\frac{Dλ^{α}}{Dτ} = \frac{dλ^{α}}{dτ} + Γ_{βγ}^{α}λ^{β} \frac{dx^{γ}}{dτ} = 0

the tangent vector is λ^{α} = dx^{α}/dτ [1 mark] so its absolute derivative gives the geodesic condition

\frac{d^{2}x^{α}}{dτ^{2}} + Γ_{βγ}^{α} \frac{dx^{β}}{dτ} \frac{dx^{γ}}{dτ} = 0

[1 mark]

U_{;β}^{α}U^{β} = 0 = (\frac{∂U^{α}}{∂x^{β}} + Γ_{βγ}^{α}U^{γ})U^{β}

[1 mark]

\frac{dx^{β}}{dτ} \frac{∂U^{α}}{∂x^{β}} + Γ_{βγ}^{α} \frac{dx^{γ}}{dτ} \frac{dx^{β}}{dτ} = 0

[1 mark]

\frac{dU^{α}}{dτ} + Γ_{βγ}^{α} \frac{dx^{γ}}{dτ} \frac{dx^{β}}{dτ} = 0

[1 mark]

\frac{d^{2}x^{α}}{dτ^{2}} + Γ_{βγ}^{α} \frac{dx^{γ}}{dτ} \frac{dx^{β}}{dτ} = 0

which is the same as the geodesic condition [1 mark] N_{,α}^{α} = 0 in SR, so in GR the derivative changes to covariant derivative giving N_{;α}^{α} = 0 [1 mark] The stress energy tensor for dust T^{αβ} = N^{α}U^{β}. So

T _{;α}^{αβ} = 0 = N_{;α}^{α}U^{β} + N^{α}U_{;α}^{β}

[1 mark]

But from above N_{;α}^{α} = 0 so N^{α}U_{;α}^{β} = 0 [1 mark] hence ρ_{0}U^{α}U_{;α}^{β} = 0 i.e. U^{α}U_{;α}^{β} = U^{β}U_{;β}^{α} = 0 which is the geodesic condition, so dust particles follow geodesic paths [1 mark]