8.022 (E&M) – Lecture 21
Topics:
Energy and momentum carried by EM waves
Poynting vector
Transmission lines
Scattering of light and sunset demo…
Last time
Solution of Maxwell’s equations in vacuum
ˆ)
Solution of wave equation f (r ± ckt
combination of plane waves:
∇2E =
1 ∂2E
c 2 ∂t 2
can be expressed as linear
Properties of plane waves: E = E 0 sin(k ir − ωt ); B = B 0 sin(k ir − ωt )
They travel at the speed of light // to k (wave vector)
E, B and k are always perpendicular to each other
Amplitude of E and B are the same in cgs
Polarization of EM waves
Linear: when the direction of E0 is constant in time
Circular: when the vector E0 describes a circle over time
Elliptical: all the situations in between these 2 cases
Today we will complete the study of these properties…
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8.022 – Lecture 21
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EM Energy
EM radiation carries energy
Obvious if you think about the fact that is the light from the sun
that keeps us warm…
How does this energy propagate?
Consider a volume V of surface A containing E and B
Energy density: u=
V
E
B
energy
1
=
(E iE + B iB )
volume 8π
Total energy: U = ∫ udV =
V
.
1
(E iE + B iB )dV
8π ∫V
Surface A
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8.022 – Lecture 21
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The Poynting vector
How does total derivative change over time?
∂U
1 ∂
=
∂t
8π ∂t
∫ (E iE
V
+ B iB )dV =
1
∂E
∂B
iB )dV
( iE +
4π ∫V ∂t
∂t
Remembering that in vacuum: ∇ × E=⇒
1 ∂B
1 ∂E
and ∇ × B=
c ∂t
c ∂t
c
∂U
=
(∇ × B iE − ∇ × EiB )dV
∂t
4π ∫V
Remembering that ∇i(E × B ) = -E i(∇ × B ) + B i(∇ × E )
⇒
c
∂U
=−
∂t
4π
∫V ∇i(B × E ) dV ≡ − ∫V ∇iS
.
dV
where we defined the Poynting vector as S ≡
G. Sciolla – MIT
8.022 – Lecture 21
c
B ×E
4π
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Interpretation of Poynting vector
Given:
Stokes
∂U
∂U
= − ∫ ∇iS dV →
= − ∫ S ida = −ΦS ( A )
V
A
∂t
∂t
The rate of change of EM energy in the volume V is given
by the flux of the Poynting vector S through the surface A
Minus sign: dA points outward
U increases when S is opposite to dA
Interpretation of Poynting vector:
S ≡
c
B ×E
4π
points in the direction of the EM energy flow
Remember that E 0 × B 0 = E 0
2
kˆ
The flux of S through a surface gives the power through A
Power through A: ∫ S ida
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A
8.022 – Lecture 21
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Poynting vector: dimensional analysis
What are the units of the Poynting vector?
cgs
2
⎡ c
⎤
⎡ ⎤
⎤⎡B ⎦⎣
⎤⎡E ⎦⎤ = ⎣⎡c ⎦⎣
⎤⎡E
⎦⎤
⎣ S ⎦ = ⎢⎣ 4 π E × B ⎥⎦ = ⎣⎡c ⎦⎣
Lenght
⎡⎣c ⎤⎦ =
T im e 2
1
E n e rg y F ro m u =
( E iE + B iB ) ⇒ ⎡⎣E ⎤⎦ =
8π
V o lu m e
L e n g h t E n e rg y
E n e rg y
Pow er
⇒ ⎡⎣ S ⎤⎦ =
=
=
T im e A re a
A re a
T im e V o lu m e
Expected if the flux of S is the power through area A
In cgs: [S]=erg s-1 cm-2
NB: Magnitude of S is know as Intensity I
Intense source of radiations emit a lot of power per unit area
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8.022 – Lecture 21
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Applications: plane waves
Consider a linearly polarized plane wave:
⎧⎪ E = E 0 cos( kz − ω t ) xˆ
⎨
⎪⎩ B = B0 cos( kz − ω t ) yˆ
Poynting vector associated with it:
c
c 2 2
S ≡
E 0 sin (kz − wt )kˆ
E×B =
4π
4π
This can be compared to the energy density of the wave:
1
1
E 0 2 sin2 (kz − ωt )
u=
(E iE + B iB ) =
8π
4π
⇒ S = uc = uckˆ
This is similar to J = ρv
another way to show that S tells us about the flow of energy!
