Appendix: All Solutions to the Problems
759
=
S
[n · a∇ · b + (n × a) · ∇ × b]dS
Transposing a and b of the above expression, then doing the subtraction of the two
expressions, and using the parallelepipedal product formula of trivector, to obtain
˚
V
(a · b − b · a)dV =
S
(a∇ · b + a × ∇ × b − b∇ · a − b × ∇ × a) · ndS
=
S
[n · a∇ · b − n · b∇ · a + (n × a) · ∇ × b − (n × b) · ∇ × a]dS
Quod erat demonstrandum.
1.19 Given a second order tensor T i j is
T i j =
⎡
⎣
T 11 T 12 T 13
T 21 T 22 T 23
T 31 T 32 T 33
⎤
⎦ =
⎡
⎣
1 1 0
1 2 2
0 2 3
⎤
⎦
Find (1) T ii ; (2) T i j T i j ; (3) T i j T ji .
Solution
(1) Expanding T ii , to get T ii = T 11 + T 22 + T 33 = 1 + 2 + 3 = 6;
(2) Expanding T i j T i j , it can be obtained
T i j T i j = T 1 j T 1 j + T 2 j T 2 j + T 3 j T 3 j
= T 11 T 11 + T 12 T 12 + T 13 T 13 + T 21 T 21 + T 22 T 22 + T 23 T 23 + T 31 T 31
+ T 32 T 32 + T 33 T 33 = 1 + 1 + 0 + 1 + 4 + 4 + 0 + 4 + 9 = 24
(3) Because T i j is the second-order symmetric tensor, T i j T ji = T i j T i j = 24.
1.20 Suppose that the second order tensor is
T i j =
⎡
⎣
2 4 −2
0 3 2
4 2 5
⎤
⎦
Decompose T i j into the sum of a symmetric tensor and an antisymmetric tensor.
Solution The conjugate tensor of T i j is
T
c
i j =
⎡
⎣
2 0 4
4 3 2
−2 2 5
⎤
⎦