Wave Phenomena Physics 15c

Lecture 16 Radiation of EM Waves

What We Did Last Time EM waves in conductors are confined within the skin depth   Higher frequencies thinner

Studied reflection and refraction   Derived Snell’s law three times

  Using Huygens’ principle   Using Fermat’s principle   Solving boundary-condition problem

  Fresnel coefficients intensities of reflected/refracted light   Brewster’s angle Reflection is polarized

Let’s wrap up what was left…

sinθ2

sinθ1

=n1

n2

Fresnel Coefficients

  For air glass, β = 1.5

β ≡

Z1

Z2 α ≡

cosθ2

cosθ1

ERV

EIV =

α − βα + β

ETV

EIV =

2α + β

ERH

EIH =

αβ −1αβ +1

ETH

EIH =

2αβ +1

Vertical polarization Horizontal polarization

ETV

EIV

ERV

EIV

ERH

EIH

ETH

EIH

No reflection at this angle

θ1 θ1

Reflectivity and Transmittivity Intensity is

  Reflectivity R is simply

  Transmittivity is a little more subtle   Consider the power going through

unit area on the boundary surface

R =

ER

EI

⎛

⎝⎜⎞

⎠⎟

2

Acosθ2A

Acosθ1

S = E 2 Z

PI =

EI2

Z1

Acosθ1

PT =

ET2

Z2

Acosθ2

T =

PT

PI

=ET

2

EI2

Z1

Z2

cosθ2

cosθ1

=ET

EI

⎛

⎝⎜⎞

⎠⎟

2

αβ

Reflectivity and Transmittivity

  R + T = 1 in both cases   For air glass, β = 1.5

RV =

α − βα + β

⎛⎝⎜

⎞⎠⎟

2

TV =

4αβ(α + β)2

RH =

αβ −1αβ +1

⎛⎝⎜

⎞⎠⎟

2

T H =

4αβ(αβ +1)2

Vertical polarization Horizontal polarization

No reflection around here

RV

TV

RH

T H

θ1 θ1

Brewster’s Angle Look at the condition α = β   Assume µ1 = µ2

  A little trig gives us

At Brewster’s angle, the reflected light and the refracted light are perpendicular to each other   Is this a coincidence?   If not, what is the significance?

β ≡

ε2µ1

ε1µ2

=ε2

ε1

=n2

n1

=sinθ1

sinθ2 α ≡

cosθ2

cosθ1

cosθ2

cosθ1

=sinθ1

sinθ2

sin2θ1 = sin2θ2 θ1 = θ2 θ1 +θ2 =

π2

or

θ1 θ1

θ2

Slow Fast Transition What if n1 > n2?   For glass air, β = 1/1.5

  R = 1, T = 0 beyond the critical angle

Total internal reflection

Total internal reflection

RV

TV

RH

T H

Total Internal Reflection Angle of refraction is determined by the difference between wavelengths   Wavelength along the surface is

What if we choose θ2 so that is shorter than λ1?

  Can’t match the wavefronts no matter what θ1 we use

  No EM waves above water

λs =

λ1

sinθ1

=λ2

sinθ2

λs = λ2 sinθ2

Boundary Condition Examine boundary conditions   No EM Field above water

Only solution is   No waves anywhere   Something went wrong…

EI

BI

BR

E1 = E2

ε1E1⊥ = ε2E2

⊥

H1 = H2

µ1H1⊥ = µ2H2

⊥

(EI + ER )cosθ = 0

ε2(EI − ER )sinθ = 0

HI − HR = 0

Nothing

EI = ER = 0

ε1,µ1

ε2,µ2

Imaginary Waves There must be EM field above water   Boundary conditions cannot be satisfied without it   But it cannot be usual EM waves either

Solution: above water

  Wave equation

  Wavelength along the surface is linked to

E = E0ei (kx+ iκz−ω t ) = E0e

−κzei (kx−ω t )

∇2E =

n12

c2

∂2E∂t 2

k 2 −κ 2 =

n12

c2 ω2

λs = 2π k

λs =

λ2

sinθ=

2πcωn2 sinθ

k =ωn2 sinθ

c

κ =

ωc

n22 sin2θ − n1

2

Evanescent Waves EM “waves” above water satisfies wave equation and boundary conditions   It shrinks as e–κ z Not traveling waves

  “Range” of the imaginary waves is

  Comparable to

Total internal reflection creates “leakage” or evanescent waves that extends a few wavelengths   You can detect it by having two

boundaries close enough to each other

E = E0e−κzei (kx−ω t )

1κ=

c

ω n22 sin2θ − n1

2

λ1 =

2πcωn1

Today’s Goals How to create EM waves   What is happening in the antennas, light bulbs, etc? Moving charge (electrons)!

