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Electromagnetism Physics 15b

Lecture #19 Electromagnetic Waves

Purcell 9.4–9.7

What We Did Last Time Complete Maxwell’s equations

  Displacement current needed for mathematical consistency

Electromagnetic waves   Maxwell’s eqns. in vacuum

  Solutions are waves propagating with speed of light … which is light itself

∇ ⋅E = 4πρ ∇ ⋅B = 0

∇ ×E = −1c∂B∂t

∇ ×B =4πc

J+1c∂E∂t

Jd =

14π

∂E∂t

∇2E =

1c2

∂2E∂t 2 ∇2B =

1c2

∂2B∂t 2

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Today’s Goals Examine electromagnetic waves in more details   Plane waves and their completeness   Wave number k and angular frequency ω   Relationship between E and B   Polarization

Consider the energy carried by electromagnetic waves   Poynting vector   Power density and momentum density

Sinusoidal Wave Solution Let’s consider sinusoidal E waves traveling along +x

  k : wave number [unit = radian/cm] is related to the wave length λ by kλ = 2π

  Just as ω [unit = radian/s] is related to the period T by ωT = 2π

In order to satisfy the wave equation

  With this condition,

E = E0 sin(kx −ω t)

k =

2πλ

∇2E =

1c2

∂2E∂t 2

∂2

∂x2 sin(kx −ω t) = 1c2

∂2

∂t 2 sin(kx −ω t)

k 2 =ω 2

c2 ω = ck

sin(kx −ω t) = sink(x − ct)

Expected from general soln.

f(x ± ct)

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Sinusoidal Wave Solution There are more equations to satisfy

  E is perpendicular to x, the direction of wave propagation   Let’s take the direction of E0 as the y axis

The last equation to satisfy

  Integrating by t, we find

  B is perpendicular to both E and x,

and |B| = |E|

∇ ⋅E = 0 ∇ ⋅E0 sin(kx −ω t) = x ⋅E0k cos(kx −ω t) = 0

∂B∂t

= −c∇ × yE0 sin(kx −ω t) = −zE0ck cos(kx −ω t)

E = yE0 sin(kx −ω t)

∇ ×E = −

1c∂B∂t

B = zE0 sin(kx −ω t)

x

E

B

Plane Waves Plane sinusoidal EM waves are described by 3 vectors:   E0 : amplitude and direction of E field   B0 : amplitude and direction of B field   k : wave number and direction of propagation

Generally   E0, B0 and k are perpendicular and right-handed   |E0| = |B0|

k = kxE0 = E0yB0 = E0z

⎧

⎨⎪

⎩⎪ +x

+y

+z

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General Plane Waves General form of plane EM waves are

with conditions   ... plus the corresponding cosine terms

Fourier transform:   Any (repetitive) function of space can be represented by an infinite

sum of sines and cosines of the form

  Understanding plane waves is understanding all possible waves

E = E0 sin(k ⋅r −ω t) = E0 sin(kxx + ky y + kzz −ω t)

B = B0 sin(k ⋅r −ω t) = B0 sin(kxx + ky y + kzz −ω t)

⎧⎨⎪

⎩⎪

ω = kc, E0 = B0 , k = E × B

Ak sin(k ⋅r) + Bk cos(k ⋅r)⎡⎣ ⎤⎦

k∑

Polarization For a given k, there are two possible directions of E   E is in the y- or z-plane   Direction of E is called

the polarization

They are 2 independent solutions of the wave eqn   Linear combinations make

all the possibilities

Polarizing filters (a.k.a. Polaroids) can selectively absorb one polarization   Used in photography, sun

glasses, LCD, etc.

EB

E

B

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Energy We know light carries energy with it   Sun light brings (or brought) all the energy we use through vacuum

between Sun and Earth

It contains nothing but E and B fields   Energy density [erg/cm3] of E and B fields are   For plane waves

u =dUdv

=E 2 + B2

8π

E = yE0 sin(kx −ω t)B = zE0 sin(kx −ω t)

⎧⎨⎪

⎩⎪ u =

E02

4πsin2(kx −ω t)

This energy is moving with speed c

How much power arrives in a unit area?

Poynting Vector Consider the energy inside a volume V

  Time derivative

  Use Maxwell

  Divergence theo.

