390 Gravitational lensing
(a) nI≥ 1
(b) ntot<∞
(c) nI+nIII= 1 +nII
(d) for|β|sufficiently large ntot=nI= 1.It thus follows from (c) that the total number of images ntot= 1 + 2 nIIis odd.
The number of images with positive parity (nI+nIII) exceeds by one those with
negative parity (nII); nII≥nIIIand ntot> 1 if and only if nII≥ 1. The number of
images is odd; however, in practice some images may be very faint or be covered
by the lens itself and are thus not observable.
Theorem 14.2.The image of the source which will appear first to the observer
is of type I and it is at least as bright as the unlensed source would appear
(μ(θ 1 )≥ 1 ).
For a proof of the two theorems we refer to [4]. The second theorem is a
consequence of the fact that the surface mass densitykis a positive quantity.
14.3 Simple lens models
14.3.1 Axially symmetric lenses
Let us consider a lens with an axially symmetric surface mass density, that is
&(ξ)=&(|ξ|), in which case the lens equation reduces to a one-dimensional
equation. By symmetry we can restrict the impact vectorθto be on the positive
θ 1 -axis, thus we haveθ=(θ, 0 )withθ>0. We can then use polar coordinates:
θ′ =θ′(cosφ,sinφ)(thus d^2 θ′=θ′dθ′dφ). Withk(θ)=k(θ)we get for
equation (14.32)
α 1 (θ)=1
π∫∞
0θ′dθ′k(θ′)∫ 2 π0dφθ−θ′cosφ
θ^2 +θ′^2 − 2 θθ′cosφ, (14.51)
α 2 (θ)=1
π∫∞
0θ′dθ′k(θ′)∫ 2 π0dφ−θ′sinφ
θ^2 +θ′^2 − 2 θθ′cosφ. (14.52)
Due to symmetry,αis parallel toθand with equation (14.52) we getα 2 (θ)=0.
Only the mass inside the disc of radiusθaround the centre of the lens contributes
to the light deflection, therefore from equation (14.51) one finds
α(θ)≡α 1 (θ)=2
θ∫θ0θ′dθ′k(θ′)≡m(θ)
θ. (14.53)
This way we can write the lens equation as
β=θ−α(θ)=θ−m(θ)
θ