Feb 16, 2011

Finish PolarizationThe Radiation Spectrum

HW & Questions

LCD Displays

Bees can see polarized light polarization of blue sky enables them to navigate

Humans: Haidinger’s Brush

Vikings: Iceland Spar?

Many invertebrates can see polarization, e.g. the Octopus

Not to navigate (they don’t go far)Perhaps they can see transparent jellyfish better?

unpolarized polarized

Polarization and Stress Tests

In a transparent object, each wavelength of light is polarized by a different angle. Passing unpolarized light through a polarizer,then the object, then another polarizer results in a colorful pattern which changes as one of the polarizers is turned.

CD cover seen in polarized light from monitor

3D movies

Polarization is also used in the entertainment industry to produce and show 3-D movies. Three-dimensional movies are actually two movies being shown at the same time through two projectors. The two movies are filmed from two slightly different camera locations. Each individual movie is then projected from different sides of the audience onto a metal screen. The movies are projected through a polarizing filter. The polarizing filter used for the projector on the left may have its polarization axis aligned horizontally while the polarizing filter used for the projector on the right would have its polarization axis aligned vertically. Consequently, there are two slightly different movies being projected onto a screen. Each movie is cast by light which is polarized with an orientation perpendicular to the other movie. The audience then wears glasses which have two Polaroid filters. Each filter has a different polarization axis - one is horizontal and the other is vertical. The result of this arrangement of projectors and filters is that the left eye sees the movie which is projected from the right projector while the right eye sees the movie which is projected from the left projector. This gives the viewer a perception of depth.

2121

2121

22

21

22

21

sin2

cos2

V

U

Q

I

)cos( and )cos( 2211 tEtE yx

Stokes Parameters:

2121

2121

22

21

22

21

sin2

cos2

V

U

Q

I

)cos( and )cos( 2211 tEtE yx

Linear Polarized:

€

I

Q

U

V

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

= Io

1

1

0

0

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

2121

2121

22

21

22

21

sin2

cos2

V

U

Q

I

)cos( and )cos( 2211 tEtE yx

Linear Polarized:

€

I

Q

U

V

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

= Io

1

−1

0

0

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

2121

2121

22

21

22

21

sin2

cos2

V

U

Q

I

)cos( and )cos( 2211 tEtE yx

Linear Polarized (45 deg):

€

I

Q

U

V

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

= Io

1

0

1

0

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

2121

2121

22

21

22

21

sin2

cos2

V

U

Q

I

)cos( and )cos( 2211 tEtE yx

Linear Polarized (- 45 degrees):

€

I

Q

U

V

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

= Io

1

0

−1

0

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

2121

2121

22

21

22

21

sin2

cos2

V

U

Q

I

)cos( and )cos( 2211 tEtE yx

Left-handCircular:

€

I

Q

U

V

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

= Io

1

0

0

1

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

2121

2121

22

21

22

21

sin2

cos2

V

U

Q

I

)cos( and )cos( 2211 tEtE yx

Right-handCircular:

€

I

Q

U

V

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

= Io

1

0

0

−1

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

2121

2121

22

21

22

21

sin2

cos2

V

U

Q

I

)cos( and )cos( 2211 tEtE yx

Unpolarized:

€

I

Q

U

V

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

= Io

1

0

0

0

⎛

⎝

⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟

The Radiation Spectrum Rybicki & Lightman, Section 2.3

Consider

€

rE (t) Transverse E-field

What is the spectrum?

€

W (ω) Energy / time as a function of frequency

Note:

€

=2π

tRadians / sec

Define the Fourier Transform of

€

E(t) ≡ ˆ E (ω)

€

ˆ E (ω) =1

2πE(t)e iωtdt

−∞

∞

∫

€

ˆ E (ω) =1

2πE(t)e iωtdt

−∞

∞

∫

The inverse is

€

E(t) =1

2πˆ E (ω)e−iωtdt

−∞

∞

∫

Eqn. 1

€

E(t) is real; ˆ E (ω) is complex

Eqn. 2

Take complex conjugate of Eqn. (1):

€

What is E(t)e iωt[ ]

*?

€

ˆ E (ω)* =1

2πE(t)e iωt

[ ]*dt

−∞

∞

∫

€

What is E(t)e iωt[ ]

*?

