hohmov的博客

电磁场原理题与答案 1010101.1 根据算符∇ 的微分性与矢量性,推导下列公式： ∇(A⋅ B) = B× (∇ × A) + (B ⋅ ∇)A + A× (∇ × B) + (A⋅ ∇)B A A A (A )A 2 × (∇ × ) = 1 ∇ 2 - ⋅ ∇ 【解】记∇(A⋅ B) = ∇ (A⋅ B) + ∇ (A⋅ B) A B , A ∇ 是作用于A 的算符, B ∇ 是作用于B 的算符,利用a × (b × c) = b(c ⋅ a) - c(a ⋅ b) ,有 ∇ (A⋅ B) = B× (∇ × A) + (B ⋅ ∇)A A ∇ (A⋅ B) = A× (∇ × B) + (A⋅ ∇)B B ∇(A⋅ B) = B× (∇ × A) + (B ⋅ ∇)A + A× (∇ × B) + (A⋅ ∇)B 在上式中令B = A,即得 A A A (A )A