Fundamentals of electromagnetic theory
Reference:
Slides of EE4C05, TUD
Jin J M. Theory and computation of electromagnetic fields
Content
- Electromagnetic quantities (forces and fields)
- Maxwell's equations in time domain
- Constitutive relations
- Poynting vector and Poynting theorem in time domain
- Maxwell's equations in frequency domain
- Maxwell's equations in phasor domain
- Complex permittivity, permeability
- Poynting theorem in phasor domain
- Uniqueness theorem
- Boundary conditions
Electromagnetic quantities (forces and fields)
Before fields were introduced: force
Coulomb observed electric force:
f
⃗
e
=
k
e
q
1
q
2
r
2
r
^
\vec{f}_e=k_e\frac{q_1q_2}{r^2}\hat{r}
f
e=ker2q1q2r^
Define electric field intensity
e
⃗
\vec{e}
e
as the force on the charge
q
q
q when
q
=
1
C
q=1\rm C
q=1C, i.e., force per unit charge (newtons/coulomb):
e
⃗
=
f
⃗
e
q
=
1
4
π
ε
0
Q
r
2
r
^
\vec{e}=\frac{\vec{f}_e}{q}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}
e
=qf
e=4πε01r2Qr^
Magnetic force (empirical relation):
f
⃗
m
=
q
v
⃗
×
b
⃗
\vec{f}_m=q\vec v \times \vec b
f
m=qv
×b
The magnetic flux density b ⃗ ( Newtons C ⋅ m / s = T ( T e s l a ) ) \vec{b}(\frac{\text{Newtons}}{\rm C \cdot m/s}=\rm T(Tesla)) b (C⋅m/sNewtons=T(Tesla)) is defined from the equation above.
If there is both electric and magnetic fields, the total electromagnetic force is
f
⃗
=
f
⃗
e
+
f
⃗
m
=
q
e
⃗
+
q
v
⃗
×
b
⃗
(Lorentz force)
\vec f=\vec f_e+\vec f_m=q\vec e+q\vec v \times \vec b\qquad\text{(Lorentz force)}
f
=f
e+f
m=qe
+qv
×b
(Lorentz force)
| f ⃗ e \vec f_e f e | f ⃗ m \vec f_m f m |
|---|---|
| Always in the direction of e ⃗ \vec e e | Always perpendicular to b ⃗ \vec b b |
| Acts on a charge whenever or not it is moving | Acts on a charge only if it is moving |
| Expends energy when displacing a charge: d W = f ⃗ e ⋅ d l ⃗ dW=\vec f_e \cdot d\vec l dW=f e⋅dl | Expends no energy when displacing a charge: d W = f ⃗ m ⋅ d l ⃗ = ( f ⃗ m ⋅ v ⃗ ) d t = 0 dW=\vec f_m\cdot d\vec l=(\vec f_m \cdot \vec v)dt=0 dW=f m⋅dl =(f m⋅v )dt=0 ( b ⃗ \vec b b can’t change the speed of a charge but only its direction) |
Introducing volume charge density ρ \rho ρ and current density j ⃗ \vec j j :
The kinetic energy of a moving particle (charge):
E
k
=
1
2
m
v
2
=
1
2
m
v
⃗
⋅
v
⃗
E_k=\frac{1}{2}mv^2=\frac{1}{2}m\vec v \cdot \vec v
Ek=21mv2=21mv
⋅v
Power (time-rate of work done by the force on the charge) is the time derivative of the kinetic energy:
P
(
t
)
=
d
E
k
i
n
d
t
=
1
2
m
d
d
t
v
⃗
⋅
v
⃗
=
m
v
⃗
⋅
d
v
⃗
d
t
=
a
f
⃗
⋅
v
⃗
=
b
(
q
e
⃗
+
q
v
⃗
×
b
⃗
)
⋅
v
⃗
=
c
q
e
⃗
⋅
v
⃗
,
\begin{aligned} P(t)&=\frac{dE_{kin}}{dt}=\frac{1}{2}m\frac{d}{dt}\vec v \cdot \vec v=m\vec v\cdot \frac{d\vec v}{dt}\stackrel{a}=\vec f \cdot \vec v\\ &\stackrel{b}=(q\vec e+q\vec v \times \vec b)\cdot \vec v\stackrel{c}=q\vec e \cdot \vec v, \end{aligned}
P(t)=dtdEkin=21mdtdv
⋅v
=mv
⋅dtdv
=af
⋅v
=b(qe
+qv
×b
)⋅v
=cqe
⋅v
,
where
=
a
\stackrel{a}=
=a is due to Newton’s equation
m
d
v
⃗
d
t
=
f
⃗
m\frac{d\vec v}{dt}=\vec f
mdtdv
=f
;
=
b
\stackrel{b}=
=b is due to Lorentz force
f
⃗
=
q
e
⃗
+
q
v
⃗
×
b
⃗
\vec f=q\vec e+q\vec v \times \vec b
f
=qe
+qv
×b
;
=
c
\stackrel{c}=
=c is due to
v
⃗
⋅
(
v
⃗
×
b
⃗
)
=
0
\vec v \cdot(\vec v \times \vec b)=0
v
⋅(v
×b
)=0.
Introducing volume charge density ρ \rho ρ and current density j ⃗ = ρ v ⃗ \vec j=\rho \vec v j =ρv
Power per unit volume:
P
(
t
)
=
ρ
e
⃗
⋅
v
⃗
P(t)=\rho \vec e \cdot \vec v
P(t)=ρe
⋅v
Total power:
P
=
q
e
⃗
⋅
v
⃗
=
∭
V
e
⃗
⋅
ρ
v
⃗
d
V
=
∭
V
e
⃗
⋅
j
⃗
d
V
P=q\vec e\cdot \vec v=\iiint \limits_{V} \vec e\cdot \rho\vec v d V=\iiint \limits_{V} \vec e \cdot \vec j dV
P=qe
⋅v
=V∭e
⋅ρv
dV=V∭e
⋅j
dV
power expended by (or lost from) the fields and converted into kinetic energy of the charges, or heat
| Electromagnetic quantities | Symbol | Units |
|---|---|---|
| Electric field | e ⃗ ( r ⃗ , t ) \vec e(\vec r,t) e (r ,t) | [ V / m ] volt per meter [\rm{V/m}]\text{ volt per meter} [V/m] volt per meter |
| Magnetic field | h ⃗ ( r ⃗ , t ) \vec h(\vec r,t) h (r ,t) | [ A / m ] ampere per meter [\rm{A/m}]\text{ ampere per meter} [A/m] ampere per meter |
| Electric flux density | d ⃗ ( r ⃗ , t ) \vec d(\vec r,t) d (r ,t) | [ C / m 2 ] coulomb per square meter [\rm{C/m^2}]\text{ coulomb per square meter} [C/m2] coulomb per square meter |
| Magnetic flux density | b ⃗ ( r ⃗ , t ) \vec b(\vec r,t) b (r ,t) | [ W e b e r / m 2 ] weber per square meter [\rm{Weber/m^2}]\text{ weber per square meter} [Weber/m2] weber per square meter |
| Volume charge density | ρ ( r ⃗ , t ) = lim Δ V → 0 Charge q in Δ V Δ V \rho(\vec r,t)=\lim \limits_{\Delta V\to 0}\frac{\text{Charge }q\text{ in }\Delta V}{\Delta V} ρ(r ,t)=ΔV→0limΔVCharge q in ΔV | [ C / m 3 ] coulomb per cube meter [\rm{C/m^3}]\text{ coulomb per cube meter} [C/m3] coulomb per cube meter |
| Current density | j ⃗ ( r ⃗ , t ) = lim Δ S → 0 Flux of current through Δ S Δ S \vec j(\vec r,t)=\lim \limits_{\Delta S\to 0}\frac{\text{Flux of current through }\Delta S}{\Delta S} j (r ,t)=ΔS→0limΔSFlux of current through ΔS | [ A / m 2 ] ampere per square meter [\rm{A/m^2}]\text{ ampere per square meter} [A/m2] ampere per square meter |
The two definition of j ⃗ \vec j j ( j ⃗ = ρ v ⃗ \vec j=\rho \vec v j =ρv and j ⃗ ( r ⃗ , t ) = lim Δ S → 0 Flux of current through Δ S Δ S \vec j(\vec r,t)=\lim \limits_{\Delta S\to 0}\frac{\text{Flux of current through }\Delta S}{\Delta S} j (r ,t)=ΔS→0limΔSFlux of current through ΔS) are equivalent:
Since q = ∭ V ρ d V = ∫ t ( ∬ S ρ v ⃗ d S ⃗ ) d t q=\iiint_{V} \rho dV=\int_t(\iint_S\rho \vec vd\vec S)dt q=∭VρdV=∫t(∬Sρv dS )dt, we have I = d q d t = ∬ S ρ v ⃗ d S ⃗ = ∬ S j ⃗ d S ⃗ I=\frac{dq}{dt}=\iint_S\rho \vec vd\vec S=\iint_S \vec jd\vec S I=dtdq=∬Sρv dS =∬Sj dS , which is equivalent to j ⃗ ( r ⃗ , t ) = lim Δ S → 0 Flux of current through Δ S Δ S \vec j(\vec r,t)=\lim \limits_{\Delta S\to 0}\frac{\text{Flux of current through }\Delta S}{\Delta S} j (r ,t)=ΔS→0limΔSFlux of current through ΔS.
