⊥^{ν}_{α} \nabla_{μ} T^{μ}_{ν} = 0,

and substitute the decomposition from part a. The

ρ n^{μ} n_{ν}

term contributes

ρ a_{α}

. The two momentum terms combine into

L_{n} j_{α} - K j_{α},

because the two contractions with the full extrinsic curvature cancel and

\nabla_{μ} n^{μ} = - K .

Finally, part b gives

⊥^{ν}_{α} \nabla_{μ} S^{μ}_{ν} = D_{μ} S^{μ}_{α} + S^{μ}_{α} a_{μ} .

Thus

ρ a_{α} + L_{n} j_{α} - K j_{α} + D_{μ} S^{μ}_{α} + S^{μ}_{α} a_{μ} = 0.

L_{n} j_{α} = - D_{μ} S^{μ}_{α} - S^{μ}_{α} a_{μ} + K j_{α} - ρ a_{α} .

Therefore

c_{1} = - 1, c_{2} = - 1, c_{3} = 1, c_{4} = - 1 .

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