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Advanced Electromagnetic Theory
Advanced Electromagnetic Theory
Lecture 2: Wave Equation and its Solutions
Dimitri PeroulisSchool of Electrical and Computer Engineering
Birck Nanotechnology Center, Tel: 765 494 3491, [email protected]
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Constructing Solutionsin Source-Free Media
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2 ~ E = ~ M i + j ~ J i + 1 q ve + j ~ E 2 ~ E
Reminder of Wave EquationWave equation for Electric field (from Maxwells eqns):
Magneticcurrent source
Electriccurrent source
Electriccharge
Source-free wave equation for Electric field:
2 ~ E = j ~ E 2 ~ E
2 = j ~ E 2 ~ E = j ( + j ) = + j Propagation constant: Attenuation constant
Phase constant
exp( j t)Assumed timedependence
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A. Lossless media
Source-free and lossless wave equation for Electric field:
2 ~ E + 2 ~ E = 0 Helmholtz equation
2 = k2
General solution method: Separation of Variables
Rectangular Cylindrical SphericalUsually:
Focus of this class
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1. Rectangular Coordinates~ E (x,y,z ) = ~ax E x (x,y,z ) + ~ay E y (x,y,z ) + ~az E z (x,y,z )
2
~ E + 2
~ E = 0 and by using vector identity
2 ~ E = 2 (~ax E x + ~ay E y + ~az E z ) =
= ~ax 2E x + ~ay 2E y + ~az 2E z
only true for rectangular coordinates
We get:
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1. Rectangular Coordinates
2E x (x,y,z ) + 2E x (x,y,z ) = 0 2E y (x,y,z ) + 2E y (x,y,z ) = 0 2E z (x,y,z ) + 2E z (x,y,z ) = 0
Three scalar Helmholtz equations
Explicitly:
E x x
+ E x y
+ E x z
+ 2E x = 0
Assuming: E x (x,y,z ) = f (x)g(y)h(z)
Key step: Separation of Variables
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1. Rectangular Coordinates
1f
d2 f dx2
+ 1g
d2gdy2
+ 1h
d2hdz2
+ 2 = 0
Only a function of x
and:
Therefore:
2x
+ 2y
+ 2z
= 2
1f
d2 f dx2 =
2x
Only a function
of x
1g
d2gdy2
= 2yOnly a function
of y
1h
d2hdz2 =
2z
Only a function
of z
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1. Rectangular CoordinatesSolution (for x, similarly for others):
f (x) = C 1 cos( x x) + C 2 sin( x x)
or
f (x) = C 1 exp( j x x) + C 2 exp( j x x)
Standing waves
Wave propagating along positive x axis
BC BC
BC BC exp( j t)Assumed timedependenceOpposite for
exp(
j t)
Wave propagating along negative x axis
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B. Lossy MediaSource-free and lossy wave equation for Electric field:
2 ~ E 2 ~ E = 0 with
2
= j ~ E 2 ~ E = j ( + j ) = + j
Propagation constant: Attenuation constant
Phase constant
= p j ( + j )Choose the sign of the
square root that yields zerowave amplitude at infinity
Separation of Variables: 2x
+ 2y
+ 2z
= 2
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B. Lossy MediaSolution (for x, similarly for others):
or
Standing waves
Wave propagating along positive x axis
f (x) = C 1 cosh( x x) + C 2 sinh( x x)
f (x) = C 1 exp( x x) + C 2 exp( x x)
Wave propagating along negative x axis
BC BC
BC BC exp( j t)Assumed timedependenceOpposite for
exp(
j t)
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Cylindrical Coordinates: Lossless media
2 ~ E + 2 ~ E = 0
Helmholtz equation:
