I was going over the derivation of the dimensionless Navier-Stokes equation, which is explained in this answer. By introducing dimensionless variables in the NS equation, one gets

$$ \frac{D\mathbf u}{Dt} = - \nabla P + \frac 1 {\text{Re}} \nabla^2 \mathbf u $$

where $\text{Re} \equiv \rho UL/\eta$ is the Reynolds number. For $\text{Re} \gg 1$ (high Reynolds number), the viscosity (the one with $\nabla^2$) term is negligible and thus

$$ \frac{D\mathbf u}{Dt} = - \nabla P \ \ \ (\text{Re} \gg 1) $$

For $\text{Re} \ll 1$, instead, the viscosity term dominates and we get

$$ \nabla P = \frac 1 {\text{Re}} \nabla^2 \mathbf u \ \ \ (\text{Re} \ll 1) $$

My question is: In all derivations I've seen, people keep the pressure gradient term $\nabla P$ for $\text{Re} \ll 1$. Why can't we also get rid of this term?


1 Answer 1


Notice how when we derive the non-dimensionalization of the Navier-Stokes equations there is no Re related to the gradient pressure term

$$\nabla p \longrightarrow\textrm{non-dimensionalization} \longrightarrow (\nabla p)^{*}, \ \textrm{where * stands for dimensionless quantities}$$

No Re appears when we perform the non-dimensionalization, so the pressure term should not let us say, "suffer influence" if either we have a Re $\ggg 1$ or if Re $\lll 1$.

Besides that, if even though you still wanted to get rid of the pressure term, doing so doesn't give many advantages since pressure is one of the three fundamental physical quantities that we are interested in when modelling fluid flow.

The main reason why we do non-dimensionalization is that we want to change the fundamental question when encountering the Navier-Stokes equation. Usually, the question that we look for is how the distribution velocity flow ($\mathbf{u}$) looks under certain conditions. But when we perform the non-dimensionalization the question changes to what is the Reynolds number since a high Re implies very low viscosity effects and future turbulent flow, and if we have a low Re, as you said, the viscosity becomes dominant and there is no way of forming a turbulent flow, but independently of any of these situations, pressure still is a quantity of our interest.

A zero difference in pressure could imply in a fluid at rest, which doesn't have a "WOW" behavior

Obs: the three fundamental quantities are velocity, density, and pressure

  • $\begingroup$ Thanks for your answer. Doesn't $\frac 1 \rho \nabla p$ go to $(\nabla p)^*$ i.e. we lose a $\rho^{-1}$ factor? In any case, I thought about it and kinda convinced myself of the fact that people are just redefining an "effective" dimensionless pressure $(p^*)'=\text{Re}\cdot P$... $\endgroup$
    – valerio
    Nov 25, 2021 at 14:01
  • $\begingroup$ oh yeah, you are completely right! I forgot about the inverse of the density term, but besides that, that's it! $\endgroup$
    – Caio Cesar
    Nov 25, 2021 at 17:49

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