# Derivation of viscosity using basic kinetic theory?

This question has been asked in part before in the question Kinetic theory derivation of viscosity of a gas although the given accepted answer does not give the required detail for the part of the question I am most interested in.

I will first give a derivation of viscosity as I have learned it (this is the same method given in 'Concepts in Thermal Physics' by Blundell, Blundell and a numerous other sources).

Derivation of viscosity

Assuming we have a gradient of the $x$ component of the mean velocity, $u_x$, in the $z$ direction, such that: $$\frac{\partial u_x}{\partial z}\ne 0$$ The momentum flux (of the $x$ momentum in the $z$ direction) is given by: $$\Pi_{xy}=-\eta \frac{\partial u_x}{\partial z}$$ Where $\eta$ is by definition the viscosity of the gas. The particle flux through the surface $z=z_0$ (that is number of particles per unit area per unit time) is given by: $$\mathrm{d}\Omega=nv_z f(v_z)\mathrm{d}v_z$$ We can assume these particles have travailed on average a distance $\lambda_{mfp}$ before crossing the plane and their last collision meaning they started at a position $z=z-\lambda \cos(\theta)$ and thus carry with them an excess in momentum of: \begin{align}\Delta p&=m(u_x(z-\Delta z)-u_x(z)) \\&\approx m\lambda \cos(\theta) \frac{\partial u_x}{\partial z}\end{align} Meaning the momentum flux through the surface is given by: \begin{align}\Pi_{xy}&= \int \Delta p \mathrm{d} \Omega\\ &=\int m\lambda \cos(\theta) \frac{\partial u_x}{\partial z} \mathrm{d} \Omega\end{align} ...

I have several related problems with this derivation, none of which are explained clearly in the above linked question.

1. Why have we assumed that the particle has the value of $u_x$ at the position of its last collision.

2. Why do we assume the particle has travailed a distance $\lambda_{fmp}$ between its last collision and crossing the boundary. Particles that do not collide at the boundary will also carry a momentum through the boundary and therefore contribute to the momentum flux.

3. Why are we using the 'excess' momentum $\Delta p$ the particle carries across the boundary rather then the total momentum $p$. If a particle crosses the boundary it contributes a momentum $p$ to the total flux across the boundary not $\Delta p$.

Please can someone give an (intuitive) explanation of these points.

• I think the initial position is $z=z_0-\lambda\cos(\theta)$, right? I'm not sure if I understood your questions, but: 1- it's just a label, 2-how is it possible for a particle to carry momentum through the boundary if it doesn't reach the boundary in first place? 3- you're wrong, if a particle crosses the boundary it contributes ITS momentum to the total flux, and its momentum on the boundary is given by the excess momentum, i.e., its momentum on the boundary minus its initial momentum. I could elaborate more as an answer, but I have to be sure what you're not understanding. Dec 17 '15 at 14:57

1) There are more carefully derived values assuming a distribution of particles (I do not have the reference here), but instead of considering all particles as distributed between different initial positions and velocities, lets us take them all equal to the same "typical" particle: the one with the typical velocity, let us say the median velocity $u_x$; the mean instead of the median could have also been used. They differ only in a factor, which is not really important (see answer (2)).

2) The typical particle has moved the typical distance before crossing the boundary. It is about $\lambda_{fmp}$, or, perhaps better, $\lambda \cos{\theta}$. But this is a very crude approximation. I have seen derivations where it is considered to be $1/2 \lambda$ (assuming that molecules hitting the unit area come from all distances between 0 and λ; equally distributed), or, calculated in more detail, $1/3 \lambda$ (F. Reif, Statistical and Thermal Physics (McGraw-Hill), Ch. 12: http://physics.bu.edu/~redner/542/refs/reif-chap12.pdf.). Even this last factor is not very accurate because, as the author states: "Our calculation has been very simplified and careless about the exact way various quantities ought to be averaged. Hence the factor 1/3 is not to be trusted too much".

3) You should not use the total momentum, but only the difference. You can see this intuitively by noticing that from the other side of the surface there will be the same number of particles crossing back with momentum $p-\Delta p$. Thus the net transfer is only $\Delta p$

Let's first take a broad perspective; what is this question asking? It's asking "suppose that the fluid has a velocity gradient across it, $u_x(z)$ not constant, how does that velocity gradient create a force on the plane of the fluid beneath?" The fact that $u_x(z)$ is not constant is very important, as is the fact that it varies only linearly over a distance of several mean free paths.

First, why do we use the velocity at some distance away from the boundary? And it's because we're approximating this fluid with very short-range interactions (e.g. the billiard-ball model) where the momentum-vectors of particles (including each component, hence $u_x$) are not changing substantially between interactions. We're saying: Collisions change momenta! But other than that we're going to pretend that the momentum doesn't change! So when a particle comes through this surface, we "backtrack" to the position it had before.

