I am currently dealing with multiphase flows and have to use the non-dimensional form of the Navier-Stokes equations (NSE). In the scientific literature I found various formulations (and almost no explaining o how they get them) but I wanted to get to the point by myself as well. Although the procedure should be simple, the results do not lign up and I cannot wrap around my head on why. NSE can be written for incompressible fluids as: \begin{equation} \frac{\mathrm{D}\mathbb{u}}{\mathrm{D}t}=-\frac{1}{\rho}\mathbb{\nabla}p+\mathbb{g}+\frac{\mu}{\rho}\Delta\mathbb{u} \end{equation} where the ratio $\frac{\mathrm{D}}{\mathrm{D}t}$ represent the material derivative, $\mathbb{u}$ is the fluid velocity, $\rho$ the density, $\mathbb{g}$ the gravitational acceleration and $\mu$ the dynamic viscosity. Also, $\Delta$ is the Laplacian operator.

With surface tension, the equation becomes (CSF formulation of the surface tension) \begin{equation} \frac{\mathrm{D}\mathbb{u}}{\mathrm{D}t}=-\frac{1}{\rho}\mathbb{\nabla}p+\mathbb{g}+\frac{\mu}{\rho}\Delta\mathbb{u}+\frac{\sigma\kappa\hat{\mathbb{n}}\delta_{\varepsilon}}{\rho} \end{equation} where $\sigma$ is the coefficient of surface tension, $\kappa$ is the curvature, $\hat{\mathbb{n}}$ is the normal unit vector to the interface surface and $\delta_{\varepsilon}$ a Dirac delta function (smeared out version).

Introducing the following dimensionless variables (marked with $'$): \begin{equation} x'L=\mathbb{x} \hspace{5mm} u'U=\mathbb{u} \hspace{5mm} g'g=\mathbb{g} \hspace{5mm} \nabla'=\mathbb{\nabla}L \hspace{5mm} \Delta'=\Delta L^2 \hspace{5mm} p=p'\rho U^2 \hspace{5mm} t'=tU/L \end{equation} I get \begin{equation} \frac{U^2}{L} \frac{\mathrm{D}u'}{\mathrm{D}t'}=-\frac{U^2}{\rho}\frac{\nabla'}{L}p'+g'g+\frac{\mu}{\rho}\frac{U}{L^2}\Delta' u'+\frac{\sigma\kappa\hat{\mathbb{n}}\delta_{\varepsilon}}{\rho} \end{equation} and, after multiplying both sides of the equation for $L/U^2$, \begin{equation} \frac{\mathrm{D}u'}{\mathrm{D}t'}=-\frac{1}{\rho}\nabla'p'+\frac{g'gL}{U^2}+\frac{\mu}{\rho}\frac{1}{LU}\Delta' u'+\frac{\sigma\kappa\hat{\mathbb{n}}\delta_{\varepsilon}}{\rho}\frac{L}{U^2} \end{equation} Now, after introducing the well-known Reynolds and Froude numbers \begin{equation} \mathrm{Re}=\frac{\rho LU}{\mu} \hspace{10mm} \mathrm{Fr}=\frac{U}{\sqrt{gL}} \end{equation} it follows that: \begin{equation} \frac{\mathrm{D}u'}{\mathrm{D}t'}=-\frac{1}{\rho}\nabla'p'+\frac{1}{\mathrm{Fr}^2}g'+\frac{1}{\mathrm{Re}}\Delta' u'+\frac{\sigma\kappa\hat{\mathbb{n}}\delta_{\varepsilon}}{\rho}\frac{L}{U^2} \end{equation} Until here no problem.

However, for the manipulation of the surface tension (SF) form, some authors (e.g. "An improved level set method for incompressible two-phase flows". Sussman et al., 1998) use the Weber number: \begin{equation} \mathrm{We}=\frac{\rho LU^2}{\sigma} \end{equation} Plugging this latter in the SF term though, I get: \begin{equation} SF'=\frac{\kappa\hat{\mathbb{n}}\delta_{\varepsilon}L^2}{\mathrm{We}} \end{equation} instead of \begin{equation} SF'=\frac{\kappa\hat{\mathbb{n}}\delta_{\varepsilon}}{\mathrm{We}} \end{equation}

I dare to think that this latter formulation is incorrect due to the incontrovertible principle of dimensional homogeneity.

$\kappa$ is measured (I hereinafter use the S.I. units for simplicity) in [1/m], $\hat{\mathbb{n}}$ is dimensionless [-] (unit vector) and the delta function has as unit of measure the inverse of its argument (which is a distance function) [1/m]. Thus \begin{equation} \mathrm{dimensionless \hspace{3mm}number}[-] \neq \frac{[1/m][-][1/m]}{\mathrm{dimensionless \hspace{3mm}number}[-]}=[1/m^2] \end{equation}

If someone can explain to me if I made some mistake I would greatly appreciate it.

Note: \begin{equation} \hat{\mathbb{n}}=\frac{\mathbb{\nabla}{d}}{|\mathbb{\nabla}{d}|}=[-] \end{equation} \begin{equation} \kappa=\mathbb{\nabla}\cdot\hat{\mathbb{n}}=[1/m] \end{equation}

  • $\begingroup$ Since when is a surface tension term included in the Navier Stokes equations? Surface tension occurs at a free surface, and is treated as a boundary condition rather than a term in the momentum balance. $\endgroup$ Sep 3, 2018 at 13:07
  • $\begingroup$ As I wrote in the question title, I'm dealing with multiphase flows; to model the interfaces between the phases, surface tension must be taken into account. $\endgroup$ Sep 3, 2018 at 13:13
  • $\begingroup$ I'm voting to close this question as off-topic because check-my-work questions are generally not appropriate for physics SE. See physics.meta.stackexchange.com/questions/6093/… $\endgroup$
    – user191954
    Sep 7, 2018 at 7:37

1 Answer 1


As you pointed out, your $\kappa$ and $\delta_\varepsilon$ are still dimensional quantities, both in the units of [1/m]. To make them dimensionless, you just need to multiply by characteristic length $L$:

$\kappa' = \kappa L$, $\delta'_\varepsilon=\delta_\varepsilon L$

Then, $SF' = \frac{\kappa'\hat{n}\delta'_\varepsilon}{\rm We}$ is a dimensionless quantity.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.