Talk:Navier–Stokes equations/Archive 2

Who is writing this article? Give the formula in the very first section and define the variables in it before going for the derivation and other bs. Someone looking for the formula can never find from this article which one is the Navier–Stokes equation. Damn it. 122.176.58.109 (talk) 14:34, 20 September 2011 (UTC)

There are more than one equations that can be used 98.114.196.197 (talk) 17:10, 11 March 2012 (UTC)

The three dimensional exact solution in this article might not be an incompressible flow. The divergence of the velocity vector is non-zero. — Preceding unsigned comment added by 68taileddragon (talk • contribs) 21:54, 1 April 2012 (UTC)

There has been recent revival of interest in using sans serif here, I replaced \mathbb with \mathsf for the various tensors in this article since this seems to be what the literature uses (sans serif and not bold)... Let’s see the reaction... Maschen (talk) 10:34, 7 May 2012 (UTC)

Dear Participants of discussion!

Here http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations#Pressure-free_velocity_formulation we can read: “This result follows from the Helmholtz Theorem (also known as the fundamental theorem of vector calculus)”. However well-known http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics/Archive/2012/Mar#Helmholtz_decomposition_is_wrong that Helmholtz decomposition is total wrong. Fundamental theorem of vector calculus can be written as follows:

${\displaystyle {\vec {F}}=\operatorname {grad} \varphi +\operatorname {rot} \operatorname {rot} {\vec {A}}}$ . (2*)

This formula completely corresponds to transformed Navier–Stokes equations(NSE) for incompressible fluids (${\displaystyle \operatorname {div} {\dot {\vec {u}}}=0}$ )

${\displaystyle \rho {\vec {F}}-\operatorname {grad} p+\mu \nabla ^{2}{\dot {\vec {u}}}=\rho {\ddot {\vec {u}}}\Rightarrow \rho ({\vec {F}}-{\ddot {\vec {u}}})=\operatorname {grad} p+\operatorname {rotrot} \mu {\dot {\vec {u}}}.}$ (3*)

Here,

${\displaystyle {\vec {F}}={\vec {F}}_{1}+{\vec {F}}_{2}+\cdots }$ vectors sum of a given, externally applied forces (e.g. gravity ${\displaystyle {\vec {F}}_{1}}$ , magnetic ${\displaystyle {\vec {F}}_{2}}$ and other), ${\displaystyle p}$ - pressure (scalar function), ${\displaystyle {\dot {\vec {u}}}}$ - velocity vector, ${\displaystyle {\ddot {\vec {u}}}=d{\dot {\vec {u}}}/dt}$ - acceleration vector, ${\displaystyle \rho }$ - density (const), ${\displaystyle \mu }$ - viscosity (const), ${\displaystyle \nabla ^{2}}$ - Laplace operator.

Equations (3) and (2) are consistent. Hence there is no reason to say that the theory requiring (2) must be false. As we can see from NSE the sum - ${\displaystyle \operatorname {grad} p+\mu \nabla ^{2}{\dot {\vec {u}}}=-(\operatorname {grad} p+\operatorname {rotrot} \mu {\dot {\vec {u}}})}$ forms the vector field.

Note that we will receive the formula (2) also after similar transformation of the Navier–Stokes for a compressible fluid and after transformation of the Lame equations for an elastic media.

Therefore let's try to formulate the text for editing of this article.

