6
$\begingroup$

I have followed derivations of the Navier-Stokes equations and I can see how the various terms arise in the "main equation", the momentum conservation equation.

However I don't understand why the mass conservation equation has no diffusion term. There is a diffusion type term in the conservation of momentum equation, so why do we not have one in the conservation of mass equation?

$\endgroup$
  • $\begingroup$ What do you mean by diffusion type term? $\endgroup$ – Ján Lalinský Jul 19 '15 at 3:36
  • $\begingroup$ Well in the NS conservation of momentusm equation we have a term $-\mu \nabla^2 \mathbf{u}$ which accounts for diffusion due to viscous stress. But there is no counterpart for this in the conservation of mass equation. Yet there is an advection term in both equations. So why is there an advection term in both conservation of mass and momentum equations, but the diffusion term is only present in the conservation of momentum equation? $\endgroup$ – csss Jul 19 '15 at 12:30
  • $\begingroup$ Did you try to derive the equation starting from Kinetic theory? It is straightforward. $\endgroup$ – falematte Jul 19 '15 at 13:25
  • $\begingroup$ I have seen kinetic theory used to describe conservation of momentum...the momentum of particles in different "layers" of the fluid at different velocities come into contact leading to a diffusion of momentum. Why can we not say the same for the mass of the particles? $\endgroup$ – csss Jul 19 '15 at 13:36
  • 1
    $\begingroup$ @csss You can definitely do that. But in a pure fluid, there is no differences in mass. Water molecules all weigh the same, so there's no "layers" of different masses. I left an answer explaining that reasoning. $\endgroup$ – tpg2114 Jul 19 '15 at 17:04
6
$\begingroup$

For a single-component fluid, the conservation of mass follows $$ \left(\begin{array}{c}\text{mass of fluid } \\ \text{in volume }\Delta V\end{array}\right)=\left(\begin{array}{c}\text{flux of fluid } \\ \text{in/out of volume }\Delta V\end{array}\right)+\left(\begin{array}{c}\text{sources or} \\ \text{sinks in }\Delta V\end{array}\right) $$ In terms of a cube of volume $\Delta V=\Delta x\Delta y\Delta z$, this is \begin{align} \frac{\partial}{\partial t}\rho\Delta V&=\rho v_x\Delta y\Delta z\vert_{x}-\rho v_x\Delta y\Delta z\vert_{x+\Delta x} \\ &+\rho v_y\Delta x\Delta z\vert_{y}-\rho v_y\Delta x\Delta z\vert_{y+\Delta y} \\ &+\rho v_z\Delta x\Delta y\vert_{z}-\rho v_z\Delta x\Delta y\vert_{z+\Delta z} \\ &+r\Delta V \end{align} The flux here is defined as $\rho v_i$: the mass that flows outwards, $\rho$, must flow out at the speed of the fluid in the cell volume, $v_i$. If a microscopic diffusion were to take place, we would not be able to tell because the molecular masses are identical, so we would not be able to tell state 1 from state 2.

Then by dividing both sides by $\Delta V$ and taking the limit $\Delta x\to0$, we end up with the PDE $$ \frac{\partial \rho}{\partial t}=-\frac{\partial\rho v_x}{\partial x}-\frac{\partial\rho v_y}{\partial y}-\frac{\partial\rho v_z}{\partial z}+r $$ which reduces to the commonly seen continuity equation $$ \frac{\partial\rho}{\partial t}+\nabla\cdot\rho\mathbf{v}=0 $$ with no sources/sinks ($r=0$).

