Let us consider a fluid, with spacetime translation symmetry, and one internal $U(1)$ symmetry. Corresponding to the spatial translation symmetry, we can write down a momentum conservation equation for momentum density $G_i$ as : $\partial_t G_i + \nabla_j(P\delta_{ji} + v_j G_i) = 0$ ; where $P$ is the Pressure (assuming no dissipation)
Similarly we can write, for a conservation law corresponding to the $U(1)$ symmetry: $\partial_t n + \nabla_j (n_j) = 0$ ; where, $n$ is the "charge" density, and $n_j$ is $j$th component of the charge current.
We can interpret $\frac{n_j}{n} = v_j$ , where $v$ is the "fluid velocity". This is the same $v$ as appears in the momentum conservation equation, and is the velocity with which the fluid advects conserved quantities.
If we have an isotropic fluid, we know that $G_i = \rho v_i$, (basically, because the only object that transforms under rotations like we expect $G$ to, is $v$, hence the two must be proportional) where we can take $\rho$ to be some arbitrary function (scalar under rotation) of the hydrodynamic variables ; $\rho = \rho(n,|v|...)$ .
Now comes the part I do not understand : Apparently, the requirement that the above equations must be Galilean invariant (ie, must retain their form under Galilean Transforms) forces us to choose $\rho(n,|v|...) = \alpha n$, where $\alpha$ is some constant. That is, not only is the direction of $G$ fixed by the direction of $v$, but also, the magnitude of $G$ now scales linearly with the magnitude of $v$.
My question is, what goes wrong if we do not assume this? (I can see how this works when we start off with a "fluid" composed of point particles of some "mass", and then invoke the mechanical definition of momentum of a single particle to work out the momentum density of the fluid. What I am looking for is a proof using only the fact of Galilean invariance).
Addendum : the exchange of comments below made me think where I have seen any derivation of conservation of mass in the non-relativistic limit. As far as I can see, I can think of two places:
$(1)$ : $\S 133$, Fluid Mechanics, Landau (Relativistic Fluid mechanics). Towards the end of this section, Landau makes the following statement : "The non-relativistic case is that of small velocities...of the internal(microscopic) motion of fluid particles. In passing to this limit it must be borne in mind that relativistic internal energy $e$ includes the rest energy $nmc^2$ of the fluid particles ($m$ being the rest mass of $1$ particle) "
$(2)$ Weinberg (I), page 62, where he is discussing how to obtain Galilean Algebra as low velocity limit of Poincare Algebra : "..For a system of typical particle mass $m$, velocity $v$, the momentum operator ... is expected to be $P \approx mv$..."
I understand $(1)$ a lot better than $(2)$, but what is clear is that in both cases, we have an additional input, that is, a particle of mass $m$.
So, I reiterate my question:
What does this construction of a single particle of mass $m$ have to do with Galilean invariance? (And actually, what IS Galilean invariant about say, Navier Stokes equation ? )