Usually the oscillation is very fast (e.g.: visible~1014 Hz)
the average energy density <S> and intensity <I>:
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S =
all that matters is
ckˆ 2
c
E0 ;
I =
E 02
π
8π
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8.022 – Lecture 21
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Application 2: Dipole radiation
Radiation emitted by a dipole oriented along the z axis in spherical
coordinates:
2
⎧
ω p
sin ( kr − ω t ) ˆ
θ
⎪⎪ E = c 2 sin θ
r
⎨
2
⎪ B = ω p sin θ sin ( kr − ω t ) φˆ
⎪⎩
c2
r
This is 8.07 stuff:
just trust me for the moment
z
+
NB: this solution is only valid for r >>λ=2π/k
This is the Radiation propagates radially, some angular dependence too
2
Poynting vector: S = c E × B = 1 ω 4 p 2 sin2 θ sin (kr − ωt ) rˆ
4π
4π c 3
r2
⇒ S =
-
ω4p 2
sin2 θ rˆ
8π r 2c 3
NB: Poynting vector (and I) falls as 1/r2: this should be intuitive. Why?
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8.022 – Lecture 21
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Dipole radiation: cont.
Draw a sphere of radius R around the dipole centered in origin:
NB: R >> d
Compute power radiated through the sphere:
∂U
∂t
=
∫R
S ida = ∫
R
ω4p 2
sin2 θ rˆida
8π r 2c 3
+
-
Since da = R sin θ d φ rˆ :
2
∂U
∂t
=
Since
∫
π
0
z
2π
π
ω4p 2
R 2 ∫ d φ ∫ sin3 θ d θ
0
0
8π R 2c 3
sin3 θ d θ =
4
∂U
⇒
3
∂t
=
ω4p 2
(Larmor formula)
3c 3
NB: power through sphere does not depend on R
Why? S falls as 1/r2, area increases as r2
Power through S (flux through S) is constant: Energy is conserved
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Application 3: capacitor
The Poynting vector applies to ANY situation in which both E and B appear,
not just when we have radiation
Example: charging capacitor
E=−
4π Q
4Q
zˆ = − 2 zˆ
A
a
From generalized Ampere law: B ( r ) =
2 Ir ˆ
φ
ca 2
Calculate Poynting vector:
c
c 4Q 2Ir
2IQr
(−rˆ)
E ×B =
zˆ × φˆ =
4π
4π a 2 ca 2
πa 4
NB: what is important here is the direction of S:
S =
S points into the center of the capacitor as it should: the plates are
charging up!
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8.022 – Lecture 21
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Momentum carried by EM wave
Since EM carry energy it’s not surprising that they carry momentum
as well
In relativity, E and p are related by E 2 = p 2 c 2 + m 2 c 4
For EM radiation, m=0:
E2 = p 2 c 2 ⇒ p =
Remember that
U
c
Power
Energy
S Energy/c
Momentum
S=
=
⇒ =
=
Area Time Area
c Time Area
Time Area
Dimensional analysis will also tell us that:
S Momentum Force
=
= Pressure
=
c Time Area
Area
Demo
Radiation exerts pressure
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8.022 – Lecture 21
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Summary on Poynting vector Energy flux: Energy / area / unit time
Energy density u: Energy / unit volume
Momentum flux: Momentum / area / unit time
Momentum density: Momentum/ unit volume
Flux
X/(Area sec)
Density
X/Volume
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Energy
Momentum
c
S
(same as pressure)
c
S ≡
4π
B ×E
S
S
c
c2
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Transmission line
Transmission line = a pair of (twisted) cables used to transmit a signal
Current flows in one direction on one cable and comes back on the other cable
If terminated correctly, Z is purely real: Z~Rtermination
Find R when capacitance per unit length=C’ and inductance per unit length=L’
In theory:
L’
C’
R
In practice infinite sum of infinitesimal elements C and L:
…
R
…
…
Calculate Z of the last piece and impose that it’s purely real.
−1
.
