When we non-dimensionalize NS equation, we typically take some characteristic scales of length, velocity etc, say the size of the box, average speed of the fluid in the flow direction etc. Once we have chosen the length and velocity scales, we chose the time scale as the ratio of the two. Is it necessary to do so? What mathematical constraint forces me to do so? For instance- $$x=x_\text{ref} \tilde{x}\\ u=u_\text{ref} \tilde{u}\\ $$ where the quantities with tilde are dimensionless while the 'ref' quantities are dimensionful.

The only constraint I have is $u=\frac{dx}{dt}$ and unless I demand $\tilde{u}=\frac{d\tilde{x}}{d\tilde{t}}$, I don't have to say $u_\text{ref}=\frac{x_\text{ref}}{t_\text{ref}}$. However, why should I demand the latter condition at all? Solutions can still be worked out if I solve the non-dimensionalized equation and then put the dimensions back.

Further, if I assume that I have the above mentioned freedom, that would make all the reference variables free. Is it appropriate to use this freedom to normalize (not just non-dimensionalize, but bring the non-dimensional quantities between 0 and 1) the NS equations since non-dimensionalization may not necessarily normalize the equation as one of the quantities can be found by integrating the NS equations and this quantity may be very large.


1 Answer 1


It is probably more proper to start with the NS equations and write all variables as the product of a dimensionful and dimensionless number. For example, the continuity equation becomes, $$\frac{\partial\rho}{\partial t}+\nabla\cdot\rho\mathbf v=0\to\frac{\rho_0}{t_0}\frac{\partial\rho'}{\partial t'}+\frac{\rho_0v_0}{\ell_0}\nabla'\cdot\rho'\mathbf{v}'=0\tag{1}$$ where $\rho=\rho_0\rho'$ with $\rho_0$ the dimensionful quantity and $\rho'$ the dimensionless.

After doing this, you first want to isolate at least one term by multiplying/dividing appropriately (usually you want the highest-order derivative term to have a unit constant). It is at this point that you can then pick your characteristic scales and define the remaining scales as appropriately to satisfy the equations with the goal of making as many coefficients as possible equal to 1.

As an example, we can multiply Eq (1) by $t_0/\rho_0$ to get, $$\frac{\partial\rho}{\partial t}+\nabla\cdot\rho\mathbf v=0\to\frac{\partial\rho'}{\partial t'}+\frac{t_0v_0}{\ell_0}\nabla'\cdot\rho'\mathbf{v}'=0$$ which suggests that if we define any pair of $\ell_0$, $t_0$ and $v_0$ then we can also get the third one for free (using $v_0=\ell_0/t_0$). Note that the density scale is inconsequential in this equation, but almost surely will be defined via the momentum equation.

If you chose to make $v_0=\eta\ell_0/t_0$ for some $\eta\neq1$, then your equations would no longer be identical in form as the dimensionful case (which is the ideal scenario when nondimensionalizing a system of equations): $$\frac{\partial\rho}{\partial t}+\nabla\cdot\rho\mathbf v=0\not\leftrightarrow\frac{\partial\rho'}{\partial t'}+\eta\nabla'\cdot\rho'\mathbf{v}'=0$$

Then you can add to this that there are some variables, such as the pressure scale factor $p_0\sim\rho_0v_0^2$ or the gravity scale factor $g_0\sim v_0/t_0$, that then carry additional multiplicative factors that must be carried about in the momentum equations that make using inconsistent scales even less tenable a scenario.

So to answer your questions: yes, you are absolutely constrained to have the velocity scale be the same as the ratio of length and time scales, otherwise your formula are not consistent in the dimensionful and dimensionless cases.

For further reading, see the Wikipedia entry on nondimensionalization. See also the Physics StackExchange posts (both of which I answered) on similar subject matter

  • $\begingroup$ Thanks for the answer. What would go wrong if the non-dimensional equation has a different form? The equation with $\eta\neq1$ can still be solved and I can still get back the solutions by multiplying the non-dimensional solutions I obtain from the equation and then multiplying the non-dimensional variable with their dimensionful counterpart, right? $\endgroup$
    – James
    Commented Jun 3, 2023 at 4:14
  • $\begingroup$ The ideal goal of nondimensionalization is to have an identical formula after removing the scales; having $\eta\neq1$ defeats that ideal. In principle, you could handle that in your numerical solvers or analytic system, but why would you carry an additional term that you can set to 1 and not carry it? $\endgroup$
    – Kyle Kanos
    Commented Jun 3, 2023 at 15:49
  • $\begingroup$ I thought of it because I was wondering if doing that can allow me to bring more quantities between 0 and 1. $\endgroup$
    – James
    Commented Jun 3, 2023 at 18:32
  • $\begingroup$ I doubt you'd be able to find a case where your system of equations allows for more reduced variables at the cost of inconsistent scales. It's a lot easier to just fully eliminate the scales. $\endgroup$
    – Kyle Kanos
    Commented Jun 4, 2023 at 13:42

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