By Jens Clever
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The chain rule of diﬀerential calculus, Eq. 54), now requires the replacements B(A · C) = (C · A)B → B(∇ · C) + (C · ∇)B −C(A · B) = −(B · A)C → −C(∇ · B) − (B · ∇)C, with two terms appearing for each term of the BAC rule. Thus Eq. 60) obtains. We shall refer to this procedure as the operator form of the BAC rule. 1 Derive the following identities: (a) ∇ · (A × B) = B · (∇ × A) − A · (∇ × B); (b) ∇ × (∇ × A) = ∇(∇ · A) − ∇2 A; (c) ∇(A · B) = (B · ∇)A + (A · ∇)B + B × (∇ × A) + A × (∇ × B). 2 If A(r) is irrotational, show that A × r is solenoidal.
Some of the most interesting vector ﬁelds in physics satisfy the property that the results of such vector diﬀerential operations vanish everywhere in space. Such special ﬁelds are given special names: If ∇ · V(r) = 0, V is said to be solenoidal (or divergence-free). If ∇ × V(r) = 0, V is said to be irrotational. 3 Show that the gradient of any scalar ﬁeld Φ(r) is irrotational and that the curl of any vector ﬁeld V(r) is solenoidal. i j k ∇ × (∇Φ) = ∂ ∂x ∂ ∂x ∂ ∂y ∂ ∂y ∂ ∂z ∂ ∂z Φ(x, y, z) = 0 because there are two identical rows in the determinant.
4 Operator identity Gauss’s theorem, Eq. 85), can be applied to the vector ﬁeld V(r) = ei Φ(r) to give dσi Φ(r) = Ω S dτ ∂ Φ(r). ∂xi Since this is true for any i, it follows that dσΦ(r) = S = dσi Φ(r) ei S i Ω dτ∇Φ(r). 93) for operations on any ﬁeld in space. It gives rise to other integral theorems such as the following: dσ × A(r) = S dσ · (u∇v − v∇u) = S = Ω Ω Ω dτ∇ × A(r), dτ∇ · (u∇v − v∇u) dτ(u∇2 v − v∇2 u). The last identity is called Green’s theorem. 2 Show that dσ · r = r2 dτ 1 . 94) r 1 = 2, 2 r r a result that can be demonstrated readily in rectangular coordinates.