Talk:Derivation of the Navier–Stokes equations

Notation?

Does it change from Q_i to b_i for a source/sink of momentum? —Preceding unsigned comment added by 140.184.21.115 (talk) 13:40, 19 September 2007 (UTC)

Q is used for the generic conservation, b is used in the context of momentum. —Preceding unsigned comment added by Ben pcc (talk • contribs) 23:11, 24 October 2007 (UTC)

The bounding surface of the control volume is denoted $\partial\Omega$, but later the boundaries are defined as $\Gamma$ within the section ‘Continuity Equation’. Seems to be a mistake? — Preceding unsigned comment added by S. M. Peyres (talk • contribs) 19:04, 7 December 2023 (UTC)

Reynolds Transport Theorem

I'm surprised that, in a derivation of the Navier-Stokes equations, that there is no mention of the Reynolds Transport Theorem.

--71.98.78.28 04:12, 11 June 2007 (UTC)

Good point. -Ben pcc 02:23, 28 June 2007 (UTC)

There is something fishy about how Leibniz's rule is applied just after Reynold's transport theorem is mentioned. In addition, the sign on Q looks wrong in the latter portions of that section. Worth looking into. 128.83.68.26 (talk) 18:01, 6 October 2008 (UTC)

Shouldn't it be extensive, instead of intensive property? —Preceding unsigned comment added by 85.24.185.34 (talk) 14:55, 2 September 2009 (UTC)

It defenitely should if these examples are correct Intensive_and_extensive_properties#Examples_2 — Preceding unsigned comment added by 46.242.70.122 (talk) 20:29, 3 May 2016 (UTC)

Outer Product?

Would the relations for $\mathbb {T} _{ij}$ in the discussion on Newtonian fluids be equivalent to saying $\mathbb {T} =\mu (\nabla \otimes \mathbf {u} +(\nabla \otimes \mathbf {u} )^{T})+(\lambda \nabla \cdot \mathbf {v} )\mathbb {I}$, where $\mathbb {I}$ is the identity matrix? --Zemyla^t 17:57, 25 October 2007 (UTC)

I think so. I just added something that looks a lot like that minus the outer products, I think they're equivalent. I'm not using the outer product notation because my sources (MIT OCW "Surface tension module" and my fluid mechanics teacher) don't either. — Ben pcc 02:30, 3 November 2007 (UTC)

Scratch that. Using the outer product is more mathematically sound and there is zero ambiguity. Thank you! — Ben pcc 17:30, 4 November 2007 (UTC)

I think this section is needlessly confusing and there is no reason to introduce the $\nabla \otimes \mathbf {u}$ notation at all. Please correct me if my algebra is wrong but why not have the large expansion and then "and, more compactly, in vector form"

$\rho \left({\frac {\partial \mathbf {v} }{\partial t}}+\mathbf {v} \cdot \nabla \mathbf {v} \right)=-\nabla p+\mu (\nabla ^{2}\mathbf {v} )+\mu \nabla (\nabla \cdot \mathbf {v} )+\lambda \nabla (\nabla \cdot \mathbf {v} )+\rho \mathbf {g}$

This clearly shows how the incompressability reduces the stress contribution to $\mu (\nabla ^{2}\mathbf {v} )$ and also why there is no contribution from bulk viscocity. — Jon —Preceding unsigned comment added by 131.227.66.187 (talk) 13:27, 8 September 2010 (UTC)

See Volume viscosity for an example of this expansion already in use. — Jon —Preceding unsigned comment added by 131.227.66.187 (talk) 11:17, 9 September 2010 (UTC)

Are they correct?

