Talk:Derivation of the Navier–Stokes equations

message: 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)
thread_title: Notation?

message: 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)
thread_title:  Reynolds Transport Theorem 

message: 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)
thread_title:  Messed up notation 

message: 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)
thread_title:  Stress tensor for incompressible Newtonian fluid 

message: 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)
replies: text: Can't agree more, it seems an unnecessary duplication and source of confusion. Gpsanimator (talk) 07:52, 22 April 2024 (UTC)
thread_title:  This article should be combined with the main article on the Navier Stokes equation 

message: 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)
thread_title:  Stress tensors 

message: 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)
thread_title:  Momentum equations 

message: 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)
thread_title:  Energy equation/power balance 

message: 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_xdr + dt\frac{\partial v_x}{\partial t}
v_y\text{(r+dr,t+dt)}=v_y +\nabla v_ydr + dt\frac{\partial v_y}{\partial t}
v_z\text{(r+dr,t+dt)}=v_z +\nabla v_zdr + 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{N m}{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^2_E}{2 \mu}\quad
which after the separation of variables and integration gives:
\int\frac{dv_E}{v^2_E} = -\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)
thread_title: LET'S ASK SIR ISAAC NEWTON ABOUT N-S

message: This article appears to imply these two words are synonyms. Are they?
Darcourse (talk) 11:50, 12 July 2023 (UTC)
thread_title:  Continuity/conservation