Michael Scalora U.S. Army Research, Development, and Engineering Center Redstone Arsenal, Alabama,...

Michael Scalora

U.S. Army Research, Development, and Engineering CenterRedstone Arsenal, Alabama, 35898-5000

&Universita' di Roma "La Sapienza"

Dipartimento di Energetica

OPTICS BY THE NUMBERS

L’Ottica Attraverso i Numeri

Rome, April-May 2004

Soluzione Numeriche di Equazioni Nonlineari Accoppiate Usando il Predictor-Corrector Algorithm:

More Examples

2

0 0 0 0( ) ( ) ( ) ( )2

tx t t x t x t t x t

…is just a Taylor expansion for ANY function 0( )x t t

0 00 0

[ ( )] [ ( )]( ) ( )

2predictedf x t f x t t

x t t x t t

…always finds a second order accurate solution to the generic differential equation

( )[ ( )]

dx tf x t

dt

Keep in mind that an nth order differential equation can always be reduced to n coupled equations of first order.

Example 1: can be rewritten as

the system:

2

20

1 40x x x

…with appropriate initial conditions.

2

20

( )( )

( ) 1 4( ) ( )

dx ty t

dt

dy ty t x t

dt

24

( )( )

( ) ( )4 ( )

10

dx ty t

dtdy t y t

x tdt

0 0

40 0

1; 2 ;

10 ( )

(0) 1; (0) 0x x Con le condizioni iniziali:

Significa: …da cui…(0) (0) 0x y

24

( )( )

( ) ( )4 ( )

10

dx ty t

dtdy t y t

x tdt

0 0 0

4 20 0 0 0

( ) ( ) ( )

( ) ( ) 10 ( ) 4 (

predicted

predicted

x t t x t y t t

y t t y t y t x t t

0 00 0

0 0 0 04 20 0

( ) ( )( ) ( )

2

( ) ( ) ( ) ( )( ) ( ) 10 4

2 2

predicted

predicted predicted

y t y t tx t t x t t

y t y t t x t x t ty t t y t t t

Ricordiamo la Soluzione esatta:

/(2 ) 1 (0)( ) (0)cos (0) sin

2t x

x t e x t x t

1/ 2

0 2 20

11

4

0 0

40 0

1; 2 ;

10 ( )

(0) 1; (0) 0x x

-1.0

-0.5

0

0.5

1.0

900 905 910

t

x(t)

-1.0

-0.5

0

0.5

1.0

900.5 900.7 900.9 901.1 901.3 901.5

t

x(t)

( ) ( ) ( )

( ) ( ) ( )

x t x t y t

y t y t x t

( ) ( ) ( ) ( )x t x t x t y t

( ) ( ) ( ) ( )y t y t y t x t

Oscillatori smorzati accoppiati

Example 2:

damping restoring force

driving force

Numerical solution

0 0 0 0( ) ( ) ( ) ( )px t t x t x t y t t

0 0 0 0( ) ( ) ( ) ( )py t t y t y t x t t

0 0

0 00 0

[ ( ) ( )]( ) ( ) / 2

[ ( ) ( )]

p

p

x t x t tx t t x t t

y t y t t

0 0

0 00 0

[ ( ) ( )]( ) ( ) / 2

[ ( ) ( )]

p

p

y t y t ty t t y t t

x t x t t

-0.5

0

0.5

1.0

0 2 4 6 8 10

y(t)x(t)

t

==0.75; =2; =-2

-2

-1

0

1

2

0 2 4 6 8 10

y(t)x(t)

t

=0.1; =0.3; =2; =-0.5

Calculation of numerical derivatives

0 0 00

0

( ) ( ) ( )( ) lim

t

dx t x t t x tx t

dt t

However, in reality t cannot be zero, and so one incurs into an error:

0 0 00

( ) ( ) ( )( )

dx t x t t x tx t Error

dt t

?

23

0 0 0 0( ) ( ) ( ) ( ) ( ) ...2

tx t t x t x t t x t t

Consider the Taylor Expansion:

Then:

20 00 0

( ) ( )( ) ( ) ( ) ...

2

x t t x t tx t x t t

t

Calculating the numerical derivative as:

0 00

( ) ( )( )

x t t x tx t

t

Implies an error of order t …

0( )2

tError x t

Usando la matematica del buon senso

La derivata al punto puo essere definita in almeno due modi:

0t t0t 0t t

0 00

0 00

( ) ( )( )

( ) ( )( )

x t t x tx t

tx t x t t

x tt

0t

0 0 0 00

( ) ( ) ( ) ( )( )

2 2

x t x t x t t x t tx t

t

mediando le due soluzioni…

.. .