However, we can have a diffusion component to the continuity equation if we are considering different chemical species that can interact. For an arbitrary volume of some chemical species $i$, the mass-balance is $$ \left(\begin{array}{c}\text{mass of species }i \\ \text{in volume }\Delta V\end{array}\right)=\left(\begin{array}{c}\text{flux of species }i \\ \text{in/out of volume }\Delta V\end{array}\right)+\left(\begin{array}{c}\text{mass produced} \\ \text{by reactions}\end{array}\right) $$ which is really, $$ \frac{\partial c_i}{\partial t}+\nabla\cdot\mathbf{n}=r\tag{1} $$ where $\mathbf{n}$ is the flux of species $c_i$, and $r$ the source term. This is, of course, our common continuity equation with a source term. In the case of the multi-component fluid here, the diffusion of particles will change the states, so the initial state is no longer equivalent to the final state.

For stationary flows, the flux for mass-transfer is $$ \mathbf{n}=-D\nabla c_i $$ to give us Fick's law. For a moving flow, however, the flux has a diffusion and a convection/advection component, $$ \mathbf{n}=-D\nabla c_i+c_i\mathbf{v} $$ which would then allow (1) to be $$ \frac{\partial c_i}{\partial t}+\nabla \cdot c_i\mathbf{v}=\nabla\cdot\left(D\nabla c_i\right)+r\tag{2} $$ which is a convection-diffusion equation.

With the momentum equation, however, we have some extra terms that are associated with the changing shape of the control volume $\Delta V$: $$ \frac{\partial}{\partial t}\iiint_V\rho \mathbf u\,dV=-\oint_S\left(\rho \mathbf u\,d\mathbf S\right)\mathbf u-\oint_S p\,d\mathbf S+\iiint_V\rho\mathbf f_{body}\,dV+\mathbf F_{surf} $$ namely the body forces, $\mathbf{f}_{body}$, and surface forces, $\mathbf{F}_{surf}$. It is the surface force that generates the diffusion term, as it is related to the stress tensor which provides the $\nu\nabla^2\mathbf{v}$ term in the Navier-Stokes equation.

$\endgroup$
  • $\begingroup$ I think this answer could be a little more focused on the question, and maybe add a little more detail than "I cannot see how it would be possible for the fluid to diffuse." I think the point is that for a pure fluid, any flux in density is $\rho {\bf v}$, right? (I'm having trouble making this make sense, which is why this isn't a full answer). $\endgroup$ – AJK Jul 19 '15 at 3:16
  • $\begingroup$ The momentum of the fluid can diffuse...why can't the mass of the fluid diffuse too? $\endgroup$ – csss Jul 19 '15 at 13:19
  • $\begingroup$ @AJK: Correct, the mass flux is defined as $\rho\mathbf{v}$. I've updated my answer to reflect your suggestion. $\endgroup$ – Kyle Kanos Jul 19 '15 at 15:23
  • 1
    $\begingroup$ @csss: I've updated my answer to include the reasons why the momentum density can diffuse and why the total mass density doesn't. $\endgroup$ – Kyle Kanos Jul 19 '15 at 15:24
  • 1
    $\begingroup$ @tpg2114: I was thinking about expanding the Fick's law bit to cover that reasoning, so I probably will paraphrase/reformulate your entire comment to cover it ;) $\endgroup$ – Kyle Kanos Jul 19 '15 at 16:57
4
$\begingroup$

I left this as a comment but I'll expand it here since it provides another viewpoint. Imagine you have a box of gas molecules bouncing around in them. Every molecule is identical so they have the same mass, temperature and pressure. Let's also say this box has a diaphragm in the middle separating the box into two.

You now remove the diaphragm and start looking for changes in the gas in the box. But you see nothing happening because for every molecule that started on the left of the divider that moves to the right of the divider, a molecule from the right moves to the left. But they are the exact same masses, pressures, and temperatures so there is no change in the actual state of the box.

Now imagine that you had the same box, split in half, but this time you put a light molecule on the left and a heavy one on the right, again all with the same pressure and temperature. Now when you remove the diaphragm, when a heavy molecule moves into the light molecule side, several light molecules will move to the heavy side. And if you look at this over time, the sharp interface will diffuse out as these molecules bounce off one another and mix. Eventually, it will become homogeneous and you won't see any further changes in the system.

