Lecture 3
Constitutive Relations, Wave
Equation, Electrostatics, and
Static Green’s Function
3.1 Simple Constitutive Relations
The constitution relation between D and E in free space is
D = ε0E (3.1.1)
When material medium is present, one has to add the contribution to D by the polarization
density P which is a dipole density.1 Then [29, 31, 36]
D = ε0E + P (3.1.2)
The second term above is the contribution to the electric flux due to the polarization density
of the medium. It is due to the little dipole contribution due to the polar nature of the atoms
or molecules that make up a medium.
By the same token, the first term ε0E can be thought of as the polarization density
contribution of vacuum. Vacuum, though represents nothingness, has electrons and positrons,
or electron-positron pairs lurking in it [37]. Electron is matter, whereas positron is anti-
matter. In the quiescent state, they represent nothingness, but they can be polarized by an
electric field E. That also explains why electromagnetic wave can propagate through vacuum.
For many media, it can be assumed to be a linear media. Then P = ε0χ0E
D = ε0E + ε0χ0E
= ε0(1 + χ0)E = εE (3.1.3)
1Note that a dipole moment is given by Q` where Q is its charge in coulomb and ` is its length in m. Hence,
dipole density, or polarization density as dimension of coulomb/m2, which is the same as that of electric flux
D.
25
26 Electromagnetic Field Theory
where χ0 is the electric susceptibility. In other words, for linear material media, one can
replace the vacuum permittivity ε0 with an effective permittivity ε.
In free space:
ε = ε0 = 8.854 × 10−12 ≈
10−8
36π
F/m (3.1.4)
The constitutive relation between magnetic flux B and magnetic field H is given as
B = µH, µ = permeability H/m (3.1.5)
In free space,
µ = µ0 = 4π × 10−7 H/m (3.1.6)
As shall be explained later, this is an assigned value. In other materials, the permeability can
be written as
µ = µ0µr (3.1.7)
Similarly, the permittivity for electric field can be written as
ε = ε0εr (3.1.8)
In the above, µr and εr are relative permeability and relative permeability.
3.2 Emergence of Wave Phenomenon, Triumph of Maxwell’s
Equations
One of the major triumphs of Maxwell’s equations is the prediction of the wave phenomenon.
This was experimentally verified by Heinrich Hertz in 1888 [18], some 23 years after the com-
pletion of Maxwell’s theory [17]. Then it was realized that electromagnetic wave propagates
at a tremendous velocity which is the velocity of light. This was also the defining moment
which revealed that the field of electricity and magnetism and the field of optics were both
described by Maxwell’s equations or electromagnetic theory.
To see this, we consider the first two Maxwell’s equations for time-varying fields in vacuum
or a source-free medium.2 They are
∇ × E = −µ0
∂H
∂t (3.2.1)
∇ × H = −ε0
∂E
∂t (3.2.2)
Taking the curl of (3.2.1), we have
∇ × ∇ × E = −µ0
∂
∂t∇ × H (3.2.3)
2Since the third and the fourth Maxwell’s equations are derivable from the first two.
Constitutive Relations, Wave Equation, Electrostatics, and Static Green’s Function 27
It is understood that in the above, the double curl operator implies ∇×(∇×E). Substituting
(3.2.2) into (3.2.3), we have
∇ × ∇ × E = −µ0ε0
∂
2
∂t2
E (3.2.4)
In the above, the left-hand side can be simplified by using the identity that a × (b × c) =
b(a · c) − c(a · b),3 but be mindful that the operator ∇ has to operate on a function to its
right. Therefore, we arrive at the identity that
∇ × ∇ × E = ∇∇ · E − ∇2E (3.2.5)
and that ∇ · E = 0 in a source-free medium, we have
∇2E − µ0ε0
∂
2
∂t2
E = 0 (3.2.6)
where
∇2 = ∇ · ∇ =
∂
2
∂x2
∂
2
∂x2
+
∂
2
∂y2
+
∂
2
∂z2
Here, (3.2.6) is the wave equation in three space dimensions [31, 38].
To see the simplest form of wave emerging in the above, we can let E = ˆxEx(z, t) so that
∇ · E = 0 satisfying the source-free condition. Then (3.2.6) becomes
∂
2
∂z2
Ex(z, t) − µ0ε0
∂
2
∂t2
Ex(z, t) = 0 (3.2.7)
Eq. (3.2.7) is known mathematically as the wave equation in one space dimension. It can
also be written as
∂
2
∂z2
f(z, t) −
1
c
2
0
∂
2
∂t2
f(z, t) = 0 (3.2.8)
where c
2
0 = (µ0ε0)
−1
. Eq. (3.2.8) can also be factorized as
∂
∂z −
1
c0
∂
∂t ∂
∂z +
1
c0
∂
∂t
f(z, t) = 0 (3.2.9)
or
∂
∂z +
1
c0
∂
∂t ∂
∂z −
1
c0
∂
∂t
f(z, t) = 0 (3.2.10)
The above implies that we have
∂
∂z +
1
c0
∂
∂t
f+(z, t) = 0 (3.2.11)
3For mnemonics, this formula is also known as the “back-of-the-cab” formula.
