Past exam of the astrophysics course of the University of Oxford / 2018 / 5

Interpret ${\sigma }_{\mathrm{DM}}$ as a constant one-dimensional, isotropic velocity dispersion. The effective pressure is $p=\rho {\sigma }_{\mathrm{DM}}^2$, corresponding to the stated gas-temperature analogy. For $\rho =A/r^2$, the spherical enclosed mass isThe two sides of the hydrostatic equilibrium equation are thenEquating them fixes the singular isothermal sphere:For collisionless dark matter this is an isotropic Jeans-equation construction, not a claim that the particles thermalize by gas collisions. The central singularity and infinite total mass show that the untruncated profile is an idealization.

Use $R$ for the physical projected radius, or impact parameter, in the lens plane. Although the source calls it angular, a physical length is required in the mass and deflection equations; its relation to an angle is $R=D_L|\theta |$. Integrate along physical line-of-sight coordinate $z$:Consequently the projected cylindrical mass inside the image radius isSubstituting a givesIf an angular radius $\vartheta$ is used instead, write $\Sigma (D_L\vartheta )={\sigma }_{\mathrm{DM}}^2/(2G{D}_L\vartheta )$ and $M_{2D}(<\vartheta )=\pi {\sigma }_{\mathrm{DM}}^2{D}_L\vartheta /G$. The projected mass differs from the spherical mass $2{\sigma }_{\mathrm{DM}}^{2}R/G$; material at three-dimensional radii larger than $R$ still projects into the cylinder.

In the weak-field thin-lens limit, circular symmetry makes the physical deflection depend on the projected enclosed mass. With the mass calculated in b,For ${\sigma }_{\mathrm{DM}}=155\mathrm{km}{\mathrm{s}}^{-1}$, this is $3.36\times 10^{-6}\mathrm{rad}$, orIt is independent of impact radius in this ideal profile.

Enclosed-mass qualification. The source's appeal to Birkhoff's theorem must not be read as permission to use the mass of a three-dimensional sphere of radius $R$ here. Deflection accumulates along the light path. A concrete counterexample is a thin spherical shell of radius $r_s>R$: its spherical enclosed mass at $R$ is zero, but its projected mass is $M[1-\sqrt{1-R^2/r_s^2}]>0$, and it produces a nonzero deflection. The formula above is the axisymmetric projected-mass lens result; it is consistent with b and is not that incorrect spherical-mass interpretation.

Let $\alpha$ be the reduced, observed angular displacement between source and image. In the small-angle lens geometry, the physical bending angle is weighted by the lens-to-source and observer-to-source angular diameter distances:In a Euclidean diagram this follows because the transverse displacement at the source plane is $D_{LS}\xi$, whose observed angle is that displacement divided by $D_S$. In cosmology the same thin-lens expression uses the corresponding angular diameter distances. For a signed one-dimensional coordinate, the direction is supplied by the image's side of the lens.

The general thin gravitational lens equation is a two-component angular relation,For a spherical lens, image, source and lens centre lie in one plane containing the line of sight. Choose its sky direction as a signed axis. Then $\alpha ,\theta ,\beta$ can be signed scalars andThey are not fundamentally scalar quantities for a general lens; the reduction uses this symmetry and axis choice. An exactly aligned source requires the full ring geometry, not a unique preferred sky axis.

In general, $D_S\ne D_L+D_{LS}$. For example, in a spatially flat expanding cosmology, comoving radial distances ${\chi }_L,{\chi }_S$ giveThus $D_S-D_{LS}={\chi }_L/(1+z_S)$ differs from $D_L$ when the redshifts differ. Ordinary Euclidean addition applies only when the same static physical-distance convention is appropriate.

The Einstein angular radius is the image-ring radius of an exactly aligned source. For the singular isothermal sphere lens, set $\beta =0$ in ii:Choose $\beta >0$ without loss of generality. The positive branch always gives ${\theta }_{+}=\beta +{\theta }_E$. A negative-side image would have ${\theta }_{-}=\beta -{\theta }_E$; this branch is admissible only when its result is negative. ThereforeAt $\beta =0$, symmetry produces an Einstein ring of angular radius ${\theta }_E$. At $\beta ={\theta }_E$, the negative branch reaches the singular origin and is not an ordinary additional finite image; the one finite image is at $2{\theta }_E$. Reversing the source sign reverses the signed image coordinates, so the multiplicity criterion uses $|\beta |$. These features rely on the singular isothermal profile; a smooth finite-density core changes the central-image behaviour.