Derive EM waves emitted by an accelerating charge   You may have done this in 153, but let’s do it again   Intensity, angular distribution, polarization…   Qualitative explanation for Brewster’s angle   Rayleigh scattering

Creating EM Waves Look at the “full” version of Maxwell’s equations

  We need electric charge to create EM waves

The charge must be moving   Stationary charge create only static E field   Constant current create only static B field

∇ ×E = −

∂B∂t

∇ ⋅B = 0 ∇ ⋅E =

ρε0

∇ ×B = ε0µ0

∂E∂t

+ µ0J

SI

Charge density Current density

Let’s move a point charge +q with constant velocity v   We know the directions of E and B fields

  Poynting vector is   No power is radiated outward

Constant velocity No radiation

We shouldn’t be surprised   Think about Relativity   There is an inertial frame in which the charge is at rest

  No energy is radiated by EM waves in that frame   Same should be true in any coordinate system

Moving Charge

v +q

E B

S E r B v × r

S = E ×H→ S ⊥ r

E&M Puzzle Charge +q creates E at position r   Exactly which direction does E point?

  Parallel to r, of course…?

EM field travels at speed c   Field at r is not determined by

where +q is now, but T = r /c ago

  In the meantime, +q moved by   So…?

Which point does the tail of E point?   Where +q is now, or where it was at t = −T ?

v +q

E

r

vT

vT =

vrc now t = −T

E&M Puzzle Answer: where +q is now   See your E&M textbook for a real explanation

Sloppy explanation:   Consider two charges +q and −q

moving in parallel   Let –q fall toward +q

v +q

v −q

E

E

“now”

“past”

Hit

Miss

  Watch this in a frame where v = 0 They will collide

  “Now” hypothesis is correct

v +q

E

r

How Does It Know? But how does E at r know where the charge is now?   It can’t — Special Relativity!

  Sounds odd, but that’s the only Relativistically-Correct way

E doesn’t necessarily point back to the actual location of the charge if it accelerated (or decelerated)   We are on to something now…

E(r, t) is determined by what q was doing at t – r/c, in such a way that it points back to where q should be now if v has been constant v

+q

E

r

Accelerating Charge Consider this picture:   Charge +q is at rest until t = 0   It accelerates for Δt by a = dv/dt   Keep constant velocity v after Δt

Want to know E at large distance r   Easy for t < 0 and t > Δt   What happens during the

acceleration? +q v

t = 0 t = Δt

r

E

Before/After Acceleration

A B C

A : t = 0B : t = ΔtC : t = Δt +T

accelerate between

now

Radius c(Δt + T) from A   Outside this circle,

acceleration has not happened

  E points outward from A

Radius cT from B   Inside this circle, the

acceleration has finished   E points outward from C

How can we connect them?

During Acceleration E field lines must be continuous   They start and end

only where there is charge

  So we just connect them like

Acceleration causes “kinks” in E field   Let’s looks at it

more closely

A B C

Radiated Field We are interested in E at large distance r   Radial component is

the usual Coulomb field

  Geometry gives us the ratio

  ET is proportional to the acceleration

A B C t = 0 Δt Δt +T

θ

E0 ET

E0 =

q4πε0r

2

ET

E0

≈vT sinθ

cΔt v c

Δt Tassuming

ET =q

4πε0r2

vT sinθcΔt

=qasinθ4πε0c

2r

cΔt vT sinθ

a =

vΔt

T =

rc

Radiated Field

At large r,   i.e., E becomes transverse   This is the EM radiation!

