What is SUCCESSIVE APPLICATIONS OF ∇ in World of Mathematics ?

SUCCESSIVE APPLICATIONS OF ∇ :

We have now defined gradient, divergence, and curl to obtain vector, scalar, and vector

quantities, respectively. Letting ∇ operate on each of these quantities, we obtain

(a) ∇ · ∇ϕ (b) ∇ ×∇ϕ (c) ∇∇ · V

(d) ∇ · ∇ ×V (e) ∇ ×(∇ ×V)

50 Chapter 1 Vector Analysis

all five expressions involving second derivatives and all five appearing in the second-order

differential equations of mathematical physics, particularly in electromagnetic theory.

The first expression, ∇·∇ϕ, the divergence of the gradient, is named the Laplacian of ϕ.

We have

∇ · ∇ϕ =

ˆx

∂

∂x + ˆy

∂

∂y + ˆz

∂

∂z

·

ˆx

∂ϕ

∂x + ˆy

∂ϕ

∂y + ˆz

∂ϕ

∂z

=

∂2ϕ

∂x2 +

∂2ϕ

∂y2 +

∂2ϕ

∂z2 . (1.81a)

When ϕ is the electrostatic potential, we have

∇ · ∇ϕ = 0 (1.81b)

at points where the charge density vanishes, which is Laplace’s equation of electrostatics.

Often the combination ∇ · ∇ is written ∇2, or in the European literature.

Example 1.9.1 LAPLACIAN OF A POTENTIAL

Calculate ∇ · ∇V (r).

Referring to Examples 1.6.1 and 1.7.2,

∇ · ∇V (r) = ∇ · ˆr

dV

dr =

2

r

dV

dr +

d2V

dr2 ,

replacing f (r) in Example 1.7.2 by 1/r · dV/dr. If V (r) = rn, this reduces to

∇ · ∇rn = n(n+ 1)rn−2.

This vanishes for n = 0 [V (r) = constant] and for n=−1; that is, V (r) = 1/r is a solution

of Laplace’s equation, ∇2V (r) = 0. This is for r 

= 0. At r = 0, a Dirac delta function is

involved (see Eq. (1.169) and Section 9.7).

Expression (b) may be written

∇ ×∇ϕ =

ˆx ˆy ˆz

∂

∂x

∂

∂y

∂

∂z

∂ϕ

∂x

∂ϕ

∂y

∂ϕ

∂z

.

By expanding the determinant, we obtain

∇ ×∇ϕ = ˆx

∂2ϕ

∂y ∂z −

∂2ϕ

∂z ∂y

+ ˆy

∂2ϕ

∂z ∂x −

∂2ϕ

∂x ∂z

+ ˆz

∂2ϕ

∂x ∂y −

∂2ϕ

∂y ∂x

= 0, (1.82)

assuming that the order of partial differentiation may be interchanged. This is true as long

as these second partial derivatives of ϕ are continuous functions. Then, from Eq. (1.82),

the curl of a gradient is identically zero. All gradients, therefore, are irrotational. Note that

1.9 Successive Applications of ∇ 51

the zero in Eq. (1.82) comes as a mathematical identity, independent of any physics. The

zero in Eq. (1.81b) is a consequence of physics.

Expression (d) is a triple scalar product that may be written

∇ · ∇ ×V =

∂

∂x

∂

∂y

∂

∂z

∂

∂x

∂

∂y

∂

∂z

Vx Vy Vz

. (1.83)

Again, assuming continuity so that the order of differentiation is immaterial, we obtain

∇ · ∇ ×V = 0. (1.84)

The divergence of a curl vanishes or all curls are solenoidal. In Section 1.16 we shall see

that vectors may be resolved into solenoidal and irrotational parts by Helmholtz’s theorem.

The two remaining expressions satisfy a relation

∇ ×(∇ ×V) = ∇∇ · V− ∇ · ∇V, (1.85)

valid in Cartesian coordinates (but not in curved coordinates). This follows immediately

from Eq. (1.55), the BAC–CAB rule, which we rewrite so that C appears at the extreme

right of each term. The term ∇ · ∇V was not included in our list, but it may be defined by

Eq. (1.85).

Example 1.9.2 ELECTROMAGNETIC WAVE EQUATION

One important application of this vector relation (Eq. (1.85)) is in the derivation of the

electromagnetic wave equation. In vacuum Maxwell’s equations become

∇ · B = 0, (1.86a)

∇ · E = 0, (1.86b)

∇ ×B = ε0μ0

∂E

∂t

, (1.86c)

∇ ×E=−

∂B

∂t

. (1.86d)

Here E is the electric field, B is the magnetic induction, ε0 is the electric permittivity,

and μ0 is the magnetic permeability (SI units), so ε0μ0 = 1/c2, c being the velocity of

light. The relation has important consequences. Because ε0, μ0 can be measured in any

frame, the velocity of light is the same in any frame.

Suppose we eliminate B from Eqs. (1.86c) and (1.86d). We may do this by taking the

curl of both sides of Eq. (1.86d) and the time derivative of both sides of Eq. (1.86c). Since

the space and time derivatives commute,

∂

∂t

∇ ×B = ∇ ×

∂B

∂t

,

and we obtain

∇ ×(∇ ×E)=−ε0μ0

∂2E

∂t2 .

52 Chapter 1 Vector Analysis

Application of Eqs. (1.85) and (1.86b) yields

∇ · ∇E = ε0μ0

∂2E

∂t2 , (1.87)

the electromagnetic vector wave equation. Again, if E is expressed in Cartesian coordinates,

Eq. (1.87) separates into three scalar wave equations, each involving the scalar

Laplacian.