Second, yes that's sort of a mathematically incorrect statement, but it turns out to be correct. So, during a time interval $dt$, some amount $dn$ of particles come through this surface. Now suppose we trace them back, we find that if the situation is homogeneous enough that the probability of interaction between $\ell$ and $\ell + d\ell$ is the same as the probability of interaction between that and $\ell + 2d\ell,$ then we should have an exponential random variable $dp_{\text{interaction at distance }\ell} = \lambda^{-1} ~\exp(-\ell/\lambda)~d\ell.$ Notice that $\int \ell dp(\ell) = \lambda$ is the mean free path, it turns out that you can develop any probability distribution here.

And that's because if $\lambda$ is small enough that we can linearize $u_x(z - \ell) \approx u_x(z) - u_x'(z)~\ell$ for most of this probability-density, then the typical $u_x$ carried through the barrier is going to be $int (u_x(z) - u_x'(z)~\ell) dp(\ell) = u_x(z) - u_x'(z)~\lambda.$ The first term follows because the probability-density is normalized and integrates to 1; the second term follows because integrating $\ell$ over the probability density gives the mean free path.

So it follows from the fact that we're linearizing $u_x(z)$ for $0 > z > -10\lambda$, say, so that the only assumption about $p(\ell)$ is that it decays more rapidly than the velocity does for $\ell \gg \lambda.$

Third, because we want to know how much extra momentum we're absorbing in a certain time! Let me just totally cop-out of this one thusly: There is a reference frame where the $u_x$ at this point on the boundary is zero. We can use this reference frame to calculate the force by seeing how its velocity is inclined to change over a unit of time! Here the velocity is changing due to convective transport of particles with some x-velocity-component $u_x(z-\ell) \ne u_z(z)$ into our control volume, lowering/raising the total $x$-velocity in that space.

# Separate, quicker derivation

Suppose an object is moving at some speed $v$ through a fluid; I like to think of slowly cutting through the fluid with a knife blade. We can imagine that until it starts to "collect" particles on its front surface, the particles hit its sides with some characteristic small time scale $\tau,$ and that each time a particle hits it, it will give that particle some forward velocity proportional to $v$. Therefore in a time $dt$ there will be some number $dt/\tau$ particles which have been given some momentum $\alpha v$ corresponding to a reaction force due to conservation of momentum looking like $-\alpha v / \tau.$ What this means is that the force is proportional to the velocity of the particle, "linear drag", characteristic of a normal viscous interaction.

My issue with this bit is, why don't we just find the total momentum rather than the amount by which it exceeds the momentum of that layer?

Your question is really interesting, and you need to go down to very basics. Please note that in the Kinetic theory it's assumed that it's all just collisions;

Except during collisions, the interactions among molecules are negligible. (That is, they exert no forces on one another.) This implies:

1. Relativistic effects are negligible.
2. Quantum-mechanical effects are negligible. This means that the inter-particle distance is much larger than the thermal de Broglie wavelength and the molecules are treated as classical objects.
3. Because of the above two, their dynamics can be treated classically. This means, the equations of motion of the molecules are time-reversible.

But if you just think viscosity, you will immediately note, that it's much more than just a collision. Just think spoon in a honey or syrup, and how it follows the spoon when you lift it out from the pot. This attraction is Quantum-mechanical, it's very similar like the magnets pulling each other; and these forces are not negligible! So your basic problem is that you actually can't truly explain viscosity through kinetic theory. You can have usable results so far your fluid is fully turbulent, but that's simply because the Turbulence is collisions, und thus it just fits to the the Kinetic theory. -The math works, but it doesn't describe the true physics behind the phenomenon.

To conclude this, though Kinetic theory predicts quite accurately the viscous-like behaviour of gases, the idea behind this thery is not quite correct, and thus the answers to your questions;

1. Why have we assumed that the particle has the value of $u_x$ at the position of its last collision.
2. Why do we assume the particle has travailed a distance $λ_{fmp}$ between its last collision and crossing the boundary.
3. Why are we using the 'excess' momentum $Δp$ the particle carries across the boundary rather then the total momentum $p$.

are; Just because these simplification works with reasonable accuracy in typical real life range. But if you study over wide range, you note easily how the simplifications stops working; ie. the Pressure does have an influence in the viscosity. The Kinetic theory is only a theoretical, simplified model which truly works only with a theoretical gas. It's ability to explain real phenomenons (like viscosity), is thus quite limited.

I hope this help's you forward. Here's some thoughts about things; Would a solution to the Navier-Stokes Millennium Problem have any practical consequences?