In conservation of momentum, why is the following derivation correct: 1)$\nabla (\rho )\mathbf {v} \cdot \mathbf {v} =\mathbf {v} \mathbf {v} \cdot \nabla (\rho )$; 2)$\rho \nabla (\mathbf {v} )\cdot \mathbf {v} =\rho \mathbf {v} \cdot \nabla \mathbf {v}$. The $\nabla \mathbf {v} \cdot \mathbf {v}$ is not equal to $\mathbf {v} \cdot \nabla \mathbf {v}$ —Preceding unsigned comment added by Run Jiang (talk • contribs) 10:23, 14 July 2008 (UTC)

This is most easily seen in Cartesian coordinates, by using the summation convention to write the momentum equation with body forces as

${\frac {\partial (\rho v_{i})}{\partial t}}+{\frac {\partial (\rho v_{j}v_{i})}{\partial x_{j}}}=b_{i},$

summing the second term over all coordinate directions j, so j from 1 to 2 in two spatial dimensions and j from 1 to 3 in 3D. Here v_j are the components of the vector v in the respective coordinate directions associated with x_j. Then, by using the chain rule:

$v_{i}{\frac {\partial \rho }{\partial t}}+\rho {\frac {\partial v_{i}}{\partial t}}+v_{i}v_{j}{\frac {\partial \rho }{\partial x_{j}}}+\rho v_{i}{\frac {\partial v_{j}}{\partial x_{j}}}+\rho v_{j}{\frac {\partial v_{i}}{\partial x_{j}}}=b_{i}.$

Which is equal to the last equation in the expansion. The third of the three rules, as mentioned by you above, has to be read as (∇v)•v=v•(∇v) and is also an identity (with ∇v the tensor derivative of vector v). The same holds in curvilinear coordinate systems, provided care is taken with respect to using contravariant or covariant vector component representations, and by using the covariant derivative instead of a simple partial derivative. In vector notation, as in the article, it is independent of the coordinate system used. -- Crowsnest (talk) 21:48, 15 July 2008 (UTC)

While the simple method shown in the above response is correct, it does not answer the original question posted by Run Jiang. The answer to that question is that the first line of the derivation on the main page is, in fact, incorrect and so is the second one. It is only by magic that the third line (which we could have arrived at using Crowsnest's choice, namely the derivative of products) is correct again.

If we wish to do this on a step-by-step basis, as in the main article, these lines should do as follows

$\nabla \cdot (\rho \mathbf {v} \mathbf {v} )=\nabla \cdot (\rho \mathbf {v} )\mathbf {v} +\rho \mathbf {v} \cdot \nabla \mathbf {v}$

Which would greatly simplify the rest of the derivation, since in this form the terms of the continuity equation are already pulled together. Czigi (talk) 17:17, 13 May 2010 (UTC)

Please correct the derivation on the main page, if you find mistakes (e.g. "first line of the derivation on the main page is, in fact, incorrect and so is the second one" by Czigi). Bkocsis (talk) 23:38, 4 June 2010 (UTC)

In Stream function formulation the derivation seems to assume that the $\mathbf {f}$ body forces are conservative, but this is not stated. To fix this, I suggest to insert $\nabla \times \mathbf {f}$ to the right hand side of the equation of motion of the stream function, and then note that this term drops out if $\mathbf {f} =\nabla \phi$. However, I am not sure what the analogous equation is for the 2D flow in orthogonal coordinates. Bkocsis (talk) 23:38, 4 June 2010 (UTC)

Messed up notation

In the beginning of the section "General form..." the very first equation incorrectly takes the divergence of the stress tensor components, not the tensor itself. I suggest replacing $\sigma _{ij}$ with ${\boldsymbol {\sigma }}$ —Preceding unsigned comment added by Kallikanzarid (talk • contribs) 19:42, 28 July 2009 (UTC)

Stress tensor for incompressible Newtonian fluid

Does the simplified stress tensor look like this $\mathbb {T} _{ij}=\mu \left({\frac {\partial u_{i}}{\partial x_{j}}}\right)$ or this $\mathbb {T} _{ij}=\mu \left({\frac {\partial u_{j}}{\partial x_{i}}}\right)$. If i denotes rows and j denotes columns then I think the second one, right? Thanks. --kupirijo (talk) 16:02, 22 September 2010 (UTC)

Stress tensor for compressible Newtonian fluid

I am fairly certain that the general expression for the stress tensor (i.e. compressible) is wrong. Instead

