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.

I've been studying fluid mechanics and I have some doubts on the mathematical model of a fluid. First we let $D\subset \mathbb{R}^3$ be a region in three-space filled with a fluid. This region has as many points as the real line. Since in Classical Mechanics when studying systems with a finite number of particles we consider each particle as a point in $\mathbb{R}^3$ I thought that each point of this region $D$ should be considered a particle of fluid. In that setting the fluid would be composed of uncountably many particles.

With this logic, I thought on labeling each particle by it's position at time $t = 0$, so that a state of the fluid would be a bijection from $D$ onto $D$. The time evolution would be then $t \mapsto \varphi_t : D \to D$ and we could summarize this with a map $\varphi : D\times \mathbb{R}\to D$ given by $\varphi(a,t) = \varphi_t(a)$. This is exactly the Lagrangian description of the fluid, so I thought I was right in considering each particle of the fluid as a point in $D$.

Now, I've been told it's wrong to think it this way, because if it were so, each particle would have zero mass. I've also been told that the right way to think is to imagine each particle as a infinitesimal volume $dV$ containing some real molecules of fluid with mass density $\rho : D\times \mathbb{R}\to \mathbb{R}$ so that each particle would have mass $\rho dV$. Since each piece of volume is infinitesimal we could still identify them with points.

My doubt is, when modeling a fluid with math which viewpoint is the standard one and why? The view point as in Newtonian and Lagrangian Mechanics of considering the fluid particles as points of $D$ or the view point of considering the fluid particles as infinitesimal chunks of fluid with volume $dV$ one at each point of $D$?

share|improve this question
    
There are no particles in a continuum mechanics. Texts and journal papers do however typically use a concept of differential volumes. Is it mathematically rigorous? No. It's "physics math." Those books definitely aren't using non-standard analysis. –  David Hammen Aug 28 at 0:45
    
@DavidHammen, the book I'm reading talks all the time of "fluid particles", then what this really means is a chunk of fluid with volume $dV$ associated with some point in the region of the fluid? –  user1620696 Aug 28 at 1:23

1 Answer 1

up vote 3 down vote accepted

The latter is definitely more standard because suppose you have a collection of different particles (e.g., $H_2O$ and $H_2O_2$). The masses of the individual molecules are different, but over an infinitesimal volume, the density could be taken as an average value.

I suppose, since $dV$ is an infinitesimal, then $\rho dV=m$ and the two pictures are equivalent, but the latter image gives a bit more intuitive sense of the fluid picture since we don't deal with particles in this model.

So, rather than using $$ m\frac{d\mathbf v}{dt}=\mathbf F \\ \frac{d\mathbf x}{dt}=m\mathbf v $$ to model $n$ particles, we can simply use the continuity + Navier-Stokes equations*: $$ \partial_t\rho+\nabla\cdot\rho\mathbf v=0 \\ \rho\left(\frac{\partial\mathbf v}{\partial t}+\mathbf v\cdot\nabla\mathbf v\right)=-\nabla p+\nabla\cdot\mathsf T+\mathbf f $$ and get our fluid model exactly from the properties we started off assuming: a roughly constant density, $\rho$, over an infinitesimal volume element, $dV$ (combined with the conservation law that flow in = flow out)--similarly for the velocity, $\mathbf v$, to get the Navier-Stokes equations.

Computationally speaking, the first method can only work for $10^{10}$ particles before cluster admins will be yelling at you, whereas the fluid picture is looking at way more particles than that.


Alternatively, you can use the Euler equations if you can assume zero viscosity.

share|improve this answer
    
Your first treatment is Lagrangian while the second is Eulerian in frame. It is possible to treat fluids in a Lagrangian fashion using particle methods such as Smoothed Particle Hydrodynamics. But the particles aren't atoms/molecules/etc.. –  tpg2114 Aug 28 at 0:56
    
@tpg2114: I was thinking more along the lines of the PIC method for the first set, rather than the Lagrangian frame. –  Kyle Kanos Aug 28 at 0:58
    
Thanks @KyleKanos for your answer. Just one doubt: when we pick a point in $D$ and consider the velocity field at the point at some time, we are really considering the velocity of a point of reference in a three-dimensional infinitesimal parcel of fluid? Because if points are not particles, the velocity field is exactly the velocity of what? –  user1620696 Aug 28 at 1:06
1  
In order to do fluid mechanics, you need to stop thinking about individual particles and think about bulk properties/effects; you have a fluid, not a collection of particles. The velocity is simply the velocity of the fluid element in the infinitesimal volume $dV$. –  Kyle Kanos Aug 28 at 1:10

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.