€

E(t)e iωt[ ]

*= E(t)* e iωt

( )*

€

Since E(t) is real, E(t)* = E(t)

€

e iωt = cosωt + isinωt

e iωt( )

*= cosωt − isinωt

= cos(−ωt) + isin(−ωt)

= e−iωt

€

E(t)e iωt[ ]

*= E(t)e−iωtSo….

and

€

ˆ E (ω)* =1

2πE(t)e iωt

[ ]*dt

−∞

∞

∫ =1

2πE(t)e−iωtdt

−∞

∞

∫

But since

€

ˆ E (ω) =1

2πE(t)e iωtdt

−∞

∞

∫ We have

€

ˆ E (−ω) =1

2πE(t)e−iωtdt

−∞

∞

∫

So

€

ˆ E ω( )*

= ˆ E −ω( ) eqn. (a) Also

€

ˆ E −ω( )*

= ˆ E ω( ) eqn. (b)

Now

€

ˆ E ω( )2

= ˆ E ω( ) ˆ E * ω( )

€

ˆ E −ω( )2

= ˆ E −ω( ) ˆ E * −ω( )

= ˆ E ω( )* ˆ E ω( )

= ˆ E ω( )2

From Eqn. (a), (b)

€

ˆ E −ω( )2

= ˆ E ω( )2 So…

Parseval’s Theorem for Fourier Transforms:

€

E 2(t)dt =−∞

∞

∫ 2π ˆ E (ω)2dω

−∞

∞

∫

Proof:

€

E 2(t)dt =−∞

∞

∫ dt E(t)−∞

∞

∫ dω ˆ E (ω)−∞

∞

∫ e−iωt

= dω ˆ E (ω)−∞

∞

∫ dt E(t)−∞

∞

∫ e−iωt

= dω ˆ E (ω)−∞

∞

∫ 2π ˆ E * ω( )

= 2π ˆ E (ω)2

−∞

∞

∫ dω

€

E 2(t)dt =−∞

∞

∫ 2π ˆ E (ω)2

−∞

∞

∫ dω

= 2π ˆ E (ω)2

0

∞

∫ dω + 2π ˆ E (ω)2

−∞

0

∫ dω

= 2π ˆ E (ω)2

0

∞

∫ dω + 2π ˆ E (−ω)2

0

∞

∫ dω

= 4π ˆ E (ω)2

0

∞

∫ dω

Poynting Theorem: Energy /time/area

€

dW

dtdA=

c

4πE 2(t)

Integrate over pulse:

€

dW

dA=

c

4π −∞

∞

∫ E 2(t)dt

= c0

∞

∫ ˆ E ω( )2dω

So…

€

dW

dAdω= c ˆ E ω( )

2

Electromagnetic Potentials

Rybicki & Lightman, Chapter 3

Electromagnetic Potentials

Instead of worrying about E and B, we can use the scalar and vector potentials

),( PotentialVector

),( PotentialScalar

trA

tr

Simpler: 1 scalar and 1 vector quantity instead of 2 vector quantities. Relativistic treatment is simpler.

AAcurlBBBdiv

0

t

A

c

t

B

cEEcurl

1

1

01

t

A

cE

t

A

cE

1

t

A

cE

1

So

Therefore, there exists a φ such that

or

(1)

(2)

Equations (1) & (2) already satisfy 2 of Maxwell’s Equations – what about the others?

4 E

1, D

E

41

41

2

Atc

t

A

c

becomes

For reasons which will become clear in a minute, we re-write thislast equation as

4111

2

2

22

tc

Atctc

(3)

The 4th Maxwell equation

t

D

cj

cH

14ED

HB

becomes

jct

A

ctcA

411

AAA

2Since

we get

jctc

At

A

cA

4112

2

22

(4)

GUAGE TRANSFORMATIONS

(1)-(4) do not determine A and φ uniquely:one can add the gradient of an arbitrary scalar ψ to A and leave B unchanged

BBAA

Likewise E will be unchanged if you add

EEtc

1

These are called GUAGE TRANSFORMATIONS

For the LORENTZ GUAGE we take

01

tc

A

so that (3) and (4) simplify to

jt

A

cA

tc

41

41

2

2

22

2

2

22

(5)

(6)

RETARDED POTENTIALS

It turns out that the solutions to (5) and (6) can be expressedas integrals over sources of charge, provided you properlytake into account the fact that changes in the E and B fieldscan move no faster than the speed of light.

RETARDED POTENTIALS

At point r =(x,y,z), integrate over charges at positions r’

rd

rr

j

ctrA

rdrr

tr

3

3

1),(

),( (7)

(8)

where [ρ] evaluate ρ at retarded time:

rr

ctr

1, Similar for [j]