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Electromagnetic Waves and
Polarization
Physics 2415 Lecture 30
Michael Fowler, UVa
Today’s Topics
• Measuring the speed of light
• Wave energy and power: the Poynting vector
• Light momentum and radiation pressure
• Radio transmission
• Polarization
First Measurement of the Speed of Light
Roemer, in 1676, noticed that eclipses of Jupiter’s satellite Io
gradually fell behind a regular schedule for about half the year,
then just as gradually caught back up.
He figured it was because the Earth was going away from Jupiter
for a little over half a year, then approaching again: it’s a Doppler
effect! He got a reasonable estimate of the speed of light.
First Measurement on Earth
Light goes out between teeth of rotating wheel, reflects off distant mirror,
by the time it gets back, a tooth may be blocking its path, depending on
wheel speed: at certain speeds, the observer sees nothing. Knowing the
wheel rotation rate, the speed of light can be figured out. (1849)
6 miles
(720 teeth!)
Best Early Measurements of the Speed
of Light (1879)…
• were by a physics instructor, Albert Michelson, at
the Naval College in Annapolis—his rich father-
in-law (who had a castle in Scarsdale) bankrolled
the experiment. (It was within about 50 km/sec.)
• The velocity of light c is precisely:
c = 2.99792458 x 108 m/sec.
• This is exact, because it’s the definition of the
meter. (The second is defined as a precise
number of oscillations of a particular atom.)
Wave Energy
• Recall we found the energy/meter in a
vibrating string, then multiplying that by the
wave speed gave power delivered by the wave.
• The same analysis works for electromagnetic
waves: for a harmonic wave
and (from the previous lecture) .
• The energy density u (E, B mean rms values!) is
 
0 sin
E E kx t

 
E vB cB
 
2 2 2
1 1
0 0 0 0
2 2 /
u E B E c EB
   
   
(using to see magnetic energy = electric).
0 0
/ , 1/
B E c c  
 
Wave Power
• The energy density in the harmonic wave is
equal to joules/m3. ( .)
• The power intensity—the energy delivered
across one square meter perpendicular to the
direction of the wave—equals the energy in a
volume with a base of 1 sq m, a length of c =
3x108 meters.
• Intensity
• These terms all represent power/m2 in watts.
0
u c EB


2 2
0 0 0
/
S cu c E c EB EB
  
   
0 0
1/
c  

Poynting Vector
• The energy density in the harmonic wave is
equal to joules/m3.
• Intensity—power delivery/m2 —is
• Obviously, this power is delivered in a
particular direction (that of the wave) and can
be represented as a vector:
This is called the Poynting vector.
0
u c EB


0
/
S EB 

  
0
1/
S E B

 
Poynting Vector for Static Fields
• The Poynting vector gives energy
flow even for static fields.
• If an electric field E is driving a
steady current I along a wire, there
will be a magnetic field B = 0I/2r,
and so a Poynting flow .
inwards as shown here.
• Flow across a cylindrical surface of
radius r, length 1 meter, will be
AEB/0, A = area = 2r, so total
energy flow rate = EI. This is just
the power dissipated as heat!
• .
I1
  0
/
B
S E 
 
E
Energy flows out of the battery,
through space, down into the wire.
Light has Momentum
• Maxwell proved from his equations that a flash
of light with total energy E carries momentum
E/c.
• The proof is quite difficult, but one way to see
the result is to use Einstein’s equation E = mc2.
The energy E in a flash of a beam of light
means it has a (very tiny!) mass E/c2, moving of
course at c, so momentum p = mv = mc = E/c.
• All you need to know is the result.
Radiation Pressure
• Since light carries momentum, anything
absorbing or reflecting light feels a pressure, a
force equal to the rate of change of
momentum, from Newton’s laws.
• How can the perpendicular E and B fields
push something forwards?
• The electric field causes charged particles
(electrons) to oscillate perpendicular to the
wave direction, then the force from the
magnetic field pushes the charge forwards.
qv B