28 Electromagnetic Field Theory
or
∂
∂z −
1
c0
∂
∂t
f−(z, t) = 0 (3.2.12)
Equation (3.2.11) and (3.2.12) are known as the one-way wave equations or advective equa-
tions [39]. From the above factorization, it is seen that the solutions of these one-way wave
equations are also the solutions of the original wave equation given by (3.2.8). Their general
solutions are then
f+(z, t) = F+(z − c0t) (3.2.13)
f−(z, t) = F−(z + c0t) (3.2.14)
Eq. (3.2.13) constitutes a right-traveling wave function of any shape while (3.2.14) constitutes
a left-traveling wave function of any shape. Since Eqs. (3.2.13) and (3.2.14) are also solutions
to (3.2.8), we can write the general solution to the wave equation as
f(z, t) = F+(z − c0t) + F−(z + c0t) (3.2.15)
This is a wonderful result since F+ and F− are arbitrary functions of any shape (see Figure
3.1); they can be used to encode information for communication!
Figure 3.1: Solutions of the wave equation can be a single-valued function of any shape. In
the above, F+ travels in the positive z direction, while F− travels in the negative z direction
as t increases.
29 Electromagnetic Field Theory
Furthermore, one can calculate the velocity of this wave to be
c0 = 299, 792, 458m/s ' 3 × 108m/s (3.2.16)
where c0 =
p
1/µ0ε0.
Maxwell’s equations (3.2.1) implies that E and H are linearly proportional to each other.
Thus, there is only one independent constant in the wave equation, and the value of µ0 is
defined to be 4π × 10−7 henry m−1, while the value of ε0 has been measured to be about
8.854 × 10−12 farad m−1. Now it has been decided that the velocity of light is defined to
be the integer given in (3.2.16). A meter is defined to be the distance traveled by light
in 1/(299792458) seconds. Hence, the more accurate that unit of time or second can be
calibrated, the more accurate can we calibrate the unit of length or meter.
3.3 Static Electromagnetics
When the fields and sources are not time varying, namely that ∂/∂t = 0, we arrive at the
static Maxwell’s equations for electrostatics and magnetostatics, namely [29, 31, 40]
∇ × E = 0 (3.3.1)
∇ × H = J (3.3.2)
∇ · D = % (3.3.3)
∇ · B = 0 (3.3.4)
Notice the the electrostatic field system is decoupled from the magnetostatic field system.
However, in a resistive system where
J = σE (3.3.5)
the two systems are coupled again. This is known as resistive coupling between them. But if
σ → ∞, in the case of a perfect conductor, or superconductor, then for a finite J, E has to
be zero. The two systems are decoupled again.
Also, one can arrive at the equations above by letting µ0 → 0 and 0 → 0. In this case,
the velocity of light becomes infinite, or retardation effect is negligible. In other words, there
is no time delay for signal propagation through the system in the static approximation.
3.3.1 Electrostatics
We see that Faraday’s law in the static limit is
∇ × E = 0 (3.3.6)
One way to satisfy the above is to let E = −∇Φ because of the identity ∇ × ∇ = 0.4
Alternatively, one can assume that E is a constant. But we usually are interested in solutions
that vanish at infinity, and hence, the latter is not a viable solution. Therefore, we let
E = −∇Φ (3.3.7)
4One an easily go through the algebra to convince oneself of this.
30 Electromagnetic Field Theory
Example
Fields of a sphere of radius a with uniform charge density ρ:
Figure 3.2: Figure of a sphere with uniform charge density for the example above.
Assuming that Φ|r=∞ = 0, what is Φ at r ≤ a? And Φ at r > a?
3.3.2 Poisson’s Equation
As a consequence of the above,
∇ · D = % ⇒ ∇ · εE = % ⇒ −∇ · ε∇Φ = % (3.3.8)
In the last equation above, if ε is a constant of space, or independent of r, then one arrives
at the simple Poisson’s equation, which is a partial differential equation
∇2Φ = −
%
ε
(3.3.9)
Here,
∇2 = ∇ · ∇ =
∂
2
∂x2
+
∂
2
∂y2
+
∂
2
∂z2
For a point source, we know that
E =
q
4πεr2
rˆ = −∇Φ (3.3.10)
From the above, we deduce that5
Φ = q
4πεr
(3.3.11)
Therefore, we know the solution to Poisson’s equation (3.3.9) when the source % represents a
point source. Since this is a linear equation, we can use the principle of linear superposition
to find the solution when % is arbitrary.
