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# Talk:Navier–Stokes equations/Archive 1

I moved the following reference from the main page:

* V.Budarin.A mehod of finding equations of motion of a discountinuity flow.//Heart/Mass transfer forum, MIF-2000, V.1, P.238-241.

http://www.itmo.by/forum/forum7/index.html

Who is the author of this work. What is the //Heart/Mass transfer forum. - science conference in Lukov institute, Minsk, May 2000 y.

What does MIF stand for.- Minsk International Forum.

What is explained in this work? - Consider solution method for

mechanical fluid tasks, which according or not accoding compact(continuous)environment (method for flows with any macrostructure). More information -http://www.geocities.com/week65053/index.html

--Vit.

--AxelBoldt

Wouldn't this be agood spot to actually show the equations?

Yes, they most definitely should be here. --AxelBoldt

## Body forces?

Ack. In the Derivation and description, each term in the right hand side of the Navier-Stokes system is described as a "body force". In what sense is that right? The terms ${\displaystyle -\nabla p}$ and ${\displaystyle \nabla \cdot \mathbf {T} }$ arise from an application of the divergence theorem from the surface stress terms in the integral version of the momentum equations. Only ${\displaystyle \mathbf {f} }$ represents a body force term in the integral version of the momentum equations.

${\displaystyle {\frac {D}{Dt}}\int _{V(t)}\rho \mathbf {v} dV=\oint _{S(t)}\left(-p\mathbf {I} +\mathbf {T} \right)\cdot \mathbf {n} dS+\int _{V(t)}\mathbf {f} dV}$

--71.98.69.218 01:28, 26 July 2007 (UTC)

but those forces act on the infinitesimal volume elements, precisely because of the divergence theorem. A non-body force would be the wind acting on a free surface of water. CyrilleDunant 05:42, 26 July 2007 (UTC)
We seem to be having an issue of semantics. Any differential volume element of fluid can be accelerated or experience a change in momentum by the action of stresses applied along its surface (e.g., pressure, shear stress) or by forces acting "throughout" it (e.g., gravity, Maxwell stresses if the fluid element carries bulk charge, etc.). The former I am referring to as "surface forces" and the latter I'm referring to as "body forces". If by a body force you just mean "a force acting on a fluid element divided by the volume of that element", then I have no problem - however, this doesn't match the definition I'm used to.
This is because the equations are expressed in Lagrangian form. In a fluid problem, there is no such thing as a surface force, except for those that come from the fluid-surface interactions. All the forces are "body" forces, because they are expressed as integrals over infinitesimal volumes. Because the volumes are "arbitrary" and can be "infinitesimal" (yay continuity), the integral can be dropped, and you are left with quantities that are added to each other. As such, they need to be compatible, and a dimensional analysis will reveal them to be volume (body) forces.CyrilleDunant 20:38, 2 August 2007 (UTC)
I certainly agree that all of the terms in the Navier-Stokes equations have the dimension of force per unit volume - I believe I did so in the previous comment. This does not mean that the best way to interpret all of these terms is as "body forces", but again this appears to be a problem of semantics. I'm not going to push this any more or change the article, but I do want to clarify my position. Any fluid particle (even an infinitesimal one) has a bounding surface (albeit possibly a differential one), and a good, intuitive way to explain physically the concept of shear stress in a flow, for example, is to imagine a faster moving fluid particle "dragging" an adjacent, slower moving fluid particle - an interaction between the surfaces of these fluid particles resulting in a transfer of momentum is what I mean by a surface stress. Indeed, the solutions of very simple problems (ones that can be described as 1-dimensional flows in some orthogonal curvilinear coordinate system) are typically approached in undergraduate fluid dynamics classes by a method of shell balances (usually prior to the derivation of the Navier-Stokes equations in those classes) which explicitly describe fluid flow in such a manner. To risk belaboring my point, consider the meaning of the divergence of the viscous stress tensor, ${\displaystyle \nabla \cdot \mathbf {\tau } }$. For a given fluid element (say, in the middle of a vast expanse of fluid), the viscous stress vector at some point along its bounding surface is ${\displaystyle \mathbf {s} \equiv \mathbf {n} \cdot \mathbf {\tau } }$, where ${\displaystyle \mathbf {n} }$ is the unit normal to the surface at that point. If we integrate ${\displaystyle \mathbf {s} }$ around the bounding surface, we have the total force acting on the fluid element due to viscous stress. The divergence of ${\displaystyle \mathbf {\tau } }$ is simply what we get when we divide that integral by the volume of the fluid element and take the limit as that volume becomes infinitesimally small. It represents the resultant force per unit volume on the particle due to viscous stresses acting along its bounding surface. While its tempting to call this quantity a body force, I think it is more physically meaningful to think of it this way, and not lose sight of the fact that these are forces arising from surface stresses. Again, semantics. Keep up the good work, by the way. I'm very impressed overall with how well Wikipedia has dealt with fluid mechanics.--71.98.86.246 05:03, 3 August 2007 (UTC)
Yes, absolutely. When the problem is expressed in Eulerian form. Because continuity is assumed, there is no such thing as a "fluid particle". The essential quantity gotten from the solving of the equations is velocity (and a bunch of state variables), not position in time. I think this is an important point that is perhaps not well expounded in the article: there are no forces acting on fluid particles, because there are no fluid particles. There are forces acting on infinitesimal volumes, but they only make sense if you understand them as pervasive body forces. When you solve the problem numerically, using finite elements, it becomes obvious why those are body forces: their effect gets integrated on the element volumes. You can also solve a fluid problem by having a huge amount of particles bouncing into each other, which would relate to your understanding. This also works, but is not derived directly from the canonical equations.CyrilleDunant 06:30, 3 August 2007 (UTC)

Just because they're derived from conservative laws doesn't mean that stresses (or stress gradients in this case) don't qualify as (body) forces. f is an external force, a point source/sink of momentum. "Body force" isn't meant to imply externally applied force, it's only a force per volume, just as a surface force is force per area. Pressure is a surface force, right? No protest there? -Ben pcc 23:18, 31 July 2007 (UTC)

## Normal and shear stresses

In the Derivation and description section, there seems to be some confusion about the nature of viscous stresses and pressure. While pressure always acts normally to a given surface in a fluid, viscous stresses can also act normally. The description seems to imply otherwise.--71.98.69.218 00:53, 26 July 2007 (UTC)

Correcto! The words "shear" and "viscous" seemed to be swapped everywhere. Also, I kind of get your message concerning the surface force business. It's a matter of semantics indeed, but it'd be dishonest not to mention what those forces come from. I changed it a little, it still says "body force" but it tells a little more about the stress gradients, I hope it's ok. I think it's still important to use the phrase "body force" since many people don't see that the momentum equation is (yes, this is gross, horrible, dishonest oversimplification) kind of like Newton's law divided by volume.
Though I love wikipedia, be aware that there are some really pitiful articles. Are you into horror? Check out this article from a few months ago. -Ben pcc 20:13, 5 August 2007 (UTC)

"viscous stresses can also act normally"- it is false. Viscous stresses - it is force, acts on the unit of a surface. But this force acts into the volume of liquid. Equation of Navier-Stokes was derivated for volume, not for a surface. Any surface of liquid has a folume.There is not surface of liquid without a volume. —Preceding unsigned comment added by Dmitri Gorskin (talkcontribs) 01:23, 16 May 2008 (UTC)

## The continuity equation is curved correct

If you look carefully you will see that that the left hand side of the equation contains a time derivative of a scalar field, which is a scalar field (i.e. has a dimension of 1)

${\displaystyle {\frac {D\rho }{Dt}}}$

plus a space derivative of a vector field, which is a tensor field (i.e. has the dimension of a 3x3 matrix)

${\displaystyle \rho \nabla \mathbf {v} }$

Therefore, dimensions in the terms of the equation do not agree, i.e., it is not possible to add a scalar to a matrix. The problem seems to be that one of the authors is under the impression that the continuity equation is derived from the substantial derivative expression. The fact is that the continuity equation is derived from considering a volume fixed in space (does not move with the fluid). Therefore changes in mass inside such a volume are stated in the following way:

${\displaystyle \int _{V}{\frac {d\rho }{dt}}dV=\int _{A}\rho \mathbf {v\cdot n} dA}$

the sign of the right hand side depends on the direction of the surface normal n. normally it's because n pointing away from the surface

${\displaystyle \int _{V}{\frac {d\rho }{dt}}dV=-\int _{A}\rho \mathbf {v\cdot n} dA}$

Where V is the control volume and A its area; :${\displaystyle \rho }$ is the density of the fluid and ${\displaystyle \mathbf {v} }$ its velocity. The left hand side represents changes in density inside the volume whereas the right han side represents advective transport of mass (mass carried out of the boundary of the volume by the moving fluid). This equation states simply that the density of the fluid inside the volume must change (expand or compress) in response to either loss or addition of mass through the boundaries of the volume. Now applying the divergence theorem we have

${\displaystyle \int _{V}({\frac {d\rho }{dt}}-\nabla \cdot (\rho \mathbf {v} ))=0}$

Where the differential operator :${\displaystyle \nabla \cdot ()}$ is the divergence. Eliminating the integral, one finally finds

${\displaystyle {\frac {d\rho }{dt}}-\nabla \cdot (\rho \mathbf {v} )=0}$

This equation is the "continuity equation" for a moving fluid (see eq. 1.2 in p. 2 of Landau L.D., and Lifshitz E.M., 1987, Fluid Mechanics, 2nd Edition, Pergamon Press, pp. 539). Now, the divergence of a vector field :${\displaystyle \nabla \cdot (\rho \mathbf {v} )}$ is a scalar field. Thus dimensions agree for all terms in the equation.

If density is constant, the continuity equation is reduced to

${\displaystyle \nabla \cdot (\mathbf {v} )=0}$

The equation pretending to be the Navier-Stokes equation contains two terms of different physical dimension, rho u and rho d/dt(rho). This cannot be right.

Formula removed from article, thus:

${\displaystyle -\nabla p+\mu \left(\nabla ^{2}\mathbf {u} +{1 \over 3}\nabla (\nabla \cdot \mathbf {u} )\right)+\rho \mathbf {F} =\rho \left({\partial \mathbf {u} \over \partial t}+\mathbf {u} \cdot \nabla \mathbf {u} \right)}$

Equations without defined variables and context are useless.

Note that the article in its current form gives the impression that there is "a Navier-Stokes equation", rather than a set of them: we should either give them all, or none of them.

-- The Anome 11:05, 23 Dec 2003 (UTC)

Look, I don't want to sound nasty or anything, but
• they are defined
• the article does state that NS are a 'system' of equations and that many specific form are derived from the general form, of which there is a single form.CyrilleDunant 16:38, 25 August 2005 (UTC)

## Removed equation again

${\displaystyle -\nabla p+\mu \left(\nabla ^{2}\mathbf {u} +{1 \over 3}\nabla (\nabla \cdot \mathbf {u} )\right)+\rho \mathbf {u} =\rho \left({\partial \mathbf {u} \over \partial t}+\mathbf {u} \cdot \nabla \mathbf {u} \right)}$

where:

p is pressure
${\displaystyle \mathbf {u} }$ is fluid velocity
${\displaystyle \mu }$ is viscosity
${\displaystyle \rho }$ is density

Problems include: no sign of which one this is, no derivation or motivation, and what's with the factor of one-third? -- The Anome 19:28, 19 Jul 2004 (UTC)

## Why are the equations missing?

The equaitons are correct (yes, the units are correct) except the div(u) term is zero, because constant density is assumed for the "official" Navier-Stokes equations. Also, to address the other concern, this is not one equation, it is a set of equations. u is a vector and has three components. This set of equations should be added to the text with the variables defined.

