$$ \frac{∂u}{∂t}+u\frac{∂u}{∂x}+v\frac{∂u}{∂y}+w\frac{∂u}{∂z}=-\frac{1}{ρ}\frac{∂P}{∂x}+gx+\nu\left(\frac{∂^2u}{∂x^2}+\frac{∂^2u}{∂y^2}+\frac{∂^2u}{∂z^2}\right) $$

Why would someone use a form of this equation where both sides are not divided by the density? Are there any cases where the above equation cannot be used because the equation has been divided by the density?

  • $\begingroup$ Density is nonzero everywhere so they're entirely equivalent. $\endgroup$ May 30, 2019 at 18:50
  • $\begingroup$ Apparently if $\rho=0$. $\endgroup$
    – Qmechanic
    May 30, 2019 at 19:39
  • $\begingroup$ What are v and w? I assume u to be a component of the velocity field. $\endgroup$
    – DanielC
    May 30, 2019 at 20:06
  • 1
    $\begingroup$ @ DanielC (u, v, w) are commonly used to denote the (x, y, z) velocity components $\endgroup$
    – D. Halsey
    May 30, 2019 at 22:18
  • $\begingroup$ @Qmechanic would you mind expanding on that? $\endgroup$
    – Chris
    May 31, 2019 at 15:39

1 Answer 1


Let's take the general conservative compressible form of the mass continuity \eqref{1} and momentum equations \eqref{2}

$\frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} \frac{\partial (\rho u_j )}{\partial x_j }=0 \tag{1}\label{1}$

$\frac{\partial (\rho u_i )}{\partial t} + \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_i u_j )}{\partial x_j} = \sum\limits_{j \in \mathcal{D}} \frac{\partial \sigma_{ij}}{\partial x_j } + \rho g_i \tag{2}\label{2}$

where $\mathcal{D} = \{ x, y, z \}$ and assume a Newtonian fluid with the Stokes' hypothesis $\sigma_{ij} = - p \delta_{ij} + 2 \mu S_{ij} \tag{3}\label{3} - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} S_{kk} \delta_{ij}$

where the rate of strain tensor is given by

$$S_{ij} := \frac{1}{2} \left( \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right). \tag{4}\label{4}$$

Now let's go through what simplification have been made to arrive at your form.

First insert equation \eqref{3} and \eqref{4} into \eqref{2} and apply the chain rule

$$\underbrace{u_i \frac{\partial \rho}{\partial t} + u_i \sum\limits_{j \in \mathcal{D}}\frac{\partial (\rho u_j)}{\partial x_j}}_{u_i \cdot (1) = 0} + \rho \frac{\partial u_i}{\partial t} + \sum\limits_{j \in \mathcal{D}} (\rho u_j) \frac{\partial u_i}{\partial x_j} = - \frac{\partial p}{\partial x_i } + \sum\limits_{j \in \mathcal{D}} \frac{\partial}{\partial x_j } \Big[ \mu \big( \frac{\partial u_i}{\partial x_j} + \underbrace{ \frac{\partial u_j}{\partial x_i} }_{II}\big) \underbrace{ - \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k} \delta_{ij} }_{I} \Big] + \rho g_i.$$

The first term cancels out due to continuity equation \eqref{1} and we are left with the non-conservative form of the momentum equation. Your equation looks fairly similar but we instantly see that the shear viscosity $\mu$ is in your formula replaced by the kinematic viscosity $\nu := \frac{\mu}{\rho}$ and it stands outside the derivative - it is assumed constant.

Furthermore the term $- \frac{2 \mu}{3} \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k}$ corresponding to the volumetric dilatation is not there. This means a fluid parcel is not compressed along its way on a stream line

$$\frac{D \rho}{D t} = \frac{\partial \rho}{\partial t} + \sum\limits_{j \in \mathcal{D}} u_j \frac{\partial \rho}{\partial x_j} = 0,$$

something termed incompressible flow (it could also be slightly more strict and be an incompressible fluid). With equation \eqref{1} we can find - applying the chain rule - that in such a case the velocity field is divergence free

$$\sum\limits_{j \in \mathcal{D}} \frac{\partial u_j}{\partial x_j} = 0.$$

This is the reason why the two terms I and II are equivalent to zero and not present in your equation.

Summing up, these equations are only valid for incompressible flow of a Newtonian fluid with a constant viscosity.

  • $\begingroup$ I’m a little lost as to where the $\frac{2\mu}{3}$, the negative sign’s disappearance, and $\partial_ku_k$ are coming from. Could you elaborate a tiny bit on that part? Everything else I can follow. Maybe I should try writing it out for $ \{ i, j \} \in \{0,1,2\}$ ( or 1,2,3 if you prefer)? $\endgroup$
    – Shinaolord
    Oct 23, 2019 at 1:19
  • 1
    $\begingroup$ Oh, I forgot the $\delta_{ij}$ in $- \frac{2}{3} \mu \sum\limits_{k \in \mathcal{D}} \frac{\partial u_k}{\partial x_k} \delta_{ij}$. Was that the issue? It is simply equation (4) inserted into (3) and then the result into (2) and making use of $S_{kk} = \frac{\partial u_k}{\partial x_k}$. In case you don't understand where the last term in the strain rate tensor $S_{ij}$ physically comes from and how it can be interpreted: I wrote a post on the underlying Stokes' hypothesis recently physics.stackexchange.com/a/509518/245414 if you are interested. $\endgroup$
    – 2b-t
    Oct 23, 2019 at 10:04

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