Radiation Pressure
• The equation p = E/c means that it takes a lot
of energy to get much momentum.
• The only known successful antisatellite laser,
called MIRACL, delivered one megawatt
continuous power to an area of 200cm2.
• If all that energy hits a satellite, very
approximately what force does it exert?
A. 100N B. 10N C. 1N D. 0.1N E. 0.01N
Radiation Pressure
• If one megawatt hits a satellite, very
approximately what force does it exert?
E. 0.01N approximately-- p = E/c means 1 Mw,
or 106J/sec, is momentum 106/3x108  3x10-3
delivered per second, and force is rate of
change of momentum, so this is 3x10-3N. If
the radiation is reflected, the force is double
this.
The point is to fry the satellite, not to push it!
Radio Transmission
• The basic radio transmitter is an
oscillating dipole: at some
instant, a dipole is created, its
field propagates outwards, but it
rapidly dies to be replaced by a
dipole in the opposite
direction—the outgoing electric
field must switch direction, it
does this by looping around as
seen here. The magnetic field
lines from current up and down
the dipole antenna are circular. Some animations
Radio Reception
• Although the radio wave looks complicated near
the transmitter, far away (meaning more than a
few wavelengths) it has the familiar form shown
above, the direction of propagation being directly
away from the source.
• For the wave shown above, generated by a
vertical transmitting antenna, reception would be
best with a vertical receiving antenna. The
oscillating vertical electric field would set up
oscillating currents in a vertical wire.
Radio Reception
• A radio receiving antenna has simultaneously
many small oscillating currents, from all the
transmitters within range.
• The antenna is linked to an LC circuit with
tuneable oscillation frequency—this circuit is
driven by the antenna current with the right
frequency, the signal is then amplified.
• Frequencies vary from kHz to GHz, and f = c
gives the corresponding wavelengths.
Polarization
• The electromagnetic wave shown above
is said to be vertically polarized—
meaning the direction of the electric
field vector is vertical.
• A wave like this with the electric field
vector at an angle  to the vertical can
be represented as a sum of a vertically
polarized wave of amplitude Ecos and
a horizontally polarized wave of
amplitude Esin.

E
Polarizing Light
• Ordinary light is a random mixture of
polarizations. Certain materials, like polaroid,
only allow light polarized, say, vertically, to
pass.
• These materials have long horizontal
electrically conducting molecules, so the
horizontal component of the electric field is
absorbed driving currents in these molecules.
How Much Intensity Gets Through
Polaroid?
• For a wave polarized at  to the
vertical, only the component
Ecos gets through. This means
the intensity is down by a factor
cos2.
• For incoming light with random
polarizations, the reduction in
intensity will be .

E
2 1
2
cos  
Polarization by Reflection
• We’ll discuss this in more detail later, but just
mention here that reflected light is partially
polarized, that from a flat horizontal surface
partially horizontally polarized—polaroid
sunglasses cut this out.
• Light reflected from a surface between two
materials with different refractive indices n1,
n2 is fully polarized if reflected at Brewster’s
angle, given by .
2 1
tan /
p n n
 

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2415_web_Lec_30_EM_Waves.5234524524045040pptx

