# Viscosity and mean free path

I have troubles understanding how to derive the formula for viscosity in terms of the mean free path $$\eta\sim \rho \lambda \bar v$$ where $\bar v$ is the average molecular velocity of the gas, $\lambda$ is the mean free path and $\rho$ is the density. I was following the derivation on the wikipedia article as well as a very similar one in page 8 here.

The derivation considers a moving slab in a liquid/gas, which induces a gradient of the velocity $u_x$ between the moving slab and a stationary one. The first relation I have problems with is $\langle u_x \rangle=\frac{1}{2}\lambda du_x/dy$, where $\langle u_x \rangle$ is the average velocity in the $x$ direction (parallel with the direction of the moving slab) of particle crossing a control surface and $\lambda$ is the mean free path (the wikipedia article strangely does not have the factor of 2). This is explained by (see the second link) assuming that there are particles crossing the surface from all distances from $0$ to $\lambda$ (which makes sense). Then it is explained that the velocity changes linearly with distance (which makes sense, if the particles are to hit the control surface at the same time, particles further away must move faster). This would give $\langle u_x\rangle\sim\frac{1}{\lambda}\int_0^\lambda s ds=\frac{1}{2}\lambda$ (the first term is just the distribution of particles from 0 to $\lambda$ and the integrand is the velocity dependence).

However, the overall proprtionality factor is $du_x/dy$ and I have no idea how to get that (I also do not understand the remaining parts of the derivation either so help on the full derivation would be very much appreciated)

• What is $s$ in your integral? – Han-Kwang Nienhuys Jun 11 '16 at 18:07
• It is the distance of the particle from the control surface. The assumption is that all particles crossing the surface come from some distance $s$ from the surface and the distribution is uniform up to distance $\lambda$. Since the average velocity depends linearly on the distance (another assumption) the average velocity should be this integral. At least this is how I understood it but I might have missed something since I dont know how to get their overall constant right anyway. – Jonathan Lindgren Jun 11 '16 at 18:20
• I'd suggest that you edit your question to clarify what you meant. In any case, that equation seems to have a dimension mismatch (velocity equals length). – Han-Kwang Nienhuys Jun 11 '16 at 18:23
• viscosity = force per square meter / gradient of velocity, so $du_x/dy$ enters the definition of the viscosity. The other thing which tricky is that the viscosity force is due to vertical transfer of horizontal momentum. – Kostas Oct 5 '19 at 20:41

The gas as a whole is moving in the $x$ direction everywhere, but at different heights it is moving different speeds, so we have $u_x(y)$.

Even though the gas as a whole has velocity in the $x$ direction only, individual molecules have motion in the $y$ direction (and $z$-direction, but that doesn't matter here).

So let's suppose that at $u_x(y_0) = 2 m/s$. And let's say that $u_x$ is an increasing function of $y$. Then the molecules from higher values of $y$ will sometimes drift down past $y_0$, and they will be moving faster than $2 m/s$ in the x-direction, on average. This is because they come from higher $y$ values, where $u_x$ is higher on average.

Will they be going $2.1 m/s$ or $2.01 m/s$ in the $x$-direction on average? In your language, is $\langle u_x \rangle$ equal to $.1 m/s$ or $.01 m/s$? That depends. If one particular molecule drifts down to $y = y_0$ from $y = y_0 + \Delta y$, then on average its $x$ velocity should be $u_x(y_0 + \Delta y) \approx u_x(y_0) + \frac{\partial u_x}{\partial y}(y_0) \Delta y$.

The document you were reading was simply approximating $\Delta y \sim \lambda$, which is not exact. If you do it carefully, I believe you should actually get a factor of $1/3$, not $1/2$ as stated in the document you linked.

Then it is explained that the velocity changes linearly with distance (which makes sense, if the particles are to hit the control surface at the same time, particles further away must move faster).

This is not what it means at all. It says that the excess $x$ velocity that a molecule has, on average, above $u_x(y_0)$, increases linearly in $y$. This has nothing to do with particles' motion in the $y$ direction.

• Thank you, that was a much clearer explanation (my confusion was partly from that I thought the control surface was in the y-z plane) – Jonathan Lindgren Jun 11 '16 at 20:30
• I have one question though, how come there can be momentum flow between the layers? I thought the whole system is stationary, if there is momentum flow shouldnt some layers start accelerating? – Jonathan Lindgren Jun 11 '16 at 20:34
• Acceleration will depend on the divergence of the flux of momentum. If there is a flux of momentum that is constant, then the same amount of momentum flows in and out of an small volume of fluid and there is no acceleration. Eventually, this momentum will go into a wall or barrier of some kind at the edge of the pipe / whatever the fluid is flowing through. – Mark Eichenlaub Jun 11 '16 at 23:44