 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-67/4/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

ξ

M_{proj} (< ξ) = 2 π \int_{0}^{ξ} \frac{σ ^{2}}{2 G ξ ^{'}} ξ^{'} d ξ^{'} = \frac{π σ ^{2} ξ}{G} .

With

ξ = D_{d} θ

this gives the reduced deflection magnitude

\frac{D _{d s}}{D _{s} D _{d}} \frac{4 G M _{proj} ( < D _{d} θ )}{c ^{2} θ} = 4 π \frac{σ ^{2}}{c ^{2}} \frac{D _{d s}}{D _{s}},

independent of

θ

θ_{E} = 4 π (\frac{σ}{c})^{2} \frac{D _{d s}}{D _{s}} .

The spherical three-dimensional mass

2 σ^{2} r / G

would give the wrong coefficient here; the thin gravitational lens equation needs the projected cylindrical mass.

Using the same

θ_{E} = 3 \times 1 0^{- 4}

and

D_{d s} / D_{s} = 0.5

, the one-dimensional velocity dispersion is

σ = c \frac{θ _{E}}{4 π ( D _{d s} / D _{s} )} = 3 \times 1 0^{8} \frac{3 \times 1 0 ^{- 4}}{2 π} m s^{- 1} ≃ 2.1 \times 1 0^{6} m s^{- 1} .

Thus $σ ≃ 2100 km s^{- 1}$ within the idealized SIS model.

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