--- title: "Navier-Stokes equations" sort_title: "Navier-Stokes equations" date: 2021-04-12 categories: - Physics - Fluid mechanics - Fluid dynamics layout: "concept" --- While the [Euler equations](/know/concept/euler-equations/) govern *ideal* "dry" fluids, the **Navier-Stokes equations** govern *nonideal* "wet" fluids, i.e. fluids with nonzero [viscosity](/know/concept/viscosity/). ## Incompressible fluid First of all, we can reuse the incompressibility condition for ideal fluids, without modifications: $$\begin{aligned} \boxed{ \nabla \cdot \va{v} = 0 } \end{aligned}$$ Furthermore, from the derivation of the Euler equations, we know that Newton's second law can be written as follows, for an infinitesimal particle of the fluid: $$\begin{aligned} \rho \frac{\mathrm{D} \va{v}}{\mathrm{D} t} = \va{f^*} \end{aligned}$$ $$\mathrm{D}/\mathrm{D}t$$ is the [material derivative](/know/concept/material-derivative/), $$\rho$$ is the density, and $$\va{f^*}$$ is the effective force density, expressed in terms of an external body force $$\va{f}$$ (e.g. gravity) and the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{\sigma}$$: $$\begin{aligned} \va{f^*} = \va{f} + \nabla \cdot \hat{\sigma}^\top \end{aligned}$$ From the definition of viscosity, the stress tensor's elements are like so for a Newtonian fluid: $$\begin{aligned} \sigma_{ij} = - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i) \end{aligned}$$ Where $$\eta$$ is the dynamic viscosity. Inserting this, we calculate $$\nabla \cdot \hat{\sigma}^\top$$ in index notation: $$\begin{aligned} \big( \nabla \cdot \hat{\sigma}^\top \big)_i = \sum_{j} \nabla_j \sigma_{ij} &= \sum_{j} \Big( \!-\! \delta_{ij} \nabla_j p + \eta (\nabla_i \nabla_j v_j + \nabla_j^2 v_i) \Big) \\ &= - \nabla_i p + \eta \nabla_i \sum_{j} \nabla_j v_j + \eta \sum_{j} \nabla_j^2 v_i \end{aligned}$$ Thanks to incompressibility $$\nabla \cdot \va{v} = 0$$, the middle term vanishes, leaving us with: $$\begin{aligned} \va{f^*} = \va{f} - \nabla p + \eta \nabla^2 \va{v} \end{aligned}$$ We assume that the only body force is gravity $$\va{f} = \rho \va{g}$$. Newton's second law then becomes: $$\begin{aligned} \rho \frac{\mathrm{D} \va{v}}{\mathrm{D} t} = \rho \va{g} - \nabla p + \eta \nabla^2 \va{v} \end{aligned}$$ Dividing by $$\rho$$, and replacing $$\eta$$ with the kinematic viscosity $$\nu = \eta/\rho$$, yields the main equation: $$\begin{aligned} \boxed{ \frac{\mathrm{D} \va{v}}{\mathrm{D} t} = \va{g} - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} } \end{aligned}$$ Finally, we can optionally allow incompressible fluids with an inhomogeneous "lumpy" density $$\rho$$, by demanding conservation of mass, just like for the Euler equations: $$\begin{aligned} \boxed{ \frac{\mathrm{D} \rho}{\mathrm{D} t} = 0 } \end{aligned}$$ Putting it all together, the Navier-Stokes equations for an incompressible fluid are given by: $$\begin{aligned} \boxed{ \frac{\mathrm{D} \va{v}}{\mathrm{D} t} = \va{g} - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} \qquad \nabla \cdot \va{v} = 0 \qquad \frac{\mathrm{D} \rho}{\mathrm{D} t} = 0 } \end{aligned}$$ Due to the definition of viscosity $$\nu$$ as the molecular "stickiness", we have boundary conditions for the velocity field $$\va{v}$$: at any interface, $$\va{v}$$ must be continuous. Likewise, Newton's third law demands that the normal component of stress $$\hat{\sigma} \cdot \vu{n}$$ is continuous there. ## References 1. B. Lautrup, *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, CRC Press.