…si presume con un errore piu piccolo.

5

2 3( ) ( ) ( ) ( ) ( )0 0 0 0 02 6

4( ) ( ) ..0 24

.

t tx t t x t x t t x t x t

tx t t

Consider the Taylor Expansions:

2 3

0 0 0 0 0

54

( ) ( )0 24

( ) ( ) ( ) ( ) ( )2 6

tx t t

t tx t t x t x t t x t x t

35

0 0 0 0( ) ( ) 2 ( ) ( ) ( ) ...3

tx t t x t t x t t x t t

20 0

0 0

( ) ( )( ) ( )

2 6

x t t x t t tx t x t

t

Subtract…

20 0 0 0

4( )0 12

( ) ( ) 2 ( ) ( )t

x tx t t x t t x t x t t

20 0 0

0 2( )0 12

( ) 2 ( ) ( )( )

tx t

x t t x t x t tx t

t

0 0 0 0 0

50

2 3( ) ( ) ( ) ( ) ( )

2 64

( ) ( ) ..24

.

t tx t t x t x t t x t x t

tx t t

2 3

0 0 0 0 0

54

( ) ( )0 24

( ) ( ) ( ) ( ) ( )2 6

tx t t

t tx t t x t x t t x t x t

Add…

1. Increasing precision requires more information, storage

0 00

( ) ( )( ) ( )

x t t x tx t t

t

20 0

0

( ) ( )( ) ( )

2

x t t x t tx t t

t

2. An nth order differential equation can always be reduced to n coupled equations of first order.

2

20

1 40x x x

2

20

( )( )

( ) 1 4( ) ( )

dx ty t

dt

dy ty t x t

dt

3. Calculating derivatives and integrating differential equations is more of an art than a science, and one uses whatever works. Care should be exercised when considering functions that vary rapidly in space or time.

Ex.:

In general,

2 2 2

2 2 2

( , ) ( , )E z t n E z t

z c t

We have all the elements for a simple one dimensional electromagnetic pulse propagation algorithm

0

0.2

0.4

0.6

0.8

1.0

0 0.2 0.4 0.6 0.8 1.0

z (distance)

t (t

ime)

0 0( , )z t

0 0( , )z t t

0 0( , )z t t

0 0( , )z z t 0 0( , )z z t

0t

0z

22 20 0 0 0 0 0

2 2 2 2

20 0 0 0 0 0 0 02

0 0

2

( , ) 2 ( , ) ( , )( ,

( , ) ( , ) 2 ( , ( )

)

) ,

E z t E z t E z t tn n

c t c t

E z t E z z t E z t E z z t

z z

E z t t

Combine and solve for 0 0( , )E z t t

0 0 0 0

2 2

0 0

0 0 0 0 0 02 2

2 ( , ) ( , )

( , ) 2 ( , ) , )

,

(

( ) E z t E z t t

c tE z z t E z t E z z

z

E z t

tn

t

2 20( ) /

0 0 0( , 0) cos( )z z wE z t E e k z

Initial Condition

0.2

0.6

1.0

10 20

Position in m

Fie

ld Inte

nsity

Input Intensity 00

2k

0

0.2

0.4

0.6

0.8

1.0

0 25 50 75

Position in m

Fie

ld I

nten

sity

Propagation in free Space

1 20t t t t t

0

0.2

0.4

0.6

0.8

1.0

0 10 20 30 40

position in m

Fie

ld I

nte

nsit

y

For clarity only envelopes are shown

n=1 n=2

Incident

transmitted

reflected

|E|2

2 2 2

2 2 2

( , ) ( , )E z t n E z t

z c t

Another point of view:

L’equazione di secondo grado:

Diventa due equazioni accoppiate di primo grado…

2

2

( , ) ( , )

( , ) ( , )

E z t c B z t

t n zB z t E z t

t z

…che sono le equazioni di Maxwell in una dimensione e il tempocioe il punto di partenza.

