Einstein ring 2026-10-14
An aligned source behind an axisymmetric gravitational lens produces a circular image at an Einstein radius. Its angular radius satisfies . Exact alignment gives a ring; displaced sources generally give arcs or distinct images rather than a complete ring.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 72 3 ii Solution Created 2026-10-03 Updated 2026-10-14
Substituting and into the previous geometric relation verifiesFor a point mass gravitational lens, the given physical deflection yieldsAn aligned source has , and therefore . Rotational symmetry turns these signed intersections into an Einstein ring, whose Einstein radius isFor an offset source, multiplying by gives a quadratic equation. Its two image positions areFor , lies on the source side of the lens and lies on the opposite side.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 72 3 iv b Solution Created 2026-10-03 Updated 2026-10-14
For an axisymmetric gravitational lens, the reduced deflection is . The specified mean lensing convergence therefore givesThe derivative is smallest at the centre, where . If , the mapping is monotone and an offset source has only one image. If , 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.
Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 67 4 i Solution Created 2026-10-03 Updated 2026-10-14
A complete Einstein ring requires an axisymmetric gravitational lens and exact alignment of the source with the lens centre and observer: . A finite aligned circular source gives a ring of finite width. The background source must lie behind the lens, so the lensing distance factor is positive.
With the thin gravitational lens equation givesThe Einstein radius here is angular, and the mass is projected inside the corresponding aperture. For an extended mass distribution the equation is implicit because the enclosed mass depends on .
Using the supplied approximations, and , the inferred mass isThusThis measures the projected aperture mass without requiring the singular isothermal sphere model used subsequently.
Point-mass gravitational lens 2026-10-14
A point-mass gravitational lens is the ideal gravitational lens obtained by concentrating a positive mass at one location. Use geometrical optics, the weak-field approximation, small angles in radians, and a stationary lens whose deflecting region is short compared with the observer-lens and lens-source distances. At impact parameter , require , the Schwarzschild radius, and a ray passing outside any opaque body. The physical bend has magnitude from Schwarzschild light deflection; is the gravitational constant and the speed of light. Multiple images can occur within this weak-field approximation: strong gravitational lensing does not require a ray to approach a black hole closely.[ref]
Gravitational Lensing by Matthias Bartelmann. Section 1.2, printed p. 2: geometrical optics, weak fields and slowly moving lenses; compact-object near-field propagation excluded
Let , and be angular diameter distances from observer to lens, observer to source, and lens to source, respectively. They refer to the background geometry used in the thin gravitational lens equation. In nearby Euclidean geometry, ; in cosmology, do not make that subtraction. The angular source coordinate is the direction the source would have without this lens; the observed image coordinate is . Both are measured relative to the observer-lens optical axis.[ref]
Lectures on Gravitational Lensing by Ramesh Narayan and Matthias Bartelmann. Section 2.1.3, pp. 8-9, equations (13)-(15), including the warning that lens-source angular diameter distance is not the difference of the observer distances
For the flat sketch below, extend the observer-side light ray backwards to the source plane. Its transverse position would be , whereas the actual source is at . Their separation is . Here is defined outward along the lens-plane impact vector, as in the thin gravitational lens equation; the physical ray bends toward the mass. Substituting givesThe same thin gravitational lens equation uses angular diameter distances in the cosmological setting. The factor converts the physical bend into the reduced angular deflection; these are different angles.
Schwarzschild light deflection Created 2026-09-24 Updated 2026-10-14