Now imagine if we had the same box, same divider, with identical molecules on the left and right but now the temperature of the left was higher than the right. When the divider is removed and a high temperature molecule moves to one side and a low temperature one takes it's place, you would see the temperature of the system diffuse and mix until homogeneous. You can make the same argument about momentum to get the viscous diffusion term.

So all of this is to say that the equations are all describing the same things, but the diffusion of identical mass gives an indistinguishable system from the previous state. There just aren't any differences that are observable, and so there's no diffusion term in the mass equation. Unless you have multiple species (different molecules), in which case those partial mass equations do have a diffusion term.

$\endgroup$
2
$\begingroup$

So it's actually a really simple reason, but you're going to have to think a little bit about what's going on.

The transport equation states that everything which is a "stuff" can be viewed in this way: "A small box flows downstream; the time rate of change of the stuff inside of the box is equal to the flow of stuff through the boundary of the box, plus whatever stuff is inserted into the box through some other mechanism." Of course, the fluid's own mass is a stuff, its momentum in the x-direction is a stuff, its temperature is a stuff in the form of thermal energy, etc. Pretty much anything which is conserved can be viewed as a "stuff".

Taking it by parts, the stuff is described by some concentration or density $c$; the stream by some velocity field $\vec v(\vec r, t)$. The part which says "a box flows downstream, the time rate of change of the stuff inside of the box" starts us off with:$$\frac{\partial c}{\partial t} + (\vec v \cdot \nabla) ~ c = \dots~~~.$$(If you've never seen this before: the box will at time $t + dt$ be at $\vec r + \vec v~dt;$ Taylor-expand $c(\vec r + \vec v ~dt, t + dt) - c(\vec r, t)$ to find that "convective derivative".)

The flow of $c$ is then described by a current density $\vec j$, but this only accumulates in the box with its negative divergence. Finally the "other mechanism" is just left as some term $q$ to be filled in later, hence$$\frac{\partial c}{\partial t} + (\vec v \cdot \nabla) ~ c = -\nabla \cdot \vec j + q~.$$A typical form for $\vec j$ indeed states that$$\vec j = c ~\vec v - D ~\nabla c ~.$$This says that the "stuff" is chiefly flowing downstream, but also has some effect where it doesn't flow downstream: namely it flows from a high concentration to a lower concentration by a "locally linear" flow (Fick's law, linear in the sense that twice the concentration gap locally = twice the flow). Plugging this form in gives the common form:$$\frac{\partial c}{\partial t} + \nabla \cdot (c ~\vec v) = D \nabla^2 c ~+~ (\nabla D) \cdot (\nabla c) ~+~ q~.$$

So now jump back to that expression for $\vec j$: can you see why $D = 0$ is the only appropriate choice when we're talking about the mass of the fluid itself?

Yes: it's because all of the information that we need is already in $\vec v$. The flow of the mass of the fluid itself is simply $\rho ~\vec v$, full stop, nothing else.

Put another way, If the fluid mass were flowing any other way, then $\vec v$ would be different to compensate. The fact, for example, that the fluid may be compressible is already there in the equation, hidden in the $\nabla \cdot (\rho ~ \vec v)$ term. The only thing that's not there is if fluid is coming into the stream from the outside world, but that's buried in $q$. There is no possibility for the fluid mass to self-interact outside of this mechanism without us defining a different set of particles as the "fluid" proper and following that fluid as our $\vec v$, in which case those particles' $\rho$ has the same phenomenon occurring.