0 0 0 0

2 20 0 0 0 0 0

2

0

2

0( , ) ( ,( , )

2

( , ) ( , ) ( , )

2

)E z t E z t t

t t

B z t B z

E z t

z t B z z tc c

n z n z

t

0 0

2

0 0 0 0 0 02( , ) ( , )( ) ( ,, )

c tE z t t B z z t B z z tt t

n zE z

Algebraically Solve for 0 0( , )E z t t

0 00 0 0 0

0 0 0 0 0 0

( , ) ( , )

2( , ) ( , ) ( , )

2

( , )B z t B z t t

t tE z t E z z t E

B z

z

t t

z t

z z

0 0 0 0 0 00 0( , ) ( , ) ( , ) ( , )t

B z t t E z z t EB zz z tz

t t

Algebraically Solve for 0 0( , )B z t t

1. The SVEA, the Wave Equation, and Diffraction

2. Spectral Methods and Free-Space Diffraction

Diffraction

The Bending of Light Around Corners

a

0 a

a

0 ~ aL

L

ray opticswave optics

0

0.2

0.4

0.6

0.8-40-30

-20-10

010

2030

40

transverse coordinate

intensity

a

0 a

a

0 ~ aL

ray opticswave optics

2 22

2 2

n

c t

E

E

( )ˆ ( , , ) .i kz tE x y z e c c E x

x

z

y

E

2 2 22 2

2 22 t

E E nik k E E E

z z c

2 22

2 2t x y

k n

c

This term decribes diffraction

Assumption: the beam envelpe does not vary in timeso-called CW (continuous wave) beam

2

22

E Eik

z z

( )( , ) .i kz tE E x z e c c

For simplicity, let’s assume only one transverse dimension:

Assume…

2

22

E Eik

z x

22

2t x

E(x,z) e’ un inviluppo che varia lentamente nello spazio rispetto a e nel tempo rispetto a

/c.

2

22

E i E

z k x

k nc

2 2 2 22

2 2 22

E E E nik k E E

z z x c

Apply the SVEA…

/ /z L x x L

We may also define the Fresnel number as…0

4 nLF

0E

F

F small

Il fascio non diffrange: Ray optics..

Regime di Forte diffrazzione.

2

22

E i E

z k x

2

2

E i E

F x

2 2 2

2 2 2

( , ) ( , )E z t n E z t

z c t

How to calculate diffraction using the wave equation

Confrontiamo…

Con…2

2

E i E

F x

20 0 0 0 0 0 0 02 2

( , ) ( , ) 2 ( , ) ( , )E x z E x x z E x z E x x z

x x

20 0 0 0 0 0 0 02 2

( , ) ( , ) 2 ( , ) ( , )E z t E z z t E z t E z z t

z z

0 0 0 0 0 0( , ) ( , ) ( , )

2

E x E x E x

20 0 0 0 0 0 0 02 2

( , ) ( , ) 2 ( , ) ( , )E x E x x E x E x x

x x

Combine and solve for 0 0( , )E x

0 0 0 0

0 0 0 0 0 02

( , ) ( , )

( , ) 2 ( , ) ( , )2

E x E x

E x x E x E x x

x

0

0 0

0 0

0 0

0 02

0

( , )

( , ) 2 (( ,

( ,

) )

)

,2

E x

E x

E x x E x x

E

x

x

We can proceed as follows:

0 0 00 0

0( , )

2

( ))

,( ,

E x xE x

E

Substitute and solve for 0 0( , )E x

0 0

0 0

0 0

0 0 0 0 0 02

( , )

( , ) ( , )

( , )

( , ( )2

) ,

E x

E x

E x

E x x E x E x x

x

Let…

0 0

0 0 0 0 0 02 2

0 0 ( , )

( , ) ( , ) ( , ) 2

(1 2

( , )

/ )

E x

E x x E x E

x

x

x x

E

x

0 0

0 0

0 0

0 0 0 0 0 02

( , )

( , ) ( , )

( , )

( , ( )2

) ,

E x

E x

E x

E x x E x E x x

x

Ci sono altri modi di procedere che non richiedono un’impostazione cosi onerosa dal punto di vista del numero di varibili e vettori di cui

tener conto. Ci occuperemo di metodi cosidetti spettrali.

How to calculate diffraction using the wave equation…

Useful properties of the Fourier Transform: Derivatives…

2 22

2 2

( , ) ( , ) ( , )

( , ) ( , ) ( , )

( , ) ( , ) ( , )

( , ) ( , ) ( ) ( , )n

iqx

iqx

iqx

niqx n

n n

FT E x E q E x e dx

FT E x E x e dx iqE qx x

FT E x E x e dx q E qx x

FT E x E x e dx iq E qx x

2

2

E i E

F x

21,2,3,4...

x

q j jL

xL N x

N is the number of points used to discretize the space x in units x.