$\mathbb {T} _{ij}=\mu \left({\frac {\partial u_{i}}{\partial x_{j}}}+{\frac {\partial u_{j}}{\partial x_{i}}}\right)+\delta _{ij}\lambda \nabla \cdot \mathbf {v}$

should read

$\mathbb {T} _{ij}=\mu \left({\frac {\partial u_{i}}{\partial x_{j}}}+{\frac {\partial u_{j}}{\partial x_{i}}}-{\tfrac {2}{3}}\delta _{ij}{\frac {\partial u_{k}}{\partial x_{k}}}\right)+\delta _{ij}\lambda {\frac {\partial u_{k}}{\partial x_{k}}}$

See Landau and Lifshitz, Fluid Mechanics, Second Edition: Volume 6 (Course of Theoretical Physics) page 45. — Preceding unsigned comment added by Neeson.m (talk • contribs) 02:50, 5 July 2011 (UTC)

I second that opinion. While this has been corrected, further down the page (Navier–Stokes equations for a compressible Newtonian fluid) the term making the tensor T_ij traceless is still missing. 143.210.37.186 (talk) 09:26, 22 November 2012 (UTC)

It depends on whether $\lambda$ is the volume viscosity or the second viscosity (Lame's first parameter). There is a lot of confusion over which is called bulk viscosity, volume viscosity, or second viscosity. If $\lambda$ is the second viscosity, or Lame's first parameter, then the stress tensor is:

$\mathbb {T} _{ij}=\mu \left({\frac {\partial u_{i}}{\partial x_{j}}}+{\frac {\partial u_{j}}{\partial x_{i}}}\right)+\delta _{ij}\lambda \nabla \cdot \mathbf {v}$

If $\lambda$ is the volume viscosity, or bulk modulus, then the stress tensor is:

$\mathbb {T} _{ij}=\mu \left({\frac {\partial u_{i}}{\partial x_{j}}}+{\frac {\partial u_{j}}{\partial x_{i}}}-{\tfrac {2}{3}}\delta _{ij}{\frac {\partial u_{k}}{\partial x_{k}}}\right)+\delta _{ij}\lambda {\frac {\partial u_{k}}{\partial x_{k}}}$

Consequently, the Stokes hypothesis is different for each. For the former one takes $\lambda =-{\frac {2}{3}}\mu$ and for the latter $\lambda =0$. Lucky.v.tran (talk) 21:23, 28 November 2017 (UTC)

This article should be combined with the main article on the Navier Stokes equation

It is unclear why the authors have made this a separate article when the main article on the Navier Stokes equation covers almost all of the same material. Also this article was judged to be under WikiProject Physics whereas the main article is judged to be under WikiProject Mathematics... both (or more properly one article combined) should be under the former since it is about the physics and not the applied math (singular please).Danleywolfe (talk) 17:02, 8 September 2011 (UTC)

Can't agree more, it seems an unnecessary duplication and source of confusion. Gpsanimator (talk) 07:52, 22 April 2024 (UTC)

NS is one of the very few mathematical models that is simultaneously insufferably difficult, both conceptually and mathematically, and widely applicable and even commonly used on a daily basis. similar examples would be derivations of the Boltzmann equation, Schrodinger equation, Kohn Sham system, Dirac Equation, etc. The usefulness of the derivation either laying in the examples of the historical ingenuity and/or a visual depiction of the amazingly complicated, and clunky, mathematics that lay at the heart of the various subsets of theoretical physics/math. Unrelated to the technical issue, I personally think that no wikipedia article should be excessively long. At a certain point, a very very long article detracts from some subset of its contents. In this case, I think the article might actually be in need of splitting it based upon the pure theoretical/mathematical, solutions (why? because this is a whole area of research in and of itself)/approximations/competing/equivalent theories, and maybe even an "applications" article.... or just a "lay article," to keep those people who never even saw a professor write the word "tensor" on a board, let alone personally handle them on paper, happy without too much loss of detail.184.189.220.114 (talk) 11:25, 28 March 2013 (UTC)