R
i ωL '− ω 2RL ' C '+ R impose
⎛1
⎞
=
= R
Z eq = i + ⎜ + i ωC ⎟ = i ωL +
1 + i ωRC
1 + i ωRC '
R
⎝
⎠
L
L'
i ωL − ω LCR + R = R + i ωCR 2 . Ignoring term with LC (small): ⇒ R= =
G. Sciolla – MIT
8.022 – Lecture 21
C
C'
2
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Transmission line (2)
What happens when transmission line is terminated correctly?
Z is purely real: Z~Rtermination
Z is a constant of the cable:
Z does not depend on how long the cable is!
If R ≠ L'/C :
Z will depend on how long the cable is and on the frequency of the signal
Distortions of the signal!
Example of transmission line: coaxial cable, a pair of conducting
tubes nested in one another
Homework: prove that for a cylindrical coaxial cable
Z=2 ln(b/a) /c and the velocity of propagation is c.
Typical Rtermination: 50 Ohm
G. Sciolla – MIT
8.022 – Lecture 21
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Transmission line: demo
Coaxial cable (127.4 m long)
Pulse generator: pulse duration 0.1 µsec, period 20 µsec
Scope
Ch1 Ch2
Pulse
generator
Trigger Out
Add terminals here
Simultaneously send pulse from pulse generator (splitter)
to Ch 1 of scope
to transmission line (back and forth and display on Ch 2)
Measure speed of propagation: Time difference: 656 ns
v=L/T~2/3c
What happens if:
Open: signal will bounce back but nasty reflections
Short: signal will be reversed on the same cable, nothing on the other cable
If I terminate it with 50Ω resistor: signal comes back on return cable with no reflections
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8.022 – Lecture 21
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Scattering of light
(Logically this topic belongs to last lecture, but we did not have time…)
When we send light into a medium, the light is scattered in many
directions
Example: light from Sun (unpolarized) passing through atmosphere
Propagation of light //z
We look up in x direction
What kind of light do we see?
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8.022 – Lecture 21
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Scattering of light (2)
Since light propagates //z: no polarization // z
We measure the light (with our eyes!) along the x direction: no
polarization // x
The light we see must be polarized along the y direction
This is actually not really true because the light scatters multiple times,
but it suggests the general tendency
What if the put a giant polaroid in front of the Sun?
Scattered light would be more intense in direction perpendicular to
polarization direction
Rotating the polaroid would allow us to change intensity of the light:
Max intensity when polarization direction is // y axis
Dark when polarization direction is // x axis
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8.022 – Lecture 21
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Scattering of light (3)
How is light scattered?
Light hits a molecule; the E shakes the molecule’s charges with frequency
w; the molecule re-radiates the light often changing the direction
changes in polarization
Are all frequencies scattered in the same way?
Electric fields of scattered radiation depend on acceleration of (dipole)
charges 2
∂d
E scattered ∝ 2 ∝ ω 2 if dipole moment of the shaken molecule goes as d~cosω t
∂t
Intensity of scattered radiation: I ∝ E 2scattered ∝ ω 4 ∝ λ −4
Since λ red~ 2 λ blue
Blue is scattered 16 times more than red
This explains why the sky is blue during the day and why it’s red at
sunset
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8.022 – Lecture 21
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Summary and outlook
Today:
Energy and momentum carried by EM waves
Poynting vector and some of their applications
Transmission lines
Scattering of light
What happens at sunset?
Next Thursday:
Magnetic fields through matter? Or review problems?
G. Sciolla – MIT
8.022 – Lecture 21
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Sunset experiment
Solution of distilled water and salt.
Unpolarized light is shining through it to the wall
wall
Add Na2S2O335H2O (Na thiosulfate)
Lights starts scattering: fog; light on the wall becomes red first and then dark as
all the light is scattered toward the audience (as in sunset)
What happened?
Chemical reaction creates bigger and bigger molecules that scatter more and more light.
Blue light is scattered first. Red makes it for a while but eventually scatters too.
NB: light is polarized!
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8.022 – Lecture 21
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Sugar solution experiment (T8)
Light goes through a polarizer and then through an optically active
sugar solution
wall
Polarizer
Polarizer
The first polarizer creates a linearly polarized wave, overlap of right-handed
and left-handed circularly polarized waves which propagate at different
speeds in the solution. This causes linear polarization direction to change
slowly. Since the effect depends on λ, different colors are rotated differently.
The second polarizer check polarization direction at exit
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