7.6 Solutions to Problems
193
(D) =
1
2
α
D0
pqr ((, ω 1 , ω 2 ))r s
+
1
2 ¯
h 2
r s
ω 1
g,m,n
ρ
(0)
m
g| ˆ
μ p |nn| ˆ
μ r |mm| ˆ
μ q |g
(( − ω ng )
+
g,m,n
ρ
(0)
g
g| ˆ
μ r |nn| ˆ
μ p |mm| ˆ
μ q |g
(( − ω mn )
−
g,m,n
ρ
(0)
m
g| ˆ
μ p |nn| ˆ
μ q |mm| ˆ
μ r |g
(( − ω ng )
−
g,m,n
ρ
(0)
g
g| ˆ
μ q |nn| ˆ
μ p |mm| ˆ
μ r |g
(( − ω mn )
=
1
2
α
D0
pqr ((, ω 1 , ω 2 ))r s = (C)
where all the terms in the square bracket cancel each other by properly replacing the
suffixes, g, m, n. We could also prove (B) = (E) in the same way as above.
Therefore, we derive Eq. (7.119) for F=D1 from Eq. (7.126) as
α
D1
pqrs
((, ω 1 , ω 2 ) = (A) − (B) − (C) − (D) + (E) = (A) − 2(C)
= α
D1
pqrs ((, ω 1 , ω 2 ) − α
D0
pqr ((, ω 1 , ω 2 ))r s .
We can also confirm Eq. (7.119) for F=D2 and Q in the same way.
7.6.4 Levi-Civita Tensor
[Problem 7.4] Derive the following formulas (i)-(vi) using the Levi-Civita tensor.
(A, B, C, D refer to vectors, and φ to a scalar.)
(i) (A × B) · (C × D) = (A · C) (B · D) − (A · D) (B · C)
(ii) ∇ × (∇ × A) = ∇ (∇ · A) − ∇
2 A
(iii) ∇ · (∇ × A) = 0
(iv) ∇ × (∇φ) = 0
(v) ∇ · (A × B) = B · (∇ × A) − A · (∇ × B)
(vi) ∇ × (A × B) = A (∇ · B) + (B · ∇) A − B (∇ · A) − (A · ∇) B