Talk:Stokes flow

Are the viscous forces really large in comparison with the advective inertial forces?

As the headline reads, are the viscous forces really large in comparison with the advective inertial forces? Have this actually been measured or showed somehow? Where does this claim come from? That the Reynolds number is very low is a completely different thing and is much easier to show. Has some Wikipedian made the conclusion that the two are equivalent? I'm not so sure they are, even though they might be. I have put a {{fact}} tag after the statement "Stokes flow ... is a type of fluid flow where advective inertial forces are small compared with viscous forces" until someone can find a source for the claim or show the equivalence I just mentioned. —Kri (talk) 19:34, 14 August 2012 (UTC)

I think you will find the equivalence follows trivially from the definition of Reynolds number as the ratio of inertial forces/viscous forces. Obviously when Re << 1, the numerator (inertial) must be very small relative to the denominator (viscous). Bungeh (talk) 07:20, 20 May 2013 (UTC)

I agree with Bungeh. The wording itself was take from Kirby "Micro- and Nanoscale Fluid Mechanics: Transport in Microfluidic Devices" which says "In this chapter, we discuss Stokes flow (equivalently termed creeping flow), in which case the Reynolds number is so low that viscous forces dominate over inertial forces." Fluellen7 (talk) 03:44, 5 January 2014 (UTC)

Are creeping flow and creeping motion the same as Stokes flow?

I am confused about "creeping flow" and "creeping motion". Both redirect here, and are mentioned as synonyms to Stokes flow. I actually added "creeping motion" as a synonym, because the article didn't mention "creeping motion" at all.

But I am unsure if that is correct. [1] quotes McGraw-Hill Concise Encyclopedia of Physics. © 2002 by The McGraw-Hill Companies, Inc. for saying that "In creeping flow the Reynolds number is very small (less than 1) such that the inertia effects can be ignored in comparison to the viscous resistance. Creeping flow at zero Reynolds number is called Stokes flow." That means that "Stokes flow" (Re << 1, or maybe even Re = 0) is just a special case of "creeping flow" (Re < 1). Is that true?

And what about "creeping motion"? Is that the same as "Stokes flow", or "creeping flow", or something different from both?

--Jhertel (talk) 01:04, 29 August 2012 (UTC)

http://www.andrew.cmu.edu/course/06-703/Creeping%20Flow.pdf says that "One limiting case is creeping flow which corresponds to the limit in which the Reynolds number is small (i.e. Re → 0)." So, it seems like "creeping flow" is not the same as Stokes flow. My guess is that "creeping motion" is the same as "creeping flow", but I don't know. --Jhertel (talk) 01:16, 29 August 2012 (UTC)

I did a google search for "creeping motion" and found this picture. I guess "creeping motion" is not the same thing as "Stokes flow". —Kri (talk) 11:54, 29 August 2012 (UTC)

creeping motion and stokes flow are the same, they are both derived by neglecting the inertia in the Navier-Stokes equations (i.e. imposing Re = 0, justified by the fact that Re << 1). In Kim and Karrilla's "Microhydrodynamics", they say: "The linearized Navier-Stokes equations for steady (i.e. time-independent) motion are known as the creeping motion or (inhomogeneous) Stokes equations, and these are obtained by neglecting the substantial derivative of v," (the velocity), "in the Navier-Stokes equations" Fluellen7 (talk) 03:16, 5 January 2014 (UTC)

Stokes Section confusing

Is there something wrong here? In one instance the divergence is treated as a vector and in another it is treated as a scalar. I don't know enough about tensors to be sure about things here but maybe the result of "dot producting" a tensor produces a gradient. For what it is worth, this section of the article is flagged as being unclear.

${\boldsymbol {\nabla }}\cdot \mathbb {P} +\mathbf {f} =0$

where $\scriptstyle \mathbb {P}$ is the Cauchy stress tensor representing viscous and pressure stresses, and $\scriptstyle \mathbf {f}$ an applied body force. The full Stokes equations also includes an equation for the conservation of mass, commonly written in the form:

${\frac {d\rho }{dt}}+\rho \nabla \cdot \mathbf {u} =0$ — Preceding unsigned comment added by 99.240.143.10 (talk) 17:37, 5 November 2015 (UTC)

For the record, the confusing notation ${\boldsymbol {\nabla }}\cdot \mathbb {P}$ is standard in fluid dynamics. It represents a vector. The ith entry of this vector is the divergence of the ith row of $\mathbb {P}$ (or the ith column, because $\mathbb {P}$ is symmetric). Disclaimer: I am not an expert. Mgnbar (talk) 03:10, 8 March 2018 (UTC)

References

Assessment comment

The comment(s) below were originally left at Talk:Stokes flow/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Stokes flow should be combined with the creeping motion article, as they are the same thing.Easyl 09:56, 11 January 2007 (UTC)

Last edited at 09:56, 11 January 2007 (UTC). Substituted at 07:06, 30 April 2016 (UTC)

Notation in Stokeslet section

The section on Stokeslets uses the notation r r, where r is a position vector. The result is a 2-tensor. What does this notation mean? The outer product r r^T? And is this notation standard in this subject? Mgnbar (talk) 03:10, 8 March 2018 (UTC)

So it's been 4.5 years now. Does r r mean r r^T? Mgnbar (talk) 02:20, 22 October 2022 (UTC)