The problem is that there is no single "official" version of the Navier-Stokes equation or equations. Different ways of posing the problem (streamline/turbulent, incompressible/compressible...) yield different but related sets of equations. Some people prefer to write everything out in components, others like vectors. Some like the Reynolds-approximation version, some want to do the whole thing right down to thermal terms and shearing. There are lots of different choices for variable names. Just pulling an equation out of a hat and saying "this is the Navier-Stokes equation" is not useful, because it does not provide insight for the reader into what the equation(s) mean(s). What would be useful would be a derivation from scratch, and then some special case versions for simple cases. See some of the references, particularly [1] for a good example of this.
Just for comparison, the reference above gives the (simplified) Navier-Stokes equations as:
${\displaystyle {{\partial \rho } \over {\partial t}}+\nabla \cdot (\rho \mathbf {u} )=0}$

${\displaystyle {\partial \mathbf {u} \over \partial t}+({\mathbf {u} }\cdot \nabla )\mathbf {u} =-{1 \over \rho }\mathbf {\nabla } p-\mathbf {\nabla } \phi +{\mu \over \rho }\nabla ^{2}{\mathbf {u} },}$
${\displaystyle \rho \left({\partial \varepsilon \over \partial t}+{\mathbf {u} }\cdot \nabla \varepsilon \right)-\nabla \cdot (K_{H}\nabla T)+p\nabla \cdot {\mathbf {u} }=0.}$
-- The Anome 07:45, 21 Jul 2004 (UTC)
Is a full derivation appropriate here? That seems to be more of a wikibooks thing. However, if we really want one, I'm sure I can cobble one up. -- Kaszeta 21:16, 28 Oct 2004 (UTC)

I think that the (simplified) Navier Stokes equations should be put somewhere in the beginning of the page. We can just add a small note that this is a simplified version and there are some other verisons that account for stuff like thermal gradients and magnetic fluids. I think that most of the readers to know what the formula looks like. We don't want to force these people to read the entire derivation. Therefore the equations should be placed first and the derivation (or whatever) should be placed last. (anonymous)

Most of the equations here are just plain wrong! Phys 19:26, 30 August 2005 (UTC)

Ok. What equations are wrong? I mean, the expanded forms are a bit painful to read, but they are correct...CyrilleDunant 21:38, 30 August 2005 (UTC)
I corrected most of them Phys 17:06, 31 August 2005 (UTC)
I am sorry to contradict you. You did not correct them, you just swapped a set of conventions for another (I know, I checked in a textbook which used the "other" convention.). Wich is sort of OK, just a bit pointless, but it would have been more useful to eradicate.
I disagree with the statement saying "The nature of the trace of ${\displaystyle \mathbb {P} }$ is known, it is thrice the pressure, thus:"
${\displaystyle {\frac {\sigma _{xx}+\sigma _{yy}+\sigma _{zz}}{3}}=-p}$
as opposed to the diagonal part of ${\displaystyle \nabla \mathbb {T} }$ is the gradient of pressure. Mostly because the gradient of pressure has some physical sense to it unlike the trace of the tensor...
Also, but again, this is purely conventional, the stress components are divided in shear and non-shear, symbolised by ${\displaystyle \tau }$ and ${\displaystyle \sigma }$. I am at loss to understand what ${\displaystyle \tau _{ii}}$ might be (${\displaystyle {\frac {\partial d_{i}}{\partial x_{i}}}=\tau _{ii}}$ derivative of the displacement i along axis i normal to i ??). Of course they are nil...
I will not revert brutally (because I suppose you checked what you wrote, and I you think the way it is now is better, well, why not :)) But it would be nice to discuss before making radical changes...CyrilleDunant 12:39, 1 September 2005 (UTC)
${\displaystyle \nabla \mathbb {T} }$ is a tensor of rank 3 whereas ${\displaystyle \nabla \cdot \mathbb {T} }$ is a vector. By the diagonal, I assume you mean the trace. Neither of them will give a matrix, since neither of them are of rank 2. And it simply isn't covariant to split a tensor of rank 2 like this:
${\displaystyle {\begin{pmatrix}\sigma _{xx}&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\sigma _{yy}&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\sigma _{zz}\end{pmatrix}}}$

Phys 19:15, 1 September 2005 (UTC)

The decomposition of a tensor of rank 2 into a traceless part and its trace goes as
${\displaystyle {\begin{pmatrix}\tau _{xx}&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\tau _{yy}&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\tau _{zz}\end{pmatrix}}+{\begin{pmatrix}\sigma &0&0\\0&\sigma &0\\0&0&\sigma \end{pmatrix}}}$
where ${\displaystyle \tau _{xx}+\tau _{yy}+\tau _{zz}=0}$. Phys 19:19, 1 September 2005 (UTC)
Yes, but this is not pure math: there is also some physics behind... The goal is not to get a traceless part, but to make physically meaningful terms appear. Such, for example as pressure. Or the constraints on the fluid particle. What you do is correct, but not very didactic or indeed physicall relevant at this point in the article. Again Tau_xx has _no_sense_ CyrilleDunant 20:19, 1 September 2005 (UTC)
Let's define ${\displaystyle \mathbb {T} }$ as
${\displaystyle \mathbb {T} ={\begin{pmatrix}\tau _{xx}&\tau _{xy}&\tau _{xz}\\\tau _{yx}&\tau _{yy}&\tau _{yz}\\\tau _{zx}&\tau _{zy}&\tau _{zz}\\\end{pmatrix}}}$
and ${\displaystyle \mathbb {T} '}$ as
${\displaystyle \mathbb {T} ={\begin{pmatrix}0&\tau _{xy}&\tau _{xz}\\\tau _{yx}&0&\tau _{yz}\\\tau _{zx}&\tau _{zy}&0\\\end{pmatrix}}}$
${\displaystyle \nabla \cdot \mathbb {T} }$ and ${\displaystyle \nabla \cdot \mathbb {T} '}$ are NOT equivalent. One gives
${\displaystyle {\begin{pmatrix}{\frac {\partial \tau _{xx}}{\partial x}}+{\frac {\partial \tau _{xy}}{\partial y}}+{\frac {\partial \tau _{xz}}{\partial z}}\\{\frac {\partial \tau _{yx}}{\partial x}}+{\frac {\partial \tau _{yy}}{\partial y}}+{\frac {\partial \tau _{yz}}{\partial z}}\\{\frac {\partial \tau _{zx}}{\partial x}}+{\frac {\partial \tau _{zy}}{\partial y}}+{\frac {\partial \tau _{zz}}{\partial z}}\\\end{pmatrix}}}$
and the other gives
${\displaystyle {\begin{pmatrix}{\frac {\partial \tau _{xy}}{\partial y}}+{\frac {\partial \tau _{xz}}{\partial z}}\\{\frac {\partial \tau _{yx}}{\partial x}}+{\frac {\partial \tau _{yz}}{\partial z}}\\{\frac {\partial \tau _{zx}}{\partial x}}+{\frac {\partial \tau _{zy}}{\partial y}}\\\end{pmatrix}}}$
${\displaystyle \nabla \cdot \mathbb {P} }$ is equal to
${\displaystyle -{\begin{pmatrix}{\frac {\partial p}{\partial x}}\\{\frac {\partial p}{\partial y}}\\{\frac {\partial p}{\partial z}}\\\end{pmatrix}}+{\begin{pmatrix}{\frac {\partial \tau _{xx}}{\partial x}}+{\frac {\partial \tau _{xy}}{\partial y}}+{\frac {\partial \tau _{xz}}{\partial z}}\\{\frac {\partial \tau _{yx}}{\partial x}}+{\frac {\partial \tau _{yy}}{\partial y}}+{\frac {\partial \tau _{yz}}{\partial z}}\\{\frac {\partial \tau _{zx}}{\partial x}}+{\frac {\partial \tau _{zy}}{\partial y}}+{\frac {\partial \tau _{zz}}{\partial z}}\\\end{pmatrix}}}$
but not
${\displaystyle -{\begin{pmatrix}{\frac {\partial p}{\partial x}}\\{\frac {\partial p}{\partial y}}\\{\frac {\partial p}{\partial z}}\\\end{pmatrix}}+{\begin{pmatrix}{\frac {\partial \tau _{xy}}{\partial y}}+{\frac {\partial \tau _{xz}}{\partial z}}\\{\frac {\partial \tau _{yx}}{\partial x}}+{\frac {\partial \tau _{yz}}{\partial z}}\\{\frac {\partial \tau _{zx}}{\partial x}}+{\frac {\partial \tau _{zy}}{\partial y}}\\\end{pmatrix}}}$
And yes, a symmetric tensor (which has 6 independent components) decomposes into a shear part (which is 5 components, not three!!!) and a trace (which has 1 component) if we insist upon a covariant decomposition. It does not decompose into an off-diagonal part (with three components) and a diagonal part (with three components) covariantly. And stress isn't a constraint. Phys 19:41, 30 September 2005 (UTC)

Perhaps the expanded forma ought to simply die. They are illegible, and apparently are there only to make the following point : "look! those equations are horrible !"

I would second that. All it does is drawn the physical meaning of the equations into an indigest ocean of trivial calculations. Rama 12:58, 1 September 2005 (UTC)

## Historical background and actual relevance

Outside of these mathematical concerns, I think it would be very nice to have some historical background on the equations. When were they v�� �� � > � � � �

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E t E E t E d t t t tUser:CyrilleDunant|CyrilleDunant]] 21:45, 20 November 2005 (UTC) Also, what is their significance today ? Do we have solutions to them or just approximations, and for which cases (turbulent, laminar, etc) ?

