Feb. 7, 2011 Plane EM Waves The Radiation Spectrum: Fourier Transforms.

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Feb. 7, 2011 Plane EM Waves The Radiation Spectrum: Fourier Transforms
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Transcript of Feb. 7, 2011 Plane EM Waves The Radiation Spectrum: Fourier Transforms.

Feb. 7, 2011

Plane EM WavesThe Radiation Spectrum:

Fourier Transforms

01

2

2

22

t

E

cE

01

2

2

22

t

B

cB

Vector Wave Equations for E and B:

For solutions to the 3-Dimensional wave equation, usecomplex notation

)(01ˆ

trkieEaE

)(02ˆ trkieBaB

frequencyangular

vector waveˆˆ

constantscomplex are ,

ectorposition vCartesian ),,(

rsunit vecto are ˆ,ˆ

00

21

nkk

BE

zyxr

aa

where

Before we go further, let’s review complex numbers

Argand Diagram

imaginary

real

y

x

x+iy

Complex number

zr

iyxz Complex conjugate iyxz

1ix = real part of zy = imaginary part of z

In polar coordinates sin cos ryrx

so )sin(cos iriyxz

The Euler Formula

sincos iei

implies sincos irrrez i

r = magnitude of zθ = phase angle of z

Re(z) = real part of z = rcosθIm(z) = imaginary part of z = r sinθ

Since exponentials are so easy to integrate and differentiate,it is convenient to describe waves as

)(),( tkziAetz Where A is a real constant

To get the physically meaningful quantity, which must be a real number,one solves the wave equation and then takes the REAL part of thesolution.

This is OK, since the wave equation is linear, so that the real part ofΨ and its imaginary part are each separately solutions.

So for example you can write

ctorcomplex ve a is where

Re

0Re

00

)(0

imaginaryal

tkzi

EiEE

eEE

then

rE = Re

r E 0

real + ir E 0

imaginary( ) cos(kz −ωt) − isin(kz −ωt)( )[ ]

=r E 0

real cos(kz −ωt) +r E 0

imaginary sin(kz −ωt)

vector vector

Solution to the wave equations:

)(01ˆ

trkieEaE

)(02ˆ trkieBaB

scalarscomplex are , 00 BE

nkk

vector wave

k

n

The waves travel in direction or

surfaces of constant phase travel with time in direction k

)(01ˆ

trkieEaE

)(02ˆ trkieBaB

These must also satisfy Maxwell’s Equations

z

E

y

E

x

E E zyx

∇⋅ ˆ a 1E0ei(

r k ⋅

r r −ωt )

[ ] = 0

Recall the definition of the divergence:

0E

)(01)ˆ( trki

xxx iekEa

x

E

)(01)ˆ( trki

xx eEaE

0E

0ˆ 01 Eaki

So

“Can show” the other equations are

0ˆ 01 Eaki

0ˆ 02 Baki

)(01ˆ

trkieEaE

)(02ˆ trkieBaB

0201 ˆˆ Bac

iEaki

0102 ˆˆ Ea

c

iBaki

ktransverse

BaEa

akak

n,propagatio ofdirection the to are

) is(that ˆ and ) is(that ˆboth

0ˆ and 0ˆ Since

21

21

0ˆ 01 Eaki

0ˆ 02 Baki

)(01ˆ

trkieEaE

)(02ˆ trkieBaB

0201 ˆˆ Bac

iEaki

0102 ˆˆ Ea

c

iBaki

00 Bkc

E

00 Ekc

B

Also

0

2

0 Ekc

E

222 kc

Require k>0 and ω>0 ω = c k

Hence, E0=B0

0ˆ 01 Eaki

0ˆ 02 Baki

)(01ˆ

trkieEaE

)(02ˆ trkieBaB

0201 ˆˆ Bac

iEaki

ir k × ˆ a 2B0 = −

cˆ a 1E0

othereach lar toperpendicu bemust

) is(that ˆ and ) is(that ˆ

ˆ opposite is 0ˆ and

ˆdirection in is 0ˆ Since

21

12

21

BaEa

aak

aak

othereach lar toperpendicu always

and n,propagatio ofdirection theto

larperpendicu always are and

SO

BE

Qualitative Picture: For “one” wave with one λ

In real situations, one wants to consider the superposition of many waveslike this – and the more general case where the direction of E(and hence B) is random as the wave propagates.

