In Question 3, you assumed that the lens and sources were point sources. You found that the lens projects the source into two images on opposing sides of the chord connecting the lens and source, one outside the Einstein ring (positive, or major image) and the other inside (negative, or minor image).
For sources with finite area, the images are elongated tangentially (i.e. in the axis perpendicular to the lens-source alignment) and compressed radially (i.e. in the axis parallel to this alignment). The geometry for a circular source is shown below.
(a) Referring to the picture above, reason that the factor by which each image is tangentially elongated is y/u
(b) Reason that the factor by which each image is radially compressed is dy/du.
(c) Therefore, show the area of each image is enlarged by the factor
\[A_{\pm}=\left|\frac{y_{\pm}}{u}\right| \left|\frac{dy_{\pm}}{du}\right|\]
(d) The surface brightness per area of the source is conserved in the images. This means that the total brightness of the images scales directly with its area. Consequently, the total magnification of the system during a microlensing event is found by adding the area the images and dividing by the area of the original source. That is,
\[A \equiv |A_+| + |A_-|\]
After class, feel free to convince yourself that
\[A=\frac{u^2+2}{u(u^2+4)^{1/2}}\]
Using Equation 4, what is the maximum magnification of a source crossing tangent to the Einstein ring of the lens?
(e) For an additional challenge after class, you can try to show that the maximum magnification in a microlensing event where the source crosses well within the Einstein ring of the lens (i.e. u << 1) approaches:
\[A(u \rightarrow 0) \rightarrow 1/u\]
\[A=\frac{u^2+2}{u(u^2+4)^{1/2}}\]
Using Equation 4, what is the maximum magnification of a source crossing tangent to the Einstein ring of the lens?
(e) For an additional challenge after class, you can try to show that the maximum magnification in a microlensing event where the source crosses well within the Einstein ring of the lens (i.e. u << 1) approaches:
\[A(u \rightarrow 0) \rightarrow 1/u\]
(a) Referring to the picture above, reason that the factor by which each image is tangentially elongated is y/u
First we need to define \(u\) and \(y\). \(u\) is the angle between the source and the lens measured from Earth divided by the angular radius of the Einstein ring. Basically it is a unit-less quantity that measures the visual distance between the source and the lens. in the above diagram, \(u\) is the vertical axis. \(y\) is another unit-less quantity which is the ratio of a) the angular distance between the lens and the image, and b) the Einstein ring radius. Basically, it measure the separation of the center of the ring and the upper image.
Looking at the picture, we can form a triangle with 2 legs being the distance from S to L and the third being the diameter of S. The distance from S to L is \(u\), and the distance from L to the upper image I is \(y\). These two triangles are similar, so:
\[\frac{S}{u}={I}{y}\]
We want to solve for the stretch factor, so we want:
\[\frac{I}{S}={u}{y}\]
Which checks out perfectly.
(b) Reason that the factor by which each image is radially compressed is dy/du.
This is a bit less obvious. It follows the same principle as a move, but now we need to imagine that a little bit of the sphere is put through the lens transformation at a time. This small bit follows the same reasoning as before, but because we are using radial compression with small pieces, instead of \(u\) and \(y\) we use \(du\) and \(dy\). Thus:
\[\frac{I}{S}=\frac{du}{dy}\]
(c) Therefore, show the area of each image is enlarged by the factor
\[A_{\pm}=\left|\frac{y_{\pm}}{u}\right| \left|\frac{dy_{\pm}}{du}\right|\]
Area is proportional to both length and height, so we can simply multiply our modification factors for this variables to gather to get the factor for area.
\[\boxed{A_{\pm}=\left|\frac{y_{\pm}}{u}\right| \left|\frac{dy_{\pm}}{du}\right|}\]
(d) The surface brightness per area of the source is conserved in the images. This means that the total brightness of the images scales directly with its area. Consequently, the total magnification of the system during a microlensing event is found by adding the area the images and dividing by the area of the original source. That is,
\[\frac{S}{u}={I}{y}\]
We want to solve for the stretch factor, so we want:
\[\frac{I}{S}={u}{y}\]
Which checks out perfectly.