Stress tensors

Hi, I corrected an error in what I think is a confusion in notation between the deviatoric stress tensor $\mathbb {T} _{ij}$ and the viscosity stress tensor. The latter includes the volume viscosity, but is not required to be traceless. However, in the notation used in the article, $\mathbb {T} _{ij}$ has, by definition, to be traceless for all values of λ, and not just for the provided value of $-{\frac {2}{3}}\mu$. I introduced π as the mechanical pressure, so as to preserve most of the notation in the latter parts of the article, where p corresponds to the thermodynamical pressure. Hope this improves the article coherence. Donvinzk (talk) 17:15, 24 May 2012 (UTC)

Momentum equations

The second line in the derivation of the momentum equation moves the source/sink term to the right hand side, but the sign is kept positive with no explanation why:

${\frac {\partial }{\partial t}}(\rho \mathbf {u} )+\nabla \cdot (\rho \mathbf {u} \mathbf {u} )+\mathbf {s} =0$

$\mathbf {u} {\frac {\partial \rho }{\partial t}}+\rho {\frac {\partial \mathbf {u} }{\partial t}}+\mathbf {u} \mathbf {u} \cdot \nabla \rho +\rho \mathbf {u} \cdot \nabla \mathbf {u} +\rho \mathbf {u} \nabla \cdot \mathbf {u} =\mathbf {s}$

Why is $\mathbf {s}$ not negative in the second line? — Preceding unsigned comment added by Voyvoyvoy (talk • contribs) 19:11, 26 August 2015 (UTC)

Energy equation/power balance

A section on the derivation of the energy equation/power balance is missing from the article.

The section Compressible Newtonian fluid lists a formulation of the energy equation/power balance, but does not mention that $\nabla \cdot (k\nabla T)$ is $-\nabla \cdot \mathbf {q}$ with the isotropic material model $\mathbf {q} =-k\nabla T$ applied. 2.202.211.78 (talk) 10:09, 6 June 2022 (UTC)

LET'S ASK SIR ISAAC NEWTON ABOUT N-S

Read if you wish.