Maxwell’s equations in time domain
Electromagnetism is based on a set of four equations: Maxwell’s equations
- connect fields with sources ( e ⃗ / d ⃗ \vec e/\vec d e /d with ρ \rho ρ and h ⃗ / b ⃗ \vec h/\vec b h /b with j ⃗ \vec j j )
- describe the coupling between electric and magnetic fields
| Differential form | Integral form | |
|---|---|---|
| Gauss’s law | ∇ ⋅ d ⃗ = ρ \nabla \cdot \vec d=\rho ∇⋅d =ρ | ∯ S d ⃗ ⋅ d S ⃗ = q \oiint_S\vec d \cdot d \vec S=q ∬ Sd ⋅dS =q |
| Gauss’s law for magnetism | ∇ ⋅ b ⃗ = 0 \nabla \cdot \vec b=0 ∇⋅b =0 | ∯ b ⃗ ⋅ d S ⃗ = 0 \oiint \vec b \cdot d \vec S=0 ∬ b ⋅dS =0 |
| Faraday’s law | ∇ × e ⃗ = − ∂ b ⃗ ∂ t \nabla \times \vec e=-\frac{\partial \vec b}{\partial t} ∇×e =−∂t∂b | ∮ C e ⃗ ⋅ d l ⃗ = − ∂ ∂ t ∬ S b ⃗ ⋅ d S ⃗ \oint_C \vec e \cdot d\vec l=-\frac{\partial}{\partial t}\iint_S\vec b \cdot d \vec S ∮Ce ⋅dl =−∂t∂∬Sb ⋅dS |
| Ampere - Maxwell’s law | ∇ × h ⃗ = j ⃗ + ∂ d ⃗ ∂ t \nabla \times \vec h=\vec j+\frac{\partial \vec d}{\partial t} ∇×h =j +∂t∂d | ∮ C h ⃗ ⋅ d l ⃗ = I + ∂ ∂ t ∬ S d ⃗ ⋅ d S ⃗ \oint_C \vec h \cdot d\vec l=I+\frac{\partial}{\partial t}\iint_S\vec d \cdot d \vec S ∮Ch ⋅dl =I+∂t∂∬Sd ⋅dS |
Equations above are asymmetric because of the lack of magnetic currents and charges. Although magnetic currents and charges do not exist or have not been found so far in reality, the concepts of such currents and charges are useful because sometimes we can introduce equivalent magnetic currents and charges to simplify the analysis of some electromagnetic problems. Therefore, sometimes we write Maxwell’s equations as
| Differential form | Integral form | |
|---|---|---|
| Gauss’s law | ∇ ⋅ d ⃗ = ρ e \nabla \cdot \vec d=\rho_e ∇⋅d =ρe | ∯ S d ⃗ ⋅ d S ⃗ = ∭ V ρ e d V \oiint_S\vec d \cdot d \vec S=\iiint_V\rho_edV ∬ Sd ⋅dS =∭VρedV |
| Gauss’s law for magnetism | ∇ ⋅ b ⃗ = ρ m \nabla \cdot \vec b=\rho_m ∇⋅b =ρm | ∯ b ⃗ ⋅ d S ⃗ = ∭ V ρ m d V \oiint \vec b \cdot d \vec S=\iiint_V\rho_mdV ∬ b ⋅dS =∭VρmdV |
| Faraday’s law | ∇ × e ⃗ = − m ⃗ − ∂ b ⃗ ∂ t \nabla \times \vec e=-\vec m-\frac{\partial \vec b}{\partial t} ∇×e =−m −∂t∂b | ∮ C e ⃗ ⋅ d l ⃗ = − ∬ S m ⃗ ⋅ d S ⃗ − ∂ ∂ t ∬ S b ⃗ ⋅ d S ⃗ \oint_C \vec e \cdot d\vec l=-\iint_S\vec m\cdot d \vec S-\frac{\partial}{\partial t}\iint_S\vec b \cdot d \vec S ∮Ce ⋅dl =−∬Sm ⋅dS −∂t∂∬Sb ⋅dS |
| Ampere - Maxwell’s law | ∇ × h ⃗ = j ⃗ + ∂ d ⃗ ∂ t \nabla \times \vec h=\vec j+\frac{\partial \vec d}{\partial t} ∇×h =j +∂t∂d | ∮ C h ⃗ ⋅ d l ⃗ = ∬ S j ⃗ ⋅ d S ⃗ + ∂ ∂ t ∬ S d ⃗ ⋅ d S ⃗ \oint_C \vec h \cdot d\vec l=\iint_S\vec j \cdot d \vec S+\frac{\partial}{\partial t}\iint_S\vec d \cdot d \vec S ∮Ch ⋅dl =∬Sj ⋅dS +∂t∂∬Sd ⋅dS |
To understand the meaning of those laws, we require the concept of flux.
Interpretation: flux of a vector field
flux of
e
⃗
\vec e
e
through the surface
S
S
S:
∬
S
e
⃗
⋅
d
S
⃗
\iint \limits_S \vec e \cdot d\vec S
S∬e
⋅dS
Note that the direction of the electric field line is the direction of vector e ⃗ \vec e e , and the density of electric field lines indicate ∣ e ⃗ ∣ |\vec e| ∣e ∣, flux of a vector field can describe how many field lines are coming out of a surface.
Meaning of Gauss’s law
Integral form: ∯ S d ⃗ ⋅ d S ⃗ = q \oiint_S \vec d\cdot d\vec S=q ∬ Sd ⋅dS =q
Electric charge produces an electric field, and the flux of that field passing through any closed surface is proportional to the total charge contained within that surface.
Meaning of Gauss’s law for magnetism
Integral form: ∯ S b ⃗ ⋅ d S ⃗ = 0 \oiint_S \vec b\cdot d\vec S=0 ∬ Sb ⋅dS =0
The flux of that field passing through any closed surface is always zero, so there is no single magnetic charge (monopole).
Meaning of Faraday’s law
Differential form: ∇ × e ⃗ = − ∂ b ⃗ ∂ t \nabla \times \vec e=-\frac{\partial \vec b}{\partial t} ∇×e =−∂t∂b
A time-varying magnetic field generates an electrical field.
Integral form: ∮ C e ⃗ ⋅ d l ⃗ = − ∂ ∂ t ∬ S b ⃗ ⋅ d S ⃗ \oint_C \vec e \cdot d\vec l=-\frac{\partial}{\partial t}\iint_S\vec b \cdot d \vec S ∮Ce ⋅dl =−∂t∂∬Sb ⋅dS
A magnetic flux changing with time can produce an electric current in a closed loop.
Meaning of Ampere-Maxwell’s law
Differential form: ∇ × h ⃗ = ∂ d ⃗ ∂ t + j ⃗ \nabla \times \vec h=\frac{\partial \vec d}{\partial t}+\vec j ∇×h =∂t∂d +j
The magnetic field is generated by moving charges and time-varying electrical fields.
Integral form: ∮ C h ⃗ ⋅ d l ⃗ = I + ∬ S ∂ d ⃗ ∂ t ⋅ d S ⃗ \oint_C \vec h \cdot d \vec l=I+\iint_S\frac{\partial \vec d}{\partial t}\cdot d \vec S ∮Ch ⋅dl =I+∬S∂t∂d ⋅dS
S is the surface that the total current and total electric flux pass through, C is the contour of S
Ampere’s law (first term, conduction current): A conductor wire carrying a current I I I produces a magnetic field h ⃗ \vec h h
Maxwell (second term, displacement current): Since in a capacitor there is no current, there must be a magnetic field generated by the varying E-field between the plates (equivalent to a current)
Example:
According to Gauss’s law, the charge on the plates of the capacitor is
q
=
∯
S
d
⃗
⋅
d
S
⃗
q=\oiint_S \vec d \cdot d \vec S
q=∬
Sd
⋅dS
. Therefore,
∂
q
⃗
∂
t
=
∯
S
∂
d
⃗
∂
t
⋅
d
S
⃗
=
I
d
.
\frac{\partial \vec q}{\partial t}=\oiint_S \frac{\partial\vec d}{\partial t}\cdot d \vec S=I_d.
∂t∂q
=∬
S∂t∂d
⋅dS
=Id.
This term can be interpreted as an equivalent current (displacement current).
Continuity equation
The intuitive fact that the flux of current density through the surface enclosing a volume is equal to a negative variation of charge inside the volume can be proved by Ampere - Maxwell’s law and Gauss’s law.
Ampere - Maxwell’s law: ∇ × h ⃗ = j ⃗ + ∂ d ⃗ ∂ t \nabla \times \vec h=\vec j+\frac{\partial \vec d}{\partial t} ∇×h =j +∂t∂d
Applying the divergence to both sides, and since
∇
⋅
(
∇
×
a
⃗
)
=
0
\nabla \cdot(\nabla \times \vec a)=0
∇⋅(∇×a
)=0,
∇
⋅
(
∇
×
h
⃗
)
=
∇
⋅
j
⃗
+
∇
⋅
∂
d
⃗
∂
t
=
0.
\nabla \cdot(\nabla \times \vec h)=\nabla \cdot \vec j+\nabla \cdot \frac{\partial \vec d}{\partial t}=0.
∇⋅(∇×h
)=∇⋅j
+∇⋅∂t∂d
=0.
According to Gauss’s law:
∇
⋅
d
⃗
=
ρ
\nabla \cdot \vec d=\rho
∇⋅d
=ρ, we have
∇
⋅
j
⃗
=
−
∂
ρ
∂
t
.
\nabla \cdot \vec j=-\frac{\partial\rho}{\partial t}.
∇⋅j
=−∂t∂ρ.
Applying divergence theorem, we obtain
∯
S
j
⃗
⋅
d
S
⃗
=
−
∂
q
∂
t
\oiint_S \vec j \cdot d \vec S=-\frac{\partial q}{\partial t}
∬
Sj
⋅dS
=−∂t∂q
Example:
If there is no charge, i.e., ∇ ⋅ j ⃗ = 0 , ∯ S j ⃗ ⋅ d S ⃗ = 0 \nabla \cdot \vec j=0,\oiint_S \vec j \cdot d \vec S=0 ∇⋅j =0,∬ Sj ⋅dS =0,
Constitutive relations
Maxwell equations are 8 ( 2 + 2 × 3 2+2\times3 2+2×3) scalar equations, but unknowns are 16 ( 5 × 3 + 1 5 \times 3+1 5×3+1). We need additional equations.
Flux densities d ⃗ , b ⃗ \vec d,\vec b d ,b are related to the fields e ⃗ , h ⃗ \vec e,\vec h e ,h via the so-called constitutive relations.
The mechanism behind it: see Theory and Computation of Electromagnetic Fields 1.3
For the cases involving simple media, the constitutive relations reduce the M.E., which can be rewritten with 7 unknowns:
∇
⋅
e
⃗
=
ρ
ε
∇
⋅
h
⃗
=
0
∇
×
e
⃗
=
−
μ
∂
h
⃗
∂
t
∇
×
h
⃗
=
ε
∂
e
⃗
∂
t
+
j
⃗
c
+
j
⃗
i
\begin{aligned} \nabla \cdot \vec e&=\frac{\rho}{\varepsilon}\\ \nabla \cdot \vec h&=0\\ \nabla \times \vec e&=-\mu\frac{\partial \vec h}{\partial t}\\ \nabla \times \vec h&=\varepsilon \frac{\partial\vec e}{\partial t}+\vec j_c+\vec j_i \end{aligned}
∇⋅e
∇⋅h
∇×e
∇×h
=ερ=0=−μ∂t∂h
=ε∂t∂e
+j
c+j
i
j
⃗
c
\vec j_c
j
c: Conduction currents. Currents sustained by the field
j ⃗ i : \vec j_i: j i: Impressed currents. Actual source of the field.
What is the conduction current?
In presence of external electric field, electrons move in the direction opposite to this field
Average velocity of the electrons: electron drift velocity
v
⃗
d
\vec v_d
v
d
v
⃗
d
=
−
μ
e
e
⃗
μ
e
electron mobility
[
m
2
V
s
]
\vec v_d=-\mu_e \vec e\\ \mu_e \text{ electron mobility }\left[\frac{m^2}{Vs}\right]
v
d=−μee
μe electron mobility [Vsm2]
The current is defined as the volume charge density times the velocity
j
⃗
c
=
ρ
e
v
⃗
d
=
−
ρ
e
μ
e
e
⃗
.
\vec j_c=\rho_e \vec v_d=-\rho_e \mu_e \vec e.
j
c=ρev
d=−ρeμee
.
−
μ
e
ρ
e
-\mu_e \rho_e
−μeρe is defined as conductivity
σ
\sigma
σ
j
⃗
c
=
σ
e
⃗
(Ohm’s law)
\vec j_c=\sigma \vec e \qquad \text{(Ohm's law)}
j
c=σe
(Ohm’s law)
Perfect conductor:
σ
=
∞
\sigma=\infty
σ=∞
Perfect dielectric: σ = 0 \sigma=0 σ=0
Linear medium
The electric and magnetic flux may be responding not only to the electric and magnetic fields but also to their rate of variations in time (dispersion).
These phenomena have a role in dielectric and magnetic losses.
Characterization can become tensorial and integral when the dielectric and magnetic permeability are also not constant as a function of the space.
Poynting vector and Poynting theorem in time domain
(mainly from Theory and Computation of Electromagnetic Fields 1.6)
Consider a medium characterized by permittivity
ε
\varepsilon
ε, permeability
μ
\mu
μ, and conductivity
σ
\sigma
σ. Maxwell’s equation in such a medium can be written as
∇
×
e
⃗
=
−
μ
∂
h
⃗
∂
t
−
m
⃗
i
∇
×
h
⃗
=
ε
∂
e
⃗
∂
t
+
σ
e
⃗
+
j
⃗
i
(PT.1 & PT.2)
\begin{aligned} \nabla \times \vec e&=-\mu\frac{\partial \vec h}{\partial t}-\vec m_i \\ \nabla \times \vec h&=\varepsilon \frac{\partial \vec e}{\partial t}+\sigma\vec e+\vec j_i \tag{PT.1 \& PT.2} \end{aligned}
∇×e
∇×h
=−μ∂t∂h
−m
i=ε∂t∂e
+σe
+j
i(PT.1 & PT.2)
where
J
⃗
i
\vec J_i
J
i and
M
⃗
i
\vec M_i
M
i represent the actual source of the field and are often referred to as the impressed currents.
