ECE 206: Advanced Calculus 2ece206/Lectures/slides/section10.pdf · To de ne G, we use the divergence operator again, viz. rG = r(r B 0J) = 0rJ: On the other hand, from Gauss’ law

Ampere-Maxwell Law

We are going to see that

∇×B(x, y, z, t) = µ0J(x, y, z, t)

cannot possibly be true for genuinely time varying fields.

To this end, we first apply the divergence to obtain

∇ · (∇×B) = µ0∇ · J.

Since ∇ · (∇×B) = 0, it follows that

µ0∇ · J = 0.

Then, in view of (∇ · J) + ∂ρ/∂t = 0, we obtain

∂ρ(x, y, z, t)

∂t= 0,

which contradicts our observations.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 1/1

Ampere-Maxwell Law (cont.)

To overcome the above difficulty, Maxwell’s idea was to “guess”that a correct version of Ampere’s continuity law would be

∇×B = µ0J + G,

for some time varying vector field G.

To define G, we use the divergence operator again, viz.

∇ ·G = ∇ · (∇×B− µ0J) = −µ0∇ · J.

On the other hand, from Gauss’ law for electric fields we get

ρ = ε0(∇ ·E),

which suggests∂ρ

∂t= ε0∇ ·

[∂E

∂t

].



Substituting the above qualities into the continuity equation

∇ · J +∂ρ

∂t= 0,

we obtain

ε0∇ ·[∂E

∂t

]− 1

µ0∇ ·G = 0,

that is

∇ ·[ε0∂E

∂t− 1

µ0G

]= 0,

The simplest relation between E and G consistent with the aboveis

G = ε0µ0∂E

∂t.



The resulting Ampere-Maxwell law for time varying magneticfields reads

∇×B = µ0J + ε0µ0∂E

∂t.

Note that, when the fields are time constant, the partial t-deriva-tive is identically zero, so that the law is fully consistent withAmpere’s law.

The correctness of the Ampere-Maxwell law was verified experi-mentally by Heinrich Hertz in 1887-1891.

Suppose that the current density J is identically zero, then

∇×B− ε0µ0∂E

∂t= 0.

Maxwell’s correction of Ampere’s law supplies a new physical lawwhich is a symmetric counterpart of Faraday’s law.


Maxwell’s equations

Law (Maxwell)

a) ∇ ·E =1

ε0ρ (Gauss for electric fields)

b) ∇ ·B = 0 (Gauss for magnetic fields)

c) ∇×E +∂B

∂t= 0 (Faraday)

d) ∇×B = µ0J + ε0µ0∂E

∂t(Ampere-Maxwell)

It was Maxwell who first truly understood the enormous power ofthese equations when used together.

Maxwell used the equations to discover the existence of self-sus-taining electromagnetic waves or electromagnetic radiation.


EM waves without sources

In the source free case, we have ρ = 0 and J = 0.

In this case, the Maxwell equations reduce to

a) ∇ ·E = 0,

b) ∇ ·B = 0,

c) ∇×E +∂B

∂t= 0,

d) ∇×B = ε0µ0∂E

∂t.

We get non-zero electric and magnetic fields as a consequence ofEM energy being radiated into free space by some transmittingagent (e.g., antenna).

In this case, both ρ and J are localized to the antenna, while be-ing zero everywhere else.


EM waves without sources (cont.)

Applying the curl to (c) we get

∇× (∇×E) +∇× ∂B

∂t= 0.

Since ∇× (∇×E) = ∇(∇ ·E)−∆E and ∇ ·E = 0, we get

−∆E +∂(∇×B)

∂t= 0.

Using (d), we finally obtain

c2∆E =∂2E

∂t2,

where c := 1/√ε0µ0 = 2.999 · 108 meters/seconds.

Amazingly enough, c is equal to the speed of light in vacuum.



Applying the curl to (d) we get

∇× (∇×B) = ε0µ0∇×[∂E

∂t

].

Since ∇× (∇×B) = ∇(∇ ·B)−∆B and ∇ ·B = 0, we get

−∆B = ε0µ0∂(∇×E)

∂t.

Using (c), we finally obtain

c2∆B =∂2B

∂t2,

where c := 1/√ε0µ0.



The vector fields E and B can be expressed in terms of its scalarcomponents, viz.

B = B1i +B2j +B3k, E = E1i + E2j + E3k.

In terms of the scalar components, the derived equations has theform of

c2(∆Ei) =∂2Ei∂2t

, c2(∆Bi) =∂2Bi∂2t

, i = 1, 2, 3.

Thus, we have six partial differential equations called wave equa-tions, each of which can be solved individually for the relevant Eiand Bi.


Wave equation

A wave equation is a special kind of partial differential equation,which is of the form

c2(∆u) =∂2u

∂t2,

in which c is a positive constant.