--Alexandr (talk) 11:28, 17 May 2012 (UTC)

I don't understand Helmholtz decomposition (yet), but know that it exists and people use it so it must work. Its one thing to suspect a physical law is wrong, but mathematics itself is never wrong as its founded on logic and axioms (the only "mathematical mistakes" which can ever occur are due to an individual making those mistakes in some analysis). You have raised this at least twice before: in that above link and here, and each time you have been told Helmholtz decomposition is correct while simaltaneously your calculations are in error (although rather than considering yourself as wrong you consider something else to be wrong). I suggest you stop disrupting talk pages with it, and trying to introduce OR. F = q(E+v×B) ⇄ ∑_ic_i 19:39, 19 May 2012 (UTC)

These OR are comparison of two Wikipedia articles (http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations and http://en.wikipedia.org/wiki/Helmholtz_decomposition ). In Wikipedia all contradictions must be deleted. Otherwise it is Wikipedia disrupting.--Alexandr (talk) 12:06, 22 May 2012 (UTC)

I find opaque the tensor notation used throughout the "Derivation and Description" section of the article. It seems like Einstein summation notation would be clearer. For example, the viscous term of ${\displaystyle \nabla \cdot {\boldsymbol {\mathsf {T}}}}$ is one of the following: ${\displaystyle \partial _{i}(\mu _{ij}\partial _{j}v_{k})}$ , ${\displaystyle \partial _{i}(\mu _{jk}\partial _{i}v_{k})}$ , or ${\displaystyle \partial _{i}(\mu _{jk}\partial _{j}v_{i})}$ . All three give different behavior, but it is not immediately obvious which form is intended. Comparison with the section on "incompressible flow of a Newtonian fluid" makes me think the second is what's intended, but it would be nice to have this made explicit.SMesser (talk) 20:47, 1 June 2012 (UTC)

I am curious about how the solution of this three dimentinal NS equation is from. There is reference for the two dimentinal example, but the reference is lacking for this 3D example. Is there any reference or citation? Thank you very much for your kind help. — Preceding unsigned comment added by Wanchung Hu (talk • contribs) 10:11, 27 August 2012 (UTC)

In Section 3.2, Stresses, the following is stated:"Interestingly, only the gradient of pressure matters, not the pressure itself".Maybe it should be stated that only the gradient of pressure affects the stress tensor. Currently, the statement above might lead into an impression that fluid properties are independent of absolute value of the pressure. For compressible flow, an equation of state is needed to describe the fluid. For example, a commonly used equation of state for gases is the ideal gas law, where the absolute value of pressure does matter. Correct me if I am wrong.Papow (talk) 13:02, 10 November 2012 (UTC)

I think you are right. Note that the end of the section on stresses mentions the equation of state. The absolute value of pressure also affects the viscous stresses, through its influence on viscosity. -- Crowsnest (talk) 17:04, 10 November 2012 (UTC)

I think that the 1/3 term in front of the grad div v term is erroneous. The term should be ${\displaystyle \left(\mu +\mu ^{v}\right)\nabla (\nabla \cdot \mathbf {v} )}$ . I think the error stems from the commonly used Stokes assumption ${\displaystyle \mu ^{v}=-{\frac {2}{3}}\mu }$ . 195.241.117.199 (talk) 16:34, 18 December 2013 (UTC)

I have removed the term ${\displaystyle \mu ^{v}}$ by invoking the Stokes assumption. Salih (talk) 17:22, 18 December 2013 (UTC)

It is seems this problem covered very well here: http://www.docstoc.com/docs/80173363/Navier-paradox 178.137.134.138 (talk) 13:55, 19 December 2013 (UTC)

The revised form of the equation and the first equation in the article you've pointed out are same. Salih (talk)

Compare the revised form of the equation and equations (1), (6) in the article http://www.docstoc.com/docs/80173363/Navier-paradox  !!!! Equation (6) is obtained after exact transformation of equation (1). 178.137.134.138 (talk) 19:44, 19 December 2013 (UTC)

Would be nice to have a history section. This one gives some idea, http://liwei-chen.blogspot.ca/2007/09/brief-history-of-navier-stokes-equation.html (but is short) Stephanwehner (talk) 22:11, 11 January 2014 (UTC)

The article says this problem has not been solved. But it has:

http://www.inform.kz/eng/article/2619922Psyadam (talk) 19:22, 12 January 2014 (UTC)

The paper was released a few days ago, with 20 self-citations, written by a relatively unknown mathematician, published in an obscure journal of which he is an editor..... I would wait before claiming the problem is solved.2001:638:902:2001:214:BFF:FE81:48CE (talk) 11:30, 16 January 2014 (UTC)

The derivation of the Navier–Stokes equations begins with an application of Newton's second law: conservation of momentum (often alongside mass and energy conservation) being written for an arbitrary portion of the fluid.