When external electric charge and current densities are kept as driving terms in

Maxwell’s equations, similar wave equations are valid for the electric potential and the

vector potential. To show this, we solve Eq. (1.86a) by writing B = ∇ × A as a curl of the

vector potential. This expression is substituted into Faraday’s induction law in differential

form, Eq. (1.86d), to yield ∇ ×(E+ ∂A

∂t ) = 0. The vanishing curl implies that E+ ∂A

∂t is a

gradient and, therefore, can be written as −∇ϕ, where ϕ(r, t) is defined as the (nonstatic)

electric potential. These results for the B and E fields,

B = ∇ ×A, E=−∇ϕ −

∂A

∂t

, (1.88)

solve the homogeneous Maxwell’s equations.

We now show that the inhomogeneous Maxwell’s equations,

Gauss’ law: ∇ · E = ρ/ε0, Oersted’s law: ∇ ×B−

1

c2

∂E

∂t = μ0J (1.89)

in differential form lead to wave equations for the potentials ϕ and A, provided that ∇·A is

determined by the constraint 1

c2

∂ϕ

∂t +∇ ·A = 0. This choice of fixing the divergence of the

vector potential, called the Lorentz gauge, serves to uncouple the differential equations of

both potentials. This gauge constraint is not a restriction; it has no physical effect.

Substituting our electric field solution into Gauss’ law yields

ρ

ε0 = ∇ · E=−∇2ϕ −

∂

∂t

∇ · A=−∇2ϕ +

1

c2

∂2ϕ

∂t2 , (1.90)

the wave equation for the electric potential. In the last step we have used the Lorentz

gauge to replace the divergence of the vector potential by the time derivative of the electric

potential and thus decouple ϕ from A.

Finally, we substitute B = ∇ × A into Oersted’s law and use Eq. (1.85), which expands

∇2 in terms of a longitudinal (the gradient term) and a transverse component (the curl

term). This yields

μ0J+

1

c2

∂E

∂t = ∇ ×(∇ ×A) = ∇(∇ · A)−∇2A = μ0J −

1

c2

∇

∂ϕ

∂t +

∂2A

∂t2

,

where we have used the electric field solution (Eq. (1.88)) in the last step. Now we see that

the Lorentz gauge condition eliminates the gradient terms, so the wave equation

1

c2

∂2A

∂t2 −∇2A = μ0J (1.91)

1.9 Successive Applications of ∇ 53

for the vector potential remains.

Finally, looking back at Oersted’s law, taking the divergence of Eq. (1.89), dropping

∇·(∇×B) = 0, and substituting Gauss’ law for ∇·E = ρ/ǫ0, we find μ0∇·J=− 1

ǫ0c2

∂ρ

∂t ,

where ǫ0μ0 = 1/c2, that is, the continuity equation for the current density. This step justifies

the inclusion of Maxwell’s displacement current in the generalization of Oersted’s law

to nonstationary situations.

Exercises

1.9.1 Verify Eq. (1.85),

∇ ×(∇ ×V) = ∇∇ · V −∇ · ∇V,

by direct expansion in Cartesian coordinates.

1.9.2 Show that the identity

∇ ×(∇ ×V) = ∇∇ · V−∇ · ∇V

follows from the BAC–CAB rule for a triple vector product. Justify any alteration of the

order of factors in the BAC and CAB terms.

1.9.3 Prove that ∇ ×(ϕ∇ϕ) = 0.

1.9.4 You are given that the curl of F equals the curl of G. Show that F and G may differ by

(a) a constant and (b) a gradient of a scalar function.

1.9.5 The Navier–Stokes equation of hydrodynamics contains a nonlinear term (v·∇)v. Show

that the curl of this term may be written as −∇ × [v×(∇ ×v)].

1.9.6 From the Navier–Stokes equation for the steady flow of an incompressible viscous fluid

we have the term

∇ ×

v×(∇ ×v)

,

where v is the fluid velocity. Show that this term vanishes for the special case

v = ˆxv(y, z).

1.9.7 Prove that (∇u)×(∇v) is solenoidal, where u and v are differentiable scalar functions.

1.9.8 ϕ is a scalar satisfying Laplace’s equation, ∇2ϕ = 0. Show that ∇ϕ is both solenoidal

and irrotational.

1.9.9 With ψ a scalar (wave) function, show that

(r×∇) · (r×∇)ψ = r2∇2ψ − r2 ∂2ψ

∂r2 −2r

∂ψ

∂r

.

(This can actually be shown more easily in spherical polar coordinates, Section 2.5.)

54 Chapter 1 Vector Analysis

1.9.10 In a (nonrotating) isolated mass such as a star, the condition for equilibrium is

∇P + ρ∇ϕ = 0.

Here P is the total pressure, ρ is the density, and ϕ is the gravitational potential. Show

that at any given point the normals to the surfaces of constant pressure and constant

gravitational potential are parallel.

1.9.11 In the Pauli theory of the electron, one encounters the expression

(p −eA)×(p −eA)ψ,

where ψ is a scalar (wave) function. A is the magnetic vector potential related to the

magnetic induction B by B = ∇ × A. Given that p=−i∇, show that this expression

reduces to ieBψ. Show that this leads to the orbital g-factor gL = 1 upon writing the

magnetic moment as μ = gLL in units of Bohr magnetons and L=−ir × ∇. See also

Exercise 1.13.7.

1.9.12 Show that any solution of the equation

∇ ×(∇ ×A)− k2A = 0

automatically satisfies the vector Helmholtz equation

∇2A+k2A = 0

and the solenoidal condition

∇ · A = 0.

Hint. Let ∇· operate on the first equation.

1.9.13 The theory of heat conduction leads to an equation

∇2 = k|∇ |2,

where is a potential satisfying Laplace’s equation: ∇2 = 0. Show that a solution of

this equation is

=

1

2

k 2.