45
I just wanted to give a more concrete idea of how we know these equations even though we have trouble proving analytical theorems about them.
Stuff moving in space
Consider any stuff (as in, any conserved quantity) distributed over space. We know that we can describe this with a time-dependent density field $\rho(x,y,z,t)$ such that any little volume $dV$ ...
30
None of the interesting equations in physics can be derived from simpler principles, because if they could they wouldn't give any new information. That is, those simpler principles would already fully describe the system. Any new equation, whether it's the Navier-Stokes equations, Einstein's equations, the Schrodinger equation, or whatever, must be ...
24
Turbulence is indeed an unsolved problem both in physics and mathematics. Whether it is the "greatest" might be argued but for lack of good metrics probably for a long time.
Why it is an unsolved problem from a mathematical point of view read Terry Tao (Fields medal) here.
Why it is an unsolved problem from a physical point of view, read Ruelle and Takens ...
23
They are derivable from classical mechanics using either the continuum or molecular points of view.
Starting with a continuum view, one applies conservation of mass, momentum, and energy to a control volume and the result is the Navier Stokes equations. The Navier Stokes equations, in the usual form, apply to Newtonian fluids, that is fluids whose stress ...
22
Frank White's Viscous Fluid Flow book contains a good list of these "exact" solutions. I am not sure if it is complete though. I've provided links to a few of the solutions.
Steady flow between a fixed and moving plate
Axially moving concentric cylinders
Flow between rotating concentric cylinders
Hagan-Poiseuille flow
Combined Couette-Poiseuille flow ...
19
As @QMechanic mentioned in a comment, the Navier-Stokes equations are just $F = ma$, but they look much scarier. Assuming an incompressible fluid, you have:
$$ \rho \frac{D u_i}{D t} = -\frac{\partial \sigma_{ij}}{\partial x_j} + f_b $$
where $\rho$ is the density (mass per unit volume), $D u_i /D t$ is the acceleration (written in the Lagrangian form, ...
18
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?
...
18
You are not missing anything. Rather I think you are placing too much emphasis on the scientific accuracy of something said for effect in a very chatty presentation.
The spoken words almost instantly are not repeated in the headlines, whereas the rest of the quote is. And the presenter does not develop this idea of being able to smell thioacetone almost ...
17
I once asked Putterman after a similar colloquium what he meant by this statement, and his answer was "long time tails". Long time tails are fractional powers that appear in the long time behavior of correlation functions, see, for example, here and here. These fractional powers are seen in molecular dynamics (they are more difficult to see experimentally), ...
15
If you go through the process of non-dimensionalizing the equations, the math becomes more clear. If you start with the momentum equation (ignoring viscous forces because they aren't important for the analysis):
$$
\frac{\partial u_i}{\partial t} + \frac{\partial u_i u_j}{\partial x_j} = -\frac{1}{\rho} \frac{\partial p}{\partial x_i} + g
$$
Then introduce ...
12
Dear Ondřeji, a good question but a part of the answer is that your equation for the fluid is underdetermined. It treats $p,\rho$ as independent variables. But the physical system only knows how to behave if you also substitute some equation of state, i.e. a function $p=p(\rho)$ or $p=p(\rho,\vec v)$. Note that your Ansatz for the stress-energy tensor ...
12
This usually only applies to a wall bounded flow and is normally restricted to incompressible fluids. This result usually manifests in boundary layer theory and can be obtained through order of magnitude analysis of the Navier-Stokes equations. The steady, incompressible, and constant property momentum equation in the $y$ direction takes the form,
$$ u \frac{...
12
The Navier Stokes equations are a combination of Newton's 2nd law of motion (differential form) with the 3D version of Newton's law of viscosity (i.e., the mechanical constitutive equation for a Newtonian fluid).
What you were taught in Physics at school was correct. Not many analytic solutions exist to interesting practical problems, and numerical ...
12
The NS equations do include the gravity term, see the Wikipedia entry where it's included as a body force term.
Under some conditions, it can be neglected: large Froude number (as tpg2114 showed) or hydrostatics (in which the gravity force is balanced by the pressure gradient) or horizontal flows ($g$ in $z$-direction but flow is in $x,\,y$ plane). It can ...
10
1. Why is force linear in the velocity?
On a conceptual level, the idea behind stokes flow is that it is a very overdamped system because the viscosity is very high. When I say "overdamped" I am trying to make an analogy with a damped harmonic oscillator. You gave the equation
$$\ddot{x}=-k^2 x,$$
but a damped harmonic oscillator has a drag term:
$$\ddot{x}=...
