Blog Post 11, Worksheet 4.1, Problem 2

Consider a "lens" star at a distance \(D_L\) from the Earth, which lies near the line of sign to a background "source" star at a distance \(D_S > D_L\), as shown in the figure above. If the source and lens are separated by an angle \(\beta\). The bending of light from the source will create an image that is an angle \(\theta\) away from the lens, and an angle \(\alpha\) from the source. 

(Click on image to see larger version)

(a) Convince yourself that \(\beta = \theta - \alpha '\), where \(\alpha'\), the source-image angular separation, is distinct from \(\alpha\), the deflection angle of the source, in the figure above. This trivial expression is known as the lens equation. 

From the diagram, we can see that \(\alpha' + \beta = \theta\). To check this, think about the scenario in which the telescope, the lens, and the source are all directly lined up. The angle between the source and the lens, \(\beta\), would then be zero. Then \(\theta\), the angle between the image and the lens, is the same as \(\alpha'\), which agrees with the lens equation.


(b) Next, show that \[ \alpha' = (\frac{D_S-D_L}{D_S})\alpha\] and substitute it into the lens equation.

The key to understanding this equation is to realize that all the angles given in the above diagram are extremely small. Both \(\alpha\) and \(\alpha'\) are extremely small, but are not the same. Thus, we can think of two triangles with angles so small that they can be thought of as right triangles, with their hypotenuse and longest leg having approximately the same length:

As shown in the diagram above, both triangles share a sidelength, which we can arbitrarily name length L. Now, we can use trig to show the relationship between the two triangles: \[\text{sin}(\alpha) = \frac{L}{D_S} \approx \alpha'\] \[\text{sin}(\alpha) = \frac{L}{D_S-D_L} \approx \alpha\] If we solve for L in both equations, we get: \[L = \alpha(D_S - D_L) = \alpha'D_S\] Now, we can solve for \(\alpha'\): \[ \alpha' = (\frac{D_S-D_L}{D_S})\alpha\] Substituting this into the lens equation from part (a), we get: \[\beta = \theta - (\frac{D_S-D_L}{D_S})\alpha\]

(c) Now, substitute in the correct expression for \(\alpha\) from Question 1(d) to express \(\beta\) in terms of G, \(M_L\), b, \(D_L, D_S, \theta\), and c. Remember that your result from 1(d) is a factor of two smaller than the correct expression due to our Newtonian approximations. 

The correct expressions for \(\alpha\) from Problem 1d is: \[\alpha = \frac{4M_LG}{bc^2}\] Substituting this into the equation above, we get: \[\beta = \theta - (\frac{D_S-D_L}{D_S})(\frac{4M_LG}{bc^2})\]

(d) Replace b with an expression involving \(\theta\) and \(D_L\). Rearrange algebraically to show that \[\beta = \theta - (\frac{4GM_L}{\theta c^2})(\frac{D_S-D_L}{D_SD_L})\] Take a step back and remind yourself what each variable (\(\beta, \theta\), etc.) means and how they relate to each to each other. 

From Problem 1, we know that b is the distance between the lens and a photon of light. Combining these two images, we see the following setup:

Because theta is extremely small, we can solve for theta using the small angle theorem: \[\text{tan}(\theta) = \frac{b}{D_L} \approx \theta\] \[b = \theta D_L\] Now we can substitute this expression for b into the equation from part (c): \[\beta = \theta - (\frac{D_S-D_L}{D_S})(\frac{4M_LG}{bc^2})\] \[\beta = \theta - (\frac{D_S-D_L}{D_S})(\frac{4M_LG}{\theta D_Lc^2})\] \[\beta = \theta - (\frac{(D_S-D_L)4M_LG}{D_S D_L \theta c^2})\] \[\beta = \theta - (\frac{4GM_L}{\theta c^2})(\frac{D_S-D_L}{D_SD_L})\]

(e) When the lens and source lie along the exact same sight line, \(\beta = 0\). For this case, solve for the special value of \(\theta_E\), known as the Einstein ring radius, in terms of the lens mass, \(M_L\), and the relative parallax, \(\pi_{\text{rel}} = D_L^{-1} - D_S^{-1}\)