By taking the dot product of Equation (PT.1) with
h
⃗
\vec h
h
and the dot product of (PT.2) with
e
⃗
\vec e
e
and subtracting the latter from the former, we obtain
KaTeX parse error: Undefined control sequence: \partialial at position 103: … e \cdot \frac{\̲p̲a̲r̲t̲i̲a̲l̲i̲a̲l̲ ̲\vec e}{\partia…
Using the vector identity
∇
⋅
(
e
⃗
×
h
⃗
)
=
h
⃗
⋅
(
∇
×
e
⃗
)
−
e
⃗
⋅
(
∇
×
h
⃗
)
\nabla\cdot(\vec e\times \vec h)=\vec h \cdot (\nabla \times \vec e)-\vec e\cdot(\nabla\times \vec h)
∇⋅(e
×h
)=h
⋅(∇×e
)−e
⋅(∇×h
), it can also be written as
∇
⋅
(
e
⃗
×
h
⃗
)
+
ε
e
⃗
⋅
∂
e
⃗
∂
t
+
μ
h
⃗
⋅
∂
h
⃗
∂
t
+
σ
e
⃗
⋅
e
⃗
+
e
⃗
⋅
j
⃗
i
+
h
⃗
⋅
m
⃗
i
=
0
(PT.4)
\nabla\cdot(\vec e\times \vec h)+\varepsilon \vec e \cdot\frac{\partial \vec e}{\partial t}+\mu\vec h\cdot\frac{\partial \vec h}{\partial t}+\sigma \vec e\cdot\vec e+\vec e\cdot \vec j_i+\vec h\cdot \vec m_i=0 \tag{PT.4}
∇⋅(e
×h
)+εe
⋅∂t∂e
+μh
⋅∂t∂h
+σe
⋅e
+e
⋅j
i+h
⋅m
i=0(PT.4)
To understand the physical meaning of this equation, we first integrate it over a finite volume, and by using Gauss’s theorem, we obtain
∯
S
(
e
⃗
×
h
⃗
)
⋅
d
S
⃗
+
∭
V
(
ε
e
⃗
⋅
∂
e
⃗
∂
t
+
μ
h
⃗
⋅
∂
h
⃗
∂
t
+
σ
e
⃗
⋅
e
⃗
+
e
⃗
⋅
j
⃗
i
+
h
⃗
⋅
m
⃗
i
)
d
V
=
0
(PT.5)
\oiint_S(\vec e \times \vec h)\cdot d\vec S+\iiint_V(\varepsilon\vec e\cdot \frac{\partial \vec e}{\partial t}+\mu\vec h\cdot\frac{\partial \vec h}{\partial t}+\sigma\vec e\cdot \vec e+\vec e \cdot \vec j_i+\vec h\cdot \vec m_i)dV=0 \tag{PT.5}
∬
S(e
×h
)⋅dS
+∭V(εe
⋅∂t∂e
+μh
⋅∂t∂h
+σe
⋅e
+e
⋅j
i+h
⋅m
i)dV=0(PT.5)
Then we check the unit of each term.
-
e ⃗ × h ⃗ \vec e \times \vec h e ×h has a unit of volts/meter ⋅ amperes/meter = watts/meter 2 \text{volts/meter}\cdot\text{amperes/meter}=\text{watts/meter}^2 volts/meter⋅amperes/meter=watts/meter2, which is the unit of power flux density. A dot product with d S ⃗ d\vec S dS and then integration over a closed surface S S S would yield total power exiting the surface (the direction of S ⃗ \vec S S is outward). We denote this term as P e \mathcal{P}_e Pe:
P e = ∯ S ( e ⃗ × h ⃗ ) ⋅ d S ⃗ (PT.6) \mathcal{P}_e=\oiint_S(\vec e \times \vec h)\cdot d\vec S \tag{PT.6} Pe=∬ S(e ×h )⋅dS (PT.6) -
Rewrite
ε e ⃗ ⋅ ∂ e ⃗ ∂ t = 1 2 ε ∂ e 2 ∂ t = ∂ ∂ t ( 1 2 ε e 2 ) = ∂ w e ∂ t (PT.7) \varepsilon\vec e\cdot \frac{\partial \vec e}{\partial t}=\frac{1}{2}\varepsilon\frac{\partial e^2}{\partial t}=\frac{\partial}{\partial t}(\frac{1}{2}\varepsilon e^2)=\frac{\partial \mathcal{w}_e}{\partial t} \tag{PT.7} εe ⋅∂t∂e =21ε∂t∂e2=∂t∂(21εe2)=∂t∂we(PT.7)
where w e = 1 2 ε e 2 \mathcal{w}_e=\frac{1}{2}\varepsilon e^2 we=21εe2. This quantity has a unit of farads/meter ⋅ ( volts/meter ) 2 = joules/meter 3 \text{farads/meter} \cdot(\text{volts/meter})^2= \text{joules/meter}^ 3 farads/meter⋅(volts/meter)2=joules/meter3, which represents energy density. Its integral over a volume would represent the total energy in the volume:
W e = ∭ V w e d V = 1 2 ∭ V ε e 2 d V . (PT.8) \mathcal W_e=\iiint_V\mathcal{w}_edV=\frac{1}{2}\iiint_V\varepsilon e^2dV.\tag{PT.8} We=∭VwedV=21∭Vεe2dV.(PT.8)
Since this energy is associated with the electric field, it can be termed as the electric energy.Similarly, we find that
μ h ⃗ ⋅ ∂ h ⃗ ∂ t = 1 2 μ ∂ h 2 ∂ t = ∂ ∂ t ( 1 2 μ h 2 ) = ∂ w m ∂ t (PT.9) \mu\vec h\cdot \frac{\partial \vec h}{\partial t}=\frac{1}{2}\mu\frac{\partial h^2}{\partial t}=\frac{\partial}{\partial t}(\frac{1}{2}\mu h^2)=\frac{\partial \mathcal{w}_m}{\partial t}\tag{PT.9} μh ⋅∂t∂h =21μ∂t∂h2=∂t∂(21μh2)=∂t∂wm(PT.9)
where w m = 1 2 μ h 2 \mathcal{w}_m=\frac{1}{2}\mu h^2 wm=21μh2 represents the magnetic energy density. Its integration over a volume represents the total magnetic energy in the volume:
W m = ∭ V w m d V = 1 2 ∭ V μ h 2 d V . (PT.10) \mathcal W_m=\iiint_V\mathcal{w}_mdV=\frac{1}{2}\iiint_V\mu h^2dV.\tag{PT.10} Wm=∭VwmdV=21∭Vμh2dV.(PT.10)
Then we have
∭ V ( ε e ⃗ ⋅ ∂ e ⃗ ∂ t + μ h ⃗ ⋅ ∂ h ⃗ ∂ t ) d V = d d t ( W e + W m ) (PT.11) \iiint_V(\varepsilon\vec e\cdot \frac{\partial \vec e}{\partial t}+\mu\vec h\cdot\frac{\partial \vec h}{\partial t})dV=\frac{d}{dt}(\mathcal{W}_e+\mathcal{W}_m)\tag{PT.11} ∭V(εe ⋅∂t∂e +μh ⋅∂t∂h )dV=dtd(We+Wm)(PT.11)
which is the rate of increase in the total energy in the volume. -
With the interpretations above, we can readily find that
P d = ∭ V σ e 2 d V (PT.12) \mathcal{P}_d=\iiint_V\sigma e^2dV\tag{PT.12} Pd=∭Vσe2dV(PT.12)
represents the power dissipated in the volume. -
And
P s = − ∭ V ( e ⃗ ⋅ j ⃗ i + h ⃗ ⋅ m ⃗ i ) d V (PT.13) \mathcal{P}_s=-\iiint_V(\vec e \cdot \vec j_i+\vec h\cdot \vec m_i)dV \tag{PT.13} Ps=−∭V(e ⋅j i+h ⋅m i)dV(PT.13)
represents the power supplied by the source.
By using these notations, Equation (PT.5) can be written as
P
s
=
P
e
+
P
d
+
d
d
t
(
W
e
+
W
m
)
(PT.14)
\mathcal{P}_s=\mathcal {P}_e+\mathcal{P}_d+\frac{d}{dt}(\mathcal{W}_e+\mathcal{W}_m) \tag{PT.14}
Ps=Pe+Pd+dtd(We+Wm)(PT.14)
which states that in a volume the supplied power must be equal to the sum of the exiting power, the dissipated power, and the rate of increase in the total energy in the volume. Obviously, Equation (PT.14) is the statement of the conservation of energy for electromagnetic fields, which is also known as Poynting’s theorem.
Poynting’s theorem establishes a relation between five quantities. Knowing any four quantities, the remaining quantity can be calculated easily. This can be useful in a variety of applications where the desired quantity cannot be measured directly, but can be evaluated indirectly.
As illustrated above,
e
⃗
×
h
⃗
\vec e \times \vec h
e
×h
represents the power flux density in the direction determined by the cross-product. This quantity is named the Poynting vector, defined as
s
⃗
=
e
⃗
×
h
⃗
(PT.15)
\vec s=\vec e \times \vec h \tag{PT.15}
s
=e
×h
(PT.15)
which indicates that once both the electric and magnetic fields are known at any point in space, the power flow density is determined and the power flow is perpendicular to the directions of the electric and magnetic fields.