Sir, Newton if you could read this derivation of Navier-Stokes equations what would you say? Good Lord! There is all dimensionally inhomogeneous! There are added and subtracted apples and oranges! Where there are my laws? So do you sleep on lectures of physics? Physicists WAKE UP! (catch-32) And you, too Leonhard Euler ! Immediately clean up that mess! And leave it, the ideal fluid - shame on you! OK Sir! Right away Sir, only to find my old papers from 1755 on the movement of fluids. Here: MOTION OF THE FLUID IN THE CONTINUUM (Euler's approach): (differential move in point (r) by speed (v) for the length (dr) in time (dt)): $v_{x}{\text{(r+dr,t+dt)}}=v_{x}+\nabla v_{x}dr+dt{\frac {\partial v_{x}}{\partial t}}$ $v_{y}{\text{(r+dr,t+dt)}}=v_{y}+\nabla v_{y}dr+dt{\frac {\partial v_{y}}{\partial t}}$ $v_{z}{\text{(r+dr,t+dt)}}=v_{z}+\nabla v_{z}dr+dt{\frac {\partial v_{z}}{\partial t}}{\text{ (1)}}$ Or shorter written: $dv=dr\nabla v+dt{\frac {\partial v}{\partial t}}\quad {\frac {dr}{dt}}=v{\text{ , so it will be:}}$ ${\frac {dv}{dt}}={\frac {\partial v}{\partial t}}+v\nabla v\quad {\text{(2) (2nd Newton's law in the fluid!)}}$ "Force acting on a point of the fluid by simultaneously changing the speed and energy of that point." "Force of action (F) opposing reaction forces: potential force, pressure force and viscous force": ${\frac {\partial v}{\partial t}}+v\nabla v=F+F_{G}-F_{P}-F_{\mu }\quad {\text{(3) (3rd Newton's law in the fluid!) all force per unit mass}}$ POTENTIAL FORCE: $F_{G}=-\nabla U\quad {\text{ (potential force per unit mass)}}$ PRESSURE FORCE: $P={\begin{vmatrix}p&0&0\\0&p&0\\0&0&p\end{vmatrix}}\quad {\text{ (pressure tensor)}}$ On volume of fluid: $F_{VP}=-\int Pda=-\nabla P\quad {\text{ (Gauss's theorem)}}$ $F_{P}=-{\frac {\nabla P}{\rho _{F}}}\quad {\text{ (pressure force per unit mass)}}$ VISCOUS FORCE: Since, in the fluid there are not specific layers, surfaces, even volumes, in a fluid is the simplest to define all the force per unit mass (F) and thus should be defined and the viscous force. Sir Newton gave us the law of force and the law of friction: $G={\text{mass}},i={\text{impuls}},M=Gi={\text{moment}},F_{m}={\frac {dM}{dt}}={\text{force}},$ $F={\frac {F_{m}}{G}}={\frac {di}{dt}}={\frac {dv}{dt}}{\text{ = force per unit mass.}}$ $j=\mu {\frac {dv}{dr}}\quad {\text{(flow of impulses!) = (Newton's Law of friction!)}}$ "The flow of impulses, is in fact the work per unit mass [N m/kg] at some point in the fluid." $\left[{\frac {Nm}{kg}}\right]=\left[{\frac {m^{2}}{s^{2}}}\right]{\text{ so the dimension of the coefficient }}\mu {\text{ will be }}\left[{\frac {m^{2}}{s}}\right]$ Then if we do a balance of impulses in the fluid: $Acceleration-Deceleration-Friction=0,$ or, in another way: $j_{f}=\mu _{a}{\frac {dv}{dr}}-\mu _{d}{\frac {dv}{dr}}=\mu {\frac {dv}{dr}}\quad {\text{(loss of impulse = friction!)}}$ $\mu _{a}=\mu _{d}$ = coefficient of self-diffusion of particles of fluid. $\mu {\text{ = viscosity coefficient of friction in the fluid}}\left[{\frac {m^{2}}{s}}\right]$ Vector form of the law of viscosity friction is: $j_{f}=\mu {\text{ grad }}v$ "Viscosity is work performed by a fluid in motion"! "Equation of continuity for impulse = equation of force"! ${\frac {di}{dt}}+{\text{div}}j=0\quad F={\frac {di}{dt}}.$ $F_{\mu }=-{{\text{div }}j_{f}}=-\mu {\text{ div grad }}v=-\mu \Delta v$ $F_{\mu }=-\mu \Delta v{\text{ , (viscous force per unit mass), }}F_{r}=-\left(\mu {\frac {\partial ^{2}v}{\partial r^{2}}}\right)_{r}$ The basic equation (3) for the balance of forces in the fluid will now be: ${\frac {\partial v}{\partial t}}+v\nabla v={\frac {dv}{dt}}-\nabla U+{\frac {\nabla P}{\rho _{F}}}+\mu \Delta v{\text{ (4) with:}}$ $v\nabla v={\frac {1}{2}}v^{2}-v{\text{ × rot }}v\quad {\text{(total kinetic energy!)}}$ BASIC EQUATION OF FLUID DYNAMICS (2) becomes (correct, Navier-Stokes): ${\frac {dv}{dt}}={\frac {\partial v}{\partial t}}-v{\text{ × rot }}v+\nabla \left(U-{\frac {P}{\rho _{F}}}+{\frac {v^{2}}{2}}\right)-\mu \Delta v\quad {\text{ (5) with:}}$ ${\frac {d\rho _{F}}{dt}}+{\text{ div }}\rho _{F}v=0{\text{ , (equation of continuity of mass)}}$ EQUATION OF FLUID STATICS (fluid is at rest, and acting external forces): ${\frac {dv}{dt}}=\nabla U-{\frac {\nabla P}{\rho _{F}}}\quad {\text{ (6)(First condition of the stationary flow!)