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 Incoming links: Spatial projector of a timelike congruence

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 38D / iii / Solution Created 2026-09-24 Updated 2026-09-29

First differentiate the unit normalization

X^{α} X_{α} = - 1

. Metric compatibility gives

B_{α β} X^{α} = X^{α} \nabla_{β} X_{α} = \frac{1}{2} \nabla_{β} (X^{α} X_{α}) = 0.

The timelike geodesic vector field equation gives the other contraction,

B_{α β} X^{β} = X^{β} \nabla_{β} X_{α} = 0.

For the spatial projector of a timelike congruence,

h_{α β} X^{α} = X_{β} + (X_{α} X^{α}) X_{β} = 0,

and symmetry also gives

h_{α β} X^{β} = 0

. Thus

B_{α β} X^{α} = B_{α β} X^{β} = h_{α β} X^{α} = h_{α β} X^{β} = 0.

Splitting the deformation tensor of a timelike geodesic congruence into symmetric and antisymmetric parts and substituting the definitions gives

B_{α β} = B_{(α β)} + B_{[α β]} = \frac{1}{3} Θ h_{α β} + σ_{α β} + ω_{α β} .

Moreover, the orthogonality of

B_{α β}

implies

Θ = B^{α β} h_{α β} = B^{α}_{α},

while in four dimensions

g^{α β} h_{α β} = 4 + X^{α} X_{α} = 3.

It follows that

g^{α β} σ_{α β} = B^{α}_{α} - \frac{1}{3} Θ g^{α β} h_{α β} = 0.

Hence

B_{α β} = \frac{1}{3} Θ h_{α β} + σ_{α β} + ω_{α β}, g^{α β} σ_{α β} = 0,

which is the kinematic decomposition of a timelike geodesic congruence.

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