Download Lecture 21. Energy and Momentum of

Lecture 21. Energy Transfer in Electromagnetism
Energy and Intensity of Electromagnetic Waves
Outline:
Transfer of EM Energy in Space: Poynting Formalism.
Energy Transfer by EM Waves.
Radiation Pressure
Final exam, Tuesday, December 20, 4:00 to 7:00
pm on CAC. If you have 3 exams on December 20
or have 3 exams in a row that includes Physics 227,
ASAP but not later than 5:00 pm on December 5
send your ENTIRE exam schedule to Prof. Cizewski
[email protected].
Iclicker Question
Consider the magnetic field 𝐵 = 𝑥 sin 𝑘𝑥 + 𝜔𝑡 . Can this field
represent the plane electromagnetic wave in free space?
A. sure, why not
B. no, because it’s a
sin
and not
cos
function of the phase.
C. no, because 𝐵 is directed along the direction of phase
variation.
D. no, because of a wrong dependence of the phase on 𝑡.
E. no, because of B - D.
2
Iclicker Question
Consider the magnetic field 𝐵 = 𝑥 sin 𝑘𝑥 + 𝜔𝑡 . Can this field
represent the plane electromagnetic wave in free space?
A. sure, why not
B. no, because it’s a
sin
and not
cos
function of the phase.
C. no, because 𝐵 is directed along the direction of phase
variation.
D. no, because of a wrong dependence of the phase on 𝑡.
E. no, because of B - D.
3
Energy Density in E.-M. Waves
EM waves:
𝐸 = 𝑐𝐵
This doesn’t mean that B is “weaker” than E (“apples” vs. “oranges”). The meaningful
comparison - the energy densities in the 𝐸 and 𝐵 fields that form the wave:
Energy density in any 𝐸 field:
Energy density in any 𝐵 field:
𝑢𝐸 =
𝑢𝐵 =
𝜀0 2
𝐸 𝑟, 𝑡
2
1 2
𝐵 𝑟, 𝑡
2𝜇0
(lecture 23)
𝜀0 2 𝜀0 2 2
1
1 2
2
𝑢𝐸 = 𝐸 = 𝑐 𝐵 = 𝑐 =
=
𝐵 = 𝑢𝐵
2
2
𝜀0 𝜇0
2𝜇0
Energy densities of the 𝐸 and 𝐵 fields in e.-m. waves are equal.
4
Energy Density in E.-M. Waves (cont’d)
Energy density in e.-m. waves:
𝑢𝐸𝐵
2
1
𝐵
1 2
1
2
2
=
𝜀 𝐸 +
= 𝜀0 𝐸 = 𝐵 =
𝐸𝐵
2 0
𝜇0
𝜇0
𝑐𝜇0
The time-averaged energy density in
a monochromatic e.-m. wave:
𝑐𝑜𝑠 2 𝑘𝑥 − 𝜔𝑡
𝑡
=
1 1
+ cos 2 𝑘𝑥 − 𝜔𝑡
2 2
𝐸0 2 𝑐𝑜𝑠 2 𝑘𝑥 − 𝜔𝑡
𝑢𝐸𝐵
𝑡
𝑡
=
𝑡
1
2
1
= 𝐸0 2
2
𝜀0 2
1
1
2
= 𝐸0 =
𝐵 =
𝐸 𝐵
2
2𝜇0 0
2𝑐𝜇0 0 0
𝐸0 , 𝐵0 - the amplitudes
5
Intensity of Electromagnetic Waves
Intensity: the average energy transported
by the e.-m. wave per unit area per 1s
(the flux of energy per 1m per 1s, scalar)
𝑢𝐸𝐵
𝑡
1
=
𝐸𝐵
2𝜇0 𝑐
𝐼 = 𝑐 𝑢𝐸𝐵
𝑡
𝒖𝑬 , 𝒖𝑩
1m2
𝑐
𝑐 ∙ 1𝑠
1
1
=
𝐸0 𝐵0 = 𝑐𝜀0 𝐸0 2
2𝜇0
2
Units:
W/m2
Example:
The intensity of sunlight hitting the Earth is ~ 1,400 W/m2. Find the
amplitudes of the electric and magnetic fields in the e.-m. wave.
1
1
𝐼=
𝐸0 𝐵0 =
𝐸0 2
2𝜇0
2𝜇0 𝑐
𝐸0 =
2𝐼𝜇0 𝑐 =
2800 × 3 ∙ 108 × 4𝜋 ∙ 10−7
𝐸0
103 𝑉/𝑚
−6 𝑇
𝐵0 =
=
≈
3
∙
10
𝑐
3 ∙ 108 𝑚/𝑠
𝑉
𝑉
≈ 1,000
𝑚
𝑚
(~ 1% of the Earth’s magnetic field)
6
Example
In a fairly brightly-lit room, the intensity of light is about 100 W/m2.
What is the amplitude of the electric field of the light waves in the
room? If the room’ volume is 50𝑚3 and the light intensity is fairly
uniform throughout the room, about how much energy is stored in
the room in the form of light waves?
𝐼 = 𝑐 𝑢𝐸𝐵
𝑢𝐸𝐵
Total energy:
𝑡
𝑡
𝑢𝐸𝐵
𝑡
𝐼
=
𝑐
100 𝑊/𝑚2
−7 𝐽/𝑚3
=
≈
3.3
∙
10
3 ∙ 108 𝑚/𝑠
𝑢𝐸𝐵
𝑡 ∙ 𝑣𝑜𝑙𝑢𝑚𝑒 =
𝐽
−7
3.3 ∙ 10
𝑚3
∙ 50𝑚3 ≈ 1.7 ∙ 10−5 𝐽
7
Iclicker Question
Imagine that when you switch on your lamp, you increase the
intensity of light shinning on your textbook by a factor of 16. By
what factor does the average electric field strength in this light
increase?
