Simple lens models 391forθ ≥0. Due to the axial symmetry it is enough to considerβ≥0. Since
m(θ)≥0 it follows thatθ≥β(forθ≥0). Instead of equation (14.34) we get
+(θ)= 2∫θ0θ′dθ′k(θ′)ln(
θ
θ′)
, (14.55)
whereas the Fermat potential can be written as
(θ,β)=^12 (θ−β)^2 −+(θ). (14.56)This way we get the lens equation (14.54) from
∂(θ,β)
∂θ= 0. (14.57)
To get the Jacobi matrix we write:
α(θ)=m(θ)
θ^2θ (withθ=(θ 1 ,θ 2 )andθ=|θ|)and thus
A=(
10
01
)
−
m(θ)
θ^4(
θ 22 −θ 12 − 2 θ 1 θ 2
− 2 θ 1 θ 2 θ 12 −θ 22)
−
2 k(θ)
θ^2(
θ 12 θ 1 θ 2
θ 1 θ 2 θ 22)
, (14.58)
wherewemadeuseofm′(θ)= 2 θk(θ). The determinant of the Jacobi matrix is
given by
detA=(
1 −
m
θ^2)(
1 −
d
dθ(m
θ))
=
(
1 −
m
θ^2)(
1 +
m
θ^2− 2 k)
. (14.59)
14.3.1.1 Tangential and radial critical curves
The critical curves (the points for which detA(θ)=0) are then circles of radius
θ. From equation (14.59) we see that there are two possible cases:
(1) θm 2 =1 : defined astangential critical curve;and
(2) ddθ(mθ)=1: defined asradial critical curve.For case (1) one getsm/θ=θand thus from the lens equation (14.54) we
see thatβ=0 is the corresponding caustic, which reduces to a point. If the axial
symmetric gets only slightly perturbed this degeneracy is lifted.
We can look at the critical points on theθ 1 -axis withθ=(θ, 0 ),θ >0. Then
A= 1 −
m(θ)
θ^2(
− 10
0 + 1
)
−
m′
θ(
10
00
)
(14.60)
and this matrix must have an eigenvectorXwith eigenvalue zero. For symmetry
reasons, the vector must be either tangential,X=( 0 , 1 ), or normal,X=( 1 , 0 ),