• Well, if we had solutions, they'd be in the article, I guess :). Certain cases, we can solve. As for the significance, let us say that as far as we know, the describe perfectly the behaviour of fluids, if continuity is verifiedCyrilleDunant 21:45, 20 November 2005 (UTC)

All of this I would really like to know ! What do you think ? Olivier

--82.32.65.149 16:19, 20 November 2005 (UTC)

• That a section on History would be good. But I have no clue.CyrilleDunant 21:45, 20 November 2005 (UTC)
• What I can add is that the article is by Sir George Gabriel Stokes, Fellow of Penbroke College, read april 14, 1845 (this is the correct publication date) and published in the transactions of the Cambridge Philosophical Society volume IIX, 1849 titled:"On the Theories of the internal Friction of Fluids in Motion, and of the Equilibrium and Motion of Elastic Solids" Greetings, Roger Jeurissen
• My understanding is that Maxwell and Kirchhoff actually played the critical roles in adding the missing pieces to Euler's work, and deriving the Navier-Stokes equations in full. (10:28, September 5, 2007)

1) definition & historical relevance
2) conservation principles and additional assumptions for the general equations
3) the general equations

All the details for the substantial derivative need be removed in addition to the special cases. This article is too large and the additional details are completely meaningless to someone who doesn't already know them. Rather then explaining them further, I vote to have them removed. I would like input before I remove large amounts of this article. Josh Quinnell 03:10, 2 December 2005 (UTC)

• Actually, it would be a good idea to extend the article by introducing the adimentionnal forms of the equations as well as their classification (parabolic, hyperbolic, elliptic, mixed). Also I believe it is very necessary to have the details on the substantive derivative, as it allows one to understand the equations. But then certainly those sections could be a tad more didactic.CyrilleDunant 16:22, 2 December 2005 (UTC)
• This decision to arbitrarily include more information confuses me. There are loads of contributions to NS development that are just as relevant as the substantial derivative. None of these details are pertinent to understanding the NS equations. The details should be linked to their own articles. When inserted into this article they detract from the ability for a lay person to come here and learn about the NS equations. This is not a reference for an expert in NS; we have text books for that. Josh Quinnell 21:03, 2 December 2005 (UTC)
• Should it me merged in a larger section about control volumes ? In fact, I would rather the "full (as in, very large)" version of the equations was eliminated, As it brings nothing else than a statement of the type "look, plenty of confusing signs". Arguably, this is not a vulagarisation article either. The subject is highly technical (in a mathematical sense), and the article reflects that.CyrilleDunant 08:05, 3 December 2005 (UTC)

## Too much sorry

Never too much, needs much expanding

Sorry, I'm a fresh man here,and not able to use this systems here very much. Well,I decide to apology to all men. I will try my best to learn.

For the followings,I would tell something to discuss.

《Quote:》

Original subject:

             The continuity equation is flat wrong

If you look carefully you will see that that the left hand side of the equation contains a time derivative of a scalar field, which is a scalar field (i.e. has a dimension of 1)

${\displaystyle {\frac {D\rho }{Dt}}}$

plus a space derivative of a vector field, which is a tensor field (i.e. has the dimension of a 3x3 matrix)

${\displaystyle \rho \nabla \mathbf {v} }$

Therefore, dimensions in the terms of the equation do not agree, i.e., it is not possible to add a scalar to a matrix. The problem seems to be that one of the authors is under the impression that the continuity equation is derived from the substantial derivative expression. The fact is that the continuity equation is derived from considering a volume fixed in space (does not move with the fluid). Therefore changes in mass inside such a volume are stated in the following way:

${\displaystyle \int _{V}{\frac {d\rho }{dt}}dV=\int _{A}\rho \mathbf {v\cdot n} dA}$

Where V is the control volume and A its area; :${\displaystyle \rho }$ is the density of the fluid and ${\displaystyle \mathbf {v} }$ its velocity.:

Sorry but I'd deleted the other words of your talking about. I recently just use

${\displaystyle {\frac {D}{Dt}}={\frac {\partial }{\partial t}}+{\mathbf {u} }\cdot \nabla }$

to correct to

       ${\displaystyle {\frac {D\rho }{Dt}}}$ which is a scalar field (it's a vector one)

Actually, you would see that it(the math formula) does not belong scarlar field,but vector one does. Because of

${\displaystyle {\frac {D}{Dt}}u={\frac {\partial {u}}{\partial t}}+......}$

(we just take the front terms to analyse) which is operated to acceralations. Hence it is not a scalar field.

Totally I want to say is :

          ${\displaystyle {\frac {D}{Dt}}{\rho }*u={\frac {\partial {{\rho }*u}}{\partial t}}+......}$ refers to vectors * vectors = tensors

I'm from Taiwain. If I have any mistake opinions, tell me. Thank you.

Well... The Continuity equation is
${\displaystyle {\frac {D\rho }{Dt}}={\frac {\partial \rho }{\partial t}}+\mathbf {v} \cdot \nabla \rho =0}$

in indicial notation:

${\displaystyle {\frac {\partial \rho }{\partial t}}+v_{i}{\frac {\partial \rho }{\partial x_{i}}}\|x\in \{1,2,3\}}$

which is all scalar. The original discussion point was wrong :)CyrilleDunant 20:23, 19 December 2005 (UTC)

## Math in Physics About The Substantive Derivative

         About The Substantive Derivative

We might get some simple derivations by the following descriptions.

A Control Volume is filled with fluids,P represents pressure,with P=P(x,y,z,t) (Note:Pressure can be easily to image.) First we make total differential

${\displaystyle dP={\frac {\partial {P}}{\partial {t}}}dt+{\frac {\partial {P}}{\partial {x}}}dx+{\frac {\partial {P}}{\partial {y}}}dy+{\frac {\partial {P}}{\partial {z}}}dz}$

The rate of pressure change is

${\displaystyle {\frac {dP}{dt}}={\frac {\partial {P}}{\partial {t}}}+{\frac {\partial {P}}{\partial {x}}}{\frac {dx}{dt}}+{\frac {\partial {P}}{\partial {y}}}{\frac {dy}{dt}}+{\frac {\partial {P}}{\partial {z}}}{\frac {dz}{dt}}}$

Hence,

${\displaystyle {\frac {dP}{dt}}={\frac {\partial {P}}{\partial {t}}}+V_{x}{\frac {\partial {P}}{\partial {x}}}+V_{y}{\frac {\partial {P}}{\partial {y}}}+V_{z}{\frac {\partial {P}}{\partial {z}}}}$

by

${\displaystyle {\frac {D}{Dt}}={\frac {\partial }{\partial t}}+{\mathbf {V} }\cdot \nabla }$

therefore,

${\displaystyle {\frac {dP}{dt}}={\frac {\partial {P}}{\partial {t}}}+V_{x}{\frac {\partial {P}}{\partial {x}}}+V_{y}{\frac {\partial {P}}{\partial {y}}}+V_{z}{\frac {\partial {P}}{\partial {z}}}={\frac {DP}{Dt}}}$

where ${\displaystyle {\mathbf {V} }}$ is the fluid velocity,${\displaystyle V_{(x,y,z)}}$is the fluid speed, and ${\displaystyle \nabla }$ is the differential operator del. ^^

         (I'm happy on what achived:Mathematics technology from Taiwanese educations)^^

• Reference:James R. Welty,Charles E. Wicks,Robert E. Wilson,Gregory Rorrer Foundamentals of Momentum,Heat,and Mass Transfer ISBN 0-471-38149-7
Yes, this is a useful addition which could go under a section named "Control Volume", which certainely lacks in an article about NS :). But please, pay attention to the fact that throughout the article, vector quantities are in bold and scalar not. Also, velocity is ${\displaystyle \mathbf {v} }$ and pressure ${\displaystyle p}$. A figure would be nice also :)CyrilleDunant 15:47, 23 December 2005 (UTC)
Thank you first.I like physics very much though I major in C.E. .The culture background of my country Republic of China is conservative but privacy is a little open in some public permitted,so...I talk very much. I will move that article I wrote to Control Volume.
Thanks for the correction by you.[2] Are you the Wikipedia editor here,please?

## Stars in a galaxy

I find it hard to believe that the motion of stars in a galaxy would be governed by Navier-Stokes.

Keithdunwoody 01:51, 18 February 2006 (UTC)

in the same way that a fluid is composed of particles that don't actually touch themselves, but interact through electro-magnetic forces, the stars in a galaxy interact through gravity. This creates a form of viscosity. Note that for accurate modeling, you would have to add relativistic behaviour. CyrilleDunant 09:43, 18 February 2006 (UTC)
I'd be interested to read more about this. Do you have some references? All I could find on google was either 1) Navier-Stokes for protoplanitary systems, 2) University speaker series where both Navier-Stokes and galaxies were mentioned, but not in the same talk, or 3) Non-linear PDE course webpages. Thanks, 154.20.174.63 17:04, 18 February 2006 (UTC)
Well, aside from the size differences, a proto-planetary system and a galaxy would be pretty similar, as far as dynamics are concerned :) But no, I don't have references.CyrilleDunant 18:45, 18 February 2006 (UTC)

Motion of stars in a galaxy is not governed by Navier-Stokes, of course. In fact the situations are not even similar: gravity does not dissipate energy, as viscosity does, and what is the analogy of pressure? This statement needs to be removed. Even the claim that NS models weather is dubious. Perturbationist 02:02, 12 October 2007 (UTC)

## Wilkinson equations

From the article:

The equations can be converted to Wilkinson equations for the secondary variables vorticity and stream function. Solution depends on the fluid properties (such as viscosity, specific heats, and thermal conductivity), and on the boundary conditions of the domain of study.

I tried to make a Wikilink to Wilkinson equations, but there is no Wiki-page. I tried to search wikipedia, but there is no reference to Wilkinson equations. I tried to let Google search for Wilkinson equations, but still no useful results... Is this Wilkinson important? If so, one should make a Wikipage, if not, it shouldn't be mentioned... Pie.er 09:59, 28 April 2006 (UTC)

## $1,000,000 prize problem It should perhaps be mentioned that$1,000,000 prize problem concerns the incompressible Navier-Stokes equations.

Just noticed that the article has two references to the Clay Institute Millenium prize problem, one in the intro section and another in the turbulence section. One should probably be removed. --Chetvorno 01:22, 3 October 2007 (UTC)

The main Navier-Stokes page should not be in the "Millenium Prize" cattegory.

The Millenium Prize applies to only one small aspect of the theory around the Navier Stokes Equations, it seems perverse to have links to other Millenium Prize problems rather than other fluid dynamic resources. (Even links to other "famous physical equations" or some such would be better than what is currently there.)

If no one objects I am happy to make this change, but I'm open to criticism.