Summary
Maxwell’s equations in frequency domain
Fourier Transform (FT)
A signal, function of time, can be represented via its Fourier Transform in the frequency domain
Direct Transform:
F
(
ω
)
=
∫
−
∞
∞
f
(
t
)
e
−
j
ω
t
d
t
F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt
F(ω)=∫−∞∞f(t)e−jωtdt
Inverse form:
f
(
t
)
=
1
2
π
∫
−
∞
∞
F
(
ω
)
e
j
ω
t
d
ω
f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{j\omega t}d\omega
f(t)=2π1∫−∞∞F(ω)ejωtdω
Most powerful advantage of spectral representations:
- easiness in performing derivations and integrations in the transformed domains
- the time dependence only in exponent
∂ ∂ t f ( t ) = 1 2 π ∫ − ∞ ∞ F ( ω ) ∂ ∂ t e j ω t d ω = 1 2 π ∫ − ∞ ∞ j ω F ( ω ) e j ω t d ω ∫ − ∞ t f ( t ) d t = 1 2 π ∫ − ∞ ∞ F ( ω ) ( ∫ − ∞ t e j ω t d t ) d ω = 1 2 π ∫ − ∞ ∞ F ( ω ) j ω e j ω t d ω \frac{\partial }{\partial t}f(t)=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)\frac{\partial}{\partial t}e^{j\omega t}d\omega=\frac{1}{2\pi}\int_{-\infty}^\infty j\omega F(\omega)e^{j\omega t}d\omega\\ \int_{-\infty}^t f(t)dt=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)(\int_{-\infty}^t e^{j\omega t}dt)d\omega=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{F(\omega)}{j\omega}e^{j\omega t}d\omega ∂t∂f(t)=2π1∫−∞∞F(ω)∂t∂ejωtdω=2π1∫−∞∞jωF(ω)ejωtdω∫−∞tf(t)dt=2π1∫−∞∞F(ω)(∫−∞tejωtdt)dω=2π1∫−∞∞jωF(ω)ejωtdω
time domain spectral domain ∇ ⋅ e ⃗ ( r ⃗ , t ) = ρ e ( t ) ε ∇ ⋅ E ⃗ ( r ⃗ , ω ) = ϱ e ( ω ) ε ∇ ⋅ h ⃗ ( r ⃗ , t ) = ρ m ( t ) μ ∇ ⋅ H ⃗ ( r ⃗ , ω ) = ϱ m ( ω ) ε ∇ × e ⃗ ( r ⃗ , t ) = − μ ∂ h ⃗ ( r ⃗ , t ) ∂ t − m ⃗ ( r ⃗ , t ) ∇ × E ⃗ ( r ⃗ , ω ) = − j ω μ H ⃗ ( r ⃗ , ω ) − M ⃗ ( r ⃗ , ω ) ∇ × h ⃗ ( r ⃗ , t ) = ε ∂ e ⃗ ( r ⃗ , t ) ∂ t + j ⃗ ( r ⃗ , t ) ∇ × H ⃗ ( r ⃗ , ω ) = j ω ε E ⃗ ( r ⃗ , ω ) + J ⃗ ( r ⃗ , ω ) \begin{aligned} &\text{time domain} &&\text{spectral domain} \\ &\nabla \cdot \vec e(\vec r,t)=\frac{\rho_e(t)}{\varepsilon} && \nabla \cdot \vec E(\vec r,\omega)=\frac{\varrho_e(\omega)}{\varepsilon}\\ &\nabla \cdot \vec h(\vec r,t)=\frac{\rho_m(t)}{\mu} && \nabla \cdot \vec H(\vec r,\omega)=\frac{\varrho_m(\omega)}{\varepsilon}\\ &\nabla \times \vec e(\vec r,t)=-\mu\frac{\partial \vec h(\vec r,t)}{\partial t}-\vec m(\vec r,t) && \nabla \times \vec E(\vec r,\omega)=-j\omega \mu\vec H(\vec r,\omega)-\vec{M}(\vec r,\omega) \\ &\nabla \times \vec h(\vec r,t)=\varepsilon \frac{\partial \vec e(\vec r,t)}{\partial t}+\vec j(\vec r,t) &&\nabla \times \vec H(\vec r,\omega)=j\omega\varepsilon\vec E(\vec r,\omega)+\vec J(\vec r,\omega) \end{aligned} time domain∇⋅e (r ,t)=ερe(t)∇⋅h (r ,t)=μρm(t)∇×e (r ,t)=−μ∂t∂h (r ,t)−m (r ,t)∇×h (r ,t)=ε∂t∂e (r ,t)+j (r ,t)spectral domain∇⋅E (r ,ω)=εϱe(ω)∇⋅H (r ,ω)=εϱm(ω)∇×E (r ,ω)=−jωμH (r ,ω)−M (r ,ω)∇×H (r ,ω)=jωεE (r ,ω)+J (r ,ω)
Note: ∇ \nabla ∇ operates on r ⃗ \vec r r , so the structure keeps when we do Fourier Transform.
E.g.,
3
r
d
3^{rd}
3rd eq.
∇
×
e
⃗
(
r
⃗
,
t
)
=
−
μ
∂
h
⃗
(
r
⃗
,
t
)
∂
t
−
m
⃗
(
r
⃗
,
t
)
\nabla \times \vec e(\vec r,t)=-\mu\frac{\partial \vec h(\vec r,t)}{\partial t}-\vec m(\vec r,t)
∇×e
(r
,t)=−μ∂t∂h
(r
,t)−m
(r
,t)
Applying inverse FT,
1
2
π
∫
−
∞
∞
∇
×
E
⃗
(
r
⃗
,
ω
)
e
j
ω
t
d
ω
=
−
μ
2
π
∫
−
∞
∞
j
ω
H
⃗
(
r
⃗
,
ω
)
e
j
ω
t
d
ω
−
1
2
π
∫
−
∞
∞
M
⃗
(
r
⃗
,
ω
)
e
j
ω
t
d
ω
\frac{1}{2\pi}\int_{-\infty}^\infty \nabla \times \vec E(\vec r,\omega)e^{j\omega t}d\omega=-\frac{\mu}{2\pi}\int_{-\infty}^\infty j\omega\vec H(\vec r,\omega)e^{j\omega t}d\omega-\frac{1}{2\pi}\int_{-\infty}^\infty \vec M(\vec r,\omega)e^{j\omega t}d\omega
2π1∫−∞∞∇×E
(r
,ω)ejωtdω=−2πμ∫−∞∞jωH
(r
,ω)ejωtdω−2π1∫−∞∞M
(r
,ω)ejωtdω
Note that if
∫
−
∞
∞
A
(
ω
)
e
j
ω
t
d
ω
=
∫
−
∞
∞
B
(
ω
)
e
j
ω
t
d
ω
\int_{-\infty}^\infty A(\omega)e^{j\omega t}d \omega=\int_{-\infty}^\infty B(\omega)e^{j\omega t}d \omega
∫−∞∞A(ω)ejωtdω=∫−∞∞B(ω)ejωtdω is valid for every
t
t
t, we have
A
(
ω
)
=
B
(
ω
)
A(\omega)=B(\omega)
A(ω)=B(ω). Therefore,
∇
×
E
⃗
(
r
⃗
,
ω
)
=
−
j
ω
μ
H
⃗
(
r
⃗
,
ω
)
−
M
⃗
(
r
⃗
,
ω
)
\nabla \times \vec E(\vec r,\omega)=-j\omega \mu\vec H(\vec r,\omega)-\vec{M}(\vec r,\omega)
∇×E
(r
,ω)=−jωμH
(r
,ω)−M
(r
,ω)
Maxwell’s equations in phasor domain
Let us assume the time variation is a specific one: sinusoidal
ρ
(
r
⃗
,
t
)
=
ρ
(
r
⃗
)
c
o
s
(
ω
t
)
j
(
r
⃗
,
t
)
=
j
(
r
⃗
)
c
o
s
(
ω
t
)
\rho (\vec r,t)=\rho(\vec r)cos(\omega t) \qquad j(\vec r,t)=j(\vec r)cos(\omega t)
ρ(r
,t)=ρ(r
)cos(ωt)j(r
,t)=j(r
)cos(ωt)
According to M.E., if the sources are time harmonic, all electromagnetic quantities are also time harmonic, e.g.,
e
⃗
(
r
⃗
,
t
)
=
e
x
(
r
⃗
,
t
)
x
^
+
e
y
(
r
⃗
,
t
)
y
^
+
e
z
(
r
⃗
,
t
)
z
^
e
x
(
r
⃗
,
t
)
=
e
x
(
r
⃗
)
cos
(
ω
t
+
ϕ
x
)
e
y
(
r
⃗
,
t
)
=
e
y
(
r
⃗
)
cos
(
ω
t
+
ϕ
y
)
e
z
(
r
⃗
,
t
)
=
e
z
(
r
⃗
)
cos
(
ω
t
+
ϕ
z
)
\vec e(\vec r,t)=e_x(\vec r,t)\hat x+e_y(\vec r,t)\hat y+e_z(\vec r,t)\hat z\\ e_x(\vec r,t)=e_x(\vec r)\cos (\omega t+\phi_x)\\ e_y(\vec r,t)=e_y(\vec r)\cos (\omega t+\phi_y)\\ e_z(\vec r,t)=e_z(\vec r)\cos (\omega t+\phi_z)
e
(r
,t)=ex(r
,t)x^+ey(r
,t)y^+ez(r
,t)z^ex(r
,t)=ex(r
)cos(ωt+ϕx)ey(r
,t)=ey(r
)cos(ωt+ϕy)ez(r
,t)=ez(r
)cos(ωt+ϕz)
Time harmonic is the most important case for 2 main reasons:
- The output waveform of most generators is sinusoidal or nearly so.
- Other time variations can be written and superposition of sinusoidal ones, because of the property of the Fourier transform and because the equations are linear.
For sinusoidal (i.e. harmonic) time variation, a more compact notation is introduced: Phasors
For example, the electric field can be written as:
e
⃗
(
r
⃗
,
t
)
=
e
⃗
0
(
r
⃗
)
cos
(
ω
t
+
ϕ
(
r
⃗
)
)
\vec e(\vec r,t)=\vec e_0(\vec r)\cos(\omega t+\phi(\vec r))
e
(r
,t)=e
0(r
)cos(ωt+ϕ(r
))
Using Euler’s formula
e
j
θ
=
cos
(
θ
)
+
j
sin
(
θ
)
e^{j\theta}=\cos(\theta)+j\sin (\theta)
ejθ=cos(θ)+jsin(θ), this can be written as:
e
⃗
(
r
⃗
,
t
)
=
e
⃗
0
(
r
⃗
)
ℜ
{
e
j
ω
t
+
j
ϕ
(
r
⃗
)
}
=
ℜ
{
e
⃗
0
e
j
ϕ
(
r
⃗
)
e
j
ω
t
}
\begin{aligned} \vec e(\vec r,t)&=\vec e_0(\vec r)\Re\{e^{j\omega t+j\phi(\vec r)}\}\\ &=\Re\{\vec e_0e^{j\phi(\vec r)}e^{j\omega t}\} \end{aligned}
e
(r
,t)=e
0(r
)ℜ{ejωt+jϕ(r
)}=ℜ{e
0ejϕ(r
)ejωt}
Now define a complex quantity named phasor
E
⃗
(
r
⃗
)
=
e
⃗
0
e
j
ϕ
(
r
⃗
)
\vec E(\vec r)=\vec e_0 e^{j\phi(\vec r)}
E
(r
)=e
0ejϕ(r
)
which contains both the amplitude and phase of the field and is only a spatial function.
Then
e
⃗
\vec e
e
can be written as
e
⃗
(
r
⃗
,
t
)
=
ℜ
{
E
⃗
(
r
⃗
)
e
j
ω
t
}
\vec e(\vec r,t)=\Re\{\vec E(\vec r)e^{j\omega t}\}
e
(r
,t)=ℜ{E
(r
)ejωt}
We can rewrite M.E. in phasor domain, i.e.,
3
r
d
3^{rd}
3rd eq.
∇
×
e
⃗
=
−
μ
∂
h
⃗
∂
t
−
m
⃗
\nabla \times \vec e=-\mu \frac{\partial \vec h}{\partial t}-\vec m
∇×e
=−μ∂t∂h
−m
Applying phasor definition, we have
ℜ
{
∇
×
E
⃗
(
r
⃗
)
e
j
ω
t
}
=
−
μ
∂
∂
t
ℜ
{
H
⃗
(
r
⃗
)
e
j
ω
t
}
−
ℜ
{
M
⃗
(
r
⃗
)
e
j
ω
t
}
=
−
ℜ
{
j
ω
μ
H
⃗
(
r
⃗
)
e
j
ω
t
}
−
ℜ
{
M
⃗
(
r
⃗
)
e
j
ω
t
}
\begin{aligned} \Re\{\nabla \times \vec E(\vec r)e^{j\omega t}\}&=-\mu \frac{\partial}{\partial t}\Re\{\vec H(\vec r)e^{j\omega t}\}-\Re\{\vec M(\vec r)e^{j\omega t}\}\\ &= -\Re\{j\omega \mu \vec H(\vec r)e^{j\omega t}\}-\Re\{\vec M(\vec r)e^{j\omega t}\} \end{aligned}
ℜ{∇×E
(r
)ejωt}=−μ∂t∂ℜ{H
(r
)ejωt}−ℜ{M
(r
)ejωt}=−ℜ{jωμH
(r
)ejωt}−ℜ{M
(r
)ejωt}
Since equation is valid for every
t
t
t, we can equate phasors
∇
×
E
⃗
(
r
⃗
)
=
−
j
ω
μ
H
⃗
(
r
⃗
)
−
M
⃗
(
r
⃗
)
\nabla \times \vec E(\vec r)= -j\omega \mu \vec H(\vec r)-\vec M(\vec r)
∇×E
(r
)=−jωμH
(r
)−M
(r
)
Equations for phasors same as the FT of the fields, and often we use the same notation. However the phasors are not functions of the frequency! They are complex constants.
