1.3 Fundamentals of the Theory of Field
11
Equation (1.3.9) shows that the gradient of a function ϕ is a vector function.
The modulus of gradient is
|grad ϕ| = |∇ϕ| = |G| =
∂ϕ
∂ x
2
+
∂ϕ
∂ y
2
+
∂ϕ
∂z
2
(1.3.10)
Suppose that c is a constant, ϕ, ψ, f (ϕ) and f (r ) are all the scalar functions of
point M, r is any radius vector, r is the modulus of r, r 0 is the unit vector of r, then
the basic formulae of gradient operation are as follows
∇c = 0
(1.3.11)
∇(ϕ ± ψ) = ∇ϕ ± ∇ψ
(1.3.12)
∇(cϕ) = c∇ϕ
(1.3.13)
∇(ϕψ) = ψ∇ϕ + ϕ∇ψ
(1.3.14)
∇
ϕ
ψ
=
ψ∇ϕ − ϕ∇ψ
ψ 2
(1.3.15)
∇ f (ϕ) = f
(ϕ)∇ϕ
(1.3.16)
∇ f (r ) = f
(r )∇r = f
(r )
r
r
= f
(r )r 0
(1.3.17)
Now to prove Eqs. (1.3.12)–(1.3.17).
Proof
∇(ϕ ± ψ) =
∂
∂ x
(ϕ ± ψ)i +
∂
∂ y
(ϕ ± ψ) j +
∂
∂z
(ϕ ± ψ)k
=
∂ϕ
∂ x
i +
∂ϕ
∂ y
j +
∂ϕ
∂z
k
±
∂ψ
∂ x
i +
∂ψ
∂ y
j +
∂ψ
∂z
k
= ∇ϕ ± ∇ψ
∇(ϕψ) =
∂
∂ x
(ϕψ)i +
∂
∂ y
(ϕψ) j +
∂
∂z
(ϕψ)k
=
ψ
∂ϕ
∂ x
+ ϕ
∂ψ
∂ x
i +
ψ
∂ϕ
∂ y
+ ϕ
∂ψ
∂ y
j +
ψ
∂ϕ
∂z
+ ϕ
∂ψ
∂z
k = ψ∇ϕ + ϕ∇ψ
For the above expression, let ψ = c, then there is
∇(cϕ) = c∇ϕ
∇
ϕ
ψ
=
∂
∂ x
ϕ
ψ
i +
∂
∂ y
ϕ
ψ
j +
∂
∂z
ϕ
ψ
k