Is my derivation of the Navier Stokes equation using Newton's second law valid?

[member 428835]

hey pf! can you tell me if this derivation sounds reasonable for the navier stokes equation, from Newtons second law into a partial differential equation.

i'm really just concerned with one part. specifically, i start the derivation with $\Sigma F = ma$. I am comfortable with the force term but on the acceleration term I learned there are two ways to account for a change in momentum: boundary and volumetric change. i feel good on the volume term, but on the boundary term, i believe mass may leave through the boundary, yielding $$\iint_S (\rho \vec{V}) \cdot \vec{dS} \vec{V}$$ in order to maintain dimensionality through integration so we can obtain a partial differential equation it is necessary to convert this surface integral into a volume integral using the divergence theorem, yielding: $$\iiint_v \nabla \cdot (\rho \vec{V}){dv} \vec{V}$$ and thus if i shrink volume i obtain $$\nabla \cdot (\rho \vec{V}) \vec{V}$$ but the text has, in component form, $$\nabla \cdot (\rho u_i \vec{V})$$ where $u_i$ is the $i^{th}$ component of velocity $\vec{V}$ but this seems wrong. shouldn't it be written $$\nabla \cdot (\rho \vec{V})u_i$$ please help me out here! Thanks!

 

[Achy47]

Hello, thank you for your question. Your derivation seems reasonable, but I would like to clarify a few points. Firstly, when converting the surface integral to a volume integral using the divergence theorem, the equation should be \iiint_V \nabla \cdot (\rho \vec{V}) dV, without the \vec{V} term at the end. This is because the divergence theorem states that the surface integral of a vector field is equal to the volume integral of its divergence. The \vec{V} term is already included in the vector field \rho \vec{V}.

Secondly, regarding the component form of the equation, it is correct to write \nabla \cdot (\rho u_i \vec{V}), as this represents the contribution of the i^{th} component of velocity to the overall divergence of the vector field. This is because the divergence of a vector field is a scalar quantity, so it is necessary to multiply by the velocity component to obtain the correct contribution.

I hope this helps clarify the derivation for you. Let me know if you have any further questions.