Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 72 / 3

Use the small-angle, thin gravitational lens equation geometry. An undeflected ray seen at angular position $\theta$ would have transverse source position $D_s\theta$. Bending that ray through physical angle $\hat{\alpha }$ at the lens changes its source intercept by $D_{ds}\hat{\alpha }$. Thusand the required physical deflection isAlso $\beta =\eta /D_s$. Define the reduced deflection $\alpha$ as the angular difference $\theta -\beta$. Dividing the first equation by $D_s$ givesThese are angular diameter distances; in a cosmological spacetime one must not replace $D_s$ by $D_d+D_{ds}$ as if the distances were ordinary collinear Euclidean lengths.

Substituting $\eta =D_s\beta$ and $\xi =D_d\theta$ into the previous geometric relation verifiesFor a point mass gravitational lens, the given physical deflection yieldsAn aligned source has $\beta =0$, and therefore ${\theta }^2={\theta }_E^2$. Rotational symmetry turns these signed intersections into an Einstein ring, whose Einstein radius isFor an offset source, multiplying $\beta =\theta -{\theta }_E^2/\theta$ by $\theta$ gives a quadratic equation. Its two image positions areFor $\beta >0$, ${\theta }_{+}>0$ lies on the source side of the lens and ${\theta }_{-}<0$ lies on the opposite side.

Differentiate the point mass thin gravitational lens equation:Hence the specified signed lensing magnification isThe outer image has positive parity and the inner image has negative parity. Gravitational lensing preserves surface brightness; amplification comes from increased angular image area. The observed flux magnifications are $|{\mu }_{+}|$ and $|{\mu }_{-}|$; a negative determinant reverses orientation, not flux. Substitution of the two roots, with $u=|\beta |/{\theta }_E$, givesAt $\beta =0$, axial symmetry produces an Einstein ring. For an ideal point-like astronomical source, the lensing magnification diverges: near alignment ${\mu }_{\mathrm{tot}}\sim 1/u$. Finite source size gives a finite observed flux and a ring of finite thickness.

For $0<\beta <{\theta }_E$, there are two images on opposite sides. Their separation is $\sqrt{{\beta }^2+4{\theta }_E^2}$, approaching $2{\theta }_E$ at alignment. For $\beta \ll {\theta }_E$, their positions are ${\theta }_{\pm }\simeq \pm {\theta }_E+\beta /2$ and ${\mu }_{\pm }\simeq 1/2\pm {\theta }_E/(2\beta )$: both images are bright, with the outer one slightly brighter. Moving away from alignment reduces the total amplification and makes the inner image relatively fainter.

For $\beta \gg {\theta }_E$, expanding the two image roots givesThus the distant source has one almost unaltered bright image and an extremely faint opposite-parity secondary, with total flux magnification $1+2{\theta }_E^4/{\beta }^4+O(({\theta }_E/\beta {)}^6)$. The secondary persists for every finite nonzero source offset in the point mass model.

It is useful to rewrite the local lensing convergence asFor positive core radius, the requested limits areA singular isothermal sphere lens instead has $\kappa (\theta )={\theta }_E/(2|\theta |)$, which diverges at the centre. The finite core of the softened isothermal lensing potential regularizes this central lensing convergence; at large angular radius the two models have the same leading profile.

For an axisymmetric gravitational lens, the reduced deflection is $\alpha (\theta )=\bar{\kappa }(\theta )\theta$. The specified mean lensing convergence therefore givesThe derivative is smallest at the centre, where ${\beta }^{\prime }(0)=1-{\theta }_E/{\theta }_c$. If ${\theta }_c\ge {\theta }_E$, the mapping is monotone and an offset source has only one image. If $0<{\theta }_c<{\theta }_E$, the derivative is negative near zero and positive at large radius, producing two stationary points and a range of source offsets with three images. Thus the condition for a nonempty multiple-image region isThis condition permits multiple images; a particular source must additionally lie inside the multiple-image caustic of a softened isothermal lensing potential. At equality the critical radii collapse to the origin and there is no finite three-image region.

Put $r=|\mathbf{\theta }|$ and write the vector thin gravitational lens equation as $\mathbf{\beta }=(1-\bar{\kappa }(r))\mathbf{\theta }$. A tangential displacement at fixed radius has eigenvalue ${\lambda }_t=1-\bar{\kappa }$. A radial displacement has eigenvalue ${\lambda }_r=1-{\alpha }^{\prime }(r)$. From the definition of the mean lensing convergence,Since $\alpha =r\bar{\kappa }$, this gives ${\alpha }^{\prime }=2\kappa -\bar{\kappa }$. Consequently,The two requested gravitational-lensing critical curve conditions are thereforeThey correspond respectively to a tangential critical curve of an axisymmetric lens and a radial critical curve of an axisymmetric lens.

The tangential condition $\bar{\kappa }=1$ gives $\sqrt{{\theta }_t^2+{\theta }_c^2}={\theta }_E$, henceThis is a nonzero gravitational-lensing critical curve precisely when ${\theta }_E>{\theta }_c$. It is also the radius of the Einstein ring for an aligned source in this cored model, so the strength parameter ${\theta }_E$ itself is not the cored model's ring radius.

For completeness the radial condition is equally explicit. Subtracting the mean lensing convergence from twice the local one givesThus the radial critical curve of an axisymmetric lens has radiusagain nonzero exactly when ${\theta }_E>{\theta }_c$.

The conditions are singularities of the image-plane Jacobian matrix, where the formal lensing magnification of a point-like astronomical source diverges. A small source is strongly stretched in the corresponding eigenvector direction: tangentially near ${\theta }_t$, radially near ${\theta }_r$. Their mapped loci are source-plane gravitational-lensing caustics, which determine where image counts change.

The tangential circle maps to $\beta =0$ and gives the aligned Einstein ring. To find the radial caustic, put $R={\theta }_E/{\theta }_c>1$. At ${\theta }_r$, $\sqrt{{\theta }_r^2+{\theta }_c^2}={\theta }_c{R}^{1/3}$ and ${\theta }_r={\theta }_c\sqrt{R^{2/3}-1}$. Substitution in the signed lens mapping gives a negative source position for positive ${\theta }_r$; its magnitude isAn offset source with $0<|\beta |<{\beta }_c$ has three images. For $\beta >0$, one outer image is on the positive side and two are on the negative side, as the sketch shows. The central and outer images have positive parity; the intermediate image has negative parity. On the radial caustic the inner pair merges, with a separate outer image remaining; outside it only the outer image survives. At exact alignment a central image coexists with the ring. Finite source extent smooths the formal divergences. The singular zero-core limit removes the central image, so its image-count rule differs from the finite-core case.