Talk:Rayleigh–Taylor instability

Old discussions
What the *** is tangential gravity??? 80.177.213.144 19:05, 19 Apr 2005 (UTC)

Now slightly rewritten to remove the phrase Linuxlad 13:50, 23 Apr 2005 (UTC)

Tangential gravity —Preceding unsigned comment added by Ayeroxor (talk • contribs) 19:26, 23 April 2005

Fine, but not the preferred explanation for Labscale RT, I think —Preceding unsigned comment added by Linuxlad (talk • contribs) 19:46, 23 April 2005

Shouldn't "which acts to break a cylindrical jet into a stream of droplets having the same volume but lower surface area" actually read "which acts to break a cylindrical jet into a stream of droplets having the same volume but greater surface area"? —Preceding unsigned comment added by 71.53.8.184 (talk • contribs) 13:06, 26 July 2006

No... the Rayleigh limit indicates that if a cylinderical jet (radius R) of liquid is subjected to spatially periodic vericose surface perturbation of wavelength L, the perturbation will grow and the jet will break up into drops, even if the perturbation is of infinitesimal amplitude, so long as the wavelength of the perturbation exceeds the circumference of the unperturbed cylinder (i.e., L>2*Pi*R). Its easy to show that the lateral surface area of a cylinder (i.e., excluding the area of the ends) of raduis R and length L is greater than the surface area of a sphere of equivalent volume as long as L > 4.5R. Clearly, then, the surface area of the drops formed from the Rayleigh instability is less than the surface area of the one wavelength segments of the cylinder from which the drops are formed. —Preceding unsigned comment added by 128.46.120.153 (talk • contribs) 19:47, 13 January 2007

Nuclear Weapons R-T
The first place I ever heard of R-T instability (called just "Taylor Instability" at the time) was in the subject of nuclear weapons design. R-T occurs in the pit when the implosion shockwave crosses the border between two components, most notibily the uranium tamper and berylium reflector. Future re-writes of this article might reflect this. Reference:

Also: Who were Rayleigh and Taylor. Is that the same Ted Taylor, the nuclear weapons designer.

Lord Rayleigh aka J. W. Strutt, and (Sir) Geoffrey Ingram Taylor - Taylor worked for UK MoD for  a time and developed the 'shaped charge' where the metal cone flows like a fluid and forms an armour-piercing jet - he also worked on the Manhattan Project - HTH Linuxlad 08:50, 27 July 2006 (UTC).

Inaccuracy in reference
''Note that the RT instability is not to be confused with the Rayleigh instability (or Plateau-Rayleigh instability) of a liquid jet. This latter instability, sometime called the hosepipe (or firehose) instability, occurs due to surface tension, which acts to break a cylindrical jet into a stream of droplets having the same volume but lower surface area.'' This reference: Actual images and videos of RT fingers ends with "Manifestations of Rayleigh-Taylor instability," which actually illustrates Plateau-Rayleigh instability. Something's gotta give. -AndromedaRoach 03:19, 17 December 2006 (UTC)

Rayleigh Theorem is incorrect
http://arxiv.org/abs/physics/0610082 According to secion 5 of the reference, (see page 9), some "Rayleigh theorem" is incorrect. I don't know of this is the place for that, but there can't be too many Rayleigh's in the field of Fluid Dynamics, so they are probably talking about the same man, even if it isn't quite the same instability. Crysta1c1ear 03:28, 8 January 2007 (UTC)

This reference, categorically, does not refer to either the Rayleigh-Taylor instability or the Rayleigh instability —Preceding unsigned comment added by 128.46.120.153  (talk • contribs) 19:02, 13 January 2007

First sentence has it backwards
It used to say "a dense, heavy fluid is being accelerated by light fluid" (which is correct), until an anonymous user changed it. It should say either the dense one is being accelerated by the light one, or the light one is being accelerated into the dense one. What we have now is totally backwards and doesn't describe any instability.

If you're confused, maybe it's because the gravitational field at the Earth's surface points down, which means that (by the E.P.) the Earth's surface is in a frame accelerating upwards. They go opposite ways. —Keenan Pepper 13:14, 10 July 2008 (UTC)


 * OK. I think the confusion with me stems from what is meant by acceleration. I am used to simple kitchen-sink fluid dynamics (i.e. Earth bound), so I interpreted acceleration as the gravitational pull by Earth, i.e. Earth's gravity, working on the fluids of different density. While you seem to refer to a frame of reference in which the gravitational pull is zero. What is missing in this first sentence is the frame of reference from which one handles this. I did this from a frame of reference in which the interface between the light and heavy fluids is stationary. And with body forces either due to gravity or fictitious ones caused by working in an accelerated frame of reference. Crowsnest (talk) 14:01, 10 July 2008 (UTC)


 * So, this is still totally backwards and wrong. What's up. —Keenan Pepper 15:30, 20 August 2008 (UTC)


 * Do you have suggestions for an unambiguous statement? The dense oil on water is also a quite obscure example: for an RT instability the oil has to be denser than water. I will myself also try to find better formulations, and propose them here. -- Crowsnest (talk) 20:05, 20 August 2008 (UTC)


 * I hope the edits I made to the lead section repair the factual correctness of the description of the RT instability. I decided to directly change the article itself, because, as pointed out by you, this was wrong. -- Crowsnest (talk) 09:11, 30 August 2008 (UTC)

