# What is the meaning of pressure in the Navier-Stokes equation?

I have a hard time to wrap my head around pressure in the Navier-Stokes equation! It may sounds ridiculous but still I cannot understand the true meaning of pressure in the Navier-Stokes equation. Let's do some math to explain my purpose more accurately! Let's start from basics of the physics and in my opinion that would be the first equation in the classical thermodynamics as equation of state. We assume: there is a fluid, which have a equation of state as:

$$\rho = \rho(P,T)$$

Where $\rho$ is the density of the fluid, $P$ is the pressure, and $T$ is the temperature. Let's take a derivative from this equation to have:

$$d\rho = (\frac{\partial \rho}{\partial P})_{T} dP + (\frac{\partial \rho}{\partial T})_{P} dT$$

Let's assume that our fluid is in the thermal equilibrium and its temperature will not change, as a result: $d T = 0$

So, we have:

$$d \rho = (\frac{\partial \rho}{\partial P})_{T} dP$$

I know it's a lot of assumption but again let's assume that density change because of pressure change is not nonlinear and our fluid in fact behaves like a ideal gas. As a result, I call $(\frac{\partial \rho}{\partial P})_{T}$ the inverse square of the speed of sound, which is a constant number, as:

$$(\frac{\partial \rho}{\partial P})_{T} = c_{s}^{-2}$$

So, finally we have:

$$d \rho = c_{s}^{-2} d P$$

Or:

$$\Delta \rho = c_{s}^{-2} \Delta P$$

Or again:

$$(\rho - \rho_{f}) = c_{s}^{-2} (P - P_{0})$$

Where $\rho_{f}$ is the density of the fluid at the rest or reference, which is a tabulated value for each fluid, and $P_{0}$ is the reference pressure.

Now, I would assume my fluid is an incompressible fluid and it means (density is constant and it is really constant!):

$$\rho = \rho_{f}$$

As a result, because, every fluid regardless of its compressibility or incompressibility has a finite speed of sound, I would argue that:

$$P = P_{0}$$

Or in other word, strictly speaking pressure should be equal to the reference pressure.

Now, I proved that for an incompressible fluid as long as density is constant, pressure should also be a constant. So in incompressible Navier-Stokes equation we have:

$$\rho_{f} \frac{\partial \mathbf{u}}{\partial t} + \rho_{f} (\mathbf{u} \cdot \nabla)\mathbf{u} = -\nabla P + \nabla \cdot \tau$$

And I showed that for incompressible fluid, P is just constant, so: $\nabla P = 0$!

As a result, I could simplify the Navier-Stokes equation as:

$$\rho_{f} \frac{\partial \mathbf{u}}{\partial t} + \rho_{f} (\mathbf{u} \cdot \nabla)\mathbf{u} = \nabla \cdot \tau$$

Now let's back to my original question:

Based on these calculations I would say that pressure in the incompressible Navier-Stokes equation is just a dummy variable, which does not have any physical meaning! I appreciate if someone could explain this to me!

• there are relevant cases in which $dT$is not neglectible even for incompressible fluids. This is the case for e.g. describing air movement on large scales. So P as a function of T is relevant in this case and $\nabla P$ is not necessary 0 Sep 5, 2018 at 21:40
• @Katermickie I know what you mean but let's just limit our analysis to those cases where $d T = 0$, which are not rare in the literature! I didn't say that my analysis could cover all the cases but as long as my assumptions are hold true, I'm interested to study the consequences. I'm truely interested in these assumptions because the application that I use Navier-Stokes for, is completely match to these assumptions!
– GGG
Sep 5, 2018 at 21:44
• @Katermickie And also please read the question more carefully because your example when air moves on large scales probably should be studied by using compressible Navier-Stokes equation and you need to couple thermal energy balance equation (i.e. heat transfer equation) to the Navier-Stokes! As a result, your example is not relevant to my assumptions!
– GGG
Sep 5, 2018 at 21:47
• Related, if not dupe of, physics.stackexchange.com/q/319577/25301 Sep 5, 2018 at 21:49
• Why are you saying that the pressure is constant for an incompressible fluid? Sep 6, 2018 at 0:26

There are two pressures: Thermodynamic pressure $p_\text{thermo}$, and Mechanical pressure $p_\text{mech}$. Thermodynamic pressure, a concept from equilibrium thermodynamics and therefore applicable only to a static fluid, is given by an equation of state: $p_\text{thermo}=f(\rho,T)$, where $\rho$ is fluid density and $T$ its temperature. A moving fluid is not in equilibrium and its $p_\text{thermo}$ is not defined. Mechanical pressure is the isotropic part of the stress tensor and is defined for a moving fluid too; $p_\text{mech}$ appears in the Navier-Stokes equation.