--cfp 16:00, 17 July 2006 (UTC)

Well. I would tend to agree. But even better would be a fluid dynamics project/category...CyrilleDunant 16:16, 17 July 2006 (UTC)
The new Navier stokes Millennium prize page is at: Navier-Stokes existence and smoothness
The new navigation infobox is at Template:Continuum mechanics, and pages should also have Template:Physics-footer as a footer.
Hope everyone approves.
--cfp 16:26, 18 July 2006 (UTC)
It is an improvement. Good stuff. :) Nephron  T|C 21:18, 28 July 2006 (UTC)
Still on the same topic, I find the text in the introduction misleading: "Even though turbulence is an everyday experience it is extremely difficult to find solutions for this class of problems. A $1,000,000 prize (...) in the understanding of this phenomenon." The problem is not just about turbulence... ## New Paper I just stumbled over a brand new paper: I dont even understand the problem, but im sure someone here will be able to check if its relevant. MillKa 10:27, 6 October 2006 (UTC) This article under the references section has two citations for the Penny Smith fiasco, but has no text explaining it. Either dump the refs or add a section capturing the brouhaha. Actually a general section on attempts to solve the problem, the millennium prize, etc. would be of general interest. This article, while having lots of nice PDEs, does not give enough "encyclopedia article" description for the general user. Remember there are lots of technical people or general educated people who come across the term Navier Stokes and want some general impression of what that means, but who can not follow the math or don't need to.TCO 17:14, 8 April 2007 (UTC) I just found another paper from a German guy who explains, why the equations could not be solved. I think this link should be added to the external Link sektion. I think we should stick to peer-reviewed papers. User A1 (talk) 14:06, 26 April 2010 (UTC) ## How we can classify Nav.. eqn as parabolic or hyperbolic or elliptic? will somebody explain°140.158.49.62 17:41, 30 January 2007 (UTC) Shoeb This is still (AFAIK) a not very well defined problem. Basically, depending on various parameters (Reynolds' number, Mach number), various terms of the equations become negligible under certain assumptions. The resulting simplified equations can then be classified as parabolic, hyperbolic or elliptic. Those considerations are important as they define the type of boundary conditions required for the solving of the problem, and thus the type of numerical code needed, for example.17:05, 2 February 2007 (UTC) I think the answer to this question is simple: The Navier-Stokes equation is neither a hyperbolic, nor a parabolic, nor an elliptic equation. It is a first-order partial differential equation. Parabolic/hyperbolic/elliptic is only defined for second-order PDE's. Greetings,--Roger Jeurissen 11:54, 26 March 2007 (UTC) The second order is "hidden" in the expression of stress. Stress is a function of strain, which derived spatially from displacement. The divergence of stress appears, making it a second order term in space. CyrilleDunant 15:45, 26 March 2007 (UTC) No, the NS equations are second order PDEs. They can't be classified as hyperbolic/parabolic/whatever because they're nonlinear. Like the other guys said, kill the right terms and you can get all kinds of linear equations which you can classify by the conventions. -Ben pcc 16:00, 26 March 2007 (UTC) I wld have to disagree - non-linearity has nothing to do with the classification of PDEs. e.g. the 2D advection equation is hyperbolic but can be (and mostly is) non-linear. N-S is hyperbolic/parabolic/elliptical depending on the relative magnitude of the terms. Physically, its all about the significant way in which information propagates at a particular coordinate & instant. However, apart from a stokes flow when inertia is negligible, the existence of the advection term in NS means that it is nearly always non-linear.86.159.211.92 (talk) 01:56, 3 April 2008 (UTC) I've seen different conventions; for example based on the elliptic operator which is by definition linear. One way or another, the complete NS equation can't be generally classified. — Ben pcc (talk) 23:05, 9 April 2008 (UTC) The important thing to appreciate is that in a (subsonic) flow past a body, the NS eqns are parabolic in the boundary layer and hyperbolic elsewhere. This as implications for the type of numerical scheme used in CFD codes (e.g. centered-differencing or up-winding). Tt261 (talk) 13:29, 13 May 2008 (UTC) ## Inertia Quoting the Inertia article: Physics and mathematics appear to be less inclined to use the original concept of inertia as "a tendency to maintain momentum" and instead favor the mathematically useful definition of inertia as the measure of a body's resistance to changes in momentum or simply a body's inertial mass. [...] This meaning of a body's inertia therefore is altered from the original meaning as "a tendency to maintain momentum" to a description of the measure of how difficult it is to change the momentum of a body.  A "resistance to changes in momentum" is exactly what the left side of the NS eqns contains. I think that it's ok that inertia isn't technically a quantity, the left side of the eqn represents inertia just fine. I can't think of a better phrase and I don't think anyone else can, hence absurdities like "force density" (which is something else) or "quantity of motion" (which I've tried looking up and am now very confident that it was made up on the spot). In science we use non-quantities all the time when it's logical to do so. For example, ${\displaystyle \lim _{t\to \infty }t=\infty }$ is technically very incorrect since infinity is not a number (though it can be considered an "element". See extended real numbers). Everyone uses it anyway since it'd be impractical and confusing to do something else. I've asked three fluid mechanics junkies in the old ME department about this. They say it's called inertia. -Ben pcc 16:29, 12 April 2007 (UTC) I resent that, it was not made on the spot, it came from a bad translation :) . It was, however, wrong. The quantity on the left has indeed the dimension of inertia (a tensor of Newtons per metre cube) -- inertia is actually a quantity, a very useful one at that . I apologise for my unfortunate edit. CyrilleDunant 17:06, 12 April 2007 (UTC) Aww, don't apologize :->. Ben pcc 06:51, 14 April 2007 (UTC) ## Article splitted Alot of people have been suggesting that this page be split. So it's done. I've made a subarticle concerning the derivation and other "non encyclopedic" details which I think are important but indeed out of place in the main article. Compare this to mathematic proofs, which use this cool template: Template:Proof. I had structured it like that a long time ago, and it seemed a good idea. The logic behind it was that it is not so much a derivation (well, ok the part about the derivatives and control volume is) as a description of the different component of the equations (all the conservation laws). Otherwise, you are left with a system of equation and a bunch of simplified forms. On the other hand, if you feel it is better split, well, I am not going to fight you :)CyrilleDunant 07:06, 14 April 2007 (UTC) Also I noticed that someone recently moved the substansive derivative article into convective derivative. I think that this is a bad idea but I'm not sure; the problem is that there's more then convection happening in that derivative. Any thoughts? I think it is a terrible idea: the substantive derivative is really the time derivative in lagrangian coordinates -- in any case it is central to the derivation and understanding of NS.CyrilleDunant 07:06, 14 April 2007 (UTC) I do intend to add more content/adjust both articles (specially sample usage and the like), so please don't freak out just yet. -Ben pcc 06:51, 14 April 2007 (UTC) OK, have fun CyrilleDunant 07:06, 14 April 2007 (UTC) I think that the "See Also" section should be properly expanded, there are many other related articles on this matter. Anyway, I find introduction very nice.Xerenio (talk) 18:58, 23 May 2008 (UTC) ## Classical mechanics vs Solid mechanics Solid mechanics does not dictate position, please stop saying that it does. It dictates displacement; the difference is important. Wikipedia doesn't have a lot of info on the equations of solid mechanics, but have look in any introductory (structural) finite element analysis text; you'll find equations describing displacement. Solid mechanics, as with fluid mechanics and rheology, can't appropriately describe position (well, it can, but it sucks, and that's no different for fluid mechanics); even the wave equation honestly describes displacement. On the other hand, look up classical mechanics, for example this, this, this, or this; these include Newtonian, Hamiltonian, and Lagrangian formulations of classical mechanics and they almost always lead to finding the position. It's true that NS is classic, but it's not true that classical mechanics MUST deal with position, or that "solid mechanics" can substitute "classic". —Preceding unsigned comment added by Ben pcc (talkcontribs) 19:27, 19 September 2007 (UTC) ## Convection Vs Advection Strictly speaking, when any scalar quantity is carried by the fluid flow (including its own inertia), it is called advection. Convection is reserved for the special case of heat transfer from a surface to a fluid or the fluid flow due to the buoyancy created by the heat itself (ie convective currents). However, if heat is transport by the bulk movement of a fluid (say in a hot water pipe), that is advection of heat. I think we need to change all the terms convection here to advection. 86.159.212.126 (talk) 14:06, 3 March 2008 (UTC) I'm not so sure. I have heard of the phrase advection, and the Wikipedia article supports what you say. However, none of my fluid mechanics books nor math books call it advection, they call it convection, as do all three fluid mechanics teachers I've had; also, a quick Google books search suggests that convection is the more popular term. Further, there's a lot more reference to "convection" as fluid acceleration on Wikipedia other than that one little bit you've changed in this article. Please either change it back, or justify and change everything. -Ben pcc (talk) 16:49, 3 March 2008 (UTC) I've looked it up, and it seems that, generally, advection refers to the transport of a scalar (heat, mass, concentration, etc), while convection ties more with changes of currents and velocities in a material. So I'm changing it back to convection. —Ben pcc (talk) 22:40, 18 March 2008 (UTC) I think you misunderstand the term scalar - a scalar can be anything at all e.g. x-momentum, as in the case of the N-S equations. I grant you that one does see the term convection used in texts but my point is that this is inaccurate and slightly misleading colloquial speak (hence this discussion!). Also, the articles on Wikipedia about convection and advection more than support my view: “In the context of heat and mass transfer, the term convection is used to refer to the sum of advective and diffusive transfer”. The diffusion term in the N-S equations is quite distinct from the advective term and so calling the latter the convective term is surely incorrect?! Furthermore, I've studied Fluid Mechanics at both the University of Cambridge and Imperial College, London and while the term convection was used colloquially by some lecturers, they always stressed that, strictly speaking, it was called advection. 86.159.211.92 (talk) 00:59, 3 April 2008 (UTC) A component of the momentum is indeed a scalar, but it's very important to understand that, despite the appearance of each component being advected, momentum is a vector by concept. A really good example of why this matters is 1D flow: in 1D flow, there aren't any convection terms and the left side of NS consists of the time derivative only. So while it looks like individual components are advected, it turns out that the vectorial nature of momentum is important, so it's not the simple advection of a scalar. Concerning diffusion, it seems that convection is colloquially used in heat transfer with a different meaning. Indeed, diffusion is unrelated to the fluid acceleration we're talking about (though it's worth mention that in creeping flow the fluid can by all means change velocity over position). One of my teachers told me (I'm in the USA) that in Europe the term "advection" sees more use than here. That might explain some things. — Ben pcc (talk) 00:35, 10 April 2008 (UTC) The advection term in the 1D NS mom-eqn is only zero in an incompressible flow due to continuity, just like in the 1D scalar transport equation. In a compressible 1D flow the advective acceleration remains in NS, and in the scalar transport equation. The only mathematical difference between the equations is that one is linear and the other isn’t. Are you suggesting that this difference in linearity requires a different name for the two terms? Tt261 (talk) 13:57, 5 May 2008 (UTC) I'm not sure what you mean by "just like in the 1D scalar transport equation". I'm not suggesting that linearity has anything to do with it, I'm suggesting that the hard vectorial nature of momentum transport is the reason. Concerning continuity, that's correct, the term is dropped due to continuity; however, if you want to approach this issue not considering continuity, then you'll find several extra steady acceleration terms in your NS eqn. See the derivation article, there are two ugly terms that disappear thanks to continuity (compressible or otherwise). What I'm saying is that if you ignore continuity, your advection/convection takes on several more terms; if you retain continuity, you lose a term in 1D. So, either way momentum transport is not 1D advection of individual components. It may seem like it is with compressible continuity, but the ideas must hold for all cases, shouldn't they? — Ben pcc (talk) 21:52, 7 May 2008 (UTC) The compressible NS momentum equation is more general than the incompressible eqn (which is an approximation and never exactly holds for any fluid). So anything that holds for the compressible eqn, must hold for the incompressible eqn and NOT vice versa. We have advection in the (more general) compressible 1D NS eqn and only in the special case of incompressible flow (which never actually occurs in practice) does this term disappear. Advective acceleration therefore has nothing at all to do with the fact that momentum is a vector as it occurs in the 1D (scalar), 2D and 3D NS momentum equations. Besides, the 3D incompressible NS momentum eqn (for a Newtonian fluid and ignoring body forces) in the x-dirn is: ${\displaystyle \rho \left({\frac {\partial u}{\partial t}}+u{\frac {\partial u}{\partial x}}+v{\frac {\partial u}{\partial y}}+w{\frac {\partial u}{\partial z}}\right)=-{\frac {\partial p}{\partial x}}+\mu \left({\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}+{\frac {\partial ^{2}u}{\partial z^{2}}}\right)}$ The 3D incompressible scalar advection equation is: ${\displaystyle \rho \left({\frac {\partial \phi }{\partial t}}+u{\frac {\partial \phi }{\partial x}}+v{\frac {\partial \phi }{\partial y}}+w{\frac {\partial \phi }{\partial z}}\right)=\rho S+\Gamma \left({\frac {\partial ^{2}\phi }{\partial x^{2}}}+{\frac {\partial ^{2}\phi }{\partial y^{2}}}+{\frac {\partial ^{2}\phi }{\partial z^{2}}}\right)}$ where ${\displaystyle \phi }$ is the concentration of a scalar, S is a source/sink term and ${\displaystyle \Gamma }$ is the molecular diffusivity. Can you spot the difference between the equations? You talk about the NS momentum equations having a “hard vectorial