Ast 8110 (LSS & GL)
Problem Set #3
Finding positions of images using two methods, (a) and (b).
All quantities below are assumed to be projected on the plane of the sky.
The mass distribution in the lens is axisymmetric, centered on (0,0), and its radial density
profile is given by
κ = κ0
|
~θ| + θ0
θ0
m
,
where κ0 = 30, θ0 = 1, m = −3.2. The density profile is truncated at θmax = 5, so κ = 0 outside
this radius.
Let us place a source on the positive θx-axis, at a distance θx = β = 2 from the origin. Because
of the symmetry, all the images will be located on the θx-axis.
(a) Time delay surface. Given this density distribution numerically calculate the crosssection of the time delay surface along the θx-axis. Use
T(θx) ∝
1
2
(θx − β)
2 −
1
π
Z
dθ0
x dθ0
y κ(θ
0
x
, θ0
y
) ln q
(θx − θ
0
x
)
2 + (θ
0
y
)
2.
Plot this cross-section of T as a function of θx. Locate the images as the local extrema. You don’t
need to calculate their θx values; just mark the images on the plot as accurately as you can. Label
the images by the type of extrema: maxima, minima, or saddle points.
(b) Lens equation. Lens equation can be written as
~θ − β~ = ~α.
In an axisymmetric case the contribution to the deflection angle at any location ~θ in the lens plane
comes only from the mass interior to |
~θ|. Therefore the magnitude of the deflection angle can be
calculated as
α =
M(≤ θ)
π θ .
Plot two lines on the same plot:
(i) θ − β vs. θ, and
(ii) α vs. θ.
According to the lens equation, images form where the two lines intersect. Indicate the images on
your plot. Are the positions you found using the lens equation consistent with those found using
the time delay surface?