2E +
E 2
22
E +
2E = 0
In cylindrical coordinates becomes:
2E +
E 2 +
22
E +
2E = 0
2E z + 2E z = 0 Start from here
coupled
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2. Cylindrical Coordinates
2E z 2
+1
E z
+12
2E z 2
+ 2E z z2
+ 2E z = 0
Assuming: E z (, , z) = f ()g()h(z)
Key step: Separation of Variables
1f
d2 f d2 +
1f
1
df d +
1g
12
d2gd2 +
1h
d2hdz2 +
2
= 0
Only a function of z
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2. Cylindrical Coordinates
Only a function of
1
h
d2h
dz2 =
2z
Thus:
ord2h
dz2 =
2z h
Putting this back into previous equation:
2
f d2 f d2 +
f
df d +
1g
d2gd2 + 2 2z 2 = 0
2
1g
d2gd2
= m2
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2. Cylindrical CoordinatesThus the last equation becomes:
2d2 f
d2+
df
d+
2
2
m2
f = 0
Bessel differential equation
With the additional equations being:
d2hdz2
= 2z h
d2gd2
= m2gConstraint equation
2z + 2 = 2
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2. Cylindrical CoordinatesFor h:
or
Standing waves
Wave propagating along positive z axis
h(z) = C 1 cos( z z) + C 2 sin( z z)
h(z) = C 1 exp( j z z) + C 2 exp( j z z)
Wave propagating along negative z axis
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2. Cylindrical CoordinatesFor g:
or
periodic waves
Wave propagating along positive axis
g() = C 1 cos(m) + C 2 sin(m)
g() = C 1 exp( jm ) + C 2 exp( jm )
Wave propagating along negative axis
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2. Cylindrical CoordinatesFor f:
or
standing waves (Bessel functions 1 st and 2 nd kind)
Wave propagating along positive axis
f () = C 1J m ( ) + C 2Y m ( )
Wave propagating along negative axis
f () = C 1H (2)m ( ) + C 2H (1)m ( )
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2. Cylindrical CoordinatesReminder:
H (1)m ( ) = J m ( ) + jY m ( )
H (2)m ( ) = J m ( ) jY m ( )
For lossy media the solutions are modified accordingly:
H (1)m ( ) = H (1)m ( + j )
J m ( ) = J m ( + j )etc. etc.
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Constructing SolutionsWhen Sources are Known
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Constructing Solutions
1. Elementary approach: Vector Potentials
2. Greens functions
a. ~ A and ~ F
b. ~ e = j ~ A and ~ h = j ~ F
sources FieldsDirect integrationGreens functions
Integration Potentials Differentiation
Hertz Potentials
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Vector PotentialsReminder:
2 ~ E = ~ M + j ~ J + 1 q ve + j ~ E 2 ~ E
2 ~ E +
2 ~ E =
~ M + j
~ J +
1
q ve
which for lossless media becomes:
0
and similarly for the magnetic field:
2 ~ H + 2 ~ H = ~ J + j ~ M + 1 q vm
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Vector PotentialsWe will solve these by superposition:
1. First, assume that no magnetic sources exist
and find the solution. We will call these fields:
2. Second, assume that no electric sources existand find the solution. We will call these fields:
~ E A and ~ H A
~ E F and ~ H F
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Vector Potentials: No magnetic sources1. No magnetic sources exist:
~ BA = 0 thus ~ BA = ~ A
Reminder: Both the curl and divergence need to bedefined for a unique definition of the magnetic vector
potential. This will be done shortly.