THIS SHOULD BE FIXED!I GET WHAT THEY'RE TRYING TO SAY BUT THE WAY IT'S WORDED MAKES THE AUTHORS SOUND PRETTY DUMB.

Please let us know what is your proposal to rewrite. Salih (talk) 05:18, 15 October 2013 (UTC)

A very common form of the incompressible equations, the non-dimentionalized form with only the Reynolds number as parameter,${\displaystyle {\frac {\partial v}{\partial t}}+v\cdot \nabla v=-\nabla p+{\frac {1}{Re}}\Delta v}$ ,does not appear in the article. It should, at least for the theoretical reasons that solutions to NS are characterized only on one parameter. Gummif (talk) 01:31, 5 April 2014 (UTC)

I was just looking for a quick info on what is, or are, Navier–Stokes equation(s). After glancing the article I still have no idea. Maybe I am reading too fast (I am...), but anyway, it looks confusing. To the point:

• The first equation we get to see, and the only one in the lede, is explicitly NOT a Navier–Stokes equation, quote: «Navier–Stokes equations [...] cannot be put into the quasilinear homogeneous form: ${\displaystyle \mathbf {y} _{t}+\mathbf {A} (\mathbf {y} )\mathbf {y} _{x}=0}$  »; not much useful starting to show it is not, right? It is not a whole lot of other things...
• Then we get a long(ish) derivation, with many variously colourful boxes: «Cauchy momentum equation», «Linear stress constitutive equation», ... «Navier-Stokes momentum equation», aha! probably this is one, but by now, who knows?

The derivation may be fine, but it needed some introduction. It is kind of if the aeroplane article said something like "An aeroplane is something that flies. Here are instructions on building one: ..." before we knew what it looks like at all. Hopefully someone may improve the lede? Thank you! - Nabla (talk) 19:04, 7 February 2015 (UTC)

The incompressible equations appear in the relevant section of the article, but that is fairly far down the page. What's missing here is not just a statement in the lead (which would be ideal), but a clear first section stating the equations and describing various forms of them. The derivation is of comparatively less importance, and I would like to see that whole section moved further down. At the moment, the article is practically unreadable. Sławomir Biały (talk) 19:22, 8 February 2015 (UTC)

Yes, it needs just that. - Nabla (talk) 10:28, 14 February 2015 (UTC)

No words talk about relativistic hydrodynamics 111.121.12.141 (talk) 14:36, 11 April 2015 (UTC)

There are still some errors and misconceptions in this section in my opinion.

Maybe the problem is also it lacks of some concepts :

• differences between "total pressure" and "equilibrium pressure".
• correlation and differences between first lamé parameter and bulk viscosity.

I wuold write something like:

The Cauchy stress tensor can be decomposed into its isotropic and deviatoric parts as follow:

${\displaystyle {\boldsymbol {\sigma }}=-p{\boldsymbol {I}}+{\boldsymbol {\tau }}}$

Where the pressure ${\displaystyle p}$ is defined as minus the mean normal stress and is an osservable quantity.This quantity can be thought as the sum of the equilibrium pressure ${\displaystyle p_{e}}$ that is equal to total pressure ${\displaystyle p}$ in absence of any relative motion of the fluid and an argument that depend on the fluid velocity field gradient, it can be written as^[1]:

