 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-311/2/iv/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

G_{ab} = 8 π T_{ab}

twice with

n^{a}

. Since

n^{a} n_{a} = - 1

and

ρ = T_{ab} n^{a} n^{b}

R + 2 R_{ab} n^{a} n^{b} = 16 π ρ .

K_{ab}

is proportional to

h_{ab}

in three dimensions, tracing gives

K_{ab} = (K /3) h_{ab}

and

K_{ab} K^{ab} = K^{2} /3

^{(3)} R = 16 π ρ - \frac{2}{3} K^{2} = \frac{2}{3} (24 π ρ - K^{2}) .

Therefore

24 π ρ - K^{2} \geq 0 ⟹^{(3)} R \geq 0 .

 New to topics? Read the docs here!