# Why to write the Navier-Stokes equation with dimensionless quantities?

The Navier-Stokes equation is

$$\rho \dfrac{D\mathbf{u}}{Dt} = -\nabla p+(\lambda+\mu)\nabla(\nabla\cdot\mathbf{u})+\mu\nabla^2\mathbf{u}$$

Then if the flow is incomprresible, and the fluid is homogenous ($\rho$ is constant in space) then $\rho$ will be constant in time and $\rho = \rho_0$ so that we can write this as

$$\rho_0 \dfrac{D\mathbf{u}}{Dt} = - \nabla p + \mu\nabla^2\mathbf{u}$$

From this we can introduce numbers $U$ and $L$ in order to write $\bar{x}_i = x_i/L$ and $\bar{\mathbf{u}} = \mathbf{u}/U$ so that $\bar{t} = t/T$ where $T = L/U$. Then we obtain the equation

$$\dfrac{\partial \bar{\mathbf{u}}}{\partial \bar{t}} + (\bar{\mathbf{u}}\cdot \nabla)\bar{\mathbf{u}} = -\nabla \bar{p} + \dfrac{1}{R}\nabla^2\bar{\mathbf{u}}$$

Where $R = LU/\nu$ is the Reynolds number. That's fine, but what's the motivation to write the equation like that? What do we gain from just changing the coordinates by rescaling them? Many books says that the importance of the Reynolds number comes from writing the equation with dimensionless quantities, but why is that?

• I don't have time to write up a full answer right now, but when you write an equation in terms of dimensionless quantities, the relative importance of each part of the equation becomes immediately clear. For example if $R\gg1$, then the second term on the right hand side becomes exceedingly insignificant as compared to the other terms. – Wouter Oct 27 '14 at 13:29
• Dimensionless quantities make it easier to define scales in some cases, such as this one, which allows us to derive physical meaning more easily. Also, they are especially suited for numerical simulations if the quantities have extreme values, because fast computer arithmetic has finite precision. – auxsvr Oct 27 '14 at 13:41
• My answer here might be useful: physics.stackexchange.com/questions/138598/… – tpg2114 Nov 24 '14 at 0:00

In $\Pi$-theorem you can also derive relationships based on units of quantities.