${\displaystyle p\equiv -{\tfrac {1}{3}}tr({\boldsymbol {\sigma }})=p_{e}-\zeta \nabla \cdot \mathbf {u} }$

where

${\displaystyle \zeta }$ is the bulk viscosity

The bulk viscosity is related to the first lamé parameter ${\displaystyle \lambda }$ (another coefficient commonly used to describe the stress tensor):

${\displaystyle \zeta =\lambda +{\tfrac {2}{3}}\mu }$

By expanding also the material derivative the Cauchy momentum equation becomes explicitly:

${\displaystyle \rho \left({\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \cdot \nabla \mathbf {u} \right)=-\nabla p+\nabla \cdot {\boldsymbol {\tau }}+\rho \mathbf {g} }$

Assuming I said true things, there are any ideas to explain this in a better way?

These misconceptions lead to inaccuracies in the following steps so the whole section should be fixed.

For example:

${\displaystyle \nabla \cdot (2\mu {\boldsymbol {\varepsilon }})=\mu \nabla \cdot (2\varepsilon )=\mu \nabla \cdot ((\nabla \mathbf {u} )+(\nabla \mathbf {u} )^{\text{T}})=\mu \nabla ^{2}\mathbf {u} }$

Seems algebraically wrong to me, It should be something like:

${\displaystyle \nabla \cdot (2\mu {\boldsymbol {\varepsilon }})=\mu \nabla \cdot (2\varepsilon )=\mu \nabla \cdot ((\nabla \mathbf {u} )+(\nabla \mathbf {u} )^{\text{T}})=\mu (\nabla ^{2}\mathbf {u} +\nabla (\nabla \cdot \mathbf {u} ))}$

--Tom.sartor (talk) 17:54, 25 January 2015 (UTC)

I also found the sam mistake. I just added minor correction: in the last red bordered equation I have changed ${\displaystyle -2/3\mu }$ to 1/3 with a note about a divergence of tensor ${\displaystyle \nabla \mathbf {u} }$ .

I also don't like the whole structure of the artile. The most general case should come first. Now we have the derivation for incompressible flow as first, :followed with more general where at the end special case of incompressible flow is mentioned. — Preceding unsigned comment added by En odveč (talk • contribs) 17:15, :15 February 2015 (UTC)

Edit: I have deleted my changes. Why? Because compressible flow comes later, so ${\displaystyle \nabla \cdot \mathbf {u} }$ is zero and therefore divergence of tensor ${\displaystyle \nabla \mathbf {u} }$ is ${\displaystyle \nabla ^{2}\mathbf {u} }$ together with deviatoric part of ${\displaystyle p}$ . But obviously article isn't clearly written.

Yes I'm agree with you, and yes I notice that now that formula is correct but at time i wrote this comment the formula was used in the general expression of N-S eq. , anyway I think that also in the Incompressible version should be pointed out that ${\displaystyle \nabla \cdot \mathbf {u} =0}$ because of Incompressible hypothesis.

Something like:

${\displaystyle \mu \nabla \cdot (2\varepsilon )=\mu \nabla \cdot ((\nabla \mathbf {u} )+(\nabla \mathbf {u} )^{\text{T}})=\mu (\nabla ^{2}\mathbf {u} +\nabla (\nabla \cdot \mathbf {u} ))=\mu \nabla ^{2}\mathbf {u} \quad \quad {\text{Where}}\,\,\nabla \cdot \mathbf {u} =0\,\,{\text{for incompressible hypothesis.}}}$

and also:

${\displaystyle \nabla \cdot (2\mu {\boldsymbol {\varepsilon }})=\mu \nabla \cdot (2\varepsilon )\quad \quad {\text{For constant}}\,\mu \,{\text{hypothesis.}}}$

But maybe this is too verbose for someone.

--Tom.sartor (talk) 18:45, 15 February 2015 (UTC)

Hey, somebody exchanged my assumptions. I did really not mean assuptions. — Preceding unsigned comment added by 179.67.157.6 (talk) 22:08, 13 September 2015 (UTC)