We like to write it this way:   E is transverse   Proportional to acceleration a   Direction opposite to a   Maximum radiation at 90°   Decreases with 1/r

E0 =

q4πε0r

2

ET =

qasinθ4πε0c

2r

ET E0

ET =

q4πε0c

2

sinθr

a(t − rc )θ̂

Delayed by r/c due to propagation speed

a

Radiated Power Find BT from Maxwell

  Integrate with time:

Poynting vector

  EM radiation spreads outward   It goes with square of q, square of a

  Deceleration radiates just the same   It goes with 1/r2

  As it should for expanding waves in 3D   It’s not isotropic, but ∝ sin2θ

∂BT

∂t= −∇ ×ET = −

1r∂∂r

rET( ) φ̂ = −q sinθ

4πε0c2r

∂a(t − rc )

∂rφ̂ =

q sinθ4πε0c

3r∂a(t − r

c )∂t

φ̂

S =

ET ×BT

µ0

=q2 sin2θ

16π 2ε0c3r 2 a2(t − r

c ) r̂

a

ET =

q4πε0c

2

sinθr

a(t − rc )θ̂

BT =

q sinθ4πε0c

3ra(t − r

c )φ̂

EM Waves in Matter Matter is made of charged particles   Electrons and protons, in particular

EM waves in matter make them oscillate   They radiate EM waves   EM waves in matter = original + radiated waves   This is why EM waves propagate differently in matter

Incoming Radiated

Brewster’s Angle Why is there reflection in the first place?   Because Z1 ≠ Z2

  Electrons in medium 2 must be responsible Which way do the electrons oscillate?   Parallel to ET

  They radiate EM waves as

  At Brewster’s angle, reflection is parallel to ET θ = 0 No reflection!

θ1 θ1

θ2

EI

BI ER

BR

ET

BT

ET S =

q2 sin2θ16π 2ε0c

3r 2 a2(t − rc )r̂

Angle of radiation relative to ET

Larmor Formula What’s the total power radiated by a moving charge?   Integrate S over the surface of a sphere at r

Accelerating charge loses its energy at a rate proportional to acceleration squared   Think about an atom as electrons circling around the nucleus

  Circular motion acceleration radiation loss of energy electrons fall into the nucleus

  This paradox stimulated early development of QM

Joseph Larmor (1857–1942)

P = dφ Sr 2 sinθ dθ

0

π

∫0

2π

∫ =q2a2

8πε0c3 sin3 θ dθ

0

π

∫ =q2a2

6πε0c3

S =

q2 sin2θ16π 2ε0c

3r 2 a2(t − rc )r̂

Larmor formula

Oscillating Charge Now we consider an oscillating charge

  Transverse component of E at (r,θ) is

  Calculate the Poynting vector

Total power radiated is

  Increases with (frequency)4

x = x0 cosω t q

a = x = −ω 2x0 cosω t θ

ET ET =

q4πε0c

2

sinθr

a(t − rc )θ̂ = −

qω 2x0

4πε0c2

sinθr

cos ω(t − rc )( ) θ̂

S =

q2ω 4x02

16π 2ε0c3

sin2θr 2 cos2 ω t − r

c( )( ) r̂ S =

q2ω 4x02

32π 2ε0c3

sin2θr 2 r̂

P = dφ S r 2 sinθ dθ

0

π

∫0

2π

∫ =q2ω 4x0

2

16πε 2c3 sin3 θ dθ0

π

∫ =q2ω 4x0

2

12πε0c3

average

Rayleigh Scattering Sunlight passing through the air makes air molecules oscillate   Incoming light has a broad spectrum

  All frequencies are more or less equal

Molecules radiate power according to

  More power is absorbed and re-emitted at higher frequencies

  This is why the sky looks blue, and why the sky turns red at sunset

E

−q

F = qE

P =

q2ω 4x02

12πε0c3

Polarization Sunlight contains two polarizations

  Only one causes radiation that reaches the observer

  Scattered light (what you see in the “blue” sky) is polarized   Photographers use polarizing filters to deepen the color of the sky

Viewed from right

Summary Reviewed reflection and refraction   Total internal reflection is more subtle than it looks   Evanescent waves extend a few λ beyond the surface

How to create EM waves   Accelerated charge radiates EM waves   Power given by Larmor formula

  Proportional to (acceleration)2

  Instability of atoms QM   Polarization parallel to the acceleration

  Explains Brewster’s angle   Rayleigh scattering

Next: why light travels slower in matter?

ET =

q4πε0c

2

sinθr

a(t − rc )θ̂

P =

q2a2

6πε0c3

S =

q2 sin2θ16π 2ε0c

3r 2 a2(t − rc )r̂