}}$ EQUATION OF FLUID STATICS (v = 0, dv/dt = 0, ∂v/∂t = 0): $\nabla P=\rho _{F}\nabla U\quad {\text{ (7)}}$ So what do you say Sir Newton? Well, now everything is all right, all my laws are there! Wonderful! But now, you have to explain to people how to use this equation. Believe me it's very important, I know; I gave the people the laws but did not give them, instructions for use, and now see the mess that people create on the planet. OK, let's try to explain steady laminar fluid flow: STATIONARY FLOW OF IRROTATIONAL FLUID: Fluid flows only if an external force in it is induced pressure tensor, which will provide a fluid motion energy: kinetic energy and the energy needed to overcome internal friction. The fluid is in a state of steady flow when the work of external forces is equal to the sum of kinetic energy and energy losses due to friction during motion. ${\frac {\partial v}{\partial t}}=0,\quad v\times {\text{rot }}v=0\quad {\text{(Zeroth conditions of laminar and stationary flow!)}}$ ${\frac {dv}{dt}}=\nabla \left(U-{\frac {P}{\rho _{F}}}+{\frac {v^{2}}{2}}\right)-\mu \Delta v\quad {\text{ (8)(laminar flow equation!)}}$ ${\frac {dv}{dt}}-\nabla U+{\frac {\nabla P}{\rho _{F}}}=0\quad {\text{ (1st condition of the stationary flow!)}}$ $\nabla \left(U-{\frac {P}{\rho _{F}}}+{\frac {v^{2}}{2}}\right)=0\quad {\text{ (2nd condition of the stationary flow!)}}$ $U-{\frac {P}{\rho _{F}}}+{\frac {v^{2}}{2}}=Constant,\quad {\text{ (Bernoulli's theorem)}}$ After meeting all the conditions in equations (5) and (8), a fluid flowing, must still meet the remaining, speed members of vector equations of the stationary fluid flow: $\mu \Delta v-\nabla {\frac {v^{2}}{2}}=0\quad {\text{ (9)(3rd condition of the stationary flow!)}}$ "The force of the internal (viscous) friction is proportional to the gradient of the kinetic energy of the fluid"! $\nabla \left(\mu \nabla v-{\frac {v^{2}}{2}}\right)=0\quad {\text{ (the same as (9)) which implies:}}$ The Law of distribution of speed, at stationary flow of fluid: $\mu \nabla v-{\frac {v^{2}}{2}}=Constant=-{\frac {v_{max}^{2}}{2}}\quad {\text{(10)}}$ All these equations (8), (9) and (10) are very difficult to solve. Sir Newton, I'm stuck, I do not know further, what now? So we need to apply the rule that we professors do not like to hear: "Some of our students are better than us." Let's ask the advice of someone who knows mathematics and physics better than the two of us, for example, from Albert Einstein. Mr. Einstein, what do you think about our derivation of the N-S equation and of its use, what next? Very interesting, correct, nice, simple, and yet complicated, especially this last equation. I would simplify it even more! For example, if to all these points with the same speed in a vector field, we assign the same coordinate, we will create a one-dimensional problem of the equation (10), whose differential equation is: ${\frac {dv}{dr}}=-{\frac {(v_{max}^{2}-v^{2})}{2\mu }}$ this equation, we know to solve, but because, in the fluid, must be applied, and the principle of relativity, let us introduce to our problem: rE-relative distance and vE-relative speed between two points in the fluid, and the same equation becomes: ${\frac {dv_{E}}{dr}}=-{\frac {v_{E}^{2}}{2\mu }}\quad$ which after the separation of variables and integration gives: $\int {\frac {dv_{E}}{v_{E}^{2}}}=-{\frac {1}{2\mu }}\int dr,$ $-{\frac {1}{v_{E}}}=-{\frac {r}{2\mu }}+C,\quad {\frac {R}{2\mu }}=C,\quad {\frac {1}{v_{E}}}={\frac {r_{E}}{2\mu }}$ and general definition of viscosity and laminar flow read as follows: $\mu ={\frac {v_{E}r_{E}}{2}}\quad {\text{(in one-­dimensional space!)}}$ "A fluid flows such, that the product of the relative speed and distance is constant (of viscosity)". ? OK, Thank you! Sir Newton you are always right! Sorry about the ideal fluid. We professors now have to move on to the real fluids - let's go on beer. Let's have a beer in honor of Albert Einstein, Leonhard Euler and Sir Isaac Newton! Cheers! (Vjekoslav Brkić, Osijek).213.202.80.195 (talk) 11:28, 7 October 2015 (UTC)

Continuity/conservation

This article appears to imply these two words are synonyms. Are they?

Darcourse (talk) 11:50, 12 July 2023 (UTC)