So for a simple (linear, isotropic and homogeneous)medium we get:
~ H A =1
~ BA =1 ~ A
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Vector Potentials: No magnetic sources
We get:
h~ E A + j ~ Ai= 0Can be written as: ~ E A + j ~ A = e
So going into Faradays law:
~ E A = j ~ H A
~ H A =1 ~ A
~ E A = e j ~ A
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Vector Potentials: No magnetic sourcesIn order to find the two potentials we need the otherequation (Ampere-Maxwell)
~ H A = ~ J + j ~ E A
1 ~ A= ~ J + j ~ E APlug in the magnetic field
Which for simple media becomes:
~ A = ~ J + j ~ E A ~ A = ( ~ A) 2 ~ A
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Vector Potentials: No magnetic sources
This reduces to:
2 ~ A + 2 ~ A = ~ J +
h ~ A + j e
iWe can define the divergence freely (Lorentz gauge)
0
Thus the equation that needs to be solved is:
2 ~ A + 2 ~ A = ~ J
Furthermore:
~ E A = e j ~ A = j ~ A j1
( ~ A)
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Vector Potentials: No magnetic sources
In order to solve this consideran infinitesimal point sourcein V around the origin
~ J = J z~az
For any point except for the origin (source):
2Az (r ) + 2Az (r ) = 0
Due to symmetry:
~ A = Az (r )~az and / = 0 , / = 0
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Vector Potentials: No magnetic sources
which by taking into account the symmetry argumentsreduces to
d2
Az (r )dr 2 + 2r dAz (r )dr +
2Az (r ) = 0
This equation has two independent solutions:
Az 1(r ) = C 1 exp( j r )/r
Az 2(r ) = C 2 exp( j r )/rWe will learn how to find them onlywhen we do Greens functions
Outward wave
Inward wave
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Vector Potentials: No magnetic sources
Reminder:
2
=
q/ =1 Z V
q 4 r dv
0
2Az = J z Az =
Z V J z
4
r
dv0
When = 0 ( statics) the equation and solutions become:
2Az = J z , Az (r ) = C 1 /rC 1 /r
Instead of
C 1 exp( j r )/r
Thus by analogy: Az = Z V J zexp( j r )
4 rdv0
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Vector Potentials: No magnetic sources
If we do the same along the x and y directions:
Ax =
Z V J x
exp( j r )
4 rdv0
Ay =
Z V J y
exp( j r )
4 rdv0
Thus for an arbitrary point source at the origin:
~ A = Z V ~ J exp( j r )
4 rdv0
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Vector Potentials: No magnetic sources
The truth is that although this solution is correct, it isfull of hand-weaving arguments. When we learn Greensfunctions we will find out how to properly solve it.
Intuitively if the sourceis not at the origin:
~ A = Z V ~ J (r 0) exp( j |r r 0|)4 | r r 0| dv0
Later identify asfree-space Greensfunction in 3D
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Vector Potentials: No magnetic sources
To recap, for cases with no magnetic sources
~ A = Z V ~ J (r 0) exp( j |r r0
|)4 |r r 0| dv0
~ H A =1
~ A
~ E A = j ~ A j1
( ~ A) or ~ E A =
1 j
~ H A
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Vector Potentials: No electric sources
2. No magnetic sources exist:
~ D F = 0 thus ~ D F = ~ F
Reminder: Both the curl and divergence need to bedefined for a unique definition of the magnetic vector
potential. This will be done shortly.So for a simple (linear, isotropic and homogeneous)medium we get:
~ E F = 1 ~ F
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Vector Potentials: No electric sources
So going into Ampere-Maxwells law:
~ H F = j ~ E F ~ E F = 1
~ F We get:
h~ H F + j ~ F i= 0Can be written as: ~ H F + j ~ F = m
~ H F = m j ~ F
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Vector Potentials: No electric sources
In order to find the two potentials we need the otherequation (Faraday)
~ E F = ~ M j ~ H F
1 ~ F = ~ M j ~ H F Plug in the electric field
Which for simple media becomes:
~ F = ~ M + j ~ H F ~ F = ( ~ F ) 2 ~ F
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Vector Potentials: No electric sources
This reduces to:
2 ~ F + 2 ~ F = ~ M +
h ~ F + j m
iWe can define the divergence freely (Lorentz gauge)
0
Thus the equation that needs to be solved is:
2 ~ F + 2 ~ F = ~ M
Furthermore:
~ H F = m j ~ F = j ~ F j1
( ~ F )
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Vector Potentials: No electric sources
As we did for the magnetic vector potential:
~ F = Z V
~ M (r 0)exp( j |r r 0|)
4 | r r 0|dv0
~ E F = 1 ~ F
~ H F = j
~ F j
1 (
~ F ) or
~ H F =
1 j
~ E F
And therefore the total fields are:
~ E = ~ E A + ~ E F ~ H = ~ H A + ~ H F
i h G i i l
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With Greens Functions not rigorously