  • 1. Electromagnetic Waves and Polarization Physics 2415 Lecture 30 Michael Fowler, UVa
  • 2. Today’s Topics • Measuring the speed of light • Wave energy and power: the Poynting vector • Light momentum and radiation pressure • Radio transmission • Polarization
  • 3. First Measurement of the Speed of Light Roemer, in 1676, noticed that eclipses of Jupiter’s satellite Io gradually fell behind a regular schedule for about half the year, then just as gradually caught back up. He figured it was because the Earth was going away from Jupiter for a little over half a year, then approaching again: it’s a Doppler effect! He got a reasonable estimate of the speed of light.
  • 4. First Measurement on Earth Light goes out between teeth of rotating wheel, reflects off distant mirror, by the time it gets back, a tooth may be blocking its path, depending on wheel speed: at certain speeds, the observer sees nothing. Knowing the wheel rotation rate, the speed of light can be figured out. (1849) 6 miles (720 teeth!)
  • 5. Best Early Measurements of the Speed of Light (1879)… • were by a physics instructor, Albert Michelson, at the Naval College in Annapolis—his rich father- in-law (who had a castle in Scarsdale) bankrolled the experiment. (It was within about 50 km/sec.) • The velocity of light c is precisely: c = 2.99792458 x 108 m/sec. • This is exact, because it’s the definition of the meter. (The second is defined as a precise number of oscillations of a particular atom.)
  • 6. Wave Energy • Recall we found the energy/meter in a vibrating string, then multiplying that by the wave speed gave power delivered by the wave. • The same analysis works for electromagnetic waves: for a harmonic wave and (from the previous lecture) . • The energy density u (E, B mean rms values!) is   0 sin E E kx t    E vB cB   2 2 2 1 1 0 0 0 0 2 2 / u E B E c EB         (using to see magnetic energy = electric). 0 0 / , 1/ B E c c    
  • 7. Wave Power • The energy density in the harmonic wave is equal to joules/m3. ( .) • The power intensity—the energy delivered across one square meter perpendicular to the direction of the wave—equals the energy in a volume with a base of 1 sq m, a length of c = 3x108 meters. • Intensity • These terms all represent power/m2 in watts. 0 u c EB   2 2 0 0 0 / S cu c E c EB EB        0 0 1/ c   
  • 8. Poynting Vector • The energy density in the harmonic wave is equal to joules/m3. • Intensity—power delivery/m2 —is • Obviously, this power is delivered in a particular direction (that of the wave) and can be represented as a vector: This is called the Poynting vector. 0 u c EB   0 / S EB      0 1/ S E B   
  • 9. Poynting Vector for Static Fields • The Poynting vector gives energy flow even for static fields. • If an electric field E is driving a steady current I along a wire, there will be a magnetic field B = 0I/2r, and so a Poynting flow . inwards as shown here. • Flow across a cylindrical surface of radius r, length 1 meter, will be AEB/0, A = area = 2r, so total energy flow rate = EI. This is just the power dissipated as heat! • . I1   0 / B S E    E Energy flows out of the battery, through space, down into the wire.
  • 10. Light has Momentum • Maxwell proved from his equations that a flash of light with total energy E carries momentum E/c. • The proof is quite difficult, but one way to see the result is to use Einstein’s equation E = mc2. The energy E in a flash of a beam of light means it has a (very tiny!) mass E/c2, moving of course at c, so momentum p = mv = mc = E/c. • All you need to know is the result.
  • 11. Radiation Pressure • Since light carries momentum, anything absorbing or reflecting light feels a pressure, a force equal to the rate of change of momentum, from Newton’s laws. • How can the perpendicular E and B fields push something forwards? • The electric field causes charged particles (electrons) to oscillate perpendicular to the wave direction, then the force from the magnetic field pushes the charge forwards. qv B 
  • 12. Radiation Pressure • The equation p = E/c means that it takes a lot of energy to get much momentum. • The only known successful antisatellite laser, called MIRACL, delivered one megawatt continuous power to an area of 200cm2. • If all that energy hits a satellite, very approximately what force does it exert? A. 100N B. 10N C. 1N D. 0.1N E. 0.01N
  • 13. Radiation Pressure • If one megawatt hits a satellite, very approximately what force does it exert? E. 0.01N approximately-- p = E/c means 1 Mw, or 106J/sec, is momentum 106/3x108  3x10-3 delivered per second, and force is rate of change of momentum, so this is 3x10-3N. If the radiation is reflected, the force is double this. The point is to fry the satellite, not to push it!
  • 14. Radio Transmission • The basic radio transmitter is an oscillating dipole: at some instant, a dipole is created, its field propagates outwards, but it rapidly dies to be replaced by a dipole in the opposite direction—the outgoing electric field must switch direction, it does this by looping around as seen here. The magnetic field lines from current up and down the dipole antenna are circular. Some animations
  • 15. Radio Reception • Although the radio wave looks complicated near the transmitter, far away (meaning more than a few wavelengths) it has the familiar form shown above, the direction of propagation being directly away from the source. • For the wave shown above, generated by a vertical transmitting antenna, reception would be best with a vertical receiving antenna. The oscillating vertical electric field would set up oscillating currents in a vertical wire.
  • 16. Radio Reception • A radio receiving antenna has simultaneously many small oscillating currents, from all the transmitters within range. • The antenna is linked to an LC circuit with tuneable oscillation frequency—this circuit is driven by the antenna current with the right frequency, the signal is then amplified. • Frequencies vary from kHz to GHz, and f = c gives the corresponding wavelengths.
  • 17. Polarization • The electromagnetic wave shown above is said to be vertically polarized— meaning the direction of the electric field vector is vertical. • A wave like this with the electric field vector at an angle  to the vertical can be represented as a sum of a vertically polarized wave of amplitude Ecos and a horizontally polarized wave of amplitude Esin.  E
  • 18. Polarizing Light • Ordinary light is a random mixture of polarizations. Certain materials, like polaroid, only allow light polarized, say, vertically, to pass. • These materials have long horizontal electrically conducting molecules, so the horizontal component of the electric field is absorbed driving currents in these molecules.
  • 19. How Much Intensity Gets Through Polaroid? • For a wave polarized at  to the vertical, only the component Ecos gets through. This means the intensity is down by a factor cos2. • For incoming light with random polarizations, the reduction in intensity will be .  E 2 1 2 cos  
  • 20. Polarization by Reflection • We’ll discuss this in more detail later, but just mention here that reflected light is partially polarized, that from a flat horizontal surface partially horizontally polarized—polaroid sunglasses cut this out. • Light reflected from a surface between two materials with different refractive indices n1, n2 is fully polarized if reflected at Brewster’s angle, given by . 2 1 tan / p n n  