nature” but ultimately they constitute 3 separate (scalar) equations for the transport of the 3 components of momentum. The “hard vectorial nature” as a concept has no basis in maths nor physics. I think you also have a misunderstanding of what the material derivative that encompasses the advective acceleration term is. The term arises simply because we are applying conservation principles to a fluid lump and this lump gets transported by any velocity field acting. The fact that the momentum is of interest and momentum is a function of velocity makes the term non-linear but implies no conceptual difference. All I can suggest is that you try deriving the incompressible NS equation from first principles (ie applying Newton’s 2nd law) by considering the pressure and viscous forces acting on a fluid lump of constant density, mass and volume. …I remember that helping me understand the material derivative. The term “convective acceleration” seems to be preferred amongst those working in turbulence (the biggest research area in fluid mechanics by far) because it is the non-linearity of the advection term that leads to the Reynolds stresses and hence turbulence. Turbulence constitutes vorticity which can be thought of as “self-advecting” (i.e. “convecting”). However, this isn’t really sensical mathematically as a vortex does not advect itself, but a group of vortices advect each other and so turbulence (which is a bunch of vorticity) can be thought of as convecting. However, it remains that “advective acceleration” is a more appropriate and less ambiguous phrase to describe the spatially dependant material acceleration of a fluid lump. This is especially the case in flows involving heat transfer where there is advection and convection of heat and hence advective and convective acceleration of the fluid transporting the heat. Tt261 (talk) 19:19, 11 May 2008 (UTC) When I say "hard vectorial nature", I am invoking physical origin. Advection/convection are physical concepts. Pure mathematics contains many physical concepts; eg, why is there such a thing as a tensor in pure mathematics? Why do those who study differential geometry (a pretty advanced mathematics topic) know tensors well? How many of them know that the word "tensor" originated from "tension" because stress is a tensor? We're inspiring math with physics here, and that's essential. We can ditch assumptions of continuity and compressibility if you like and look at just the NS eqns. The fact that the vectorial nature of momentum is important isn't seen in the Cartesian eqns is because the Cartesian eqns are a special case of the vector representation of NS, a lot like how if we say F = ma we have to completely revise the statement if mass is variable. The individual Cartesian eqns just happen to look like advection, but try it in a different system, say 3D cylindric, r direction: ${\displaystyle \rho \left({\frac {\partial u_{r}}{\partial t}}+u_{r}{\frac {\partial u_{r}}{\partial r}}+{\frac {u_{\theta }}{r}}{\frac {\partial u_{r}}{\partial \theta }}+u_{z}{\frac {\partial u_{r}}{\partial z}}-{\frac {u_{\theta }^{2}}{r}}\right)=-{\frac {\partial p}{\partial r}}+\mu \left[{\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial u_{r}}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}u_{r}}{\partial \theta ^{2}}}+{\frac {\partial ^{2}u_{r}}{\partial z^{2}}}-{\frac {u_{r}}{r^{2}}}-{\frac {2}{r^{2}}}{\frac {\partial u_{\theta }}{\partial \theta }}\right]}$ ${\displaystyle \rho \left({\frac {\partial \phi }{\partial t}}+u_{r}{\frac {\partial \phi }{\partial r}}+{\frac {u_{\theta }}{r}}{\frac {\partial \phi }{\partial \theta }}+u_{z}{\frac {\partial \phi }{\partial z}}\right)=\rho S+\Gamma \left[{\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial \phi }{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}\phi }{\partial \theta ^{2}}}+{\frac {\partial ^{2}\phi }{\partial z^{2}}}\right]}$ where the first eqn is NS and the second is advection of a scalar. How about now? See a difference? Yes, and the reason for the difference is that momentum is a vector, its derivatives are tensorial, its scalar components are tied. The real difference is in the vector representation. The gradient (seen in advection equation), for example, becomes a tensor derivative. This is why I always revert ${\displaystyle (\mathbf {v} \cdot \nabla )\mathbf {v} }$ to ${\displaystyle \mathbf {v} \cdot \nabla \mathbf {v} }$ when someone changes, the first case is true (by coincidence) only in cartesian coordinates and involves no tensorial ideas, in the second case ${\displaystyle \nabla \mathbf {v} }$ is a tensor derivative and that is an important fact. You can't get the (correct) cylindrical representation of momentum conservation by using the cylindrical representation of the del operator. Try it, it'll come out wrong, it'll look just like the advection equation above; however if you properly use the tensor derivatives (which rely on the vectorial nature of momentum) you'll get the proper NS relations in any coordinate system you like. — Ben pcc (talk) 00:37, 14 May 2008 (UTC) I'm glad we're talking about this: I looked over some things I wrote over a year ago and they're slightly wrong due to component-wise treatment. I'll have to fix. — Ben pcc (talk) 00:42, 14 May 2008 (UTC) The differences in the scalar-transport and radial component of the momentum equations is due to the fact that you, besides the coordinate transformation, also change the vectorial representation of velocity from their Cartesian one ${\displaystyle (u_{x},u_{y},u_{z})\,}$ to the one ${\displaystyle (u_{r},u_{\theta },u_{z})\,}$ in the cylindrical coordinate system. The Cartesian components ${\displaystyle (u_{x},u_{y},u_{z})\,}$ satisfy an equation of similar form as the scalar ${\displaystyle \phi \,}$ (just as in the Cartesian coordinate system). In general, I think there is no consensus in the literature on what is to be called convection and what advection, both being used for momentum transport as well as scalar transport. For myself, I like the definition as I learned it (see my contribution from April 3, unfortunately I have no references). Since this is a very interesting discussion of general interest, perhaps a section on this subject can be added to the article (or another one), explaining there is a controversy, and describing the nomenclature as often used in the literature. Crowsnest (talk) 16:03, 14 May 2008 (UTC) An article on this seems sensible but I'm not sure which references would be appropriate - save perhaps a Latin dictionary. I think also, we should mention in the NS article that convective/convection is interchangeable with advective/advection and not to get the former confused with flows involving heat transfer. Shall I go ahead and make the necessary changes? Tt261 (talk) 14:20, 28 May 2008 (UTC) As far as I am concerned: Go ahead! Crowsnest (talk) 17:07, 28 May 2008 (UTC) I think you are mis-interpreting NS in cylindrical polar coordinates. Maybe rewriting the equations as follows will explain what I mean: ${\displaystyle \rho \left({\frac {Du_{r}}{Dt}}-{\frac {u_{\theta }^{2}}{r}}\right)=-{\frac {\partial p}{\partial r}}+\mu \left[{\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial u_{r}}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}u_{r}}{\partial \theta ^{2}}}+{\frac {\partial ^{2}u_{r}}{\partial z^{2}}}-{\frac {u_{r}}{r^{2}}}-{\frac {2}{r^{2}}}{\frac {\partial u_{\theta }}{\partial \theta }}\right]}$ ${\displaystyle \rho {\frac {D\phi }{Dt}}=\rho S+\Gamma \left[{\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial \phi }{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}\phi }{\partial \theta ^{2}}}+{\frac {\partial ^{2}\phi }{\partial z^{2}}}\right]}$ The last term on the LHS of the radial momentum equation is the centrifugal effect caused by the fact that the fluid lump is in a rotating frame of reference. Hence, its is effectively a source of momentum and has nothing to do with the advective acceleration or even the material acceleration ${\displaystyle {\frac {Du_{r}}{Dt}}}$. Likewise, the tangential momentum equation has a Coriolis term as a source of momentum but the axial momentum component is identical to the Cartesian form (as the vector in the axial direction is stationary). So, both scalar advection and NS equations take the form: Material derivative = Source/Sink + Diffusion. Tt261 (talk) 21:46, 25 May 2008 (UTC) As pointed out by Tt261, fictitious forces will occur in any non-inertial reference frame. These so-called tensor-derivatives will in general not be of help, especially not when using non-orthogonal coordinate systems, where one has to choose whether to use contra-variant or co-variant derivatives and velocity representations, and Christoffel symbols (which are not tensors) appear. So a preference of tensor derivatives ${\displaystyle \mathbf {v} \cdot \mathbf {\nabla } \mathbf {v} }$ over the common notation ${\displaystyle \left(\mathbf {v} \cdot \mathbf {\nabla } \right)\mathbf {v} }$ is a house of cards which (apart from cases of orthogonal coordinate systems) will not work. The more, since it is not pointed out under which conditions this concept may be valid, and (reliable) references to this preferred tensor-derivative notation are lacking. — Crowsnest (talk) 20:21, 26 May 2008 (UTC) Well, there are a few things wrong here. First of all, the cylindric equation is just that, cylindric. The equation is not written in a rotating system, write it in a rotating system and separate Coriolis and centrifuge terms appear. That one term is a centrifuge effect but not an effect of writing the equation in rotating system. The other problem is non-orthonogonal systems... first of all, a cylindric system is orthogonal, as is spheric and most other common systems. Further, the very reason tensors are studied is because they allow you to not worry about coordinate systems, indeed they work for any system, even one with non-orthogonal non-unit non-constant basis vectors. I'm not sure why you brought up rotating systems and non-orthogonal coordinates. You want reference? For tensor math, see Tensor Analysis by Lebedev. For a comment on NS, see Analytical fluid dynamics by George Emanuel. He explicitly says that: ${\displaystyle \mathbf {v} \cdot \nabla \mathbf {v} =\nabla \left({\frac {|\mathbf {v} |^{2}}{2}}\right)+\left(\nabla \times \mathbf {v} \right)\times \mathbf {v} }$ which is tensor math on the left and vector math on the right, and he says that this form works in any coordinate system without introduction of extra terms. The material derivative includes that one centrifuge term pointed out if you use this definition. If you consider scalar advection and separately add that term, you're in big trouble because you then can't have a single vector equation that works in any system! The beauty of retaining tensor derivatives and not confusing convection with scalar advection is that you can write a vector equation that contains everything you need. In many texts, such as White's Viscous Fluid Flow, tensor math is left out to make it easy, and in the appendix you'll find that he uses your (Tt261) approach, using the scalar advection operator and defining the material derivative (incorrectly) with it. The result? No tensors (easier for people who don't understand them) but the introduction of these mysterious extra terms; the best you can do with that thinking is supress your questions as to why those extra terms are there and where they come from (until you pay 60 USD for Tensor Analysis and spend an hour on Google books). Generality is always preferred, using scalar advection and writing a new NS for every coordinate system is a terrible idea because writing NS in some coordinate system will generally be wrong, wil be missing terms. Try it. Load up Maxima (or something), use ${\displaystyle \nabla \left({\frac {|\mathbf {v} |^{2}}{2}}\right)+\left(\nabla \times \mathbf {v} \right)\times \mathbf {v} }$, and derive NS in any coordinate system you like. It'll come out right. It'll come out wrong if you use scalar advection, and you will have no other way to aquire the missing terms then to look them up and suppress questions as to where they arise from. Though "convection" is more common specifically for momentum and matches the dictionary definition ("currents") I really don't care about semantics, you can call this acceleration "the poison dart frog effect" for all I care (though why?). But the argument that scalar advection works in NS is 100% wrong, except in Cartesian coordinates. — Ben pcc (talk) 23:37, 6 June 2008 (UTC) Your reference, Emanuel (Analytical fluid dynamics) states on page 6: "We shall utilize a notation, first introduced by George Stokes, to define the operator DDt = ∂t + w•∇ which is called the substantial or material derivative. This definition is independent of any specific coordinate system." And on page 7: "The dot product on the right hand side can be interpreted as w•(∇w), which involves the the dyadic ∇w, or as (w•∇)w, which does not involve a dyadic. With tensor analysis one can show that both interpretations yield the same result; the second one is usually preferred because of its greater simplicity." So this contradicts your statement that (v•∇)v does not produce the correct results in curvilinear coordinates, while v•(∇v) does. I cannot find a reference for this claim, and neither can I reproduce it. The operator ∇ just becomes expressed in terms of the covariant derivative in curvilinear coordinates, which is of different form for scalar, vector and other tensor fields (as rightly pointed out by you before; and provided the vectors and tensors are also represented in terms of the transformed coordinates instead of the original ones). For instance, for contravariant velocity components Vk the elements of v•∇v, either way, become: Vj (Vk),j. So the operator v•∇ can be regarded in terms of contravariant velocity components and covariant derivatives. I have no doubts regarding the vector identity v•∇v = ½∇(|v|2) + (∇×vv, either interpreting v•∇v as (v•∇)v or v•(∇v). With regards to the extra terms turning up in cylindrical coordinates: I was wrong in calling these fictitious forces, which is commonly reserved for non-inertial frames of reference. I was confusing coordinate systems with frames of reference. -- Crowsnest (talk) 19:28, 29 June 2008 (UTC) Right, you're correct. I'm sorry, I avoid using (v•∇)v because it is often misinterpreted as an elementwise application of the (v•∇) part. Both are equivalent if the tensor math is carried out, and a great reference is Wolfram. -Ben pcc (talk) 02:10, 30 June 2008 (UTC) But is a coordinate system not a frame of reference, and the cylindrical coordinate system not a non-inertial frame of reference? (ref: Frame_of_reference) The reason I initially brought this up is that NS is conceptually the same as the scalar advection equation, i.e. <unsteady change> + <material change> = <source/sink> + <diffusion>. This is true regardless of coordinate system (e.g. Cartesian & cylindrical as shown above). Yes, tensor notation is more useful because it is universal but these physical concepts (i.e. Newton’s laws applied to fluids) were around long before tensors were invented (do I need to look up a reference for this?). Hence, the argument that advection of momentum is a tensor concept is just not based on any mathematical or physical arguments. Tt261 (talk) 11:34, 22 July 2008 (UTC) Yes, conceptually they're the same, exactly the same in fact, only velocity is transported instead of something else. But you know, "tensor notation" wasn't invented to compact notation, it was invented because of the clear need for flexibility, to allow a law written