The stress-energy tensor representing "dust" takes the form$$T_{ab} = \rho u_au_b$$where $u^a$ is a unit timelike vector field, i.e., $u^au_a = -1$. Does it necessarily follow that in any solution to Einstein's equation with dust matter (with $\rho > 0$ everywhere), that the "flow lines" of the dust, i.e., the integral curves of $u^a$, are geodesics?

  • $\begingroup$ Yes, every freely moving object will follow a geodesic through spacetime, no matter what form the stress-energy tensor takes. $\endgroup$ – Alex S Apr 6 '16 at 5:51

Start with the fact that the Einstein tensor (and thus $T$) is divergence-free:

$$\nabla_a T^{ab} = \nabla_a(\rho u^a)u^b + \rho u^a \nabla_a u^b = 0. \tag{1}$$

The product rule expansion I've chosen is quite suggestive: in the first term we see the divergence term that appears in the continuity equation, while in the second we see the covariant acceleration of the integral curves of $u$. Physically we expect both of these terms to be zero - let's see why it's true. Contracting this equation with $u_b$ we get (using $u^a u_a = -1$)

$$ -\nabla_a (\rho u^a) + \rho u^a u_b \nabla_a u^b = 0.$$

Now note that $$u_b \nabla_a u^b = \frac12 \nabla_a (u_b u^b) = \frac12 \nabla_a(-1) = 0;$$ so we obtain the continuity equation $$\nabla_a(\rho u^a)=0.$$ Substituting this back in to $(1)$ we get $\rho u^a \nabla_a u^b = 0,$ which implies the geodesic equation $u^a \nabla_a u^b = 0$ wherever $\rho \ne 0$.


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