In general a Greens function characterizes the responseof a system due to a point source
Point source Greens function
Actual source Superposition of point sources
The Greens function can be scalar, vector or dyadic
Wi h G F i i l
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With Greens Functions not rigorously
A dyad is a juxtaposition of two vectors
D = ~ A ~ B =
Ax Bx Ax By Ax B zAy Bx Ay By Ay BzAz Bx Az By Az B z
~ A = Ax~ax + Ay~ay + Az~az~ B = Bx~ax + By~ay + Bz~az
Then
Ifand
which can be thought as a matrix as well
Wi h G F i i l
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With Greens Functions not rigorously
For example look at the following vector identity:
~ B ( ~ A ~ C ) = ~ A ~ B ~ C ~ B ~ A ~ C
number number
Or:
~ B ( ~ A ~ C ) = ~ A ~ B ~ C ~ B ~ A ~ C
dyad dyad
With G F ti t i l
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With Greens Functions not rigorously
Similarly:
~ E = ~ E 2 ~ E
divergence
gradient of divergenceOr:
~ E = ~ E 2
~ E
dyadic operator
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With Greens Functions not rigorously
The solution can be thought as
~ E (~r) = j
R V G(~r, ~r
0) ~ J (~r 0) dv0
Dyadic Greens function
Plug this in to the wave equation:
j
Z V G(~r, ~r 0 ) ~ J (~r 0 ) dv0
2
j
ZV
G(~r, ~r 0 ) ~ J (~r 0 ) dv0 = j ~ J
Source point
Observation point
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With Green s Functions not rigorously
The current on the right-hand side can be written as
~ J (~r) = Z V I ~ J (~r 0) (~r ~r 0) dv0
Plugging this in to the previous equation we get:
Unit dyad: I =
1 0 0
0 1 00 0 1
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With Green s Functions not rigorously
j Z V G(~r, ~r 0) ~ J (~r 0) dv0 2 j Z V G(~r, ~r 0) ~ J (~r 0) dv0
= j Z V I ~ J (~r 0) (~r ~r 0) dv0If we interchange differentiation with integration,which is valid only outside of the source regionwhich is valid only outside of the source region, we get:
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With Green s Functions not rigorously
G(~r, ~r 0 ) 2 G(~r, ~r 0 ) = I (~r ~r 0 )
Reminder: observation point is away from the source
We write the dyadic Greens function in terms of a scalarone:
G(~r, ~r 0 ) =
I +
1
2
g(~r, ~r 0 )
and plug it in the above equation:
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With Green s Functions not rigorously
2 + 2
g(~r, ~r 0 ) = (~r ~r 0 )
We obtain:
Scalar 3D Greens function equation
To solve this we follow an approach very similar towhat we have already seen for the magnetic potential:
First set: ~r0
= 0and observe that is spherically symmetricg(~r )
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With Green s Functions not rigorously
There are two exponential solutions to this equationbut we only accept the one representing the outgoingwave:
g(r ) = C exp(
j r )r
The constant is found by integrating the differentialequation on an infinitesimally small sphere around theorigin:
Z V 2g(~r ) dv +
2
Z V g(~r ) dv = Z V (~r ) dvV 0with
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Z V
2g(~r) dv =
Z S g(~r) ~ dS
= 4 a2
dg(r )
dr r = a= 4 C as a 0
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The first integral becomes with Divergence Theorem:
radius of the sphere
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With Green s Functions not rigorously
The second integral becomes:
Z V g(~r) dv = Z a
0 4 r2g(r ) dr = 0 as a 0
The third integral becomes:
Z V (~r ) dv = 1
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W g y
Putting all together we find:
C =1
4 and g(r ) =exp( j r )
4 r
Since r is the distance between the source and theobservation points, if the source is not at the originwe get:
g(r, r 0) = exp( j |r r0|)
4 |r r 0|Reminder: observationpoint is away fromthe source region
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g y
So now we can find the electric field:
~ E (~r) = j Z V G(~r, ~r 0) ~ J (~r 0) dv0= j I +
1 2 Z V g(~r, ~r 0) ~ J (~r 0) dv0
= j
I +
1
2
Z V e( j | r r
0 | )
4 |r r 0|~ J (~r 0) dv0
to emphasize thatintegration is overprimed coordinates
Del operators operate on unprimed coordinates
Example: Cerenkov Radiation
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p
Advanced Test Reactor inIdaho National Laboratory (250 MW) Cobalt-60 (medical) Plutonium-238 (space probes)
Reed Research Reactor inReed College, Portland, OR(250 kW)
From wikipedia
Electromagnetic radiation emitted when a charged particle(such as an electron) passes through an insulator at a speedgreater than the speed of light in that medium.