Any function u(x, y, z, t) which satisfies the wave equation is bestthought of as a time varying scalar field.

Note that u(x, y, z, t) = 0 always satisfies the equation for each tand (x, y, z).

Let n = n1i + n2j + n3k be some unit vector and ψ : R→ R besome C2 function.

Define the time varying scalar field

u(x, y, z, t) := ψ(n · (xi+yj+ zk)− c t) = ψ(n1x+n2y+n3z− c t),

which is a solution of the wave equation for all (x, y, z, t), as wellas for any n and ψ.


Wave equation (cont.)

In the case when n = i, the function u simplifies to

u(x, y, z, t) = ψ(x− c t), ∀(x, y, z, t).



One can see that the wave moves to the right at speed c.

Since, at each fixed t, the “wave function” u(x, y, z, t) is constantfor all (x, y, z) throughout Pα, the wave is called planar.



In a general case, a generic point (x, y, z) ∈ R3 is associated withvector

r = xi + yj + zk.

For each scalar α, the equation

n · r = α

defines a plane Pα.

Thus we obtain that, for each fixed instant t,

u(x, y, z, t) = ψ(α− c t)

for each (x, y, z) in Pα.

In this case, the wave propagates in the direction of n, at a speedc.

Moreover, for each t, the scalar field u(x, y, z, t) has the constantvalue ψ(α− c t) in the plane Pα.


Special cases

Particularly important special case is when

u(t, x, y, z) = η(x, y, z) cos(ωt),

where ω is a constant angular frequency and η : R3 → R is someC2-function.

Such separated functions with sinusoidal dependence on time toccur very naturally in applications.

Using Eulers formula, one can represent the sinusoid in an expo-mental form

u(t, x, y, z) = η(x, y, z)e−ωt.


Special cases (cont.)

By substitution, we see that

∆u(t, x, y, z) = ∆η(x, y, z)e−ωt,

while∂2u(t, x, y, z)

∂t2= −ω2η(x, y, z)e−ωt.

In this case, the wave equation reads[c2∆η(x, y, z) + ω2η(x, y, z)

]e−ωt = 0,

so that∆η(x, y, z) + k2η(x, y, z) = 0,

where k := ω/c.

Now we must determine η(x, y, z).



To determine η(x, y, z), we fix an arbitrary constant γ = Aeθ

and define

η(x, y, z) = γ exp{kn·(xi+yj+zk)} = γ exp{k(xn1+yn2+zn3)},

with n being an arbitrary direction.

Then, it is straightforward to check that such η(x, y, z) indeedsatisfies

∆η(x, y, z) + k2η(x, y, z) = 0.

Thus, the special form of the solutions is

u(x, y, z, t) = γ exp {k(n1x+ n2y + n3z)} e−ωt =

= γ exp { [k(n1x+ n2y + n3z)− ωt]} =

= γ exp {k [(n1x+ n2y + n3z)− c t]} =

= γ exp {k [n · (xi + yj + zk)− c t]} .



The real part of the solution has the form of

u(x, y, z, t) = A cos {k [n · (xi + yj + zk)− c t] + θ} .

The above solution corresponds to

ψ(α) := A cos{kα+ θ}.

Notice that ψ(α) is a periodic function of α with period given by

2π

k=

2πc

ω

and called the wavelength of the wave.

This describes the spatial periodicity of the solution in the direc-tion of n for every fixed t.

At every fixed (x, y, z), the time periodicity is described by 2π/ω.


EM waves

All the above properties of wave equations can be carried over tothe equations for Ei and Bi, i = 1, 2, 3.

In particular, each of the scalar components Ei and Bi propagatesthrough space as a wave moving at the speed of light!

One can show that when the voltage source for an antenna is si-nusoidal, that is

v(t) = V cos(ωt),

the solutions for Ei and Bi have the special separated form, e.g.

Ei(x, y, z, t) = A cos {k [n · (xi + yj + zk)− c t] + θ} .

In this case, A, n and θ are functions of V and the geometry ofthe antenna.


EM waves with sources

In the presence of sources we have

a) ∇ ·E =1

ε0ρ

b) ∇ ·B = 0

c) ∇×E +∂B

∂t= 0

d) ∇×B = µ0J + ε0µ0∂E

∂t

For example, these equations allow us to understand the relationbetween E and B outside the antenna and the source fields ρ andJ created inside the antenna by the voltage source.


EM waves with sources (cont.)

The magnetic field B can be expressed as

B = ∇×A,

where A is the magnetic vector potential of B.

Moreover, we have

∂B

∂t=∂(∇×A)

∂t= ∇×

[∂A

∂t

].