Phase Velocity v. Group Velocity

The speed at which the sine moves is the phase velocity

ckphase

v

The group velocity is

kg

v

This is usually discussed when you have several waves superimposed,which make a modulated wave: the modulation envelope travels with the group velocity

In a dispersive medium ω=ω(k) so ynecessaril vkg

However, in a vacuum, vgroup= c

Group and Phase Velocities

The Radiation Spectrum

Joseph Fourier

The Radiation Spectrum

The spectrum depends on the time variation of the electric field (or, equivalently, the magnetic field)

It is impossible to know what the spectrum is, if the electric field is only specified at a single instant of time. One needs to record the electric field for some sufficiently long time.

The spectrum (energy as a function of frequency) is related to the E-field (as a function of frequency) through the Poynting Vector.

The E-field (as a function of frequency) is related to the E-field (as a function of time) through the Fourier Transform

dtetEE ti

)(

2

1)(

deEtE ti)(2

1)(

Likewise,

ω = angular frequency

Fourier Transforms

dxexFf ix 2)()(

A function’s Fourier Transform is a specification of the amplitudes and phases of sinusoidals, which, when added together, reproduce the function

Given a function F(x)The Fourier Transform of F(x) is f(σ)

dxefxF ix 2)()(

see Bracewell’s book:FT and Its Applications

The inverse transform is

note change in sign

Not all functions have Fourier Transforms.

F.T. sometimes called the “power spectrum” e.g. search for periods in a variable star

Visualizing the Fourier Transform:

Visualizing the F.T.

)()()( xiFxFxF IR

xixeix sincos

dxxxFdxxxFi

dxxxFidxxxFf

IR

IR

)2sin()()2sin()(

)2cos()()2cos()()(

Suppose you have a complex function:

Recall Euler’s formula:

FT(F(x)) =

Notes:When F(x) is real (FI=0) the fourier transform f(σ) can still be complex.

For fixed σ, these integrals involve multiplying F by a sine (or cosine) with period 1/σ and summing the area underneath the result.Changing the frequency of the sines and cosines and repeating the process gives f(σ) at a second value of σ, and so on.

Some examples

22

for 1

2 and

2for 0)(

x

xxxB

(1) F.T. of box function

b(σ ) = B(x)e2πixσ dx−∞

+∞

= e2πixσ dx−ω

2

ω2

=e2πixσ

2πixσ

⎣ ⎢

⎦ ⎥

−ω2

ω2

=1

2πixσe2πi(ω

2 )σ − e−2πi(ω2 )σ

[ ]

= ωsin(πωσ )

πωσ= ω sinc(πωσ )

“Ringing” -- sharp discontinuity ripples in spectrum

When ω is large, the F.T. is narrow: first zero at

1

other zeros at

1

(2) Gaussian

F.T. of gaussian is a gaussian with narrower width

22 /1)(

xexG

222

)( egFT

Dispersion of G(x) β

Dispersion of g(σ)

1

(3) delta- function

11 for 0)( xxxx

Note:

1)( dxx

)(

)()()()(

1

111

xF

dxxxxFdxxFxx

x

)( 1xx

x1

1

1

2

12

21

)(

)()(

:function delta theof

ix

ix

ix

e

dxxxe

dxexxf

FT

Amplitude of F.T. of delta function = 1 (constant with sigma)Phase = 2πxiσ linear function of sigma

(4) )()()( 11 xxxxxF

)2cos(2

)(

1

22 11

x

eefFT ixix

So, cosine with wavelength1

1

x

transforms to delta functions at +/ x1

x

0-x1 +x1

)()()( 11 xxxxxF (5)

)2sin(2)( 1 xifFT

x-x1

x10

Summary of Fourier Transforms