Divergence and Curl of a Vector Field
Recall that a vector field F on R3 can be expressed as F(x, y, z) =< F1 (x, y, z), F2 (x, y, z), F3 (x, y, z) >, where the component functions are scalar functions from R3 to R. For the sake of brevity, we will often use x to denote the vector < x, y, z > and write F(x) =< F1 (x), F2 (x), F3 (x) > or even drop the variables altogether and write F =< F1 , F2 , F3 >. Let F =< F1 , F2 , F3 > : R3 → R3 be a differentiable vector field. The divergence of F is the scalar function div F = and the curl of F is the vector field curl F = ∂F3 ∂F2 − ∂y ∂z i+ ∂F1 ∂F3 − ∂z ∂x j+ ∂F2 ∂F1 − ∂x ∂y k ∂F1 ∂F2 ∂F3 + + . ∂x ∂y ∂z

Note that the divergence can be defined for any vector field F : Rn → Rn , while the curl is only defined on vector fields in 3-space. There is a nice second way to deal with these functions. Define the operator , called the “del” operator, by = ∂ ∂ ∂ i+ j+ k= ∂x ∂y ∂z ∂ ∂ ∂ , , ∂x ∂y ∂z

The action of the del operator on a scalar function f : R3 → R is just the gradient: f= ∂f ∂f ∂f i+ j+ k= ∂x ∂y ∂z ∂f ∂f ∂f , , ∂x ∂y ∂z = grad(f )

The divergence and the curl of a vector field are given by the two vector products that we have: the curl is the dot product and the curl is the cross product. div(F) = ·F curl(F ) = × F.

Theorem 1 Let f be a differentiable scalar function and F be a differentiable vector field on R3 . 1. curl(grad f ) = 0 or 2. div(curl F) = 0 or × ( f ) = 0. ·( × F) = 0.

The Laplacian operator on a scalar function is defined to be f = div(grad f ) = 1 · f=
2

f.

The Laplacian of a vector field F is F= F1 , F2 , F3 .

Geometrically, the curl F is a measurment of the tendency of a fluid to swirl around an axis. A field F with curl F = 0 at a point P is called irrotational at P . The divergence of a vector field measures the incompressibility of a vector field and we can often interpret the divergence of a vector field as the total outflow per unit area. The Laplacian operator helps to describe the diffusion process that is going on in that particular vector field, such as the concentration of a liquid changing as a chemical is dissolved in it or the heat of a solid diffusing from warmer to cooler regions. Theorem 2 The operators described above are linear operators: grad(f + g) div(F + G) curl(F + G)( (f + g) (F + G) = grad f + grad g, = div F + div G, = curl F + curl G, = (f ) + (g), = (F) + (G), grad(cf ) = c grad f div(cF0 = c div F curl(cF) = c curl F (cf ) = c (f ) (cF) = c (F)

There are some very intriquing properties of the operators. We see a few in the following theorems. The first theorem we have already covered. Theorem 3 grad(f g)(x) = g(x) grad(f )(x) + f (x) grad(g)(x) g(x) grad(f )(x) − g(x) grad(f )(x) f grad( )(x) = g g(x)2 Given two vector fields F and G, we will let the expression (F · field whose components are (F · Theorem 4 grad(F · G) div(f F) div(F × G) curl(f F) curl(F × G) (f g) (F · G) = = = = = = = (F · )(G) + (G · )(F) + F × curl G + G × curl F f div(F) + F · grad(f ) G · curl(F) − F · curl(G) f curl(F) + grad(f ) × F F div(G) − G div(F) + (G · )(F) − (F · )(G) g f + f g + 2 grad(f ) · grad(g) G · F + F G + 2 grad F · grad G 2 (3) (4) (5) (6) (7) (8) (9) (1) if g(x) = 0 (2)

)(G) denote the vecor

)(G) =< F · grad G1 , F · grad G2 , F · grad G3 > .

where
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grad F · grad G = i=1 grad(Fi ) · grad(Gi ).

How do these differential operators behave in different curvilinear coordinate systems? I will just present the results here. It is not particularly hard, but time-consuming to work throught the details. Let’s consider only cylindrical and spherical coordinates. In cylindrical coordinate systems, with F = Fr er + Fθ eθ + Fz ez we have grad f = div(F) = 1 ∂f ∂f ∂f er + eθ + ez ∂r r ∂θ ∂z 1 ∂(rFr ) 1 ∂Fθ ∂Fz + + r ∂r r ∂θ ∂z er reθ ez 1 ∂ ∂ ∂ r ∂r ∂θ ∂z Fr rFθ Fz 1 ∂ ∂f ∂ 1 ∂f ∂ r + + r ∂r ∂r ∂θ r ∂θ ∂z 2 2 2 ∂ f 1 ∂f 1∂ f ∂ f + + 2 2 + 2 2 ∂r r ∂r r ∂θ ∂z

curl(F) =

f = =

r

∂f ∂z

For spherical coordinates, the expressions for the different differential operators are (with F = Fρ eρ + Fθ eθ + Fφ eφ ): 1 ∂f 1 ∂f ∂f eρ + eθ + eφ ∂ρ ρ sin(φ) ∂θ ρ ∂φ 1 ∂ 2 ∂ ∂ div(F) = 2 (ρ sin(φ)Fρ ) + (ρFθ ) + (ρ sin(φ)Fφ ) ρ sin(φ) ∂ρ ∂θ ∂φ 1 ∂ 1 ∂ 1 ∂ = 2 (ρ2 Fρ ) + (Fθ ) + (sin(φ)Fφ ) ρ ∂ρ ρ sin φ ∂θ ρ sin φ ∂φ eρ ρeθ ρ sin φeφ ∂ ∂ ∂ 1 curl(F) = 2 ∂θ ρ sin φ ∂ρ ∂φ Fρ ρFφ ρ sin φFθ grad f = f = 1 ∂ ∂f ∂ 1 ∂f ∂ ρ2 sin φ + + sin φ ∂ρ ∂ρ ∂θ sin φ ∂θ ∂φ 2 2 2 ∂ f ∂ f 2 ∂f 1 1 ∂ f cot φ ∂f = + + 2 2 + 2 2+ 2 2 2 ∂ρ ρ ∂ρ ρ sin φ ∂θ ρ ∂φ ρ ∂φ ρ2 sin φ ∂f ∂φ

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