Page 147 - Vector Analysis

§4.7 The Stokes Theorem 143

as a (constant multiple of) measurement of the speed of rotation. The limit of the quantity
above, as r Ñ 0, is then a good measurement of the rotation speed of u at the point a about
the axis in the direction N.

We start from the case that N = e3 so that P be parallel to the x1x2-plane. With
u1, u2, u3 denoting respectively the first, the second and the third components of u, by the
change of variable ds = rdθ and the L’Hôspital rule (to obtain the second “=”) we find that

lim 1 ¿ u ¨ T ds
rÑ0 2πr Cr r

= lim 1 ż 2π [( + (r cos θ, r sin θ, ) cos θ ´ ( + (r cos θ, r sin θ, ) sin ] dθ
2πr u2 a 0) u1 a 0) θ
rÑ0 0

= 1 dˇ ż 2π [( + ) ´ ( )]
2π ˇ u2 a (r cos θ, r sin θ, 0) cos θ u1 a + (r cos θ, r sin θ, 0) sin θ dθ
0
dr ˇr=0

= 1 ż 2π [ B u2 (a) cos2 θ + B u2 (a) cos θ sin θ ´ B u1 (a) cos θ sin θ ´ B u1 (a) sin2 ]
2π θ dθ
0 B x1 B x2 B x1 B x2

1 [ B u2 (a) B u1 ] 1 2 B uj (a) . (4.21)
2 B x1 (a)
ÿ B xi

= ´ B x2 = 2 ε3ij

i,j=1

Now suppose the general case that N ‰ e3. Let pe3 = N and choose pe1 and pe2 so that
␣pe1, pe2, pe3( is an orthonormal basis following the right-hand rule (that is, pe1 ˆ pe2 = pe3).
Then the vector field u has two representations

u = u1e1 + u2e2 + u3e3 = v1pe1 + v2pe2 + v3pe3 . (4.22)

Let O = [pe1... pe2 ... pe3 ] and introduce a new Cartesian coordinate system y = OTx. Note
,

that y is the coordinate with coordinate axis parallel to the basis ␣pe1, pe2, pe3(. In this new

Cartesian coordinate system, (4.21) implies that

lim 1¿ u¨T ds = 1 [ B v2 (b) ´ B v1 ] ,
2πr Cr r 2 B y1 (b)
rÑ0 B y2

where b = OTa.

Now we transform the result above back to the original coordinate system (so that the

limit is in terms of derivatives of uj w.r.t. xi). Note that (4.22) implies that v = OTu so
that vj = pej ¨ u. Moreover, with ejk denoting the k-th component (w.r.t. the ordered basis
te1, e2, e3u) of pej; that is, pej = ej1e1 + ej2e2 + ej3e3, the chain rule provides that

B = e11 B + e12 B + e13 B and B = e21 B + e22 B + e23 B ;
B y1 B x1 B x2 B x3 B y2 B x1 B x2 B x3