9
I don't know a good answer to your first question (I'd be interested in a good text for that myself), but I can answer the second.
It's easier to explain if we temporarily imagine $\phi$ represents the concentration of some dye made up of little particles suspended in the fluid. The convective term (aka advective term) is transport of $\phi$ due to the ...
8
It comes from the notion of the boundary layer and whether it stays attached to the wall or not. If you consider the momentum equation normal to the wall, the only way there can be a pressure gradient normal to the wall is if there is a velocity or acceleration normal to the wall. If that is the case, then the boundary layer is no longer attached.
This ...
8
The traditional derivation of the Navier-Stokes equations starts by looking at a fluid parcel and the different fluxes over the surface in the integral form. The integral form is preferred as it is more general than the differential form: For the latter one has to assume differentiability and thus it is not valid for flow discontinuities such as shocks in ...
7
Turbulence is not one of the great unsolved problems in physics. Physics tells us exactly how turbulence emerges as a direct consequence of local mass and momentum conservation. We can create multiparticle computer models such as lattice gas automata that generate turbulence at large length and time scales. We can write down the equations that govern ...
7
It's not just a simplifying assumption. There are good reasons as to why the approximation is used for incompressible flows: Vorticity is introduced into incompressible flows only through surface boundaries. A flow that is initially irrotational and and does not interact with a surface will remain irrotational. You can see this from the equation for the ...
7
There are known solutions to the Navier-Stokes equations. A simple example would be laminar shear-driven flow between two moving plates. Just as in the case of Einstein's equations, the known solutions regard simple situations with particular boundary conditions; a general solution that covers all possible cases is not known in either case. One should not ...
7
For a single-component fluid, the conservation of mass follows
$$
\left(\begin{array}{c}\text{mass of fluid } \\ \text{in volume }\Delta V\end{array}\right)=\left(\begin{array}{c}\text{flux of fluid } \\ \text{in/out of volume }\Delta V\end{array}\right)+\left(\begin{array}{c}\text{sources or} \\ \text{sinks in }\Delta V\end{array}\right)
$$
In terms of a ...
7
The Reynolds number, with $\rho$ the density, $u$ the velocity magnitude, $\mu$ the viscosity and $L$ some characteristic length scale (e.g. channel height or pipe diameter) is given by
$$\text{Re}=\frac{\rho~u~L}{\mu}.$$
This is a dimensionless relation of the ratio of inertial forces ($\rho u u$) to viscous forces ($\mu\frac{u}{L}$). It therefore signifies ...
7
To add to tpg124's answer and to answer the implict question in your statetent:
How were the equations discovered in the first place if we can't solve them?
simplicity and clarity of an equation's meaning and difficulty of working out the implications of that equation are two wholly different things. For the Navier-Stokes equation, the meaning is crystal ...
6
Solutions of the form
$$ cos(x_i)e^{-x_j}$$
are common specific solutions of the Navier-Stokes equations in simplified (not simple) problems. These are however problems where inertia is ignored, which you include. (Please note that I am using index notation, with $i,j\in\{1,2,3\}$). $x_j$ is then the wall normal direction. This is actually quite well ...
5
Strictly speaking, turbulence doesn't exist in two dimensions. The energy cascade required for turbulence to develop (transfer energy from large scales to small scales) is due to the (incompressible for illustration) vorticity equation:
$\frac{D\vec{\omega}}{Dt} = \left(\vec{\omega}\cdot\nabla\right)\vec{v} + \nu\nabla^2\vec{\omega}$
specifically the ...
5
MS Mohamed et al begin with a "standard vector calculus formulation of the NS equations",
$$
\frac{∂ \bf u}{∂ t} - \mu ∆ {\bf u} + ({\bf u} \cdot \nabla) {\bf u} + \nabla p = 0
$$
$$
\nabla \cdot {\bf u} = 0
$$
and use the rotational form,
$$
\frac{∂ \bf{u}}{∂ t} + \mu \nabla \times \nabla \times {\bf u} - {\bf u}\times(\nabla \times {\bf u}) + \nabla p^d ...
5
The Navier-Stokes equation to which you refer is more generally the first moment of velocity of the Boltzmann equation. In order to get a proper connection to heating, you need a second-velocity-moment Navier-Stokes equation. The Boltzmann equation keeps track of distributions of particles. This changes the question from "What is the density and flow of a ...
Only top voted, non community-wiki answers of a minimum length are eligible
