Monday, September 28, 2015

Blog #11: WS4.1, #5

5. Let's think about the time dependence of magnification of a lensing event by considering the lens and the source stars as they move through the Galaxy. The stars have a non-zero relative, projected speed of \(v_{rel}\), which appears as a relative angular motion \(\mu_{rel}\) on the sky. A typical \(\mu_{rel}\) might be 10.5 mas/yr. As a result of this relative proper motion, the projected angular distance between the objects, u, will change with time. We'll call the projected distance of closest approach \(u_0\). It is analogous to an impact parameter. 
a) Give an equation for u(t), the instantaneous projected separation as a function of time, in terms of \(u_0, t_0\), and anything else relevant. \(t_0\) is the time at which the closest projected approach occurs, so \(u(t_0) = u_0\). Recall that u(t) is measured in units of the Einstein radius. 
The source star is moving along the arrow labeled \(\mu_{rel}\), and how far it has moved on that line is \(\mu_{rel}\) multiplied by a ∆t. We know that at \(t_0\), the source star is at its minimum distance \(u_0\), so we can use that time point to calibrate the time point we use. By using the right triangle, we can find the value of u(t). \(\theta_E\) is used to normalize \(\mu_{rel}\). 
\([u(t)]^2 = [u_0]^2+[\frac{\mu_{rel}(t-t_0)}{\theta_E}]^2\)
\(u(t) = [u_0^2+(\frac{\mu_{rel}(t-t_0)}{\theta_E})^2]^{1/2}\)

b) What might be a good timescale (\(t_E\) in this problem? Rewrite part a in terms of this timescale. Express this timescale algebraically, involving 'typical' values for \(M_L, \pi_{rel}\), and \(\mu_{rel}\). Find a typical \(t_E\) we might encounter for microlensing events in the Milky Way. 

A good timescale will simplify the expression and make it easier to tell that the units cancel out, so \(t_E = \frac{\theta_E}{\mu_{rel}}\) 
\(u(t) = [u_0^2+(\frac{t-t_0}{t_E})^2]^{1/2}\)
\(t_E = \frac{2}{c\mu_{rel}}(GM_L\pi_{rel})^{1/2}\) 
\(t_E = \frac{2[GM_{L(typ)}\pi_{rel(typ)}]^{1/2}}{c\mu_{rel(typ)}}(\frac{\mu_{rel(typ)}}{\mu_{rel}})(\frac{M_L}{M_{L(rel)}})^{1/2}(\frac{\pi_L}{\pi_{L(rel)}})^{1/2}\)
\(\theta_E = (\frac{4GM_L}{c^2}\pi_{rel})^{1/2} = (\frac{M}{10^{11.09}M_{\odot}})^{1/2}(\frac{\pi_{rel}^{-1}}{Gpc})^{-1/2} arcsec\) (with help)
If the typical values are: 
\(M_L = 0.3M_{\odot}\)
\(\pi_{rel} = \frac{D_S-D_L}{D_SD_L} = \frac{8kpc-4kpc}{(4kpc)(8kpc)} = 0.125/kpc\)
\(\mu_{rel} = 10.5 mas/yr \rightarrow 10.5*10^{-3}arcsec/yr\)
Then \(\theta_E = (\frac{0.3M_{\odot}}{10^{11.09}M_{\odot}})^{1/2}(\frac{8*10^{-6}Gpc}{Gpc})^{-1/2} arcsec = 5.52*10^{-4}arcsec\)
\(t_E = \frac{\theta_E}{\mu_{rel}} = \frac{5.52*10^{-4}arcsec}{10.5*10^{-3}arcsec/yr} = 0.0525 yr \rightarrow 19 days\)

c) Trace out a microlensing curve. How does changing the the lens mass and \(u_0\) affect the shape of the light curve? 

A is the magnification of the source, and t is the time. The peak occurs when the source is as close to being directly in line with the source as possible. As the mass of the lens increases, the Einstein radius increases (since \(\theta_E = (\frac{4GM_L}{c^2}\pi_{rel})^{1/2}\)), so the time that it takes for the source to pass across the Einstein radius increases, and the peak widens. By increasing the \(u_0\), the source will be farther out of line with the lens, and will therefore not be amplified as effectively; thus the peak will fall. 

1 comment:

  1. Yes, perfect! I see you have internalized our discussion at TALC session! Keep it up!

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