$\endgroup$
  • 1
    $\begingroup$ All of this is right, but as Kyle pointed out in his answer, is only true for a single-component flow. When you have multi-component (say, hydrogen and air), there is a diffusion term in the species mass transport equations. Although strictly speaking, the total mass density equation can be viewed as redundant because everything is defined by the partial mass equations. So everything is good, but it's important to point out that multi-component mass conservation does have diffusion because you can actually have gradients in components, unlike the pure case where all "air" is indistinguishable. $\endgroup$ – tpg2114 Jul 19 '15 at 16:52
  • $\begingroup$ Awesome point, yes: this explanation does assume that you are very clear on what $\vec v$ is, and if a fluid contains components that are not isotropic you might have a $\vec v$ which does not correspond to the $\vec j / \rho$ of either of the components. $\endgroup$ – CR Drost Jul 19 '15 at 18:42
0
$\begingroup$

The basic point has already been mentioned, but I would like to give my version of the answer, and point out a subtlety. The basic equations of fluid dynamics are the conservation of mass, momentum, and energy $$ \frac{\partial \rho}{\partial t} - \vec{\nabla}\cdot\vec{\jmath}_\rho \\ \frac{\partial \pi_i}{\partial t} = - \nabla_j\Pi_{ij}, \\ \frac{\partial {\cal E}}{\partial t} = - \vec{\nabla} \cdot\vec{\jmath}^{\;\epsilon} . $$ The Navier-Stokes theory corresponds to taking account diffusive terms in the stress tensor $\Pi_{ij}$. These terms are related to bulk and shear viscosity, $\delta\Pi_{ij}=-\eta\sigma_{ij}-\zeta\delta_{ij}\langle \sigma\rangle$ with $$ \sigma_{ij} = \nabla_i u_j +\nabla_j u_i -\frac{2}{3}\delta_{ij} \langle\sigma\rangle \, , \;\;\; \langle\sigma\rangle =\vec{\nabla}\cdot\vec{u}\, . $$ Diffusive terms also appear in the energy current $\vec{\jmath}^{\;\epsilon}$. In addition to viscosity the energy current contains the thermal conductivity. $\delta\jmath_i^{\;\epsilon}=u_j\delta\Pi_{ij}-\kappa\nabla_i T$.

Why are there no diffusive terms in the mass current $\jmath_\rho$? The correct answer is indeed that $\vec{\jmath}_\rho=\rho\vec{u}$ is used to define the fluid velocity. Other definitions are possible. In the relativistic domain we frequently define the fluid velocity using the energy current (this is called the Landau frame), and then diffusive terms do appear in the mass current.

The subtlety: In fluid dynamics we also use that the momentum density is $\vec{\pi}=\rho\vec{u}$. Since we used $\vec{\jmath}_\rho$ to define $\vec{u}$ it is not obvious why this relation is not modified by diffusive terms.

The answer is of course that $\vec{\pi}=\vec{\jmath}_\rho$ is related to a symmetry. Multiply mass conservation by $\vec{r}$ and integrate over space (this argument is due to Landau). We get $$ \frac{\partial}{\partial t}\int d^3r\, \vec{r}\rho + \int d^3r\, \vec{\jmath}_\rho =0. $$ Since the first term is the center of mass, the second term must be the total momentum. The center of mass is one of the generators of the Schroedinger group, so the symmetry is Schroedinger symmetry. A modern version of this argument was given by Jensen (http://arxiv.org/abs/1411.7024). $\vec{\pi}=\vec{\jmath}_\rho$ is a Ward identity that can be derived using Newton-Cartan geometry.

$\endgroup$
0
$\begingroup$

Navier-Stokes equations describe fluid in an approximate way which neglects diffusion of molecules altogether. In ordinary case, the velocity field is supposed to be smooth.

"Diffusion of momentum" mentioned in connection to the term proportional to $\Delta \mathbf u$ is not really a diffusion in the molecular sense. It is rather kind of metaphor to describe the transfer of momentum due to viscous forces in the fluid. No actual diffusion is being described at this macroscopic level. In the Euler and Navier-Stokes equations, all motion of matter is purely convective, given by smooth velocity field. It is just that viscous forces (that do have connection to actual diffusion but not at this level of theory) lead to evolution of momentum distribution that reminds of ordinary diffusion.

$\endgroup$

protected by Qmechanic Jul 19 '15 at 18:57

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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