in vector form to be eventually written in component form in any system of coordinates. You can exclude any tensor notation; in fact, I personally think that in the study of coordinate systems, it's best to not mention tensors until the reader has a clear understanding of, say orthogonal coordinate systems which are easiest to understand. Wilfred Kaplan's Advanced Calculus takes this approach, first discussing orthogonal coordinates and later tensors, and it works splendid. So you can by all means study NS without tensor concepts, all you need is very clean, properly implemented vector algebra/calculus. So neither does it really matter that these eqns preceded tensor math. But that doesn't say much. I can't really reply to what you wrote because I'm not sure where you're going with it, especially the first sentence. It's not clear that NS isn't componentwise application of a scalar eqn? That's my mathematic/physical argument: a physical law is objective (independent of coordinates, or anything like that), however we've clearly seen that you can't generally obtain NS in any coordinate system by looking at scalar eqns. — Ben pcc (talk) 16:28, 22 July 2008 (UTC) As far as I know, from Latin: ad is: by (passive), and vehere is: to carry, so advection is carried by the flow, i.e. passive transport of tracers moving with the flow. While con is: with, together. Convection is therefore active transport of quantities interacting with the flow, like e.g. momentum. In the case of heat transport, this can be called advection if the heat does not have a pronounced effect on the flow dynamics, so acting merely as a tracer. While in case of a strong interaction of the buoyancy effects with the flow, convection is more appropriate. Crowsnest (talk) 07:59, 3 April 2008 (UTC) That's good knowledge, thanks. In fact it makes perfect sense because convection is bulk fluid motion driven by the effects of the heat that is itself carried/transferred (due to buoyancy effects in convective currents and due to the temp gradients in natural convection from a surface to a fluid). However, advection is the transport of a scalar by an "external" bulk fluid motion. It seems to confuse a lot of peeps that a component of momentum is a scalar. This idea of "quantities interacting with the flow, like e.g. momentum" misses the point. The flow is not driven by the momentum that is carried in the advective term - how can it be? The NS equations has the forces acting on a fluid lump on the RHS (ie pressure, viscosity & body forces) and the effect of that on the LHS. Hence the advective acceleration is caused by imposed forces such as the pressure gradients resulting from changes of geometry. Physically, the term arises because when a fluid lump is carried through a (steady) pressure/velocity field, it's velocity changes due to spacial variations of that field. For exactly the same reasons, a tracer's velocity would change across the field. In fact, a perfect tracer is trapped into a fluid lump and so the advective acceleration of the two is exactly the same in every respect. Neither the fluid lump nor the tracer drives this acceleration and so both accelerations should be called the same - advective acceleration.86.159.211.92 (talk) 11:37, 3 April 2008 (UTC) Just going to throw a pair of dictionary definitions in here. Advection: The transfer of heat or matter by flow of a fluid. Convection: The movement caused by the the tendency of hotter and therefore less dense material to rise and colder more dense material to sink under the influence of gravity, which consequently results in a transfer of heat. The convection definition is a bit narrow IMHO, but the advection sounds spot on. User A1 (talk) 12:53, 3 April 2008 (UTC) Is that a layman's dictionary? I think the best definition I've heard from technical sources for advection is: "The transport of a scalar by bulk fluid motion." Tt261 (talk) 13:03, 3 April 2008 (UTC) So do we at least agree that advection is transport by fluid motion and convection is a special case of advection where the transport/fluid motion is "caused by itself"? Tt261 (talk) 13:11, 7 May 2008 (UTC) ## Convective Flow What is this convective flow that is mentioned in this article (e.g. in the badly written section on non-linearity)? ...and is it differnet from any fluid flow? Tt261 (talk) 12:54, 5 April 2008 (UTC) Convective flow means that momentum convection has a significant influence on the dynamics of the flow. In, for instance, acoustics the convective terms may in most cases be negligible, compared to temporal acceleration and pressure gradients. Crowsnest (talk) 23:19, 5 April 2008 (UTC) Stokes flow is another limit of the Navier-Stokes equations where convection may be neglected, so that is another example of a fluid flow which is not a convective flow. Etc. Crowsnest (talk) 23:22, 5 April 2008 (UTC) Ah, thanks. Perhaps "very low Re flow" wld make it clearer. Is it not also better to call it "advective flow" to make it clear it does not refer specifacally to flows with thermal convection currents? 86.159.211.92 (talk) 12:29, 8 April 2008 (UTC) Actually, in wave motion it is not the Reynolds number which is determining the relative importance of convection terms as compared to temporal acceleration, but the Keulegan-Carpenter number (no WP article here, yet): KC = UT/L, with U a characteristic velocity amplitude, T the oscillation period and L a characteristic length scale. This dimensionless number is in the context of oscillatory flow around bluff bodies: I am not aware of other names of such a dimensionless number for other types of oscillatory flow. Even at large Reynolds numbers, e.g. O(106), but low-enough KC number, convective terms may be neglected. Crowsnest (talk) 12:41, 8 April 2008 (UTC) "Very low Re"? I think you mean very high. High Re flows are called inertial flows. The phrase "convective flow" refers to a flow where the inertial change in velocity is important, even if it isn't very high Re. The radial flow at the bottom of the article is an example of this. Even for Re barely above zero, the convective acceleration becomes important. — Ben pcc (talk) 00:35, 10 April 2008 (UTC) Is "barely" not just another phrase for "very low"? Crowsnest (talk) 07:52, 10 April 2008 (UTC) Yep. — Ben pcc (talk) 17:56, 10 April 2008 (UTC) ## Amazing! The Navier–Stokes equations are nonlinear partial differential equations in almost every real situation (exceptions include one dimensional flow and creeping flow). The nonlinearity makes most problems difficult or impossible to solve and is part of the cause of turbulence. Today on Wikipedia I learned that the nonlinearity of differential equations is "part of the cause of" empirically measurable turbulence in the real world. Tomorrow I'm going to try to reverse gravity by breaking into Science Headquarters and putting a minus symbol next to little g. Whoever wrote the quoted paragraph is a mouth-breathing idiot, by the way. --75.49.223.247 (talk) 13:58, 28 June 2008 (UTC) Dude, CHILL. It's no secret that I'm an idiot, but you shouldn't conclude that because of one poorly writ sentence. I fixed it, and it would've been easier for you if you just fixed it yourself instead. But what does "mouth-breathing" mean? -Ben pcc (talk) 02:07, 30 June 2008 (UTC) ## phi and theta Exchanged phi and theta in the spherical and cylindrical equations. It is more common to write phi as the azimuthal angle and theta as the polar angle than vice-versa. —Preceding unsigned comment added by 130.243.230.54 (talk) 11:21, 2 November 2008 (UTC) It seems to me not all terms have been changed consistently. Are the equations still correct in spherical coordinates? -- Crowsnest (talk) 20:03, 2 November 2008 (UTC) ## Merge Wyld diagrams into N-S equations User:Headbomb suggested this merger. I Support it; the article is small and would be a better section of the N-S article than a standalone article, I think. Its page could be a redirect to the section of N-S after the merger. Awickert (talk) 18:08, 26 December 2008 (UTC) I am in doubt. Google Scholar finds 11 references to "Wyld diagrams". So it is a very, very specialised topic (but notable). But if included here, they would get an undue weight. Maybe someone will extend this stub in the future. So a weak oppose. -- Crowsnest (talk) 14:56, 8 March 2009 (UTC) The merge has been performed. -- Crowsnest (talk) 21:39, 28 April 2009 (UTC) ## The nonlinearity ... is the main contributor to the turbulence that the equations model. I don't understand this sentence. Is this supposed to mean that the nonlinearity is the cause of the turbulence? Shouldn't it be the other way round, that turbulence in the real world leads to nonlinearity in the model? Nczempin (talk) 14:22, 28 April 2009 (UTC) ## What are the Navier–Stokes equations In Navier–Stokes equations#Derivation and description the article states: "...the most general form of the Navier–Stokes equation ends up being: ${\displaystyle \rho \left({\frac {\partial \mathbf {v} }{\partial t}}+\mathbf {v} \cdot \nabla \mathbf {v} \right)=-\nabla p+\nabla \cdot \mathbb {T} +\mathbf {f} ,\ldots ''}$ Apart from not being in conservation form, so not being "the most general form", it gives the equations of motion before any stress model (relating fluid motion and stresses) is introduced. It is so general that it can be used, if you like to, to describe the motion of any substance either a fluid or solid or whatever other phase of matter in a single-phase continuum description. So these cannot be the Navier–Stokes equations, and no text book does call them so. According to Batchelor (1967) pp. 147-148; Landau & Lifshitz (1987), 2nd ed., pp. 44-49; Doering & Gibbon (1995) pp. 14-15, the Navier–Stokes equations are the equations resulting after the inclusion of a specific stress model for fluids: the shear stresses are proportional to the velocity gradient components ∂vi/∂xj and satisfying certain symmetry condititions (Batchelor, pp. 141-147). -- Crowsnest (talk) 09:40, 16 May 2009 (UTC) I think that this form is still useful, though I pretty much agree with Crowsnest. Maybe we should start the article with the equation for a predetermined (maybe Newtonian) rheology, and then have a derivation section where we say "Look! Equation (from above)!" then "Apply rheology - poof!". Awickert (talk) 09:51, 16 May 2009 (UTC) This equation is called by Batchelor simply the 'equation of motion' or 'momentum balance' equation, p. 137. Doering & Gibbon, p. 14 call it the 'most general form of the equation of motion'. It is just the expression of Newton's 2nd law for any continuum (apart from the cosmetic distinction between pressure and deviatoric stress). -- Crowsnest (talk) 10:11, 16 May 2009 (UTC) The main question is: which fluids are still covered by the Navier–Stokes equations. Newtonian fluid description of course is, but which non-Newtonian fluids are? Quite often, the equations of motion for fluids with some viscosity description of the deviatoric stresses are labelled as Navier–Stokes equations. Where viscosity may be isotropic or a non-isotropic tensor. -- Crowsnest (talk) 10:24, 16 May 2009 (UTC) ## Incompressibility and Stress Derivative The current last paragraph of Stresses deals with the simplification of the shear stress term in the case of an incompressible Newtonian fluid. I think that this information better belongs in the section on "incompressible flow of Newtonian fluids." I moved it there but Crowsnest reverted it and told me "incorrect: how can incompressibility avoid the need to model the shear stresses?" Since I was only moving information from one place to another, I don't see how it could be incorrect unless the original information was incorrect. Halberdo (talk) 09:51, 26 May 2009 (UTC) Your edit - not a move - did a lot more, and the present one does so again: it removes the shear stress model which is necessary to transform the (indeterminate) equations of motion into the Navier-Stokes equations. What remains is just a vague text without a mathematical formulation of the stress model in a section that promises to give a description and derivation of the Navier-Stokes equations. Further I do not see any logic in your present move: you move the incompressible description of the viscous shear stress term, while you leave the incompressible description of the pressure term in the preceding sentence unaltered (apart from a typographical correction). In my opinion, if the incompressible description of the stresses is not satisfactory, it is logical to extend them to the general case; not (re)move them. -- Crowsnest (talk) 16:07, 26 May 2009 (UTC) Another point is accessibility of the article to an as wide as possible audience, see Wikipedia:Make technical articles accessible. In this respect the incompressible form in Cartesian coordinates is easier than the compressible forms, and the curvilinear ones. Why not include such a form before - or at the start of - the Derivation and description section? -- Crowsnest (talk) 17:15, 26 May 2009 (UTC) Both edits were moves. The first edit I did included a little rewording, without changing the meaning of anything (except perhaps I should have said "In Newtonian incompressible flow..." instead of just "In incompressible flow....") The second edit was literally just a copy and paste from one section to the other. Your link to my newer edit is actually just another link to my older edit. This is the newer edit. My reasoning behind the move was, I (previously unfamiliar with Navier-Stokes) was trying to understand the section on Newtonian incompressible flow, and it took me a minute to realize that the only term that was altered from the general form was the stress term. I felt that for maximum readability for a wider audience (me), this fact needed a mention in that section. If you think that the stress section is uninformative without mentioning Newtonian incompressible flow, why not add a link from there down to the Newtonian incompressible flow section? Or even just duplicate the information. By the way... for the person desiring an introduction to Navier-Stokes, I found the article Derivation_of_the_Navier–Stokes_equations much more helpful, mainly because of its explanation of the convective derivative. -- Halberdo (talk) 18:54, 26 May 2009 (UTC) ## Forced Framing Wiki: Navier–Stokes existence and smoothness Search Wikipedia! The Navier-Stokes equations are one of the pillars of fluid mechanics. These equations describe the motion of a fluid (that is, a liquid or a gas) in space. Solutions to the Navier-Stokes equations are used in many practical applications. However, theoretical understanding of the solutions to these equations is incomplete. In particular, solutions of the Navier-Stokes equations often include turbulence, which remains one of the greatest unsolved problems in physics despite its immense importance in science and engineering. Even much more basic properties of the solutions to Navier-Stokes have never been proven. For the three-dimensional system of equations, and given some initial conditions, mathematicians have not yet proved that smooth solutions always exist, or that if they do exist they have bounded kinetic energy. This is called the Navier-Stokes existence and smoothness problem. Since understanding the Navier-Stokes equations is considered to be the first step for understanding the elusive phenomenon of turbulence, the Clay Mathematics Institute offered in May 2000 a US$1,000,000 prize, not to whoever constructs a theory of turbulence but (more modestly) to the first person providing a hint on the phenomenon of turbulence. In that spirit of ideas, the Clay Institute set a concrete mathematical problem: [1]