In my opinion the lead is improving a lot. There remain two problems which I have with the first sentence:
 * "fingering" refers to a later stage of the instability, not to its onset. It is an effect of the instability, at "mid-life" (when also other instabilities may set in, e.g. Kelvin-Helmholz, see the numerical simulation in the article) . No remaining problems, as far as I am concerned, with the 1st sentence. -- Crowsnest (talk) 21:56, 9 September 2008 (UTC)
 * "...the light fluid is pushing the heavy fluid..." is for a specific frame of reference. In case of an Earth-bound situation, with a heavy fluid on top of a light fluid, one may as well argue that the heavy fluid -- by its weight due to gravity -- is pushing on the light fluid. -- Crowsnest (talk) 08:25, 5 September 2008 (UTC)


 * My apologies for digging up an old discussion. It seems to me that the sentence "...the light fluid is pushing the heavy fluid..." is still a problem. According to Newton's 3rd Law, the light fluid pushes on the heavy fluid, and the heavy fluid pushes on the light fluid. Therefore, the statement that "...the light fluid is pushing on the heavy fluid..." is incomplete. What do you think? David Boyd Hansen (talk) 22:05, 9 January 2018 (UTC)


 * Another reference to this statement is added to the article lead. -- Crowsnest (talk) 18:18, 10 January 2018 (UTC)


 * Bold "Rayleigh–Taylor fingers" (redirect) probably solves the matter. -- Crowsnest (talk) 10:32, 5 September 2008 (UTC)


 * I'm not sure it does. It's one of many fingering instabilities, of which salt fingering and viscous fingering are two examples that currently have WP articles. Calling it a fingering instability comes directly from the source where I pulled the definition. I put “fingering” back in the first sentence, but omitted the link for now since I doubt anyone is going to start the fingering instability article anytime soon. Michael Belisle (talk) 08:47, 21 September 2008 (UTC)
 * In my view, the fingering is a later stage in the development of the RT instability, and it would be more precise to say that the RT instability develops into a fingering instability. But I agree with you this phrasing is often used. -- Crowsnest (talk) 09:25, 21 September 2008 (UTC)
 * On Google Scholar, it does not seem to be common to call the RT instability a fingering instability:
 * "Rayleigh-Taylor instability" and "fingering instability" -- about 100 hits
 * "Rayleigh-Taylor fingers" -- also about 100 hits
 * "Rayleigh-Taylor instability" -- about 10,000 hits
 * So I think we should mention "fingering instability", but not in the first sentence. -- Crowsnest (talk) 10:15, 21 September 2008 (UTC)
 * If you say so. I'm not going to fight about it. Remember, of course, that any assertions need to be verifiable (and NPOV). Presently, the introduction has one citation. I assume the rest of the content has some backup, so the editors who wrote it should cite some more sources. Michael Belisle (talk) 03:29, 23 September 2008 (UTC)
 * I am not going to fight about it either. I will look for additional references. Best regards, Crowsnest (talk) 09:25, 23 September 2008 (UTC)

Laymans' terms
I studied physics as an undergraduate not too horribly long ago, and I can't make heads or tails of the third paragraph in the introduction to this page. It seems like we're jumping into some very technical stuff without providing some sort of introduction to it or even linking to the relevant pages. Cww (talk) 04:39, 5 May 2009 (UTC)

Lava Lamp
Just an idea: It might be useful to include some mention of a lava lamp in this article, as a real-life example of an unstable fluid mechanical system. I'm not sure how appropriate an example it would be, though. Perhaps someone more expert in fluid mechanics would have a better opinion. CosineKitty (talk) 20:36, 10 August 2011 (UTC)

Details of linear stability analysis are incorrect
The curl of an irrotational velocity field (i.e., $$\nabla\times\textbf{u}'=0$$), which is also solenoidal (i.e., $$\nabla\cdot\textbf{u}'=0$$) does not, in general, satisfy $$\nabla^2\psi=0$$, where $$\psi$$ is the stream function satisfying $$\textbf{u}'=\nabla \times\psi$$. In general, $$\nabla\times(\nabla \times\psi) =\nabla(\nabla\cdot\psi)-\nabla^2\psi$$. On the other hand, for an irrotational and solenoidal vector field, the velocity potential ($$\varphi$$) can be shown to satisfy Laplace's equation, $$\nabla^2\varphi=0$$. What other assumptions were made in the actual derivation such that $$\nabla^2\psi=0$$, assuming this is correct? – Roche398 (talk) 16:19, 6 July 2013 (UTC)
 * Old post, but that use of the stream function $$\psi$$ is fine. Perhaps the only missing statement is that the flow is considered two-dimensional. Perhaps this is the additional assumption you were after. Otherwise, your objection seems to take $$\psi$$ as a vector field; it is in fact (canonically) a scalar field, so your use of the vector Laplacian, $$\nabla\times(\nabla \times\psi) =\nabla(\nabla\cdot\psi)-\nabla^2\psi$$, is not meaningful (i.e. on the RHS there both divergence and curl shows up, which is not meaningful if operating on a scalar field.) In truth there are formulations of it where it is a vector potential, but the end result comes to the same thing. The existence of a (scalar field) stream function satisfying $$\nabla^2\psi=0$$ follows at once from a two-dimensional irrotational and solenoidal velocity field: the (2d) solenoidal property allows that the scalar field exists (via continuity), and the irrotational property allows that same scalar field to satisfy Laplace's equation. Osmanthus22 (talk) 14:40, 13 December 2017 (UTC)

non-sequitur?
"This instability, sometimes called the hosepipe (or firehose) instability, occurs due to surface tension, which acts to break a cylindrical jet into a stream of droplets having the same volume but lower surface area." When an object is broken into multiple pieces, they usually have greater surface area than the original object. WilliamSommerwerck (talk) 23:18, 3 May 2014 (UTC)

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