If a static fluid is isothermal and has constant density ($\rho,T$ fixed) then $p_\text{thermo}$ is also fixed. But mechanical pressure given by hydrostatic equation varies with depth in an isothermal constant-density fluid.

Your derivation mixes up the two pressures. The relations are: \begin{align} \mathrm{d}\rho&=c_s^{-2}\,\mathrm{d}p_\text{mech}\\ \mathrm{d}\rho&=\left(\frac{\partial\rho}{\partial T}\right)\,\mathrm{d}T+\left(\frac{\partial\rho}{\partial p_\text{thermo}}\right)\,\mathrm{d}p_\text{thermo}. \end{align} The latter equation from thermodynamics is applicable only to a static fluid. The former is not a thermodynamic equation. An incompressible fluid is defined to be one whose density doesn't depend on its mechanical pressure $p_\text{mech}$; it doesn't say that $p_\text{mech}$ can't vary. Therefore as you approach the limit of an incompressible fluid, $\mathrm{d}\rho\to0$, we must necessarily have $c_s\to\infty$. It is incorrect to say "...every fluid regardless of its compressibility or incompressibility has a finite speed of sound..."; incompressible fluids don't exist, so a priori you wouldn't know what sound speed should be assigned to a hypothetical fluid; to be consistent with the definition of incompressibility however a variation in $p_\text{mech}$ must be allowed which demands that sound speed in an hypothetical incompressible fluid be infinite.

P.S. Here's article1 and article2 that may interest you (NB: both are PDFs).

• You said the equation $d \rho = c_{s}^{-2} d p_{mech}$ is not a thermodynamic equation. If it's not how you can derive it? I mean it's an ideal gas equation of state, so why there should be two definitions for pressure as thermodynamical and mechanical?! My approach to Navier-Stokes equation is a little bit different from conventional fluid dynamicists and as a result I see that equation ($d \rho = c_{s}^{-2} d p_{mech}$) everyday because I use lattice Boltzmann method for solving Navier-Stokes, and still I can't digest why ideal gas equation of state should give me pressure in Navier-Stokes!
– GGG
Sep 6, 2018 at 13:25
• @MehrdadYousefi Here's a derivation. What is needed for the derivation is the concept of bulk modulus. For an ideal gas, if you assumed that thermodynamic and mechanical pressures are the same (this assumption is often made in other contexts as well), then you can use its equation of state to explicitly compute the bulk modulus in terms of known properties of the gas. Also no matter how you derive the Navier-Stokes equation, a distinction between thermodynamic and mechanical pressure is necessary...
– Deep
Sep 7, 2018 at 5:23
• ... because a moving fluid is not in equilibrium and the concept of thermodynamic pressure is inapplicable to it, while its mechanical pressure is well defined. "My approach to Navier-Stokes equation is a little bit different from conventional fluid dynamicists ..." Could you post a relevant link? I would be interested in an alternate derivation.
– Deep
Sep 7, 2018 at 5:27
• It's a good starting point to be familiar with my approach "lattice Boltzmann method" to solve Navier-Stokes instead of using conventional FEM or FVM flow solvers: dartmouth.edu/~cushman/papers/2018-Boltzmann-to-NS.pdf
– GGG
Sep 7, 2018 at 13:24
• @MehrdadYousefi Thanks for the link. I ought to discuss with my colleagues who use LBM. In any case I think the pressure computed using LBM can't be thermodynamic pressure because a moving fluid is not in equilibrium. But applying equilibrium thermodynamic relations to a non-equilibrium situation (moving fluid) is an excellent approximation if the deviation from equilibrium is not too large. Usually this means that time scale of the flow is much larger than the molecular time scale (for eg. mean collision time).
– Deep
Sep 8, 2018 at 4:27

For an incompressible fluid, the pressure in the Navier Stokes equation represents the isotropic part of the stress tensor. It is determined up to an arbitrary constant value; that is, adding an arbitrary constant to the pressure at all location throughout the flow field still enables it to satisfy the NS equation. The arbitrariness in removed by specifying the pressure at any location on the boundary.