Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

We can describe (some of) the dynamics of many systems using fluid mechanics. Of course these include classical fluids like water, more exotic fluids like photon gases and the universe as a whole and even solid(ish) things over long times, like glasses and ice. Further still we can treat general classical systems in phase space using a phase fluid, quantum systems (e.g. Madelung equations) and with a little bit of hand waving anything that happens on a symplectic manifold (which gives us a Hamiltonian and hence a flow and Louiville's theorem).

So what is it about a system that makes it obey a fluid model? Are there systems that definitely do not fit a fluid model?

(I realise that I haven't said exactly what I mean by "fluid model", this is somewhat deliberate. If you like you can take it to mean "having equations of motion which are (almost) identical in some form to the Euler equations".)

edit: Since admittedly, the original question wasn't quite clear I'll try to clarify a bit. I'm not looking for a high-school answer, or one which describes only actual literal physical fluids, e.g. "a fluid doesn't support a shear stress", "a fluid is something you can wash your hair with". I've given some examples above of situations that fit this description, and without further explanation I think it's non-obvious what a "mean free path" or similar would mean in a generalised fluid. What I'm really looking for (and there may not be any) is some overarching physical or mathematical principle, or failing that an argument as to why there isn't one. I'd even be quite happy to be directed to a book or more appropriate forum. I apologise for not being clearer before and appreciate the answers already given.

share|improve this question
    
A fluid is nothing more than a material that "flows" (deforms continuously under shear stress). I think what you are really trying to ask is under what circumstances can a material be modeled as continuum. There are many systems that lend themselves to continuum approximation. Some of those systems deform continuously under shear stress and are therefore analogous to fluids. Regardless, the equations of motion remain the same, but the equation of state of a fluid will always be independent of deformation history. –  SimpleLikeAnEgg Mar 5 at 17:49
    
Comment to the question (v1): It would be good if OP (or somebody else?) would clarify/simplify/stress what is really being asked. –  Qmechanic Mar 5 at 22:25
    
@Qmechanic I've amended the question and hopefully the aim is somewhat clearer now –  Sean D Mar 6 at 9:19
    
@SeanD Your question is now even more confusing. So what is it about a system obey a fluid model? The answer is quite simple and I know you don't find it enlightening, but it really is as simple as if it "doesn't support shear stress." There is no other magical quality that makes a system a "fluid." –  SimpleLikeAnEgg Mar 6 at 17:12
    
@SimpleLikeAnEgg except that the concept of "shear stress" isn't (obviously) meaningful in many of the above-mentioned situations. Or are you saying there should be some generalised notion of this for each system? –  Sean D Mar 6 at 17:36
show 1 more comment

1 Answer 1

The most basic definition of a fluid is

a fluid is a substance that continually deforms (flows) under an applied shear stress.

To model a fluid using the Euler equations, you need to satisfy the condition that the mean free path of a particle, $\ell$, is significantly smaller than the typical size of the domain, $L$ (and also that viscosity and heat conduction are negligible/zero). The ratio of $\ell/L$ gives one the Knudsen number.

You can calculate the mean free path via $$ \ell=\frac{1}{n}\frac{1}{\sqrt{2}\pi\sigma^2} $$ where $n$ is the number density and $\sigma$ the mean particle size. For a hydrogen gas cloud (astrophysical context here), we expect $n\sim10\rm cm^{-3}$ with $\sigma\sim10^{-8}\,\rm cm$ (diameter of hydrogen atom) which gives $$ \ell\sim10^{14}\,\rm cm $$ There are a few common scale lengths in astrophysics, two relevant ones are (a) the AU ($10^{13}$ cm) and (b) the parsec ($3\cdot10^{18}$ cm). Clearly (a) is less than the scale length so the fluid equations won't work for modeling it on this scale, whereas (b) the fluid approach would work (and is used).

As far as the more exotic situations, I presume that the fluid description is an approximation, but I'm not sure of the justifications for doing it (outside the Knudsen relation).

share|improve this answer
    
Your answer is partial and indirect. For a start, that is not a valid definition for a fluid. A fluid is simply something that deforms continuously under shear stress. What you have given is an explanation as to when a material might be treated as a continuum. The author is indirectly asking about this as well, but you should correct your definition for a fluid. –  SimpleLikeAnEgg Mar 5 at 17:57
    
@SimpleLikeAnEgg: You are correct. I have amended my post to reflect this fact. –  Kyle Kanos Mar 5 at 18:38
    
Not sure why you had a negative 1... anyways, +1 because you mentioned the same thing that came to mind as soon as I read the question (your quote). @SimpleLikeAnEgg - another definition for a fluid that I like is - a fluid is any substance that takes the shape of the container it occupies (when acted under gravity). –  Isopycnal Oscillation Mar 6 at 6:28
    
@KyleKanos nice answer, but what does it even mean to have a mean free path for something that isn't ultimately particulate? –  Sean D Mar 6 at 9:27
1  
@SeanD Can you expand on why a hydrogen cloud would not fill the shape of a container? I see no objection for a fluid, which is what a hydrogen cloud is, to metamorphose into any shape commandeered by an eager force, since, after all, it cannot resist shear... –  Isopycnal Oscillation Mar 7 at 19:57
show 8 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.