Nobel prize1958
Example: Cerenkov Radiation
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p
Main Characteristics:Velocity of electronsmust be very large
Angles of radiationdepend on the electronvelocity
Electric fieldpolarization of emittedlight is parallel to the
plane determined by thedirection of the beamand direction of theradiation
While the electrons reachrelativistic speeds (~0.75 c ),we can explain it with what weknow so far!
Photonic shock wave!
Example: Cerenkov Radiation
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p
While the velocity decreases as a result of radiation,we will assume it constant.
Assume particle with charge q moving withvelocity v along the z direction in an isotropicmedium. This is equivalent to a current density:
~ J (~r, t ) = ~az qv (x) (y) (z vt)
Because of cylindrical symmetry it is convenient totransform this to cylindrical coordinates
Example: Cerenkov Radiation
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~ J (~r, t ) = ~az qv (x) (y) (z vt)
Due to symmetry, we would like to write it as:
~ J (~r, t ) = ~az C qv () (z vt)
The constant is determined by the requirement to havethe same total current as with the previous expression:
Example: Cerenkov Radiation
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1 =
Z +
Z +
(x) (y)dxdy =
= Z 2
0Z
0
(x) (y)dd = Z
0
(x) (y)2 d
Thus:
1 = Z
0 ()d
and since:
Example: Cerenkov Radiation
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(x) (y) = ()
2
we get:
So the current in cylindrical coordinates becomes:
~ J (~r, t ) = ~az qv1
2 () (z vt)
Example: Cerenkov Radiation
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This source is not time harmonic. So we need to usethe Fourier transform:
~ J (~r, ) =1
2 Z
~ J (~r, t )e
j t dt = ~azq
4 2 e(
j z/v ) ()
Thus the electric field wave equation becomes:
~ E (~r) 2 ~ E (~r) = j ~ J (~r)
= ~az j q 4 2
e( j z/v ) ()
Example: Cerenkov Radiation
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The most convenient way to solve this is to write theelectric field as a function of a vector Greens function:
~ E (~r ) = I + 1 2 ~g(, z)= ~g(, z) +
1
2 [ ~g(, z)]
Plugging this into the electric field vector equation
we get:
Example: Cerenkov Radiation
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2 + 2~g(, z) = ~az jq
4 2e( j z/v ) ()
Because of the azimuthal symmetry of the problem, wecan write the Greens function as:
~g(, z) = ~az g() j q
2e( j z/v )
Then the previous equation becomes:
Example: Cerenkov Radiation
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1 dd dd 2v2 + 2g() = ()2Away from the origin this equation becomes:
1 dd dd+ 2g() = 0where: = r 2 2v2
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Example: Cerenkov Radiation
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~ E (~ ) 'q
8 s 2 j h~a
v
~az
ie j ( + z/v )
Propagating plane wave provided that is real
All that Cerenkov observed is now explained:
is real if or
rest in hw
2 > 2 /v 2 v >
1
1/ 2
=c
n