The Geometry of Microlensing: Blog 10, Worksheet 4.1, Problem 2

In this part of the worksheet we’re going to do a bunch of geometry to figure out how massive objects can act as Galactic lenses and make background stars brighter through the influence of their gravitational field. The most important scale in lensing is the size of the so-called Einstein ring.

In this part of the worksheet we’re going to do a bunch of geometry to figure out how massive objects can act as Galactic lenses and make background stars brighter through the influence of their gravitational field. The most important scale in lensing is the size of the so-called Einstein ring.

Convince yourself that \( \beta = \Theta -  \alpha' \) , where α' , the source-image angular separation, is distinct from α, the deflection angle of the source, in the figure above. This trivial expression is known as the lens equation.
A) The lens equation makes sense with how each angle is defined in the diagram. I am sufficiently convinced. \[ \beta = \Theta -  \alpha' \]

Next, show that \[ \alpha' = \left( \frac{D_S - D_L }{D_S} \right) \alpha \] and plug it in.
B) Zooming in on specific triangles, we can see that (with a little fudging of distances):

\[ x = sin( \alpha) (D_s - D_L ) =  \alpha (D_s - D_L ) \]

\[ x = sin( \alpha') (D_s ) =  \alpha' (D_s ) \] \[ \alpha' (D_s ) =  \alpha (D_s - D_L )\]

 Plugging in the expression gives: \[ \beta = \Theta -  \left( \frac{D_S - D_L }{D_S} \right) \alpha  \]

Now substitute in the correct expression for α from Question 1(d) to express β in terms of G, ML, b, DL, DS, θ, and c. Remember that you result from 1(d) is a factor of two smaller than the correct expression due to our Newtonian approximations.
C) Using \[ alpha = \frac{4GM_L}{b c^2}\] from question 1, we get: \[ \beta = \Theta -  \frac{2GM_L}{b c^2} \left( \frac{D_S - D_L }{D_S} \right) \] But that if off by a factor of two, due to general relativity, so the equation is really \[ \beta = \Theta -  \frac{4GM_L}{b c^2} \left( \frac{D_S - D_L }{D_S} \right) \]

Replace b with an expression involving θ and DL. Rearrange algebraically to show that \[ \beta = \Theta -  \frac{4GM_L}{\Theta c^2} \left( \frac{D_S - D_L }{D_S D_L} \right) \]
D)

By the small angle approximation we can see that \[sin( \Theta ) = \frac{ b}{D_L} \] \[ b = \Theta D_L \] Plugging this in gives: \[ \beta = \Theta -  \frac{4GM_L}{\Theta c^2} \left( \frac{D_S - D_L }{D_S D_L} \right) \]

When the lens and source lie along the exact same sight line, β “ 0. For this case, solve for the special value of θE, known as the Einstein ring radius, in terms of the lens mass, ML, and the relative parallax, πrel = D^-1 L - D^-1.
E)  \[ \pi_{rel} = D_L^{-1} - D_S^{-1} \] Rearranging \( \pi_{rel} \) gives us a more useful expression.  \[ \pi_{rel} = \frac{D_S  }{D_S D_L} - \frac{D_L }{D_S D_L}= \frac{D_S - D_L }{D_S D_L} \] Plugging in our new information and solving gives:  \[ \beta = \Theta -  \frac{4GM_L}{\Theta c^2} \left( \frac{D_S - D_L }{D_S D_L} \right) \] \[ 0 = \Theta_E -  \frac{4GM_L}{\Theta_E  c^2} \left( \pi_{rel} \right) \] \[ \Theta_E =  \frac{4GM_L}{\Theta_E  c^2} \left( \pi_{rel} \right) \]  \[ \Theta_E = \left( \frac{4GM_L \pi_{rel} }{ c^2 } \right)^{ \frac{1}{2}} \]

For a typical lens of mass 0.3Md (an M-dwarf) at a distance DL = 4 kpc and a typical source at DS = 8 kpc, what is the size of θE in arcseconds?
F)  This is just a matter of plugging in the given information. Converting the givens to cgs gives:

• \(M = 0.3M_{\odot} = 0.6 \times 10^{33} = 6 \times 10^{32} \)
• \( D_L  = 4 \: kpc = 1.2 \times 10^{22} \: cm \)
• \( D_S = 8 \: kpc = 2.5 \times 10^{22} \: cm \)

\[ \Theta_E = \left( \frac{4GM_L  }{ c^2 }  \left( \frac{D_S - D_L }{D_S D_L} \right)  \right)^{ \frac{1}{2}} \] \[ \Theta_E = \left( \frac{4(6.67 \times 10^{-8} )(  6 \times 10^{32}) }{ (3 \times 10^{10})^2 }  \left( \frac{2.5 \times 10^{22} - 1.2 \times 10^{22} }{2.5 \times 10^{22} (1.2 \times 10^{22})} \right)  \right)^{ \frac{1}{2}} \]  \[ \Theta_E = \left( 1.78 \times 10^{5}  \left( 4.33 \times 10^{-23} \right)  \right)^{ \frac{1}{2}} \] \[ \Theta_E = 2.8 \times 10^{-9} radians\] \[ \Theta_E = 1.6 \times 10^{-7} \: degrees = 0.6 \: milli-arcseconds \]

In astronomy, it is often very useful to cast equations in terms of isolated variables measured in units of their typical order-of-magnitude values, with any power law dependencies explicitly labelled for each variable. There should be only one numerical constant out front bearing the proper units for the quantity of interest. This means calculating out all the remaining constants. In equations written in this way, not only can you directly read off the typical numerical value for your quantity of interest, but you can also see exactly how your quantity varies and by how much with changes in each variable.  Recast the equation you derived in 2(f) for θE in terms of the typical lens mass and relative parallax. Your answer should appear in the form: \[ \Theta_E = mas \left( \frac{M_{lens}}{M_{lens \: (typical)}} \right) \left( \frac{\pi_{lens}}{\pi_{lens \: (typical)}} \right) \]

G) \[ \Theta_E = mas \left( \frac{M_{lens}}{M_{lens \: (typical)}} \right) \left( \frac{\pi_{lens}}{\pi_{lens \: (typical)}} \right) \] \[ \Theta_E = 0.6 \: milli-arcseconds  \left( \frac{M_{lens}}{M_{lens \: (typical)}} \right)^{\frac{1}{2}} \left( \frac{\pi_{lens}}{\pi_{lens \: (typical)}} \right)^{\frac{1}{2}} \]  \[ \Theta_E = 0.6 \: milli-arcseconds  \left( \frac{M_{lens}}{6 \times 10^{32} g }   \right)^{\frac{1}{2}} \left( \frac{\pi_{lens}}{4.33 \times 10^{-23} \: mas } \right)^{\frac{1}{2}} \]

That is a lot of information. In the next question, we will convert some of this information into a more accessible format.