Power for time-harmonic fields
-
The power is proportional to e 2 e^2 e2 (since the dissipated power is ∭ V σ e 2 d V \iiint_V \sigma e^2 dV ∭Vσe2dV)
-
Since e ⃗ \vec e e is a function of time, the power is also the function of time e 2 ( t ) e^2(t) e2(t)
-
For time-sinusoidal fields, we are usually not interested in knowing the instantaneous power, but rather an average power in time
e a v 2 = 1 T ∫ 0 T e 2 ( t ) d t e_{av}^2=\frac{1}{T}\int_0^T e^2(t)dt eav2=T1∫0Te2(t)dt
These averages for harmonic signals are much simpler to express in the phasor domain:
e
a
v
2
=
1
T
∫
0
T
e
2
(
t
)
d
t
=
1
T
∫
0
T
e
0
2
(
r
⃗
)
cos
2
(
ω
t
+
ϕ
)
d
t
=
1
2
e
0
2
(
r
⃗
)
=
1
2
E
⃗
(
r
⃗
)
⋅
E
⃗
∗
(
r
⃗
)
=
1
2
∣
E
⃗
(
r
⃗
)
∣
2
\begin{aligned} e_{av}^2&=\frac{1}{T}\int_0^T e^2(t)dt=\frac{1}{T}\int_0^T e_0^2(\vec r)\cos^2(\omega t+\phi)dt\\ &=\frac{1}{2}e_0^2(\vec r)=\frac{1}{2} \vec E(\vec r)\cdot \vec E^*(\vec r)=\frac{1}{2}|\vec{E}(\vec r)|^2 \end{aligned}
eav2=T1∫0Te2(t)dt=T1∫0Te02(r
)cos2(ωt+ϕ)dt=21e02(r
)=21E
(r
)⋅E
∗(r
)=21∣E
(r
)∣2
Complex permittivity, permeability
M.E. in phasor domain:
∇
⋅
E
⃗
=
ρ
e
/
ε
∇
⋅
H
⃗
=
ρ
m
/
μ
∇
×
E
⃗
=
−
j
ω
μ
H
⃗
−
M
⃗
∇
×
H
⃗
=
j
ω
ε
E
⃗
+
J
⃗
\begin{aligned} & \nabla \cdot \vec E=\rho_e/\varepsilon\\ & \nabla \cdot \vec H=\rho_m/\mu \\ & \nabla \times \vec E=-j\omega \mu \vec H-\vec M\\ & \nabla \times \vec H=j\omega\varepsilon\vec E+\vec J \end{aligned}
∇⋅E
=ρe/ε∇⋅H
=ρm/μ∇×E
=−jωμH
−M
∇×H
=jωεE
+J
It can be found that
ε
\varepsilon
ε and
μ
\mu
μ are complex in general. Why?
It is related with dispersive linear media (time dispersion of materials).
Properties of fields inside dielectrics
Electric flux density ( D ⃗ \vec D D ) in the dielectric
Define a new quantity: Polarization
P
⃗
\vec P
P
(net electric dipole moment vector per unit volume, has the same unit as
D
⃗
\vec D
D
)
P
⃗
=
ε
0
χ
e
E
⃗
χ
e
Electrical susceptibility
\vec P=\varepsilon_0 \chi_e\vec E\qquad \chi_e\text{ Electrical susceptibility}
P
=ε0χeE
χe Electrical susceptibility
-
E ⃗ \vec E E is the average net field in the medium
-
P ⃗ \vec P P is the average net dipole polarization per unit volume: dipole moment per unit volume
-
Local field E ⃗ ( r ⃗ ) \vec E(\vec r) E (r ) is very complicate
Isotropic materials: The material responds equally independently on the orientation.
Linear dispersive media
Poynting theorem in phasor domain
(mainly from Theory and Computation of Electromagnetic Fields 1.7.3)
Energy conservation law in the time-average sense
Let us first consider the product between two instantaneous quantities
a
⃗
(
t
)
\vec a(t)
a
(t) and
b
⃗
(
t
)
\vec b(t)
b
(t). We can easily find that
a
⃗
(
t
)
∘
b
⃗
(
t
)
=
ℜ
{
A
⃗
e
j
ω
t
}
∘
ℜ
{
B
⃗
e
j
ω
t
}
=
1
2
ℜ
{
A
⃗
∘
B
⃗
∗
}
+
1
2
ℜ
{
A
⃗
∘
B
⃗
e
j
2
ω
t
}
(PTP.1)
\begin{aligned} \vec{a}(t)\circ \vec b(t)&=\Re\{\vec A e^{j\omega t}\} \circ \Re\{\vec Be^{j\omega t}\}\\ &=\frac{1}{2}\Re\{\vec A\circ \vec B^*\}+\frac{1}{2}\Re\{\vec A\circ \vec B e^{j2\omega t}\} \end{aligned} \tag{PTP.1}
a
(t)∘b
(t)=ℜ{A
ejωt}∘ℜ{B
ejωt}=21ℜ{A
∘B
∗}+21ℜ{A
∘B
ej2ωt}(PTP.1)
where the circle denotes that the product can be either a dot or a cross-product.
If we take the time average over one cycle, we have
a
⃗
(
t
)
∘
b
⃗
(
t
)
‾
=
1
T
∫
0
T
a
⃗
(
t
)
∘
b
⃗
(
t
)
d
t
=
1
2
ℜ
{
A
⃗
∘
B
⃗
∗
}
(PTP.2)
\overline{\vec a(t)\circ \vec b(t)}=\frac{1}{T}\int_0^T \vec a(t)\circ \vec b(t) dt=\frac{1}{2}\Re\{\vec A\circ \vec B^*\}\tag{PTP.2}
a
(t)∘b
(t)=T1∫0Ta
(t)∘b
(t)dt=21ℜ{A
∘B
∗}(PTP.2)
Using this result, we can easily relate the complex field quantities to the time-average power and energy for the time-harmonic fields. For example, taking the time average of the Poynting vector, we have
s
⃗
(
t
)
‾
=
e
⃗
(
t
)
×
h
⃗
(
t
)
‾
=
1
2
ℜ
{
E
⃗
×
H
⃗
∗
}
(PTP.3)
\overline{\vec s(t)}=\overline{\vec e(t)\times \vec h(t)}=\frac{1}{2}\Re\{\vec E \times \vec H^*\}\tag{PTP.3}
s
(t)=e
(t)×h
(t)=21ℜ{E
×H
∗}(PTP.3)
By defining the complex Poynting vector as
S
⃗
=
1
2
E
⃗
×
H
⃗
∗
(PTP.4)
\vec S=\frac{1}{2}\vec E \times \vec H^* \tag{PTP.4}
S
=21E
×H
∗(PTP.4)
we obtain
s
⃗
‾
=
ℜ
{
S
⃗
}
\overline{\vec s}=\Re\{\vec S\}
s
=ℜ{S
}.
For another example, the time-average electric energy density becomes
w
e
(
t
)
‾
=
1
2
ε
e
⃗
(
t
)
⋅
e
⃗
(
t
)
‾
=
1
4
ε
ℜ
{
E
⃗
⋅
E
⃗
∗
}
=
1
4
ε
∣
E
⃗
∣
2
(PTP.5)
\overline{w_e(t)}=\frac{1}{2}\varepsilon \overline{\vec e(t)\cdot \vec e(t)}=\frac{1}{4}\varepsilon\Re\{\vec E \cdot \vec E^*\}=\frac{1}{4}\varepsilon |\vec E|^2 \tag{PTP.5}
we(t)=21εe
(t)⋅e
(t)=41εℜ{E
⋅E
∗}=41ε∣E
∣2(PTP.5)
Now let us consider the energy conservation law for the time-harmonic fields. We take the time average of the Poynting theorem in time domain
−
(
e
⃗
⋅
j
⃗
i
+
h
⃗
⋅
m
⃗
i
)
=
∇
⋅
(
e
⃗
×
h
⃗
)
+
σ
e
⃗
⋅
e
⃗
+
ε
e
⃗
⋅
∂
e
⃗
∂
t
+
μ
h
⃗
⋅
∂
h
⃗
∂
t
(PTP.6)
-(\vec e\cdot \vec j_i+\vec h\cdot \vec m_i)=\nabla\cdot(\vec e\times \vec h)+\sigma \vec e\cdot\vec e+\varepsilon \vec e \cdot\frac{\partial \vec e}{\partial t}+\mu\vec h\cdot\frac{\partial \vec h}{\partial t} \tag{PTP.6}
−(e
⋅j
i+h
⋅m
i)=∇⋅(e
×h
)+σe
⋅e
+εe
⋅∂t∂e
+μh
⋅∂t∂h
(PTP.6)
to obtain
p
s
‾
=
p
e
‾
+
p
d
‾
+
∂
w
e
/
∂
t
‾
+
∂
w
m
/
∂
t
‾
(PTP.7)
\overline {\mathcal p_s}=\overline {\mathcal p_e}+\overline {\mathcal p_d}+\overline {\partial \mathcal w_e/\partial t}+\overline {\partial \mathcal w_m/\partial t} \tag{PTP.7}
ps=pe+pd+∂we/∂t+∂wm/∂t(PTP.7)
From (PTP.1) we can see that
w
e
=
1
2
ε
e
⃗
⋅
∂
e
⃗
∂
t
=
1
4
ε
ℜ
{
E
⃗
⋅
E
⃗
∗
}
+
1
4
ε
ℜ
{
E
⃗
⋅
E
⃗
e
j
2
ω
t
}
(PTP.8)
\mathcal w_e=\frac{1}{2}\varepsilon \vec e \cdot\frac{\partial \vec e}{\partial t}=\frac{1}{4}\varepsilon \Re\{\vec E \cdot \vec E^*\}+\frac{1}{4}\varepsilon \Re\{\vec E\cdot \vec Ee^{j2\omega t}\} \tag{PTP.8}
we=21εe
⋅∂t∂e
=41εℜ{E
⋅E
∗}+41εℜ{E
⋅E
ej2ωt}(PTP.8)
and its time derivative is
∂
w
e
∂
t
=
−
ω
2
ε
ℑ
{
E
⃗
⋅
E
⃗
e
j
2
ω
t
}
(PTP.9)
\frac{\partial \mathcal w_e}{\partial t}=-\frac{\omega}{2}\varepsilon \Im\{\vec E \cdot \vec Ee^{j2\omega t}\} \tag{PTP.9}
∂t∂we=−2ωεℑ{E
⋅E
ej2ωt}(PTP.9)
we have
∂
w
e
/
∂
t
‾
=
0
\overline{\partial \mathcal w_e / \partial t}=0
∂we/∂t=0 and similarly,
∂
w
m
/
∂
t
‾
=
0
\overline{\partial \mathcal w_m / \partial t}=0
∂wm/∂t=0. This indicates that for time-harmonic fields, although the instantaneous energy density changes, the time average of the change vanishes.
