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.
-
Why have we assumed that the particle has the value of $u_x$ at the position of its last collision.
-
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.
-
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.
Best Answer
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 $\Delta p$