Comparing the Lagrangian form of Newton's law & the Reynolds transport theorem

Question

If I was interested in deriving an equation for the conservation of momentum for a fluid, I could write down an expression for the change in momentum density of a fluid point using the Reynolds transport theorem:

$$\frac{\partial \rho \vec{v}}{\partial t} + \nabla \cdot (\rho \vec{v} \otimes \vec{v}) = \vec{f}$$

I could also try to write down Newton's law from a Lagrangian perspective for a parcel of fluid:

$$\frac{D(\rho \vec{v})}{Dt} = \vec{f}$$

Transforming this into a Eulerian perspective:

$$\frac{\partial \rho \vec{v}}{\partial t} + \nabla (\rho \vec{v})\cdot\vec{v} = \vec{f}$$

This expression differs from the expression derived from the Reynolds transport theorem by a $\rho \vec{v}(\nabla \cdot \vec{v})$ term—both expressions are the same for an incompressible flow field, but not when the flow is compressible.

Did I make a mistake in applying Newton's law, or is there a valid reason for the discrepancy?

Answer 1

If the left hand side is differentiated property using the product rule for differentiation, we obtain $$\frac{\partial \rho \vec{v}}{\partial t} + \nabla \cdot (\rho \vec{v} \otimes \vec{v}) =\rho\frac{\partial \vec{v}}{\partial t}+\vec{v}\frac{\partial \rho}{\partial t}+\vec{v}\cdot\nabla(\rho \vec{v})+\rho (\vec{v}\cdot \nabla) \vec{v}$$The middle two terms drop out because of the continuity equation, so we are left with $$\frac{\partial \rho \vec{v}}{\partial t} + \nabla \cdot (\rho \vec{v} \otimes \vec{v}) =\rho\frac{\partial \vec{v}}{\partial t}+\rho (\vec{v}\cdot \nabla) \vec{v}=\rho\frac{D\vec{v}}{dt}$$