I am working on Sean Carroll's problem 1.8     If $\partial_\nu T^{\mu \nu} = Q^\mu$, what physically does the spatial vector $Q^i$ represent? Use the dust energy momentum tensor to make your case.   I know this question has been asked and answered  and I understand that answer to a degree, but I'm trying to get it directly from dust and I'm not sure I'm doing it correctly or how to interpret what I'm getting.  The dust energy momentum tensor is  $$T^{\mu \nu}=\rho U^\mu U^\nu$$  Where $U$ is the four-velocity and $\rho$ is the energy density in the rest frame. Therefore what I'm looking at is  $$\partial_\nu T^{i \nu} = \partial_\nu \rho U^i U^\nu = \rho U^\nu \partial_\nu U^i+\rho U^i \partial_\nu U^\nu.$$  To examine the right-hand side of the above expression more simply I will choose to work in the rest frame of the dust. According to Carroll, the definition of dust is "a collection of particles at rest with respect to each other." Therefore, in the rest frame I expect all derivatives of velocity to be zero except the time derivative. Therefore I think I can simplify the above to  $$Q^i=\rho U^0 \frac{\partial U^i}{\partial t}+\rho U^i \frac{\partial U^0}{\partial t}.$$  If this is correct, I'm having some trouble understanding what each expression on the right hand side means. I'm not too sure how to proceed.
I'm trouble with exercise 1.8 of Carroll's Space-Time and Geometry:     If $\partial_\nu T^{\mu \nu} = Q^\mu$, what physically does the spatial vector $Q^i$ represent? Use the dust energy momentum tensor to make your case.    First of all, since the momentum energy tensor of dust $T^{\mu \nu}=\rho U^\mu U^\nu$ is symmetric, one has the energy-momentum conservation law $$ \partial_\nu T^{\mu \nu} = \partial_\nu T^{\nu \mu}  =0.$$ A vector $Q$ is constant in any frame related by a Lorentz transformation, while this isn't the case for the components $Q^\mu$. But here $Q=0$ and hence $Q^\mu=0$ also for every component $\mu =0,\cdots,4$?   Even if a straightforwardly apply the equation: $$ Q^i = \partial_\nu T^{i \nu} = (\partial_\nu \rho) U^i U^\nu + \rho(\partial_\nu U^i)  U^\nu + \rho U^i(\partial_\nu  U^\nu ),$$ doesn't seems to yield something...   Is my first reasoning right ? Otherwise: can someone give me a hint ?  Thanks
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