If we look more closely at the expression given for \(\pi_{\text{rel}}\), we see that it's very familiar:

\[\pi_{\text{rel}} = D_L^{-1} - D_S^{-1} = \frac{1}{D_L} - \frac{1}{D_S} = \frac{D_S}{D_SD_L} - \frac{D_L}{D_SD_L} = \frac{D_S - D_L}{D_SD_L}\]

\(\pi_{\text{rel}}\) is actually just a quantity already in the equation! With \(\beta = 0\), we are left with \[\theta_E = \pm(\frac{\pi_{\text{rel}}4M_LG}{c^2})^{1/2}\]

(f) For a typical lens of mass 0.3 \(M_\odot\) (an M-dwarf) at a distance \(D_L = 4\) kpc and a typical source at \(D_S = 8\) kpc, what is the sie of \(\theta_E\) in arcseconds?

This part just involves plugging in the given values for the equation we found in part (e) for \(\theta_E\): 

\[\theta_E = \pm(\frac{\pi_{\text{rel}}4M_LG}{c^2})^{1/2}\]

\[\pi_{\text{rel}} = \frac{D_S - D_L}{D_SD_L} = \frac{8 \text{ kpc} - 4 \text{ kpc}}{(8 \text{ kpc})(4 \text { kpc})} = \frac{4 \text{ kpc}}{32 \text{ kpc}^2} = \frac{1}{8 \text{ kpc}}\]

\[\theta_E = (\frac{(\frac{1}{8 \text{ kpc}})\:(4)\:(0.3 M_\odot * \frac{10^{33} \text{ g}}{M_\odot})\:(6.7 * 10^{-8} \text{ cm}^3 \text{ g}^{-1} \text{ s}^{-2})}{(3 * 10^{10} \text{ cm s}^{-1})^2})^{1/2} (\frac{ 4.8 * 10^{-6} \text{ arcseconds}}{ \text{1 radians}})\] \[\theta_E = \pm5.6 * 10^{-4} \text{ arcseconds}\]

(g) In astronomy, it is often very useful to case 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 see exactly how your quantity varies and by how much with changes in each variable. For example, Kepler's Law II recast in units relevant for a 'typical' stellar/planetary system might read: \[P = 1 \text{ yr } (\frac{a}{1 \text{ AU}})^{3/2} \:(\frac{M_\odot}{M_{\text{tot}}})^{1/2}\] 

You can easily see from this equation that, for a star-planet system whose semimajor axis measures 1 AU and whose stellar mass is 1 \(M_\odot\), the orbital period is 1 yr, as you would expect for the Sun-Earth system. If you wanted to find the orbital period for Mars around the Sun and you know \(a_{\text{Mars}}\) = 1.5 AU, you can readily deduce that its orbital period would be \(1.5^{3/2} * 1 \text{ yr} \approx 1.8 yr\)

Recast the equation you derived in 2(f) for \(\theta_E\) in terms of the typical lens mass and relative parallax. Your answer should appear in the form: \[\theta_E \approx \text{ mas } (\frac{M_{\text{lens}}}{M_{\text{lens (typical)}}})^x(\frac{\pi_{\text{rel}}}{\pi_{\text{rel(typical)}}})^y\] where mas is milli-arcseconds.

We are given that the mass of a normal lens, \(M_L\), is 0.3 \(M_\odot\), and the relative parallax, \(\pi_{\text{ relative}}\) is \(\frac{1}{8 \text{ kpc}}\), as calculated in the previous question. If we separate out the constants from the above equation, we are left with: \[\theta_E = \pm(\pi_{\text{rel}}^{1/2})(M_L)^{1/2}(\frac{4G}{c^2})^{1/2}\] Now, we can rearrange this formula in the form as asked:

\[\theta_E = \pm(\frac{4G}{c^2})^{1/2}(\frac{\pi_{\text{rel}}}{1/8 \text{ kpc}})^{1/2}(\frac{M_L}{0.3 M_\odot})^{1/2}\] We then calculate the final constant to get: \[\theta_E = \pm(5.6 * 10^{-1} \text{ mas})(\frac{\pi_{\text{rel}}}{1/8 \text{ kpc}})^{1/2}(\frac{M_L}{0.3 M_\odot})^{1/2}\]