(b) Reason that the factor by which each image is radially compressed is dy/du.
This is a bit less obvious. It follows the same principle as a move, but now we need to imagine that a little bit of the sphere is put through the lens transformation at a time. This small bit follows the same reasoning as before, but because we are using radial compression with small pieces, instead of \(u\) and \(y\) we use \(du\) and \(dy\). Thus:
\[\frac{I}{S}=\frac{du}{dy}\]
(c) Therefore, show the area of each image is enlarged by the factor
\[A_{\pm}=\left|\frac{y_{\pm}}{u}\right| \left|\frac{dy_{\pm}}{du}\right|\]
Area is proportional to both length and height, so we can simply multiply our modification factors for this variables to gather to get the factor for area.
\[\boxed{A_{\pm}=\left|\frac{y_{\pm}}{u}\right| \left|\frac{dy_{\pm}}{du}\right|}\]
\[A \equiv |A_+| + |A_-|\]
After class, feel free to convince yourself that
\[A=\frac{u^2+2}{u(u^2+4)^{1/2}}\]
\[A=\frac{u^2+2}{u(u^2+4)^{1/2}}\]
Using Equation 4, what is the maximum magnification of a source crossing tangent to the Einstein ring of the lens?
We are told that:
\[A=\frac{u^2+2}{u(u^2+4)^{1/2}}\]
and we are asked to maximize \(A\). Well, since there is a \(u\) in the denominator, if we bring this to 0, the equation will blow up to infinity. That sounds pretty maximized in my book.
So, \(A\) is maximized when \(u \rightarrow 0\).
(e) For an additional challenge, you can try to show that the maximum magnification in a microlensing event where the source crosses well within the Einstein ring of the lens (i.e. u << 1) approaches:
\[A(u \rightarrow 0) \rightarrow 1/u\]
Here is our equation:
\[A=\frac{u^2+2}{u(u^2+4)^{1/2}}\]
We know that \(u\) is going to approach 0, so let's make some comparisons. \(u^2\) will be insignificant when added to 2 or 4, so let's just ignore it. Why not? 1+0.00000000000001 is pretty much 1 as far as anyone is concerned.
\[A=\frac{2}{u(4)^{1/2}}\]
Now can take the easy square root.
\[A=\frac{2}{2u}\]
Let's cancel the 2's, and get our answer.\[\boxed{A=\frac{1}{u}}\]
That's not exactly rigorous mathematics, but I don't see any glaring flaws.
We are told that:
\[A=\frac{u^2+2}{u(u^2+4)^{1/2}}\]
and we are asked to maximize \(A\). Well, since there is a \(u\) in the denominator, if we bring this to 0, the equation will blow up to infinity. That sounds pretty maximized in my book.
So, \(A\) is maximized when \(u \rightarrow 0\).
(e) For an additional challenge, you can try to show that the maximum magnification in a microlensing event where the source crosses well within the Einstein ring of the lens (i.e. u << 1) approaches:
\[A(u \rightarrow 0) \rightarrow 1/u\]
Here is our equation:
\[A=\frac{u^2+2}{u(u^2+4)^{1/2}}\]
We know that \(u\) is going to approach 0, so let's make some comparisons. \(u^2\) will be insignificant when added to 2 or 4, so let's just ignore it. Why not? 1+0.00000000000001 is pretty much 1 as far as anyone is concerned.
\[A=\frac{2}{u(4)^{1/2}}\]
Now can take the easy square root.
\[A=\frac{2}{2u}\]
Let's cancel the 2's, and get our answer.\[\boxed{A=\frac{1}{u}}\]
That's not exactly rigorous mathematics, but I don't see any glaring flaws.
I worked with B. Brzycki, G. Grell, and N. James on this problem.
for (e)..it's a Taylor series
ReplyDeleteand 6... I like your explanations..
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