Maxwell's Equations

Z

0 0

q = ρ(r )dV (7)

V

0 0

3

dV = d r ρ(r )

with . Then, since does not depend on the unprimed coordinates on which

∇ operates, we have, 0

−

Z Z

r r 1

0 0

−

E(r) = dV ρ(r ) = dV ρ(r )∇ = (8)

0 0

3

− | |r − |

r r r

V V 0

Z ρ(r )

−∇

= dV 0

|r − |

r

V

1 The nonlinearity due to vacuum polarization effects produced by the continuos creation-annihilation

of electron-positron pairs, is a quantum mechanical effect, that can be here neglected.

2

It is worth noting that, under assumption of linearity, relation (8) is valid for an arbitrary

ρ

charge distribution, including also the case of discrete charges, in which can be expressed

in terms of a delta Dirac distribution, X

0 0 0

−

ρ(r ) = q δ(r r ) (9)

i i

i

2 Ampere’s Law and Magnetostatic Field

Experiments on the interaction between two small loops of electric current have shown that

they interact via a mechanical force, much the same way that electric charges interact. Let

0 0

γ γ dl dl

us consider two small loops and , with tangent line element and , respectively

0 0 0

r r γ I γ I

located at and . We suppose that carries a total current and a current .

0

γ γ

Ampere find out that the force acting on due to the presence of is, in vacuum,

0 0

−

I

I

II r r

0 ∧

∧

F = dl

dl = (10)

0 3

|r − |

c r

0

γ

γ

0

I I

II 1

0

− ∧ ∧ ∇

= dl dl (11)

0

|r − |

c r

0

γ γ

At a first glance, this celebrated equation, called Ampere’s law, may appear unsymmetric

in terms of the loops and therefore to be in contradiction with Newton’s third law of

dynamics. However, by the following properties of vector product,

∧ ∧ · − ·

a (b c) = b(a c) c(a b) (12)

we get, 0 0 0

−

I I I I

II 1 II r r

0 0

− − ·

· ∇

F = dl dl

dl dl (13)

0 0 3

|r − | |r − |

c r c r

0 0

γ γ γ γ 2

Since the integrand in the first integral is an exact differential

, due to the presence of the gradient operator, this integral vanishes and we can rewrite

the force expression in the following symmetric way,

0 0

−

I I

II r r 0

− ·

F = dl dl (15)

0 3

|r − |

c r

0

γ γ

As already seen in the case of electrostatic interaction, we may attribute the magnetostatic

B(r).

interaction to a statical vector field too, called magnetostatic field It turns out that

B

the elemental can be defined as, 0

0 −

I r r

0 ∧

dB(r) := dl (16)

0 3

|r − |

c r

2 df

We say that is an exact differential if we have,

f f

Z Z

≡

df A(x, y)dx + B(x, y)dy

∆f = (14)

i i

∆f i f

that is does not depend on the path from to but only on the boundaries. We can see that the

integrand is a 1-form, which correspond, on a pseudo-Riemannian manifold, to a 1-vector field, by duality

via the metric. Indeed, the concept of exact differential form is equivalent to the one of conservative vector

field, which is more familiar for physics applications. Hence, an exact differential 1-form is a form which

is the derivative (gradient) of a 0-form (a function) (∆f ), called the scalar potential.

3

dB r

which expresses the small element of the static magnetic field set up at the field point

0 0 0

dl I r

by a small line element of stationary current at the source point . We can generalize

J (r),

expression (15) to a steady state electric current density obtaining Biot-Savart’s law,

0

−

Z

1 r r

0 ∧

B(r) = dV J (r ) = (17)

0 3

|r − |

c r

V 0

Z J (r )

1 ∇∧ dV

= (18)

0

|r − |

c r

V

0 ∇

J (r )

where we used the fact that does not depend on coordinates on which operates.

Comparing the expression of the electrostatic field and the magnetostatic one, we see that

there exists a close analogy between them, but they differ in their vectorial characteristics,

as it will be proved in the following sections.

3 Gauss’s Flux Theorem

Gauss’s Theorem states that, given an arbitrary continuos charge distribution, the total

t)) 4πQ, Q

flux of the electric field (E(r, through any closed surface is equal to where is

the total charge within that closed surface. Explicitly,

Φ = 4πQ (19)

(E,S) S,

Reminding the definition of flux integral through a oriented-surface given the vector

F = (F , F , F ),

field then,

x y z Z Z

· × ∧

Φ = F dS := F (x(u, v)) (q q )dudv (20)

u v

S S Ω

x(u, v) q

where is a surface parametrization such that define the tangent versor to

(u,v) 3

Ω u, v-space dS

surface line coordinates at each point, is the domain in the , is an in-

finitesimal oriented-volume element. The proof of Gauss’s Theorem is trivial in the case of

R.

a simple surface like the sphere of radius The vector field generated by a point charge

q is, r

E(r) = q (21)