Prove or give a counter-example of the following statement: In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier-Stokes equations. Contents: 1. The Navier-Stokes equations 2. Two settings: unbounded and periodic space 3. Statement of the problem in the whole space 4. Statement of the periodic problem 5. Partial results 6. Notes 7. References 8. External links

Millennium Prize Problems P versus NP The Hodge conjecture The Poincaré conjecture The Riemann hypothesis Yang-Mills existence and mass gap Navier-Stokes existence and smoothness The Birch and Swinnerton-Dyer conjecture

1. The Navier-Stokes equations Main article: Navier-Stokes equations

In mathematics, it is a system of nonlinear partial differential equations for abstract vector fields of any size. In physics and engineering, it is a system of equations that models the motion of liquids or not-rarefied gases using continuum mechanics. The equations are a statement of the second law of Newton, with the forces modelled according to those in a viscous Newtonian fluid — as the sum of contributions by pressure, viscous stress and an external body force. Since the setting of the problem proposed by the Clay Mathematics Institute is in three dimensions, for an incompressible and homogeneous fluid, we will consider only that case.

Let be a 3-dimensional vector, the velocity of the fluid, and let be the pressure of the fluid. [note 1] The Navier-Stokes equations are:

where is the kinematic viscosity, the external force, is the gradient operator and is the Laplacian operator, which is also denoted by . Note that this is a vector equation, i.e. it has three scalar equations. If we write down the coordinates of the velocity and the external force

then for each we have the corresponding scalar Navier-Stokes equation:

The unknowns are the velocity and the pressure . Since in three dimensions we have three equations and four unknowns (three scalar velocities and the pressure), we need a supplementary equation. This extra equation is the continuity equation describing the incompressibility of the fluid:

Due to this last property, the solutions for the Navier-Stokes equations are searched in the set of "divergence-free" functions. For this flow of a homogeneous medium, density and viscosity are constants.

2. Two settings: unbounded and periodic space There are two different settings for the one-million-dollar-prize Navier-Stokes existence and smoothness problem. The original problem is in the whole space , which needs extra conditions on the growth behavior of the initial condition and the solutions. In order to rule out the problems at infinity, the Navier-Stokes equations can be set in a periodic framework, which implies that we are no longer working on the whole space but in the 3-dimensional torus . We will treat each case separately.

3. Statement of the problem in the whole space 3. 1. Hypotheses and growth conditions The initial condition is assumed to be a smooth and divergence-free function (see smooth function) such that, for every multi-index (see multi-index notation) and any , there exists a constant (i.e. this "constant" depends on and ) such that

for all

The external force is assumed to be a smooth function as well, and satisfies a very analogous inequality (now the multi-index includes time derivatives as well):

for all

For physically reasonable conditions, the type of solutions expected are smooth functions that do not grow large as . More precisely, the following assumptions are made:

1. 2.There exists a constant such that for all Condition 1 implies that the functions are smooth and globally defined and condition 2 means that the kinetic energy of the solution is globally bounded.

3. 2. The million-dollar-prize conjectures in the whole space (A) Existence and smoothness of the Navier-Stokes solutions in

Let . For any initial condition satisfying the above hypotheses there exist smooth and globally defined solutions to the Navier-Stokes equations, i.e. there is a velocity vector and a pressure satisfying conditions 1 and 2 above.

(B) Breakdown of the Navier-Stokes solutions in

There exists an initial condition and an external force such that there exists no solutions and satisfying conditions 1 and 2 above.

4. Statement of the periodic problem 4. 1. Hypotheses The functions we seek now are periodic in the space variables of period 1. More precisely, let be the unitary vector in the - direction:

Then is periodic in the space variables if for any we have that

Notice that we are considering the coordinates modulo 1. This allows us to work not on the whole space but on the quotient space , which turns out to be the 3-dimensional torus

We can now state the hypotheses properly. The initial condition is assumed to be a smooth and divergence-free function and the external force is assumed to be a smooth function as well. The type of solutions that are physically relevant are those who satisfy these conditions:

3.

4. There exists a constant such that for all

Just as in the previous case, condition 3 implies that the functions are smooth and globally defined and condition 4 means that the kinetic energy of the solution is globally bounded.

4. 2. The periodic million-dollar-prize theorems (C) Existence and smoothness of the Navier-Stokes solutions in

Let . For any initial condition satisfying the above hypotheses there exist smooth and globally defined solutions to the Navier-Stokes equations, i.e. there is a velocity vector and a pressure satisfying conditions 3 and 4 above.

(D) Breakdown of the Navier-Stokes solutions in

There exists an initial condition and an external force such that there exists no solutions and satisfying conditions 3 and 4 above.

