2 wave solutions

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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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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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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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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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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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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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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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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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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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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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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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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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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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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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Similarly:

~ E = ~ E 2 ~ E

divergence

gradient of divergenceOr:

~ E = ~ E 2

~ E

dyadic operator


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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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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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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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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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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


The first integral becomes with Divergence Theorem:

radius of the sphere



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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



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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!



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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



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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:



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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:



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(x) (y) = ()

2

we get:

So the current in cylindrical coordinates becomes:

~ J (~r, t ) = ~az qv1

2 () (z vt)



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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 ) ()



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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:



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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:



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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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~ 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

2 wave solutions

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