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Noting that the second term of (5.67) does not contribute to this integral, we obtain
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It is clear from (5.68) that K is to be identified as the coefficient of thermal conductivity. It is also clear that Iql is a small quantity of the first order, being of the order of AIL. For the pressure tensor Pij , only the second term of (5.67) contributes:
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u;)( V - uJ(J<Ol + g) = tJi}P + Pi) j
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To evaluate this, note that Pi) is a symmetric tensor of zero trace (i.e.,
L Pi; =
and it depends linearly on the symmetric tensor Ai}' Therefore Pi) must have the form
A ij -
~ tJij'V u)
= m\l U
where m\l u is none other than the trace of A i/
3 au L -'
aX i
and fL is a constant. It remains to calculate fL. For this purpose it suffices to calculate any component of P/j , e.g., P{z. From (5.71) we have
rm 4 P{z = - -o-AkJ d3UUPZ(Uk~ - }tJ kP2)f<Ol
With this we have
The second term is of the order of AIL. The coefficient fL turns out to be the coefficient of viscosity, as we show shortly.
- 2QQ C
Fig. 5.1 Ratio of thermal conductivity to the product of viscosity and specific heat for different dilute gases.
A comparison of (5.74) with (5.69) shows that
-P. =
= 2. c V 3
Since the unknown collision time 'T drops out in this relation, we might expect (5.76) to be of quantitative significance. A plot of some experimental data for different dilute gases in Fig. 5.1 shows that it is indeed so. Let uS put, with (5.6),
'T :::::
na 1
where a is the molecular diameter. Then we find that
To show that (5.74) is the coefficient of viscosity, We independently calculate the coefficient of viscosity using its experimental definition. Consider a gas of uniform and constant density and temperature, with an average velocity given by
= uy =
+ By
where A and B are constants. The gas may be thought of as being composed of different layers sliding over each other, as shown in Fig. 5.2. Draw any plane perpendicular to the y axis, as shown by the dotted line in Fig. 5.2. Let F' be the frictional force experienced by the gas above this plane, per unit area of the plane. Then the coefficient of viscosity Jl is eJ5perimentally defined by the relation
F' = - p . -
Ju x Jy
The gas above the plane experiences a frictional force by virtue of the fact that it
--1------------+ x
Fig. 5.2 Horizontal flow of a gas with average velocity increasing linearly with height.
suffers a net loss of "x component of momentum" to the gas below. Thus
F' == net amount of "x component of momentum" transported per sec across unit area in the y direction
The quantity being transported is m ( vx transport is n (vy - uy)' Hence we have
U x),
whereas the flux effective in the
F' = mn((vx - uJ(vy - uy )) = m 4
f d v (vx 3
uJ(vy - uy)(/{O) + g)
We easily see that the term j<0) does not contribute to the integral in (5.82). The !lrst correction g may be obtained directly from the approximate Boltzmann transport equation g V. '9/(0) =
g= - Ovy(vx - uJB/{O) = - O~Ux
where U == v - u. Thus
au x
= - __ - x Tm
ay ()
f d u U U j<0)
3 2 2
A comparison between this and (5.80) yields
f d U U}U/j< )
which is identical with (5.74). From the nature of this derivation it is possible to understand physically why Jl has the order of magnitude given by (5.78). Across the imaginary plane mentioned previously, a net transport of momentum exists, because molecules constantly cross this plane in both directions. The flux is the same in both directions, being of the order of nVkT1m. On the average, however, those that cross from above to below carry more "x component of momentum" than the opposite ones, because the average velocity U x is greater above than below. Since
most molecules that Cross the plane from above originated within a mean free path A above the plane, their u x is in excess of the local u x below the plane by the amount A(Jux/Jy). Hence the net amount of "x component of momentum" transported per second from above to below, per unit area of the plane, is
AnmV kT Ju x m Jy
vmkT Ju x a2 Jy
(5.87) It is interesting to note that according to (5.87) Jl is independent of the density for a given temperature. When Maxwell first derived this fact, he was so surprised that he put it to experimental test by observing the rate of damping of a pendulum suspended in gases of different densities. To his satisfaction, it was verified. According to (5.87) the coefficient of viscosity increases as the molecular diameter decreases, everything else being constant. This is physically easy to understand because the mean free path A increases with decreasing molecular diameter. For a given gradient Jux/Jy, the momentum transported across any plane normal to the y axis obviously increases as A increases. When A becomes so large that it is comparable to the size of the container of the gas, the whole method adopted here breaks down, and the coefficient of viscosity ceases to be a meaningful concept. As a topic related to the concept of viscosity we consider the boundary condition for a gas flowing past a wall. A gas, unlike a liquid, does not stick to the wall of its container. Rather, it slips by with an average velocity uo. To determine uo, it is necessary to know how the gas molecules interact with the wall. We make the simplifying assumption that a fraction 1 - a of the molecules striking the wall is reflected elastically while the remaining fraction a is absorbed by the wall, only to return to the gas later with thermal velocity. The number a is called the coefficient of accommodation. Suppose the wall is the xy plane, as shown in Fig. 5.3. Then the downward flux of particles is given by