image seen, the original source, and the lens can be equated, seen in the following illustration:

In another problem, $\alpha^\prime$ was also expanded on, taking into consideration the actual relativistic effects of lensing (using 4 instead of 2 in the $\delta v$ equation). The equation for $\alpha^\prime$ is : $$\alpha^\prime = \frac {4 G M_L}{\theta c^2} (\frac{D_S - D_L }{D_S D_L}) ,$$ where $D_S$ and $D_L$ are distances between the Source the observer and the distance between the Lens and the observer.   This equation is worked off the equation seen in the previous problem solved for angle $\alpha$ and the expanded lens equation given: $$\beta = \theta - \alpha^\prime ,$$ $$\beta = \theta - (\frac{D_S - D_L }{D_S D_L})\alpha ,$$ $$\beta = \theta - \frac {4 G M_L}{\theta c^2}(\frac{D_S - D_L }{D_S D_L}).$$ Now notice how all these equations are closely tied to $u = y - y^{-1}$ with the definitions for u and y given at the start of the problem. Now let us divide all by $\theta_E$: $$\frac{\beta}{\theta_E} = \frac{\theta - (\frac{D_S -D_L }{D_S D_L})\frac {4 G M_L}{\theta c^2} }{\theta_E} ,$$ and now we have: $$u = y - \frac{(\frac{D_S - D_L }{D_S D_L})\frac {4 G M_L}{\theta c^2} }{\theta_E} .$$ And $\theta_E$ is actually defined by factoring out the $\theta$ in the lens equation when $\beta$ is 0, ($\theta_E$ is the Einstein ring radius, a circle which forms around the Lens when the Source is directly behind it), which turns into: $$\theta_E = \left(\frac {4 G M_L}{ c^2} (\frac{D_S - D_L }{D_S D_L})\right)^{\frac{1}{2}} .$$ When we substitute this into the new lens equation with u and y we have: $$u = y - \frac{(\frac{D_S - D_L }{D_S D_L})\frac {4 G M_L}{\theta c^2} }{\left(\frac {4 G M_L}{ c^2} (\frac{D_S - D_L }{D_S D_L})\right)^{\frac{1}{2}}} ,$$ and many things start to cancel out after $\theta$ is taken out as a factor of the denominator:

$$u = y-\frac{\left(\frac {4 G M_L}{ c^2} (\frac{D_S - D_L }{D_S D_L})\right)^{\frac{1}{2}}} {\theta}  ,$$ which, in effect is the inverse of y, $\frac{\theta_E}{\theta} .$

(b)  This part of the problem asks us to solve for $y(u)$, which means the new lens equation must be re written again. We take the original equation: $$u = y-y^{-1} ,$$ and multiply all values by y to yield: $$yu = y^2 - 1 ,$$ $$0 = y^2 - yu -  1 ,$$ and we solve for y using the quadratic formula: $$y(u) = \frac{u \pm \sqrt{u^2 + 4}}{2} .$$ Next we use understand the values from a previous problem referenced in the question. So the lens angular separation is: $$\beta = 100 \mu as = 1\times 10^{-1} miliarcseconds ,$$ and the Einstein ring radius is: $$5.6 \times 10^{-1} mili-arcseconds .$$ The Einstein ring radius is found by plugging in $M_L = 0.3 M_\odot ,$ $D_L = 4kpc ,$ and $D_S = 8kpc$ into the $$\theta_E equation \to \left(\frac {4 G M_L}{ c^2} (\frac{D_S-D_L }{D_S D_L})\right)^{\frac{1}{2}} .$$ Now then, by dividing $\beta$ and $\theta_E$ values, we now have the value of u, which is $\approx 1.8 \times 10^1$, a unit-less value that serves to find what proportion to the Einstein radii the image is found. This value is now plugged into: $$y(u) = \frac{u \pm \sqrt{u^2 + 4}}{2} ,$$ and yields two values: $$y(1.8 \times 10^1) = 9 \times 10^{-2} \pm 1 ,$$ depending on whether the 1 is positive or negative.

The two values, 1.09 and -0.91, are the percentages of the Einstein radii where the image is found, as 0.18, the value of u, is the percentage of the radii where the original source is. This can be viewed as:

Where $I_1$ is the 1.09 value image,  $I_2$ is the -0.91 value image, L is the lens, and S is the original source.