36
1 Preliminaries
Proof Because i, j and k are all the unit vectors and perpendicular each other, there
is
i · i = j · j = k · k = 1, i · j = j · i = i · k = k · i = j · k = k · j = 0
Therefore
(∇ × a) · (∇ × a) =
∂a z
∂ y
−
∂a y
∂z
i +
∂a x
∂z
−
∂a z
∂ x
j +
∂a y
∂ x
−
∂a x
∂ y
k
·
∂a z
∂ y
−
∂a y
∂z
i +
∂a x
∂z
−
∂a z
∂ x
j +
∂a y
∂ x
−
∂a x
∂ y
k
=
∂a z
∂ y
−
∂a y
∂z
i ·
∂a z
∂ y
−
∂a y
∂z
i +
∂a x
∂z
−
∂a z
∂ x
j ·
∂a x
∂z
−
∂a z
∂ x
j
+
∂a y
∂ x
−
∂a x
∂ y
k ·
∂a y
∂ x
−
∂a x
∂ y
k =
∂a z
∂ y
−
∂a y
∂z
2
+
∂a x
∂z
−
∂a z
∂ x
2
+
∂a y
∂ x
−
∂a x
∂ y
2
= (rot x a) 2 + (rot y a) 2 + (rot z a) 2 = |∇ × a| 2 = (∇ × a) 2
Quod erat demonstrandum.
Extract square root of the two ends of Eqs. (1.3.91), (1.3.104) can be obtained.
1.3.5 The Stokes Theorem
Substituting Eq. (1.3.89) into Eq. (1.3.85), the Stokes formula expressed in the
vector form can be obtained
L
a · dL =
¨
S
rota · dS =
¨
S
rota · ndS =
¨
S
∇ × adS =
¨
S
n · ∇ × adS
(1.3.105)
The Stokes formula is also called the Stokes theorem. The Stokes theorem
revealed the transformation relations of the vector field line integral and surface integral. Thus it can be seen that the Green formula introduced before is just a special
case of the Stokes theorem on a planar domain. Of course the Stokes theorem can
also be directly proved, the proof method is similar to the one of the Gauss theorem.
A proof is given below.
Proof Dividing the surface S into n elements of surface S 1 , S 2 , …, S n , the
elements of perimeter which surround the elements of surface namely the closed
curves are L 1 , L 2 , …, L n . Taking the kth surface S k and the perimeter L k ,
from the definition of rotation, there exists the following relationship