Define Poynting vector

  S represents the amount and direction of energy flow par unit area

U =

18π

(E 2 + B2)dvV∫

∂U∂t

=1

8π2 ∂E∂t

⋅E + 2 ∂B∂t

⋅B⎛⎝⎜

⎞⎠⎟

dvV∫

=1

4πc ∇ ×B( ) ⋅E − c ∇ ×E( ) ⋅B( )dv

V∫= −

c4π

∇ ⋅ E ×B( )dvV∫

= −c

4πE ×B

⎛⎝⎜

⎞⎠⎟⋅da

S∫

S =

c4π

E ×B

This much energy leaks out from the surface S

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Poynting Vector Poynting vector represents the flux of energy   Per unit area and unit time   It points in the direction of the EM energy flow

Check the dimension

  Unit in CGS: erg/sec·cm2

  In SI, has the unit of watts/m2

Magnitude of S is also known as the intensity

S =

c4π

E ×B

S⎡⎣ ⎤⎦ =c

4πE ×B

⎡

⎣⎢

⎤

⎦⎥ = c⎡⎣ ⎤⎦ E⎡⎣ ⎤⎦ E⎡⎣ ⎤⎦ = c⎡⎣ ⎤⎦ E⎡⎣ ⎤⎦

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c⎡⎣ ⎤⎦ =lengthtime

, E⎡⎣ ⎤⎦2=

energyvolume

⇒ S⎡⎣ ⎤⎦ =length ⋅energytime ⋅volume

=powerarea

S = E ×B

Back to Plane Waves Consider a linearly polarized plane wave: Poynting vector is

  Compare with the energy density

  This makes sense: energy distribution is moving with velocity c

For most interesting cases, the oscillation is very fast   Visible light is ~1014 Hz   What matters is the average power density and average intensity

E = yE0 sin(kx −ω t)B = zE0 sin(kx −ω t)

⎧⎨⎪

⎩⎪

S =

c4π

E ×B = x c4π

E02 sin2(kx −ω t)

u =

E02

4πsin2(kx −ω t)

S = cu k = cu

S =

c4π

E 2 k =c

8πE0

2 k I =

c8π

E02

mean square E field

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Charging Capacitor Poynting vector appears whenever both E and B are there   Consider a charging capacitor

  “Extended” Ampère’s law gives

  Poynting vector is

  Integrating over the cylindrical “side” surface

E = −

4πQπa2 z = −

4Qa2 z

∂E∂t

= −4Ia2 z

B = −

2Irca2 φ

S =

c4π

−4Qa2 z

⎛⎝⎜

⎞⎠⎟

−2Irca2 φ

⎛⎝⎜

⎞⎠⎟= −

2QIrπa4 r Energy flows inward

Capacitor accumulates energy

S

side∫ ⋅da = −2QIaπa4 2πad = −QI

4da2 = −

QIC

= −VI = power into C

–Q d

I

+Q a

z

E

Momentum in EM Waves EM waves carry not only energy but also momentum   In relativity, E and p are related by   For EM radiation, m = 0

  Relation E = pv is a general feature of all waves Physics 15c

Energy flow is given by the Poynting vector

  Momentum / time is force; force / area is pressure

  EM waves exert a radiation pressure

E2 = p

2c2 + m2c4

E2 = p

2c2

p =

Ec

This is energy, not E field!

energy

time ⋅area= S =

c4π

E ×B momentumtime ⋅area

=Sc=

14π

E ×B

Sc=

14π

E ×B

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Radiation Pressure Pressure due to EM radiation was predicted by Maxwell   It was experimentally demonstrated 30 years later

Because p = E/c, the pressure is very small   A 100 megawatt laser would produce a force

It is important in extreme high temperatures   Very large stars are supported by the radiation pressure from

collapsing due to gravity

F =

Pc=

108 watts3 ×108m/s

= 0.3 newtons That’s an ounce in weight

Summary Plane wave solutions of Maxwell’s equations

  Propagates along k with speed c

  Two possible polarizations for the same k

Energy flow given by the Poynting vector

  For plane waves, average energy flow is

EM waves also carry momentum p = E/c   Radiation pressure is S/c

E = E0 sin(k ⋅r −ω t)B = B0 sin(k ⋅r −ω t)

⎧⎨⎪

⎩⎪ ω = kc, E0 = B0 , k = E × B

S =

c4π

E ×B

S =

c4π

E 2 k =c

8πE0

2 k