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(5.133)
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THERMODYNAMICS AND KINETIC THEORY
with the boundary condition that the fluid sticks to the sphere. Let us translate the coordinate system so that the sphere is at rest at the origin while the fluid at infinity flows with uniform constant velocity uo. The equations (5.133) remain invariant under the translation, whereas the boundary conditions become
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(5.134)
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Taking the divergence of both sides of the first equation of (5.133), we obtain "V 2 p = 0 (5.135) Thus the pressure, whatever it is, must be a linear superpoSItIOn of solid harmonics. A systematic way to proceed would be to write P as the most general superposition of solid harmonics and to determine the coefficient by requiring that (5.133) be satisfied. We take a short cut, however, and guess that P is, apart from an additive constant, a pure solid harmonic of order 1:
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(5.136)
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where Po and Pi are constants to be determined later. With this, the problem reduces to solving the inhomogeneous Laplace equation
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(5.138)
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A particular solution of (5.137) is
u = - Pi r 2 "V cos () = _ Pi 1 6 r2 6
3r~) 3
(5.139)
where denotes the unit vector along the z axis, which lies along uo. It is easily verified that (5.139) solves (5.137), if we note that l/r and z/r 3 are both solid harmonics. Thus,
2 = U1
:1 [- 3"V 2( :: )]
= P 1"V
(:3)
= P1"V c:s
() 2
(5.140)
The complete solution is obtained by adding an appropriate homogeneous solution to (5.139) to satisfy (5.138). By inspection we see that the complete solution is
(5.141)
TRANSPORT PHENOMENA
where we have set
(5.142)
to have 'V u = o. We now calculate the force acting on the sphere by the fluid. By definition the force per unit area acting on a surface whose normal point along the x) axis is - T) of (5.107). It follows that the force per unit area acting on a surface element of the sphere is
f = (
-T1 + -T2 + -T3 r r r
-f P
(5.143)
where f is the unit vector in the radial direction and total force experienced by the sphere is
F'=jdSf
P is given by (5.118). The
(5.144)
where dS is a surface element of the sphere and the integral extends over the entire surface of the sphere. Thus it is sufficient to calculate f for r = a. The vector f P has the components
A.... 1 1 au) ( r P).= -x.p.. = -xJ [ i)p-II. ( -ax. I r J JI r JI r
au i + - )] ax.
Xi }.t a a -P - - [ -(xu) - u + x - u ] r r ax I J J I J ax. I J
Hence
(5.145)
where P is given by (5.136) and (5.142), and u is given by (5.141). Since u = 0 when r = a, we only need to consider the first and the last terms in the bracket. The first term is zero at r = a by a straightforward calculation. At r = a the second term is found to be
1 -[(r 'V)u] r~a r
(au)
- - - -fu - r-a
3 Uo
cos ()
(5.146)
When this is substituted into (5.145), the second term exactly cancels the dipole part of fP, and we obtain
The constant Po is unknown, but it does not contribute to the force on the sphere. From (5.144) we obtain
(5.147)
which is Stokes' law.
THERMODYNAMICS AND KINETIC THEORY
The validity of (5.141) depends on the smallness of the material derivative of u as compared to /L'V 2U. Both these quantities can be computed from (5.141). It is then clear that we must require
(5.148)
Thus Stokes' law holds only for small velocities and small radii of the sphere. A more elaborate treatment shows that a more accurate formula for F' is
= 67T/LaU o
+ 8" -/L- + ...
3 puoa
(5.149)
The pure number puoa//L is called the Reynolds number. When the Reynolds number becomes large, turbulence sets in and streamline motion completely breaks down.
PROBLEMS
5.1 Make order-of-magnitude estimates for the mean free path and the collision time for
(a) Hz molecules in a hydrogen gas in standard condition (diameter of Hz (b) protons in a plasma (gas of totally ionized Hz) at T
= 2.9 A); 10 5 K, n = 1015
protons/cm3 , 0 = 'lTrz, where r = e 2 /kT; (c) protons in a plasma at the same density as (b) but at T = 10 7 K, where thermonuclear reactions occur; (d) protons in the sun's corona, which is a plasma at T = 10 6 K, n = 10 6 protons/cc; (e) slow neutrons of energy 0.5 MeV in 238U (0"" 'lTr z , r"" 10- 13 cm).