Now,

∇×[E +

∂A

∂t

]= ∇×E +∇×

[∂A

∂t

]= ∇×E +

∂B

∂t= 0,

and therefore there exists a (scalar) potential Ψ such that

E +∂A

∂t= −∇Ψ.



So we have

E = −∇Ψ− ∂A

∂t.

Notice that there are many Ψ and A for which the above wouldhold.

However, one can always find Ψ and A such that

B = ∇× A, E = −∂A∂t−∇Ψ, ∇ · A + ε0µ0

∂Ψ

∂t= 0.

For such Ψ and A we have

∇× (∇× A)︸︷︷︸∇(∇·A)−∆A

= µ0J− ε0µ0∂

∂t

[∂A

∂t+∇Ψ

].



Rearranging the terms, we obtain

∆A− ε0µ0∂2A

∂t2−∇

[∇ · A + ε0µ0

∂Ψ

∂t

]︸︷︷︸

=0

= −µ0J.

Consequently,

∆A− ε0µ0∂2A

∂t2= −µ0J.

The above relation can be reformulated in terms of the (scalar)coordinate components as

∆Ai − ε0µ0∂2Ai∂t2

= −µ0Ji, i = 1, 2, 3.

Note that in the absence of dependency on time, the above rela-tons reduce to three Poisson equations (static case).



In the time-varying case, the above equations can be solved bythe method of Green’s functions. Specifically,

Ai(x, y, z, t) =µ0

4π

∫R3

Ji

(u, v, w, t−

√ε0µ0 [(x− u)2 + (y − v)2 + (z − w)2]

)[(x− u)2 + (y − v)2 + (z − w)2]1/2

dV,

for all (x, y, z, t) and dV = du dv dw.

Having A determined, the field B is computed by taking the curl.

Note that in the static case we had

Ai(x, y, z, t) =µ0

4π

∫R3

Ji (u, v, w)

[(x− u)2 + (y − v)2 + (z − w)2]1/2dV.

Now we need to find the electric field E.



We start with

1

ε0ρ = ∇ ·E = ∇ ·

[−∂A∂t−∇Ψ

]= −∂(∇ · A)

∂t−∆Ψ.

Thus, we have∂(∇ · A)

∂t+ ∆Ψ = − 1

ε0ρ.

However, we have chosen A and Ψ so that ∇ · A = −ε0µ0∂Ψ/∂t,which suggests that

∆Ψ− ε0µ0∂2Ψ

∂t2= − 1

ε0ρ.

Note that the above relation reduces to a Poisson equation in astatic case.



As before, the solution to the above equation can be obtained bythe method of Green’s functions. Specifically,

Ψ(x, y, z, t) =1

4πε0

∫R3

ρ(u, v, w, t−

√ε0µ0 [(x− u)2 + (y − v)2 + (z − w)2]

)[(x− u)2 + (y − v)2 + (z − w)2]1/2

dV,

for all (x, y, z, t) and dV = du dv dw.

Once Ψ and A are found, the electric field E is computed from

E = −∇Ψ− ∂A

∂t.

Defining rs = (u, v, w) and rt = (x, y, z), the solution can be re-written as

Ψ(rt, t) =1

4πε0

∫R3

ρ (rs, t− ‖rt − rs‖/c)‖rt − rs‖

drs.


Lorentz gauge transformation

So, how to find Ψ and A such that

B = ∇× A, E = −∂A∂t−∇Ψ, ∇ · A + ε0µ0

∂Ψ

∂t= 0.

To this end, let’s choose A and Ψ which satisfy the first two ofthe above relations.

Now suppose that f is a time varying scalar field which satisfies

∆f − ε0µ0∂2f

∂t2= −

[∇ ·A + ε0µ0

∂Ψ

∂t

].

x and define

A := A +∇f, Ψ := Ψ− ∂f

∂t.

This transformation is called the Lorentz gauge transformation.


Lorentz gauge transformation (cont.)

We get∇× A = ∇× (A +∇f) = ∇×A = B,

and thus we have B = ∇× A.

We also get

∂A

∂t+∇Ψ =

∂[A +∇f ]

∂t+∇[Ψ− ∂f

∂t] =

∂A

∂t+∇Ψ = 0

and thus we have ∂A/∂t+∇Ψ = 0.

Finally,

∇ · A + ε0µ0∂Ψ

∂t= ∇ · [A +∇f ] + ε0µ0

∂

∂t[Ψ− ∂f

∂t] =

= ∇ ·A + ∆f + ε0µ0∂Ψ

∂t− ε0µ0

∂2f

∂t2= 0,

as required.


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ECE 206: Advanced Calculus 2ece206/Lectures/slides/section10.pdf · To de ne G, we use the divergence operator again, viz. rG = r(r B 0J) = 0rJ: On the other hand, from Gauss’ law