5. Partial results 1.The Navier-Stokes problem in two dimension has already been solved positively since the 60's: there exist smooth and globally defined solutions. [2] 2.If the initial velocity is sufficiently small then the statement is true: there are smooth and globally defined solutions to the Navier-Stokes equations. [1] 3.Given an initial velocity there exists a finite time , depending on such that the Navier-Stokes equations on have smooth solutions and . It is not known if the solutions exist beyond that "blowup time" . [1] 4.The mathematician Jean Leray in 1934 proved the existence of so called weak solutions to the Navier-Stokes equations, satisfying the equations in mean value, not pointwise. [3] 6. Notes 1.More precisely, is the pressure divided by the fluid density, and the density is constant for this incompressible and homogeneous fluid. 7. References 1.^ Official statement of the problem, Clay Mathematics Institute. 2.O. Ladyzhenskaya, The Mathematical Theory of Viscous Incompressible Flows", 2nd edition), Gordon and Breach, 1969. 3.Leray, J. (1934), "Sur le mouvement d'un liquide visqueux emplissant l'espace", Acta Mathematica 63: 193-248, doi:10.1007/BF02547354 8. External links •The Clay Mathematics Institute's Navier-Stokes equation prize •Why global regularity for Navier-Stokes is hard — Possible routes to resolution are scrutinized by Terence Tao. • d— Preceding unsigned comment added by 75.54.91.58 (talkcontribs) 02:36, 1 February 2010 (UTC)

## Just an opinion about Math tech.

If we consider that dx belongs to a position vector,therefore may exist:

         ${\displaystyle u{\frac {\partial {u}}{\partial {t}}}={\frac {\partial {x}}{\partial {t}}}*({\frac {\partial {u}}{\partial {x}}}{\frac {\partial {x}}{\partial {t}}})={\frac {\partial {u}}{\partial {x}}}{\frac {\partial ^{2}{x}}{\partial {t^{2}}}}}$

which the operations denote a tensor result.

        【${\displaystyle {\frac {\partial {u}}{\partial {x}}}}$ belongs to a tensor】

I'm the original author(Taiwainese)JhongHuá MínGuó. I now born a new idea for "reclaiming" the continued eqn. .

I'm trying to get "energy conservation" to do. We may use the physical Math something like:

          ${\displaystyle {\frac {1}{2}}{\frac {D}{Dt}}({\rho }u^{2})={\frac {1}{2}}{\frac {\partial {({\rho }u^{2})}}{\partial t}}+......}$

The official terms' phisical meaning refer to "no fluids missing from some indical systems". And in my created terms of functions, they refer to "energy in some indical systems are saved"(No matter what types of energy exist)

I'm a person of like-creation. Before I come here,I studied some articles from "Riley's Methematical Methods for Physics and Engineering". If I've made any possible mistake originally, please tell me. I want to learn Sciences from America,Britain,Canada,... — Preceding unsigned comment added by HydrogenSu (talkcontribs) 13:07, 21 December 2005 (UTC)

## Most important case missing

The Navier-Stokes equation for the case of a Newtonian incompressible fluid are used by more than 90% of researchers in the field. That this case is not given at all is likely to confuse many students and lead them to think that this article is irrelevant to their research (or aimed at irrelevant special cases).

I strongly suggest adding these simplfied forms early-on in the article.

why? the special forms are just that, special forms. And if you want the newtonian incompressible case, just add it: it is trivially the conservation of momentum plus incompressibility, expressed as:
${\displaystyle \nabla \mathbf {v} =0}$
which you would easily get from the article.
on the other hand, the compressible equations take way too much space :)CyrilleDunant 06:51, 3 November 2006 (UTC)

My point is: if 90% of the people coming to this page are needing the Newtonian incompressible form -- and that is missing -- one has a article well suited to 10% of those who come looking. Water, air, salad dressing, usually at low Mach number, these are the fluids of life and most technology. When you suggest "just add it" I'm unsure if you would like me to proceed to edit the article. -- best wishes -- User:Lathrop 21:31, 3 November 2006 (EST)

Well, this is WP! Of course you should edit the article if you think information should be added! I will just note that need is perhaps a bit of a strong word for a set of equations...CyrilleDunant 09:58, 4 November 2006 (UTC)
Hi, congratulations to all for the nice work. I just revised the "incompressible" section, perhaps that is what you wanted. But I feel that the rest of the page still needs some rewrite. I kept the full equations with all components etc... I know that students find them very useful. For Cyrille, "need" is how we teachers see some very useful pieces of math, we sense that students and users of science "need" the equations. You cannot do this type of science or engineering without them. Zaleski 23:54, 6 February 2007 (UTC)

Thank you Stéphane, that is a very useful improvement. —Preceding unsigned comment added by Lathrop (talkcontribs) 22:34, 1 December 2010 (UTC)

- dear fellow, what about the creeping/laminar flow liquid by special long chain molecules in water dissolved and called "creeping-fluid" being an additivum for slippery gliding over Navier-Stokes surfaces? This expression (creeping-fluid, contracted from: creeping chain molecules in water dissolved and constituing an extremely slippery fluid for gliding thru tubes and pipes without any rest of friction that could otherwise be caused by the surface passed by or slipped over) was coined in the eighties of the past century, it has been applied also by the New York fire brigades in the harbour and the water that contained the "creeping molecules fluid" reached a distance one-third wider than before when without that additivum of the so-called "creeping-fluid". The flexible pipe itself therefore is almost lubricated from inside, and the molecules in long chains of the water did flow, as if they would draw each other outside and they slipped without the slightest resistance, neither by the tube inner surfaces nor by themselves when advancing - under the same pressure as before - but getting a considerably higher speed. -- What ist the meaning of the scientists with regard to this, concerning the Navier-Stoke surface layer theory? - Yours sincerely, bluaMauritius. —Preceding unsigned comment added by 84.62.236.125 (talk) 08:32, 21 May 2011 (UTC)

## Maxwell's Equations Analogy

There's an analogy between the Navier–Stokes equations and Maxwell's equations. It would be good to have a brief section explaining this analogy. JKeck (talk) 05:08, 3 June 2010 (UTC)

## Jormakka

It's been solved! http://ejde.math.txstate.edu/Volumes/2010/93/jormakka.pdf --188.220.102.130 (talk) 23:08, 21 August 2010 (UTC)

The Millennium Prize problem related to these equations might have been solved. Until the paper you linked has been reviewed by others, and third-party sources have reported this, we shouldn't be reporting that the problem has been solved. In any case, we have a separate article, Navier–Stokes existence and smoothness, specifically for the Millennium Prize problem formulation, and any discussion of this should go there. 01:50, 22 August 2010 (UTC)

Convection implies that the flow is buoyancy-driven in some way, while advection refers to tracers (eg temperature/momentum) being carried with the flow. —Preceding unsigned comment added by 150.203.10.73 (talk) 06:41, 6 April 2011 (UTC)

For a previous discussion on this: see Talk:Navier–Stokes equations/Archive 1#Convection Vs Advection. -- Crowsnest (talk) 13:39, 6 April 2011 (UTC)

## Where are the equations?

This article has a good description of the uses, derivation, and applicability of the equations, but ironically does not label the equations themselves clearly. I assume the most general form is this equation in the 'Derivation and description' section:

${\displaystyle \rho \left({\frac {\partial \mathbf {v} }{\partial t}}+\mathbf {v} \cdot \nabla \mathbf {v} \right)=-\nabla p+\nabla \cdot \mathbb {T} +\mathbf {f} ,}$

except that it is not labelled the "Navier-Stokes equation" but rather the Cauchy momentum equation, Could someone please clarify this? Cheers --ChetvornoTALK 22:14, 10 April 2011 (UTC)

Because as such, it is not complete: it further requires a constitutive relationship (which will relate T, v and p). NS is the complete set of equations such that the problem can be (possibly) well-posed.CyrilleDunant (talk) 01:40, 11 April 2011 (UTC)
The Navier-Stokes equations are not complete in general anyway because they only define the momentum conservation. For incompressible flow the continuity equation is needed; and for compressible flow the continuity equation, energy equation, and an equation of state are all required. — Preceding unsigned comment added by 164.107.59.74 (talk) 23:24, 26 May 2011 (UTC)


text:
J.K. Rowling’s secret book shoots up bestseller lists after she’s outed on Twitter

Want to read a new mystery by bestselling Harry Potter author J. K. Rowling? You’re in luck, and you don’t even have to wait: She published a mystery, The Cuckoo’s Calling, under the pseudonym “Robert Galbraith” in April, and was outed by the Sunday Times of London (paywall) over the weekend.

Now that the secret’s out, Cuckoo is shooting up bestseller lists. The book, which was published by Hachette’s Little, Brown imprints in the U.S. and U.K., is now #1 on Amazon’s Kindle list and #2 on the print book list; #1 on Barnes & Noble’s print list and Nook (s BKS); and #1 in Apple’s iBookstore (s AAPL). They’re all selling it for $9.99, down from a digital list price of$12.99.

Cuckoo, the story of a war veteran turned private investigator, sold only 1,500 copies in the U.K. and 1,800 in the U.S. before Rowling was outed. As for how that happened, thank Twitter. The New York Times explains:

“It started on Thursday, said Richard Brooks, the paper’s arts editor, after one of his colleagues happened to post a tweet mentioning that she had loved ‘The Cuckoo’s Calling,’ and that it did not seem as if the book had been written by a novice.

‘After midnight she got a tweet back from an anonymous person saying it’s not a first-time novel — it was written by J. K. Rowling,’ Mr. Brooks said in an interview. ‘So my colleague tweeted back and said, “How do you know for sure?”

The person replied, ‘I just know,’ and then proceeded to delete all his (or her) tweets and to close down the Twitter account, Mr. Brooks said. ‘All traces of this person had been taken off, and we couldn’t find his name again.'”

Brooks then poked around online and found that Cuckoo and Rowling’s previous novel for adults, The Casual Vacancy, had the same agent, editor and publisher. And he sent copies of those books, plus Harry Potter and the Deathly Hallows, to “a pair of computer linguistic experts, who found significant similarities among them.” A Rowling spokesperson then admitted it.

“I hoped to keep this secret a little longer, because being Robert Galbraith has been such a liberating experience,” Rowling said in a statement. “It has been wonderful to publish without hype or expectation and pure pleasure to get feedback from publishers and readers under a different name…And to those who have asked for a sequel, Robert fully intends to keep writing the series, although he will probably continue to turn down personal appearances.”

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text:
1
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
The highest value of the calculated spin- only magnetic moment (in BM) among all the transition metal complexes is :
A
5.92
B
6.93
C
3.87
D
4.90
2
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
For emission line of atomic hydrogen from ni = 8 to nf = n, the plot of wave number $$\left( {\overline v } \right)$$ against $$\left( {{1 \over {{n^2}}}} \right)$$ will be (The Rydberg constant, RH is in wave number unit)
A
Linear with intercept $$-$$RH
B
Non linear
C
Linear with slope RH
D
Linear with slope $$-$$RH
3
JEE Main 2018 (Online) 16th April Morning Slot
+4
-1
Which of the following statements is false ?
A
Photon has momentum as well as wavelength.
B
Splitting of spectral lines in electrical field is called Stark effect.
C
Rydberg constant has unit of energy.
D
Frequency of emitted radiation from a black body goes from a lower wavelength to higher wavelength as the temperature increases.
4
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
The de-Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is :
A
0.529 $$\mathop {\rm A}\limits^ \circ$$
B
2$$\pi$$ $$\times$$ 0.529 $$\mathop {\rm A}\limits^ \circ$$
C
$${{0.529} \over {2\pi }}$$ $$\mathop {\rm A}\limits^ \circ$$
D
4 $$\times$$ 0.529 $$\mathop {\rm A}\limits^ \circ$$
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