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)

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)

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)

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)

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)