Hence (PTP.7) becomes
p
s
‾
=
p
e
‾
+
p
d
‾
(PTP.10)
\overline {\mathcal p_s}=\overline {\mathcal p_e}+\overline {\mathcal p_d} \tag{PTP.10}
ps=pe+pd(PTP.10)
Energy conservation law for the time-harmonic fields
The derivation follows a procedure similar to the one described in Poynting vector and Poynting theorem in time domain. We start with the following two M.E.:
∇
×
E
⃗
=
−
j
ω
H
⃗
−
M
⃗
i
∇
×
H
⃗
=
j
ω
ε
E
⃗
+
σ
E
⃗
+
J
⃗
i
(PTP.11& PTP.12)
\begin{aligned} \nabla \times \vec E&=-j\omega\vec H-\vec M_i \\ \nabla \times \vec H&=j\omega\varepsilon \vec E+\sigma\vec E+\vec J_i \tag{PTP.11\& PTP.12} \end{aligned}
∇×E
∇×H
=−jωH
−M
i=jωεE
+σE
+J
i(PTP.11& PTP.12)
By taking the dot product of (PTP.11) with
H
⃗
∗
\vec H^*
H
∗ and the dot product of the
complex conjugate of (PTP.12) with
E
⃗
\vec E
E
, then subtracting the latter from the former, we obtain
∇
⋅
(
E
⃗
×
H
⃗
∗
)
=
−
j
ω
μ
∣
H
⃗
∣
2
+
j
ω
ε
∣
E
⃗
∣
2
−
σ
∣
E
⃗
∣
2
−
H
⃗
∗
⋅
M
⃗
i
−
E
⃗
⋅
J
i
∗
(PTP.13)
\nabla \cdot(\vec E \times \vec H^*)=-j\omega\mu|\vec H|^2+j\omega\varepsilon|\vec E|^2-\sigma|\vec E|^2-\vec H^*\cdot \vec M_i-\vec E \cdot J_i^* \tag{PTP.13}
∇⋅(E
×H
∗)=−jωμ∣H
∣2+jωε∣E
∣2−σ∣E
∣2−H
∗⋅M
i−E
⋅Ji∗(PTP.13)
Denoting
p
e
=
1
2
∇
⋅
(
E
⃗
×
H
⃗
∗
)
p
d
‾
=
1
2
σ
∣
E
⃗
∣
2
p
s
=
−
1
2
(
H
⃗
∗
⋅
M
⃗
i
+
E
⃗
⋅
J
⃗
i
∗
)
w
e
‾
=
1
4
ε
∣
E
⃗
∣
2
w
m
‾
=
1
4
μ
∣
H
⃗
∣
2
(PTP.14-PTP.18)
\begin{aligned} & p_e=\frac{1}{2}\nabla \cdot (\vec E\times \vec H^*)\\ &\overline{ p_d}=\frac{1}{2} \sigma |\vec E|^2\\ & p_s=-\frac{1}{2}(\vec H^*\cdot \vec M_i+\vec E\cdot \vec J_i^*)\\ &\overline{ w_e}=\frac{1}{4} \varepsilon |\vec E|^2\\ &\overline{ w_m}=\frac{1}{4} \mu |\vec H|^2 \tag{PTP.14-PTP.18} \end{aligned}
pe=21∇⋅(E
×H
∗)pd=21σ∣E
∣2ps=−21(H
∗⋅M
i+E
⋅J
i∗)we=41ε∣E
∣2wm=41μ∣H
∣2(PTP.14-PTP.18)
we can write (PTP.13) as
p
s
=
p
e
+
p
d
‾
+
j
2
ω
(
w
m
‾
−
w
e
‾
)
.
(PTP.19)
p_s=p_e+\overline{ p_d}+j2\omega(\overline{w_m}-\overline{ w_e}). \tag{PTP.19}
ps=pe+pd+j2ω(wm−we).(PTP.19)
Integrating this over a finite volume and invoking Gauss’s theorem yields its integral form
P
s
=
P
e
+
P
d
‾
+
j
2
ω
(
W
m
‾
−
W
e
‾
)
(PTP.20)
P_s=P_e+\overline{ P_d}+j2\omega(\overline{W_m}-\overline{ W_e}) \tag{PTP.20}
Ps=Pe+Pd+j2ω(Wm−We)(PTP.20)
where
P
e
=
∭
V
p
e
d
V
=
1
2
∯
S
(
E
⃗
×
H
⃗
∗
)
⋅
d
S
⃗
P
d
‾
=
∭
V
p
d
‾
d
V
=
1
2
∭
V
σ
∣
E
⃗
∣
2
d
V
P
s
=
∭
V
p
s
d
V
=
−
1
2
∭
V
(
H
⃗
∗
⋅
M
⃗
i
+
E
⃗
⋅
J
⃗
i
∗
)
d
V
W
e
‾
=
∭
V
w
e
‾
d
V
=
1
4
∭
V
ε
∣
E
⃗
∣
2
d
V
W
m
‾
=
∭
V
w
m
‾
d
V
=
1
4
∭
V
μ
∣
H
⃗
∣
2
d
V
(PTP.21-PTP.25)
\begin{aligned} & P_e=\iiint _Vp_edV=\frac{1}{2}\oiint_S (\vec E\times \vec H^*)\cdot d\vec S\\ &\overline{ P_d}=\iiint_V\overline{ p_d}dV=\frac{1}{2}\iiint_V \sigma |\vec E|^2dV\\ & P_s=\iiint_Vp_sdV=-\frac{1}{2}\iiint_V(\vec H^*\cdot \vec M_i+\vec E\cdot \vec J_i^*)dV\\ &\overline{ W_e}=\iiint_V\overline{ w_e}dV=\frac{1}{4}\iiint_V \varepsilon |\vec E|^2dV\\ &\overline{ W_m}=\iiint_V\overline{ w_m}dV=\frac{1}{4}\iiint_V \mu |\vec H|^2dV \tag{PTP.21-PTP.25} \end{aligned}
Pe=∭VpedV=21∬
S(E
×H
∗)⋅dS
Pd=∭VpddV=21∭Vσ∣E
∣2dVPs=∭VpsdV=−21∭V(H
∗⋅M
i+E
⋅J
i∗)dVWe=∭VwedV=41∭Vε∣E
∣2dVWm=∭VwmdV=41∭Vμ∣H
∣2dV(PTP.21-PTP.25)
Here,
P
e
P_e
Pe is called the complex exiting power,
P
d
‾
\overline{ P_d}
Pd the time-average dissipated power,
P
s
P_s
Ps the complex supplied power, and
W
e
‾
\overline{W_e}
We and
W
m
‾
\overline{W_m}
Wm the time-average electric and magnetic energies, respectively.
Equations (PTP.19) and (PTP.20) are known as Poynting’s theorem for complex phasors. Both are complex equations.
Meaning of Poynting’s theorem for complex phasors
Real parts:
−
1
2
ℜ
{
∭
V
(
H
⃗
∗
⋅
M
⃗
i
+
E
⃗
⋅
J
⃗
i
∗
)
d
V
}
=
1
2
ℜ
{
∯
S
(
E
⃗
×
H
⃗
∗
)
⋅
d
S
⃗
}
+
1
2
∭
V
σ
∣
E
⃗
∣
2
d
V
-\frac{1}{2}\Re\{\iiint_V(\vec H^*\cdot \vec M_i+\vec E\cdot \vec J_i^*)dV\}=\frac{1}{2}\Re\{\oiint_S (\vec E\times \vec H^*)\cdot d\vec S\}+\frac{1}{2}\iiint_V \sigma |\vec E|^2dV
−21ℜ{∭V(H
∗⋅M
i+E
⋅J
i∗)dV}=21ℜ{∬
S(E
×H
∗)⋅dS
}+21∭Vσ∣E
∣2dV
Active Power is the average real power (the structure is same as the energy conservation law in the time-average sense).
− 1 2 ℜ { ∭ V ( H ⃗ ∗ ⋅ M ⃗ i + E ⃗ ⋅ J ⃗ i ∗ ) d V } -\frac{1}{2}\Re\{\iiint_V(\vec H^*\cdot \vec M_i+\vec E\cdot \vec J_i^*)dV\} −21ℜ{∭V(H ∗⋅M i+E ⋅J i∗)dV}: Active power from sources
1 2 ℜ { ∯ S ( E ⃗ × H ⃗ ∗ ) ⋅ d S ⃗ } \frac{1}{2}\Re\{\oiint_S (\vec E\times \vec H^*)\cdot d\vec S\} 21ℜ{∬ S(E ×H ∗)⋅dS }: Active power leaving the volume
1 2 ∭ V σ ∣ E ⃗ ∣ 2 d V \frac{1}{2}\iiint_V \sigma |\vec E|^2dV 21∭Vσ∣E ∣2dV: Active power dissipated in losses
Imaginary parts:
−
1
2
ℑ
{
∭
V
(
H
⃗
∗
⋅
M
⃗
i
+
E
⃗
⋅
J
⃗
i
∗
)
d
V
}
=
1
2
ℑ
{
∯
S
(
E
⃗
×
H
⃗
∗
)
⋅
d
S
⃗
}
+
ω
2
∭
V
(
μ
∣
H
⃗
∣
2
−
ε
∣
E
⃗
∣
2
)
d
V
-\frac{1}{2}\Im\{\iiint_V(\vec H^*\cdot \vec M_i+\vec E\cdot \vec J_i^*)dV\}=\frac{1}{2}\Im\{\oiint_S (\vec E\times \vec H^*)\cdot d\vec S\}+\frac{\omega}{2}\iiint_V (\mu |\vec H|^2-\varepsilon |\vec E|^2)dV
−21ℑ{∭V(H
∗⋅M
i+E
⋅J
i∗)dV}=21ℑ{∬
S(E
×H
∗)⋅dS
}+2ω∭V(μ∣H
∣2−ε∣E
∣2)dV
Reactive power is the power that is not transferred but is stored in the volume.
−
1
2
ℑ
{
∭
V
(
H
⃗
∗
⋅
M
⃗
i
+
E
⃗
⋅
J
⃗
i
∗
)
d
V
}
-\frac{1}{2}\Im\{\iiint_V(\vec H^*\cdot \vec M_i+\vec E\cdot \vec J_i^*)dV\}
−21ℑ{∭V(H
∗⋅M
i+E
⋅J
i∗)dV}: Reactive power from sources. The
power generated by the source at one moment and then taken back at the other moment within a cycle.
1 2 ℑ { ∯ S ( E ⃗ × H ⃗ ∗ ) ⋅ d S ⃗ } \frac{1}{2}\Im\{\oiint_S (\vec E\times \vec H^*)\cdot d\vec S\} 21ℑ{∬ S(E ×H ∗)⋅dS }: Reactive power leaving the volume. The power leaving the volume at one moment and then reentering at the other moment within a cycle.
ω 2 ∭ V ( μ ∣ H ⃗ ∣ 2 − ε ∣ E ⃗ ∣ 2 ) d V \frac{\omega}{2}\iiint_V (\mu |\vec H|^2-\varepsilon |\vec E|^2)dV 2ω∭V(μ∣H ∣2−ε∣E ∣2)dV: Reactive power Contained in field
This reactive power does not show up in the time-average supplied power or exiting power because it takes two round trips within each cycle, but it is clearly reflected in the difference between the time-average electric and magnetic energies.
To understand the concept of reactive power better, see Theory and Computation of Electromagnetic Fields P35.
