# Why are Navier-Stokes equations needed?

Can't we picture air or water molecules individually? Then, why are Navier-Stokes equations needed, after all? Can't we just aggregate individual ones? Or is it computationally difficult, or inefficient to generate the picture of whole flow?

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– Emilio Pisanty Dec 4 '12 at 2:33

## 3 Answers

Consider a standard volume of $1\textrm{ m}^3$ of air. This contains on the order of $10^{25}$ molecules of O2 and N2.

If you needed to simulate or explain the physics occurring in that volume of air, would you want to model $10^{25}$ molecules and all the interactions between them or, say, 100x100x100 cells based on the Navier-Stokes equations?

Theoretically, it is possible to simulate every fluid flow ever by tracking every single molecule. But direct simulation of turbulence using the Eulerian Navier-Stokes equations requires $Re^{9/4}$ grid points and is thus totally impractical for Reynolds numbers larger than a few thousand. So simulating something with $10^{25}$ things to track is completely impossible.

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Just recording the position coordinates as doubles would take 1000000000000000 GB of RAM. – user1504 Dec 4 '12 at 3:31

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A quibble, but a cubic metre contains more than a mole (6.023e23 molecules). At STP a mole of an ideal gas is 22.4 litres. – John Rennie Dec 4 '12 at 6:59
You're right, it's been a long time since I had to think about those relationships. I'll fix it – tpg2114 Dec 4 '12 at 19:55

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I don't think numerical simulation was the reason Navier and Stokes derived their equations. – Bernhard Dec 5 '12 at 11:59
Just to add to this correct answer: even if you could simulate every molecule, you would gain little by it unless for some reason you knew the initial position of every molecule to start with. Even then, it would only help you to predict the future behaviour of that particular volume of fluid at that particular time - it wouldn't tell you anything about what would happen if you repeated the same experiment again, but with different molecules that weren't in exactly the same starting places. – Nathaniel Dec 5 '12 at 13:00
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Apart from the experimental and numerical difficulties to track each particle in a macroscopic piece of matter, there are physical reasons.

Theoretically, we can show that in the hydrodynamic regime the exact equations of motion for the collection of atoms/molecules reduce to the Navier-Stokes equations. This is shown in statistical mechanics.

Due to complex molecular effects, including molecular chaos, most degrees of freedom self-cancel and only some few dynamical modes survive. Those modes are collective --i.e. they describe the fluid as a whole-- and robust --i.e., they survive in macroscopic scales of space and time--.

That is, even if you were to trace the individual motion of each particle and next average to describe the collective motion of the whole fluid, you would see that your description is practically indistinguishable [*] from that given by solving the Navier-Stokes equations.

[*] The differences are of the order of the inverse of the size of the system and vanish for a macroscopic system.

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"Theoretically, we can show that in the hydrodynamic ..." I believe this paragraph is wrong at the current time. I thought you can derive NS from the Boltzmann equation, so techinically you can derive NS to a certain extent only for gases, but not for liquids. Or do you know statistical NS derivation for liquids? – Yrogirg Dec 5 '12 at 14:47
@Yrogirg: There is a kinetic theory for dense fluids as well, but is not of the Boltzmann type, of course. Moreover, you can obtain the hydrodynamic equations without using any kinetic equation, for instance using the Mori formalism for the microscopic dynamical variables. – juanrga Dec 6 '12 at 11:32

As apparent from the other answers, the continuum approach of the Navier-Stokes equations makes life easier, by reducing the degrees of freedom. For example, Poisseuile flow (laminar flow between flat plates or in a pipe), is easily solved using the Navier-Stokes equations, but I have no clue how to solve it using just molecular behavior (on the back of an envelope that is).

If you calculate the Knudsen number of a specific problem, which is just the molecular mean free path divided by some macroscopic length scale, you can see if the continuum approach is a valid for the considered system.

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