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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-59/1/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 59 / 1 / a / Solution by

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

The normalized timelike normal has

n_{α} n^{α} = - 1

. Multiplying the linear projections gives

P^{μ}_{α} P^{α}_{δ} = (δ^{μ}_{α} + n^{μ} n_{α}) (δ^{α}_{δ} + n^{α} n_{δ}) = δ^{μ}_{δ} + 2 n^{μ} n_{δ} + n^{μ} (n_{α} n^{α}) n_{δ} = δ^{μ}_{δ} + n^{μ} n_{δ} .

(1)

Also

P^{μ}_{α} n_{μ} = n_{α} + (n_{μ} n^{μ}) n_{α} = 0

. Thus

P^{2} = P, P^{μ}_{α} n_{μ} = 0.

(2)

The operator fixes every vector tangent to the spacelike hypersurface and annihilates its normal. The repeated-index typo in the TeX is resolved by the actual PDF's matrix-composition indices.

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