SECTION IV.     A Singular Isothermal Sphere Lens With Shear (Off Origin)

    This section provides the step by step analytical solution for the image positions and magnifications in the case of an SIS lens with shear off the origin.  It then uses these calculations for comparison to demonstrate how to use gravlens to obtain such a solution.


Part 1. Analytical Analysis

    This exercise is very much like the one in the previous section except this time the source position will not be at the origin.  Thus, this case utilizes the same equations found in the previous section.  We will again assume shear of the system is:
(18) Gamma Value
and the value for the deflection angle, b, or Einstein radius is the same as that in the previous SIS examples:
(19) Einstein Radius Value
The source position, Beta, is defined as:
(20) Source Position Value
where the relation between  Beta and the Cartesian coordinate source position is defined by:
(21) Relationship Between Beta and (u, v)

We will also assume that the source lies on one of the axes (in this example, the y-axis) since the case in which the source does not lie on at least one of the axes cannot be solved analytically.  Thus, we will use:
(22) Source Position Coordinate Values

This lens model will produce four images.  From the equation for v, we find:
(23) v-Coordinate Equation
Using the equality to solve for x in the equation for u yields:
(24) u-Coordinate Solution
(25) x-Coordinate of Image Position Solution #1
We can now solve for the y coordinate:
(26) y-Coordinate of Image Position Equation
(27) y-Coordinate of Image Position Solution #1

Next we solve for x when y = 0:
(28) u-Coordinate Equation for Solution #2
(29) x-Coordinate Image Position Solution #2
Thus, we find the four images produced are:
(30) Image Coordinate Solution
or
(31) Angular Image Solutions

To find each image's magnification, we must solve for the each image's polar distance, r, and polar angle, Theta (Image Polar Angle):
(32) Polar Image Solution #1
(33) Polar Image Solution #2
(34) Polar Image Solution #3
(35) Polar Image Solution #4

This information can now be used to solve for each image's magnification:
(36) Image Magnification Solution

        Thus, analytical calculations for this lensing case yield the solution:
(37) Final Solutions


Part 2. Gravlens Analysis

        Now, we can use the gravlens software to obtain the same results. The model and its parameters are defined to be the same as those in the previous section.  The only thing that will differ in our input file is the source position. In this example, the Cartesian coordinates that define the source position are:
u = 0.070
v = 0.000

Thus, the input file for this run must contain the following information:
   startup 1 1
     alpha 0.70 0 0 0 0 0.1 0 0 0 1
     0 0 0 0 0 0 0 0 0 0
   findimg 0.070 0.000
 Now to use gravlens to perform this run, at the prompt we enter the command:
> gravlens <file>

The software will return the following:
findimg results:
7.000000e-002 0.000000e+000  # source
# 5 images:
-2.020732e-005 -3.574718e-012 8.502403e-008
-3.500000e-001 5.313487e-001 -6.518694e+000
-6.998889e-001 -2.648949e-022 1.111464e+001
-3.500000e-001 -5.313487e-001 -6.518694e+000
8.554646e-001 -2.435950e-022 3.942950e+000
These gravlens results will contain a fifth "ghost" image which can be ignored.  It will lie very close to the origin and will have an extremely low magnification.  This "image" is produced due to part of the code which attempts to smooth out the transition in finding the images.  Again, it should be noted that the analytical results will be a little off from the gravlens results due to rounding.

/\_INTRODUCTION_/\
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