3

|r|

F = k(r)e e

A general radial vector field is defined as , where is the radial versor; then,

r r

r = re

since , we can redefine (3) as,

r re 1

r

E(r) = q = q e (22)

r

3 2

|r| r

2

≡

k(r) q/r

that is a radial vector field with . A simple parametrization of the sphere is,

x(φ, θ) = (R cos φ sin θ)e + (R sin θ sin φ)e + (R cos θ)e (23)

1 2 3

hence, Z Z

· · ∧

Φ = E(r) dS = E(x(φ, θ)) (q q )dφdθ (24)

E,S φ θ

S S

3 2 3

⊃ → ∈ →

f : Ω (u, v) Ω x(u, v).

Rigorously, a surface is a vector-valued function , such that In

R R

f

general we use the term surface referring to the support of , but actually the surface is the whole function.

4

E

Since does not depend on angular coordinates, then, φ

2π Z

Z

Z 2

· ∧ dθR sin θ =

dφ

E(x(φ, θ)) (q q )dφdθ = k(R) (25)

φ θ 0

0

S 2

= k(R)R 4φ = 4πq (26)

ρ V Q

Given a general charge distribution within a region of volume , the total charge

within this closed domain will be, Z

Q = ρdV (27)

V

Then we could write Gauss’s flux theorem as follows,

Z Z

·

E(r) dS = 4π ρdV (28)

S V

4 V

For the divergence theorem identifying as the volume enclosed by the previously intro-

S,

duced surface we have, Z Z

·

Φ = (∇ E)dV = 4π ρdV (30)

(E,S) V V

hence, Z · −

(∇ E 4πρ)dV = 0 (31)

V V

Since this relation must be satisfied for any arbitrary volume , it holds,

∇ · −

E 4πρ = 0 (32)

or, ∇ · E = 4πρ (33)

that is the first Maxwell’s equation. Thus, starting from a global relation (1), we have

derived a local relation, which translates the content of Gauss’s flux theorem point by point

through the divergence differential operator. Indeed, the divergence theorem provides us

F

with a definition of divergence of a vector field ,

Z

1

∇ · ·

F = lim F dS (34)

δV

→0

δV +∂(δV )

F

that is, the divergence of is a sort of local flux density, connected to the presence of

sources or wells. We know that, in the case of electric field, sources are positive charges

and wells are negative charges.

Equation (24) can also be proved by an explicit calculation of divergence of the electric

field, reminding the definition (8). In fact, we have,

Z 1

0

∇ · ∇ · −

E(r) = dV ρ(r )∇ = (35)

0

|r − |

r

V

Z 1

0

− · ∇

= dV ρ(r )∇ = (36)

0

|r − |

r

V

Z 1

0

−

= dV ρ(r )∆ (37)

0

|r − |

r

V

4 3

V ∂V = S, F

Given a compact subset of with a piecewise smooth boundary if is a continuously

R V

differentiable vector field defined on a neighborhood of , then,

Z Z

· ·

(∇ F )dV = F dS (29)

V ∂V

5

∆

where is the Laplacian operator. We proceed with an explicit calculation of this expres-

0

−

r = r r

sion; we put and consider that,

0

1 r

0

∇ −

= (38)

3

|r | |r |

0 0

|r | 6

= 0.

when The divergence of this is,

0

r 0

∇· − =0 (39)

3

|r |

0

|r | 6

= 0,

hence, for all points for which we get,

0

1 =0

∆ (40)

|r |

0

∇ · ∇

∆ = V a,

Since if we integrate this function over a sphere of radius applying the

above mentioned divergence theorem, we get,

Z Z

1 1

∇ · ∇ ∇ ·

dV = dS = (41)

r r

0 0

V S

Z r 0

− ·

= dS (42)

3

|r |

0

S

S = ∂V e

where . The orientation of the sphere is defined by the normalized vector , then

r

the surface integral (33) is, |r |e

Z Z

r 0 0 r 2

− · − · |r |

dS = e sin θdθdφ = (43)

r 0

3 3

|r | |r |

0 0

S S 2π π

Z Z

− −4π

= dφ sin θdθ = (44)

0 0

That is, the value of the laplacian is zero everywhere except in the origin, and the integral

−4π.

over any volume containing the origin is equal to Then we could set,

1 −4πδ(r

∆ = ) (45)

0

|r |

0

0

−

r = r r

Reminding that , by a substitution of this expression in relation (28), we get,

0 Z 0 0

∇ · −

E(r) = dV ρ(r )4πδ(r r ) = 4πρ(r) (46)

V

where we have used the sifting property of the Dirac function.

Electric field is manifestly the gradient of a scalar field, or in another words, it is a

5

conservative vector field . Namely, ∇φ

E(r, t) = (47)

f

with a scalar function. Then, (15) and (16) yield

∇ &mi