Physical interpretation of $Q^i = \partial _\nu T^{i \nu}$ 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
 A: If the dust does not interact with anything, momentum and energy are conserved. Otherwise, the general equation reads $$ \partial_0 T^{i0} = - \sum_{k=1}^3\partial_k T^{ik} +Q^i\:. $$
If you integrate it in a regular volume $\Omega$ at rest with the reference frame defined by the used Minkowskian coordinates, exploiting the theorm of divergence you have:
$$\frac{d}{dt}\int_\Omega T^{i0} d^3x = - \oint_{+\partial \Omega} T^{ik} n_k dS +\int_\Omega Q^i d^3x\:.$$
You see that the momentum in $\Omega$ changes in time in view of two independent  reasons. One is the flux of momentum across the boundary of $\Omega$ due to the dust itself, the other is due to external interactions.  $Q^i$ is nothing but the source of momentum, i.e., density of external force along the $x^i$ direction caused by the external interactions. If the dust does not interact with anything, then $Q^i=0$. Otherwise it can be re-written in terms of the external interaction. 
For instance, if the dust is electrically charged, there is a charge density $\nu = e \rho$ (I am assuming that the mass is proportional to the charge, i.e., there is a unique type of charges in the continuum). In this case, up to signs and multiplicative constants depending on your conventions, you find the density of Lorentz' force:
$$Q_\mu = J^\sigma F_{\mu \sigma}$$
where $F_{\mu\sigma}$ is the electromagnetic tensor and $J^\mu = \nu V^\mu$.