5One can always take the gradient or ∇ of Φ to verify this.
31 Electromagnetic Field Theory
A point source located at r
0
is described by a charge density as
%(r) = qδ(r − r
0
) (3.3.12)
where δ(r−r
0
) is a short-hand notation for δ(x−x
0
)δ(y−y
0
)δ(z−z
0
). Therefore, from (3.3.9),
the corresponding partial differential equation for a point source is
∇2Φ(r) = −
qδ(r − r0)
ε
(3.3.13)
The solution to the above equation, from Coulomb’s law, has to be
Φ(r) = q
4πε|r − r0|
(3.3.14)
where (3.3.11) is for a point source at the origin, but (3.3.14) is for a point source located
and translated to r0.
3.3.3 Static Green’s Function
Let us define a partial differential equation given by
∇2g(r − r0) = −δ(r − r0) (3.3.15)
The above is similar to Poisson’s equation with a point source on the right-hand side as in
(3.3.13). But such a solution, a response to a point source, is called the Green’s function.6
By comparing equations (3.3.13) and (3.3.15), then making use of (3.3.14), it is deduced that
the static Green’s function is
g(r − r0) = 1
4π|r − r0| (3.3.16)
An arbitrary source can be expressed as
%(r) = ˚
V
dV 0
%(r0)δ(r − r0) (3.3.17)
The above is just the statement that an arbitrary charge distribution %(r) can be expressed
as a linear superposition of point sources δ(r − r0). Using the above in (3.3.9), we have
∇2Φ(r) = −
1
ε
˚
V
dV 0
%(r0)δ(r − r0) (3.3.18)
We can let
Φ(r) = 1
ε
˚
V
dV 0
g(r − r0)%(r0) (3.3.19)
By substituting the above into the left-hand side of (3.3.18), exchanging order of integration
and differentiation, and then making use of equation (3.3.9), it can be shown that (3.3.19)
6George Green (1793-1841), the son of a Nottingham miller, was self-taught, but his work has a profound
impact in our world.
32 Electromagnetic Field Theory
indeed satisfies (3.3.11). The above is just a convolutional integral. Hence, the potential Φ(r)
due to an arbitrary source distribution %(r) can be found by using convolution, namely,
Φ(r) = 1
4πε ˚
V
%(r0)
|r − r0|
dV 0
(3.3.20)
In a nutshell, the solution of Poisson’s equation when it is driven by an arbitrary source %, is
the convolution of the source with the static Green’s function, a point source response.
3.3.4 Laplace’s Equation
If % = 0, or if we are in a source-free region,
∇2Φ =0 (3.3.21)
which is the Laplace’s equation. Laplace’s equation is usually solved as a boundary value
problem. In such a problem, the potential Φ is stipulated on the boundary of a region, and
then the solution is sought in the intermediate region so as to match the boundary condition.
Examples of such boundary value problems are given below.
Example 1
A capacitor has two parallel plates attached to a battery, what is E field inside the capacitor?
First, one guess the electric field between the two parallel plates. Then one arrive at a
potential Φ in between the plates so as to produce the field. Then the potential is found so
as to match the boundary conditions of Φ = V in the upper plate, and Φ = 0 in the lower
plate. What is the Φ that will satisfy the requisite boundary condition?
Figure 3.3: Figure of a parallel plate capacitor. The field in between can be found by solving
Laplace’s equation as a boundary value problem [29].
Example 2
A coaxial cable has two conductors. The outer conductor is grounded and hence is at zero
potential. The inner conductor is at voltage V . What is the solution?
33 Electromagnetic Field Theory
For this, one will have to write the Laplace’s equation in cylindrical coordinates, namely,
∇2Φ = 1
ρ
∂
∂ρ
ρ
∂Φ
∂ρ
+
1
ρ
2
∂
2Φ
∂φ2
= 0 (3.3.22)
In the above, we assume that the potential is constant in the z direction, and hence, ∂/∂z = 0,
and ρ, φ, z are the cylindrical coordinates. By assuming axi-symmetry, we can let ∂/∂φ = 0
and the above becomes
∇2Φ = 1
ρ
∂
∂ρ
ρ
∂Φ
∂ρ
= 0 (3.3.23)
Show that Φ = A ln ρ + B is a general solution to Laplace’s equation in cylindrical coor-
dinates inside a coax. What is the Φ that will satisfy the requisite boundary condition?
Figure 3.4: The field in between a coaxial line can also be obtained by solving Laplace’s
equation as a boundary value problem (courtesy of Ramo, Whinnery, and Van Duzer).
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