Some Advantages of using Spectral (FT) methods

(i) calculation of all kinds of derivatives is simple(ii) derivatives are extremely accurate(iii) simple algorithm

Some Disadvantages

(i) It is slower(ii) Functions should be very smooth (which usually are)

2( , )( , )

E q iqE q

F

2

( , ) ( ,0)iq

FE q e E q

…is the solution in q-space. To find the solution in x-space, where things are observable, we need to take the inverse transform:

2

2

( , ) ( , )E x i E x

F x

Taking the FT of both sides with respect to x…

And…

2

1( , ) ( , ) ( ,0)iq

iqxFE x FT E q e E q e dq

( ,0) ( ,0)E q FT E x

2

1( , ) ( ,0)iq L

FE x L FT e FT E x

2

1( , ) ( ,0)iq L

FE x FT e FT E x

Algorithm:

(i) Trasformata di Fourier (FT) del campo iniziale

(ii) Moltiplicazione per il propagatore

(iii) Transformata inversa di tutto

Examples: single slit

0

0.2

0.4

0.6

0.8

-40 -30 -20 -10 0 10 20 30 40

transverse coordinate

inte

nsity

Direction of Propagation

-15 -5 5 15

TRANSVERSE COORDINATE

0

10

20

30

40

50

LO

NG

ITU

DIN

AL

C

OO

RD

INA

TE

0.0

1.0 1.1

1.1

Direction of Propagation

0

0.2

0.4

0.6

0.8

1.0

-60 -20 20 60

Example2: double slit

Poisson Spot

-25 -15 -5 5 15 25

TRANSVERSE COORDINATE

0

10

20

30

40

50

LO

NG

ITU

DIN

AL

C

OO

RD

INA

TE

0.0 0

.0

0.1 0.1

1.0 1.0

1.0 1.0 1.1 1.1 1

.1

1.1

1.1

1.1 1.2

1.2

1.2

1.2

1.2 1.2

1.3 1

.3

1.3 1.3

2 22

2 2t x y

2 2

2 2

( , , )( , , )

E x y iE x y

F x y

Per aperture con geometrie piu complicate, e.g.Apertura Circolare o quadrata, e’ necessario

Ritornare alle due dimensioni trasverse:

2 2( , , ) ( )( , , )x y x yx y

E q q i q qE q q

F

2 2( )

( , , ) ( , ,0)x yi q q

Fx y x yE q q e E q q

2

1 2 2 2( , , ) ( , ,0)iq L

Fx yE x y L FT e FT E x y q q q

Circular Aperture

Square Aperture

10 60 110 160 210

Y

10

60

110

160

210

X

0.0

0.0

0.0

0.0

0.0 0.0

0.1 0.1 0.1 0.1 0.1

0.1

0.1

0.1 0.1

Square Aperture

Algorithm:

(i) Trasformata di Fourier (FT) del campo iniziale

(ii) Moltiplicazione per il propagatore

(iii) Transformata inversa di tutto

2

1 2 2 2( , , ) ( , ,0)iq L

Fx yE x y L FT e FT E x y q q q

2

22

E Eik

z x

2

22

E i E

z k x

2 2 2 22

2 2 22

E E E nik k E E

z z x c

Removing the second order spatial derivative also meansmaking the Paraxial Wave Approximation:

the beam radius cannot be too smallor inconsistencies with experimental observations

may arisem since the wave tends to diffract very fast,contrary to expectations.

This problem may be partially cured as follows:

2 2 2 22

2 2 22

E E E nik k E E

z z x c

2 2 2 22

2 2 22

E E n Eik k E E

z x c z

2 2

2 22

E E Eik

z x z

2 2

2 2

E i E i E

F x F

22

2 2

i E iE

F

E

x F

2 3

2 3

2

2

E i E i E

F x F

2 2 3

2 2 3

E i E i i i E

F x F F

E

x F

2 2 32 2

2 2 2 2 3

E i E i i i E

F x F F

i E

x

i

F x F F

E

4

3

2 2 2 3

2 3 2 2 34

E i E i E i E

F x F

E

x FF x

i