Blog #11: Microlensing

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

(a) Show that the lens equation can be written as: 
\begin{align}
u \equiv y - y^{-1}
\end{align}

Let's recall that the lens equation was obtained from Problem 2 and was defined as follows:
\begin{align} \beta = \theta - \alpha \end{align} This equation was later proved to be rewritten in part 2(d) as:
\begin{align}
\beta = \theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)
\end{align}
Therefore, we know that:
\begin{align}
\alpha' = \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)
\end{align}

All of this information will come in handy in just a minute. First, let's try to rewrite the lens equation in terms of $\theta_E$:
\begin{align}
\beta &= \theta - \alpha'\\
\frac{\beta}{\theta_E} &= \frac{\theta}{\theta_E} - \frac{\alpha'}{\theta_E}\\
\end{align}

In order to get the lens equation in the form $u = y - y^{-1}$, the following three conditions need to be true:
\begin{align}
(1) && u = \frac{\beta}{\theta_E}\\
(2) && y = \frac{\theta}{\theta_E}\\
(3) && y^{-1} = \frac{\alpha'}{\theta_E}
\end{align}

Since we don't know what $\theta_E$ is, we can't be sure if this relation will hold true if the three conditions above are met. So let's solve for $\theta_t$, given that $y^{-1} = \frac{\theta_E}{\theta}$ and $\alpha = \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)$ :
\begin{align}
y^{-1} &= \frac{\alpha'}{\theta_E}\\
\frac{\theta_E}{\theta} &= \frac{\alpha'}{\theta_E}\\
\theta_E^2 &= \alpha' \theta\\
\theta_E^2 &= \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right) \times \theta\\
\theta_E &= \left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}
\end{align}

In order to solve prove that the lens equation can be rewritten as $u = y-y^{-1}$, let's substitute this equation with the lens equation and see if it holds true:
\begin{align}
u &= y - y^{-1}\\
\frac{\beta}{\theta_E} &= \frac{\theta}{\theta_E} - \frac{\alpha'}{\theta_E}\\
\frac{\theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}} &= \frac{\theta}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}} - \frac{\frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}}\\
\frac{\theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}} &= \frac{\theta - \frac{4GM_L}{\theta c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}}
\end{align}

Since both sides equal to each other means that the three conditions above held true, and the lens equation can be re-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 position of the image in a drawing. 

In order to solve for the roots of $y(u)$ in terms of $u$, we can look at the equation $u = y - y^{-1}$ and rewrite it and modify it to look like a polynomial equation as follows:
\begin{align}
u &= y - y^{-1}\\
0 &= y - y^{-1} - u\\
0(y) &= (y - y^{-1} - u) (y)\\
0 &= y^2 - uy - 1\\
\end{align}

Having the equation written in a polynomial form, we can solve for the roots of $y(u)$ in terms of $u$ using the quadratic formula as follows:
\begin{align}
y = \frac{u \pm \sqrt{u^2 + 4}}{2}
\end{align}

Now that we have an equation to get the two roots of $y$ in terms of $u$, let's try to calculate $u$. But before we try to calculate $u$, let's see what is all the information given to us:

• $\beta = 100 \mu as = 0.1$ milli-arcseconds
• $M_L = 0.3 M_{\odot}$
• $D_L = 4 kpc = 4000 \text{ pc}$
• $D_S = 8 kpc = 8000 \text{ pc}$
• $G = 4.3 \times 10^{-3} \frac{\text{pc} \cdot \text{km}^2}{M_\odot \cdot s^2}$
• $c = 3 \times 10^5 \frac{\text{km}}{s}$

We know from part (a) that $u = \frac{\beta}{\theta_E}$, and $\theta_E = \left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}$, which plug in the known values to solve for $u$:
\begin{align}
u &= \frac{\beta}{\theta_E}\\
u &= \frac{\beta}{\left[ \frac{4GM_L}{c^2}\left(\frac{D_s - D_L}{D_S D_L}\right)\right]^{\frac{1}{2}}}\\
u &= \frac{100 \mu as}{\left[ \frac{4(4.3 \times 10^{-3} \frac{\text{pc} \cdot \text{km}^2}{M_\odot \cdot s^2})(0.3 M_{\odot})}{(3 \times 10^5 \frac{\text{km}}{s})^2}\left(\frac{8000 \text{ pc} - 4000 \text{ pc}}{(8000 \text{ pc}) (4000 \text{ pc})}\right)\right]^{\frac{1}{2}}}\\
u &= \frac{100 \mu as}{\left[ \frac{4(4.3 \times 10^{-3} \frac{\text{pc} \cdot \text{km}^2}{M_\odot \cdot s^2})(0.3 M_{\odot})}{(3 \times 10^5 \frac{\text{km}}{s})^2}\left(\frac{8000 \text{ pc} - 4000 \text{ pc}}{(8000 \text{ pc}) (4000 \text{ pc})}\right)\right]^{\frac{1}{2}}}\\
u &= \frac{0.1 \text{ milli -arcseconds}}{2.7 \times 10^{-9} \text{ radians}}\\
u &= \frac{0.1 \text{ milli -arcseconds}}{0.6 \text{ milli-arcseconds}}\\
u &= 0.17 \text{ milli-arcseconds}
\end{align}

Now that we have solved for $u$, we can solve for the roots of $y(u)$ as follows:
\begin{align}
y &= \frac{u \pm \sqrt{u^2 + 4}}{2}\\
y &= \frac{0.17  \text{ milli-arcseconds}\pm \sqrt{(0.17  \text{ milli-arcseconds})^2 + 4}}{2}\\
y &= 1.09  \text{ milli-arcseconds}\\
&= -0.92 \text{ milli-arcseconds}
\end{align}

Okay, so now we have the 2 roots of $y$. One of them is a positive value and the other is a negative value. These two values have a real physical impact in how microlensing works. Look at the image below:

The positive value correlates to the larger, right side of the distorted image in the Einstein ring, and the negative value correlates to the smaller, left side of the distorted image in the Einstein ring. The positive and negatives indicates that the two images are mirror images of each other, with the positive value image being larger than the negative value image.