Using the Parseval’s theorem
−
∫
−
∞
∞
e
⃗
(
t
)
⋅
j
⃗
i
(
t
)
d
t
=
−
∫
−
∞
∞
E
⃗
(
ω
)
⋅
J
⃗
i
(
ω
)
d
ω
-\int_{-\infty}^\infty \vec e(t)\cdot \vec j_i(t)dt=-\int_{-\infty}^\infty \vec E(\omega)\cdot \vec J_i(\omega)d\omega
−∫−∞∞e
(t)⋅j
i(t)dt=−∫−∞∞E
(ω)⋅J
i(ω)dω
The total energy given by the sources can be expressed by time integration or frequency integration.
Uniqueness theorem
If we solve the Maxwell equations, under what conditions the solution is unique?
Time domain
We consider a linear medium.
The solution e ⃗ \vec e e and h ⃗ \vec h h of Maxwell equation in the volume V V V for all time t > 0 t>0 t>0 is unique if
- The sources j ⃗ \vec j j and m ⃗ \vec m m are specified for t ≥ 0 t \geq 0 t≥0
- The field is specified in every point of V V V, for t = 0 t=0 t=0 (starting of initial conditions)
- The tangential components of the electric or magnetic fields ( n ^ × e ⃗ ( r ⃗ , t ) \hat n \times \vec e(\vec r,t) n^×e (r ,t) or n ^ × h ⃗ ( r ⃗ , t ) \hat n \times \vec h(\vec r,t) n^×h (r ,t)) are specified over the boundary (surface S S S) for every t > 0 t>0 t>0
Proof
Proof by contradiction: Let us assume that there were two solutions
e
⃗
1
,
h
⃗
1
\vec e_1,\vec h_1
e
1,h
1 and
e
⃗
2
,
h
⃗
2
\vec e_2,\vec h_2
e
2,h
2 such that
∇
×
e
⃗
k
=
−
μ
∂
h
⃗
k
∂
t
−
m
⃗
k
∇
×
h
⃗
k
=
ε
∂
e
⃗
k
∂
t
+
j
⃗
k
,
k
∈
{
1
,
2
}
\begin{aligned} &\nabla \times \vec e_k=-\mu \frac{\partial \vec h_k}{\partial t}-\vec m_k\\ &\nabla \times \vec h_k=\varepsilon \frac{\partial \vec e_k}{\partial t}+\vec j_k, \qquad k\in \{1,2\} \end{aligned}
∇×e
k=−μ∂t∂h
k−m
k∇×h
k=ε∂t∂e
k+j
k,k∈{1,2}
The three hypotheses above can be presented as
HP1:
j
⃗
1
=
j
⃗
2
,
m
⃗
1
=
m
⃗
2
HP2:
e
⃗
1
(
r
⃗
,
0
)
=
e
⃗
2
(
r
⃗
,
0
)
,
h
⃗
1
(
r
⃗
,
0
)
=
h
⃗
2
(
r
⃗
,
0
)
HP3:
n
^
×
e
⃗
1
=
n
^
×
e
⃗
2
or
n
^
×
h
⃗
1
=
n
^
×
h
⃗
2
\begin{aligned} &\text{HP1:}\quad \vec j_1=\vec j_2, \vec m_1=\vec m_2\\ &\text{HP2:}\quad \vec e_1(\vec r,0)=\vec e_2(\vec r,0),\vec h_1(\vec r,0)=\vec h_2(\vec r,0)\\ &\text{HP3:}\quad \hat n \times \vec e_1=\hat n \times \vec e_2\text{ or }\hat n \times \vec h_1=\hat n \times \vec h_2 \end{aligned}
HP1:j
1=j
2,m
1=m
2HP2:e
1(r
,0)=e
2(r
,0),h
1(r
,0)=h
2(r
,0)HP3:n^×e
1=n^×e
2 or n^×h
1=n^×h
2
For linearity, also the field difference is a solution. Define
e
⃗
d
=
e
⃗
1
−
e
⃗
2
,
h
⃗
d
=
h
⃗
1
−
h
⃗
2
j
⃗
d
=
j
⃗
1
−
j
⃗
2
,
m
⃗
d
=
m
⃗
1
−
m
⃗
2
\begin{aligned} &\vec e_d=\vec e_1-\vec e_2,&&\vec h_d=\vec h_1-\vec h_2\\ &\vec j_d=\vec j_1-\vec j_2,&&\vec m_d=\vec m_1-\vec m_2 \end{aligned}
e
d=e
1−e
2,j
d=j
1−j
2,h
d=h
1−h
2m
d=m
1−m
2
The three hypotheses become
HP1:
j
⃗
d
=
0
,
m
⃗
d
=
0
HP2:
e
⃗
d
(
r
⃗
,
0
)
=
0
,
h
⃗
d
(
r
⃗
,
0
)
=
0
HP3:
n
^
×
e
⃗
d
=
0
or
n
^
×
h
⃗
d
=
0
\begin{aligned} &\text{HP1:}\quad \vec j_d=0, \vec m_d=0\\ &\text{HP2:}\quad \vec e_d(\vec r,0)=0,\vec h_d(\vec r,0)=0\\ &\text{HP3:}\quad \hat n \times \vec e_d=0\text{ or }\hat n \times \vec h_d=0 \end{aligned}
HP1:j
d=0,m
d=0HP2:e
d(r
,0)=0,h
d(r
,0)=0HP3:n^×e
d=0 or n^×h
d=0
From the Poynting theorem
∯
S
(
e
⃗
d
×
h
⃗
d
)
⋅
n
^
d
S
+
1
2
∂
∂
t
∭
V
(
ε
e
d
2
+
μ
h
d
2
)
d
V
+
∭
V
σ
e
d
2
d
V
=
−
∭
V
(
e
⃗
d
⋅
j
⃗
i
+
h
⃗
d
⋅
m
⃗
i
)
d
V
\oiint_S(\vec e_d \times \vec h_d)\cdot \hat n d S+\frac{1}{2}\frac{\partial }{\partial t}\iiint_V(\varepsilon e_d^2+\mu h_d^2)dV+\iiint_V\sigma e_d^2dV=-\iiint_V(\vec e_d \cdot \vec j_i+\vec h_d\cdot \vec m_i)dV
∬
S(e
d×h
d)⋅n^dS+21∂t∂∭V(εed2+μhd2)dV+∭Vσed2dV=−∭V(e
d⋅j
i+h
d⋅m
i)dV
By applying HP1,
−
∭
V
(
e
⃗
d
⋅
j
⃗
i
+
h
⃗
d
⋅
m
⃗
i
)
d
V
=
0
-\iiint_V(\vec e_d \cdot \vec j_i+\vec h_d\cdot \vec m_i)dV=0
−∭V(e
d⋅j
i+h
d⋅m
i)dV=0.
By applying HP3, and a ⃗ ⋅ ( b ⃗ × c ⃗ ) = b ⃗ ⋅ ( c ⃗ × a ⃗ ) = c ⃗ ⋅ ( a ⃗ × b ⃗ ) \vec{a}\cdot(\vec b\times \vec c)=\vec b\cdot(\vec c\times \vec a)=\vec c\cdot(\vec a \times \vec b) a ⋅(b ×c )=b ⋅(c ×a )=c ⋅(a ×b ), ∯ S ( e ⃗ d × h ⃗ d ) ⋅ n ^ d S = 0 \oiint_S(\vec e_d \times \vec h_d)\cdot \hat n d S=0 ∬ S(e d×h d)⋅n^dS=0.
Therefore,
1
2
∂
∂
t
∭
V
(
ε
e
d
2
+
μ
h
d
2
)
d
V
=
−
∭
V
σ
e
d
2
d
V
<
0
\frac{1}{2}\frac{\partial }{\partial t}\iiint_V(\varepsilon e_d^2+\mu h_d^2)dV=-\iiint_V\sigma e_d^2dV<0
21∂t∂∭V(εed2+μhd2)dV=−∭Vσed2dV<0
which means EM energy decreases for
t
>
0
t>0
t>0.
By applying HP2,
1
2
∭
V
(
ε
e
d
2
+
μ
h
d
2
)
d
V
=
0
for
t
=
0
\frac{1}{2}\iiint_V(\varepsilon e_d^2+\mu h_d^2)dV=0\quad \text{for }t=0
21∭V(εed2+μhd2)dV=0for t=0
which means EM energy is zero at
t
=
0
t=0
t=0.
Thus EM energy is zero for t ≥ 0 t\ge 0 t≥0, we have e ⃗ d = h ⃗ d = 0 , ∀ t ≥ 0 \vec e_d=\vec h_d=\mathbf 0,\forall t\ge 0 e d=h d=0,∀t≥0.
Frequency domain
We consider a lossy medium.
The field created by sources J ⃗ \vec J J and M ⃗ \vec M M is unique within the region if
- The sources are specified in the volume.
- The tangential component of E ⃗ \vec E E or H ⃗ \vec H H is specified on S S S ( n ^ × E ⃗ ( r ⃗ ) \hat n\times \vec E(\vec r) n^×E (r ) or n ^ × H ⃗ ( r ⃗ ) \hat n \times \vec H(\vec r) n^×H (r ))
The theorem can be demonstrated simply per absurd by assuming that for the same sources and tangent fields, there are two different sets of solutions. Then applying the Poynting’s theorem to a new situation obtained by the difference of these two sets. It results that the only possibility is that the field difference is equal to zero, if there are some losses in V V V. Therefore the two solutions are the same.
Boundary conditions
The differential form Maxwell’s equations are valid at points in a continuous medium. They cannot be applied to discontinuous fields that may occur at interfaces between different media. Fortunately, we can employ Maxwell’s equations in integral form to find the relations between the fields on the two sides of an interface. Such relations are called boundary conditions.
**Surface current density J ⃗ s \vec J_s J s & M ⃗ s \vec M_s M s **
So far, the current density J ⃗ \vec J J is actually a volume current density, which is often simply called current density. It represents the amount of current passing through a unit area normal to the direction of the current flow. Now, imagine a current flow confined in a thin layer. If the total current is kept constant while the thickness of the layer is reduced to zero, the volume current density approaches infinity, which can no longer describe the current sheet. In this case, the current distribution can be described by the surface current density, which is a vector denoted as J ⃗ s \vec J_s J s. Its value represents the amount of current passing through a unit width normal to the direction of the current flow and has a unit of amperes/meter. The surface magnetic current density M ⃗ s \vec M_s M s is defined similarly, which has a unit of volts/meter.
Surface charge density ϱ e , s \varrho_{e,s} ϱe,s & ϱ m , s \varrho_{m,s} ϱm,s
The surface charge density is defined as the amount of charge over a unit area on the surface.
