Astro 7B – Problem Set 6
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Magnification Factors in Gravitational Lensing
Figure 1 was sketched in class. The figure shows a source lensed by a mass M . The source
is in the shape of a thin circular arc of (angular) radius θs and radial width dθs θs .
The observer does not see the source, but sees instead its two lensed images, located above
and below the source. Each lensed image, like the parent source, is also a thin circular arc.
Image1 has radius θ1 and radial width dθ1 , and image2 has radius θ2 and radial width dθ2 .
Note that each point {a, b, c, d} on the source maps to a corresponding point on image1
{a0 , b0 , c0 , d0 } and image2 {a00 , b00 , c00 , d00 }.
This problem works out the area of image1 , A1 , and its relation to the source area, As —
and likewise for image2 . The total area of the two images, A1 + A2 , divided by the area of
the source, As , gives the total magnification M as a function of θs (how displaced the source
is relative to the lens).
Please recall that the Einstein ring is NOT actually illuminated unless θs = 0 (the source
is directly behind the lens). Nevertheless we can always draw the location of the Einstein
ring for reference, and this is done in Figure 1 — see the dashed circle. For this entire
problem, treat θE > 0, the angular radius of the Einstein ring, as a given known
quantity.
(a) Consider image1 . It is magnified azimuthally compared to the source by a factor Maz,1 .
In other words, the arclength b’–d’ is larger than the arclength b–d by a factor Maz,1 (the
same magnification factor relates arclengths a’–c’ and a–c).
Write down Maz,1 in terms of θ1 and θs .
(b) Rewrite the lens equation (derived in class)
θ2 − θs θ − θE2 = 0
(1)
as
θs = θ −
θE2
θ
Here θ can refer either to θ1 or θ2 .
Combine (2) with (a) to write down Maz,1 in terms of θ1 and θE .
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(2)
dθ1
Image1
b’
d’
a’
c’
dθS
θE
b
θS
a
c
θ1
d Source
Lens
θ2
d” dθ2
c”
b”
a”
Image2
Figure 1: Gravitational lensing. Deviations from perfectly circular arcs are due solely to the
instructor’s ineptitude with Keynote.
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(c) Now consider the radial magnification of image1 . The radial segment a’–b’ is larger (or
possibly smaller — you will find out) than the radial segment a–b by a factor Mrad,1 .
Write down Mrad,1 in terms of dθ1 and dθs . Then take the derivative of the rewritten lens
equation (2) to solve for Mrad,1 in terms of θ1 and θE .
(d) The total magnification M1 = Maz,1 Mrad,1 . Show that
M1 =
θ4
1 − E4
θ1
−1
(3)
Having shown this is true for M1 , simply replace subscript 1 with subscript 2 to
write down the analogous expression for M2 .
(e) Solve the original lens equation (1) for θ1 and θ2 in terms of θE and θS . Prove that
θ1 ≥ θE (image1 is always located outside the Einstein ring) and 0 > θ2 ≥ −θE
(image2 is always located inside the Einstein ring). Thereby also show that
M1 ≥ 1 (image1 is always magnified).
(Note that θ2 < 0 and M2 < 0 which simply mean that image2 is flipped relative to the
source; see Figure 1. If you get the right answers for M2 and θ2 , you should find that |M2 |
can be > 1 or < 1 — i.e., image2 can either be magnified or de-magnified depending on θs .
But you don’t have to show this.)
(f) By factoring (3) (it is the difference of two perfect squares), and combining with the
original lens equation (1) divided by θ2 , rewrite M1 as
M1 =
θ1
θS
2 −1
θ1
2 −1
θS
(4)
Also write down the analogous expression for M2 by simply swapping out subscript 1 for subscript 2.
(g) Define u ≡ θS /θE . Rewrite your answer for θ1 and θ2 in (e) to find that
!
√
2
θ1,2
1
u +4
=
1±
θs
2
u
(5)
(h) Combine (f ) and (g) to find
M1,2 =
1
u2 + 2
± √
2 2u u2 + 4
(i) Use (h) to solve for the TOTAL magnification:
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(6)
u2 + 2
M = M1 + |M2 | = √
u u2 + 4
(7)
Note the absolute value of M2 — see the parenthetical remark under part (e) above. Plot
M as function of u between u = 0 and u = ∞, and annotate on your plot the
precise values of M for u = 0.1 and u = 1.
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