Boundary conditions: normal electric flux
Let us now consider the height of the cylinder tends to
0
0
0, then the integral of the lateral part goes to
0
0
0. We have
∯
S
D
⃗
⋅
d
S
⃗
=
∬
S
1
D
⃗
1
⋅
d
S
⃗
+
∬
S
2
D
⃗
2
⋅
d
S
⃗
=
∬
S
−
D
⃗
1
⋅
n
^
d
S
+
∬
S
D
⃗
2
⋅
n
^
d
S
=
∬
S
ϱ
e
,
s
d
S
\oiint_S \vec D\cdot d\vec S=\iint_{S_1} \vec D_1\cdot d\vec S+\iint_{S_2} \vec D_2\cdot d\vec S=\iint_{S}- \vec D_1 \cdot \hat nd S+\iint_{S} \vec D_2\cdot\hat n d S=\iint_S \varrho_{e,s}dS
∬
SD
⋅dS
=∬S1D
1⋅dS
+∬S2D
2⋅dS
=∬S−D
1⋅n^dS+∬SD
2⋅n^dS=∬Sϱe,sdS
and thus
n
^
⋅
(
D
⃗
2
−
D
⃗
1
)
=
ϱ
e
,
s
\hat n\cdot(\vec D_2-\vec D_1)=\varrho_{e,s}
n^⋅(D
2−D
1)=ϱe,s
Following the same steps for the magnetic flux we have
n
^
⋅
(
B
⃗
2
−
B
⃗
1
)
=
ϱ
m
,
s
=
0
\hat n\cdot(\vec B_2-\vec B_1)=\varrho_{m,s}=0
n^⋅(B
2−B
1)=ϱm,s=0
Boundary conditions: tangential electric field
If the height tends to
0
0
0, we have
∮
C
E
⃗
⋅
d
l
⃗
=
E
⃗
1
⋅
t
^
Δ
l
−
E
⃗
2
⋅
t
^
Δ
l
=
−
∬
S
M
⃗
⋅
d
S
⃗
=
−
∫
Δ
l
M
⃗
s
⋅
b
^
d
l
=
−
M
⃗
s
⋅
b
^
Δ
l
\oint_C\vec E\cdot d\vec l=\vec E_1\cdot \hat t\Delta l-\vec E_2\cdot\hat t\Delta l=-\iint_S\vec M\cdot d\vec S=-\int_{\Delta l}\vec M_s\cdot \hat bdl=-\vec M_s\cdot \hat b\Delta l
∮CE
⋅dl
=E
1⋅t^Δl−E
2⋅t^Δl=−∬SM
⋅dS
=−∫ΔlM
s⋅b^dl=−M
s⋅b^Δl
and thus
t
^
⋅
(
E
⃗
2
−
E
⃗
1
)
=
M
⃗
s
⋅
b
^
\hat t\cdot(\vec E_2-\vec E_1)=\vec M_s\cdot \hat b
t^⋅(E
2−E
1)=M
s⋅b^
Since
t
^
⋅
(
E
⃗
2
−
E
⃗
1
)
=
(
n
^
×
b
^
)
⋅
(
E
⃗
2
−
E
⃗
1
)
=
b
^
⋅
[
(
E
⃗
2
−
E
⃗
1
)
×
n
^
]
\hat t\cdot(\vec E_2-\vec E_1)=(\hat n \times \hat b)\cdot(\vec E_2-\vec E_1)=\hat b\cdot[(\vec E_2-\vec E_1)\times \hat n]
t^⋅(E
2−E
1)=(n^×b^)⋅(E
2−E
1)=b^⋅[(E
2−E
1)×n^], we obtain
n
^
×
(
E
⃗
2
−
E
⃗
1
)
=
−
M
⃗
s
=
0
\hat n \times(\vec E_2-\vec E_1)=-\vec M_s=0
n^×(E
2−E
1)=−M
s=0
Analogously
n
^
×
(
H
⃗
2
−
H
⃗
1
)
=
J
⃗
s
\hat n \times(\vec H_2-\vec H_1)=\vec J_s
n^×(H
2−H
1)=J
s
In conclusion, we have the generic case
n
^
⋅
(
D
⃗
2
−
D
⃗
1
)
=
ϱ
e
,
s
n
^
⋅
(
B
⃗
2
−
B
⃗
1
)
=
ϱ
m
,
s
n
^
×
(
E
⃗
2
−
E
⃗
1
)
=
−
M
⃗
s
n
^
×
(
H
⃗
2
−
H
⃗
1
)
=
J
⃗
s
\begin{aligned} & \hat n\cdot(\vec D_2-\vec D_1)=\varrho_{e,s}\\ & \hat n\cdot(\vec B_2-\vec B_1)=\varrho_{m,s} \\ & \hat n \times(\vec E_2-\vec E_1)=-\vec M_s\\ & \hat n \times(\vec H_2-\vec H_1)=\vec J_s \end{aligned}
n^⋅(D
2−D
1)=ϱe,sn^⋅(B
2−B
1)=ϱm,sn^×(E
2−E
1)=−M
sn^×(H
2−H
1)=J
s
Boundary conditions: Perfect Electric Conductor (PEC) & Perfect Magnetic Conductor (PMC)
A perfect conductor is a medium full of free charges. When an electromagnetic field is applied to this medium, the free charges, being pushed by the applied field, move themselves such that they produce an opposing field that completely cancels the applied field. This causes the formation of the surface currents and charges on the surface of a perfect conductor. If it is a PEC, its surface can support a surface electric current and charge. If it is a perfect magnetic conductor (PMC), the surface can support a surface magnetic current and charge.
Removing magnetic sources, for PEC
n
^
⋅
D
⃗
2
=
ϱ
e
,
s
n
^
⋅
B
⃗
2
=
0
n
^
×
E
⃗
2
=
0
n
^
×
H
⃗
2
=
J
⃗
s
\begin{aligned} & \hat n\cdot\vec D_2=\varrho_{e,s}\\ & \hat n\cdot\vec B_2=0 \\ & \hat n \times\vec E_2=0\\ & \hat n \times\vec H_2=\vec J_s \end{aligned}
n^⋅D
2=ϱe,sn^⋅B
2=0n^×E
2=0n^×H
2=J
s
Removing electric sources, for PMC
n
^
⋅
D
⃗
2
=
0
n
^
⋅
B
⃗
2
=
ϱ
m
,
s
n
^
×
E
⃗
2
=
−
M
⃗
s
n
^
×
H
⃗
2
=
0
\begin{aligned} & \hat n\cdot\vec D_2=0\\ & \hat n\cdot\vec B_2=\varrho_{m,s} \\ & \hat n \times\vec E_2=-\vec M_s\\ & \hat n \times\vec H_2=0 \end{aligned}
n^⋅D
2=0n^⋅B
2=ϱm,sn^×E
2=−M
sn^×H
2=0
Radiation Boundary Conditions
From the real part of Poynting theorem in phasor domain:
−
1
2
ℜ
{
∭
V
(
H
⃗
∗
⋅
M
⃗
i
+
E
⃗
⋅
J
⃗
i
∗
)
d
V
}
=
1
2
ℜ
{
∯
S
(
E
⃗
×
H
⃗
∗
)
⋅
d
S
⃗
}
+
1
2
∭
V
σ
∣
E
⃗
∣
2
d
V
-\frac{1}{2}\Re\{\iiint_V(\vec H^*\cdot \vec M_i+\vec E\cdot \vec J_i^*)dV\}=\frac{1}{2}\Re\{\oiint_S (\vec E\times \vec H^*)\cdot d\vec S\}+\frac{1}{2}\iiint_V \sigma |\vec E|^2dV
−21ℜ{∭V(H
∗⋅M
i+E
⋅J
i∗)dV}=21ℜ{∬
S(E
×H
∗)⋅dS
}+21∭Vσ∣E
∣2dV
-
Let us assume, there are no losses in medium and sources are all within a volume V 0 V_0 V0
− 1 2 ℜ { ∭ V ( H ⃗ ∗ ⋅ M ⃗ i + E ⃗ ⋅ J ⃗ i ∗ ) d V } = 1 2 ℜ { ∯ S ( E ⃗ × H ⃗ ∗ ) ⋅ d S ⃗ } -\frac{1}{2}\Re\{\iiint_V(\vec H^*\cdot \vec M_i+\vec E\cdot \vec J_i^*)dV\}=\frac{1}{2}\Re\{\oiint_S (\vec E\times \vec H^*)\cdot d\vec S\} −21ℜ{∭V(H ∗⋅M i+E ⋅J i∗)dV}=21ℜ{∬ S(E ×H ∗)⋅dS } -
Let us take the surface S S S and an infinite sphere S → ∞ S\to \infty S→∞
1 2 ℜ { ∯ S ( E ⃗ × H ⃗ ∗ ) ⋅ d S ⃗ } = lim r → ∞ ℜ { ∫ 0 π ∫ 0 2 π S ⃗ ( r , θ , ϕ ) ⋅ r ^ r 2 sin θ d θ d ϕ } \frac{1}{2}\Re\{\oiint_S (\vec E\times \vec H^*)\cdot d\vec S\}=\lim _{r\to \infty}\Re\left\{\int_0^\pi\int_0^{2\pi}\vec S(r,\theta,\phi)\cdot \hat r r^2\sin\theta d\theta d\phi\right\} 21ℜ{∬ S(E ×H ∗)⋅dS }=r→∞limℜ{∫0π∫02πS (r,θ,ϕ)⋅r^r2sinθdθdϕ}
To ensure that the active power from sources does not grow when integrating over bigger and bigger spheres, we have
lim r → ∞ ℜ { S ⃗ ( r , θ , ϕ ) } = ℜ { S ⃗ ( θ , ϕ ) } r 2 H ⃗ ∗ ⋅ M ⃗ i + E ⃗ ⋅ J ⃗ i ∗ ) d V } = 1 2 ℜ { ∯ S ( E ⃗ × H ⃗ ∗ ) ⋅ d S ⃗ } + 1 2 ∭ V σ ∣ E ⃗ ∣ 2 d V \lim_{r\to \infty}\Re\{\vec S(r,\theta,\phi)\}=\frac{\Re\{\vec S(\theta,\phi)\}}{r^2} \vec H^*\cdot \vec M_i+\vec E\cdot \vec J_i^*)dV\}=\frac{1}{2}\Re\{\oiint_S (\vec E\times \vec H^*)\cdot d\vec S\}+\frac{1}{2}\iiint_V \sigma |\vec E|^2dV r→∞limℜ{S (r,θ,ϕ)}=r2ℜ{S (θ,ϕ)}H ∗⋅M i+E ⋅J i∗)dV}=21ℜ{∬ S(E ×H ∗)⋅dS }+21∭Vσ∣E ∣2dV -
Let us assume, there are no losses in medium and sources are all within a volume V 0 V_0 V0
− 1 2 ℜ { ∭ V ( H ⃗ ∗ ⋅ M ⃗ i + E ⃗ ⋅ J ⃗ i ∗ ) d V } = 1 2 ℜ { ∯ S ( E ⃗ × H ⃗ ∗ ) ⋅ d S ⃗ } -\frac{1}{2}\Re\{\iiint_V(\vec H^*\cdot \vec M_i+\vec E\cdot \vec J_i^*)dV\}=\frac{1}{2}\Re\{\oiint_S (\vec E\times \vec H^*)\cdot d\vec S\} −21ℜ{∭V(H ∗⋅M i+E ⋅J i∗)dV}=21ℜ{∬ S(E ×H ∗)⋅dS } -
Let us take the surface S S S and an infinite sphere S → ∞ S\to \infty S→∞
1 2 ℜ { ∯ S ( E ⃗ × H ⃗ ∗ ) ⋅ d S ⃗ } = lim r → ∞ ℜ { ∫ 0 π ∫ 0 2 π S ⃗ ( r , θ , ϕ ) ⋅ r ^ r 2 sin θ d θ d ϕ } \frac{1}{2}\Re\{\oiint_S (\vec E\times \vec H^*)\cdot d\vec S\}=\lim _{r\to \infty}\Re\left\{\int_0^\pi\int_0^{2\pi}\vec S(r,\theta,\phi)\cdot \hat r r^2\sin\theta d\theta d\phi\right\} 21ℜ{∬ S(E ×H ∗)⋅dS }=r→∞limℜ{∫0π∫02πS (r,θ,ϕ)⋅r^r2sinθdθdϕ}
To ensure that the active power from sources does not grow when integrating over bigger and bigger spheres, we have
lim r → ∞ ℜ { S ⃗ ( r , θ , ϕ ) } = ℜ { S ⃗ ( θ , ϕ ) } r 2 \lim_{r\to \infty}\Re\{\vec S(r,\theta,\phi)\}=\frac{\Re\{\vec S(\theta,\phi)\}}{r^2} r→∞limℜ{S (r,θ,ϕ)}=r2ℜ{S (θ,ϕ)}