A. 256
1
𝐼 = 𝑐𝜀0 𝐸0 2
2
B. 16
C. 4
D. 2
E. 1
8
Iclicker Question
Imagine that when you switch on your lamp, you increase the
intensity of light shinning on your textbook by a factor of 16. By
what factor does the average electric field strength in this light
increase?
A. 256
1
𝐼 = 𝑐𝜀0 𝐸0 2
2
B. 16
C. 4
D. 2
E. 1
9
Poynting Vector
How does the electromagnetic energy get from one place to another?
1m2
Intensity : the average energy transported
by the e.-m. wave per unit area per 1s
(scalar)
𝐼 = 𝑐 𝑢𝐸𝐵
𝑡
1
=
𝐸0 𝐵0
2𝜇0
𝒖𝑬 , 𝒖𝑩
𝑐 ∙ 1𝑠
In order to retain the direction of the
energy flow (along 𝑘), let’s try 𝐸 × 𝐵:
Poynting vector:
(energy transferred through 1 m2
per 1s in the direction of 𝑘)
𝑆
𝑡
1
=
𝐸0 𝐵0 𝑐𝑜𝑠 2 𝑘𝑥 − 𝜔𝑡
𝜇0
𝑡
1
𝑆=
𝐸×𝐵
𝜇0
1
=
𝐸 𝐵
2𝜇0 0 0
For a traveling
monochromatic e.-m. wave:
𝑐
𝑆
- the time-averaged flux of energy
𝑡
1
=
𝐸0 𝐵0 𝑘 = 𝐼𝑘
2𝜇0
10
Iclicker Question
The drawing shows a
sinusoidal electromagnetic
standing
wave.
The
average (averaged over
either x or t) Poynting
vector in this wave
A. points along the x-axis.
B. points along the y-axis.
C. points along the z-axis.
D. is zero.
E. none of the above
1
𝑆=
𝐸×𝐵
𝜇0
11
Iclicker Question
The drawing shows a
sinusoidal electromagnetic
standing
wave.
The
average (averaged over
either x or t) Poynting
vector in this wave
A. points along the x-axis.
B. points along the y-axis.
C. points along the z-axis.
D. is zero.
E. none of the above
1
𝑆=
𝐸×𝐵
𝜇0
12
Poynting Formalism (cont’d)
The concept of Poynting vector applies to all 𝐸 and 𝐵 fields, not just to the e.-m. waves.
Poynting formalism works in all situations:
1
𝑆=
𝐸×𝐵
𝜇0
(a) static
In statics 𝐸 , being a potential field, is normal to the
equipotential surfaces, thus 𝑆 is tangential to the
equipotential surfaces.
(b) dynamic
note that in this case 𝐸 is non-conservative, and the
equipotential surfaces do not make sense.
Few consequences:
(a) Whenever we have both E and B (and 𝐸 ∦ 𝐵), the e.-m. energy
flows in space! (even if we deal with static fields).
(b) The flow of energy is associated with the momentum of the e.m.
waves (remember, the energy has “mass”).
(c) The momentum transfer (absorption or reflection of e.-m. waves)
implies radiation pressure.
13
Example: Transmission Line
𝑩
We can apply the Poynting vector formalism to obtain
the well-known result: 𝑃 = 𝑉𝐼. However, the fields are
non-uniform, and the integration is difficult. Let’s
simplify the geometry.
𝑰
𝑬
R
Consider a transmission cable that consists of two flat metal ribbons of width
W, a small distance a<<W apart. The two conductors are held at a potential
difference V, and carry current I that travels uniformly down one strip and back
along the other.
V
𝑊
𝑰
𝑰
Transmitted
power to
the load R :
𝑎
𝑬
𝑩
𝑃=
𝑐𝑟𝑜𝑠𝑠
𝑠𝑒𝑐𝑡𝑖𝑜𝑛
𝑅
𝑉
𝐸=
𝑎
𝐼
𝐵 = 𝜇 0 𝐾 = 𝜇0
𝑊
both
uniform,
orthogonal
to one
another
1
1𝑉
𝐼
𝐸 × 𝐵 ∙ 𝑑𝐴 =
𝜇0
𝑎𝑊 = 𝑉𝐼
𝜇0
𝜇0 𝑎 𝑊
14
Circuits with Batteries and Resistors
𝑰
𝑬
+
𝑰
-
I𝑬
𝑩
𝑰
𝑩
Since 𝑆 is directed out of the
battery, the battery “radiates” the
e.-m. energy into the circuit. 𝑆
points not along the wires (we
assume that the wires have no
resistance). In the vicinity of the
resistor, 𝑆 points into the resistor,
and again not along the wires.
15
Example: Charging a Capacitor
B
I
h
a
𝑬
𝑺
If the process is slow (a<<), we can neglect the magnetic
field energy inside (𝑢𝑒 ≫ 𝑢𝑚 ). The capacitor is receiving
energy at a rate:
𝑑𝑈𝑒
𝑑
1
𝑑𝐸
2
2
2
=
𝜋𝑎 ∙ ℎ 𝜖0 𝐸 = 𝜋𝑎 ∙ ℎ 𝜖0 𝐸
𝑑𝑡
𝑑𝑡
2
𝑑𝑡
The energy must flow into this volume from somewhere. However, the energy
flow through the plates = 0 (𝐸 = 0 outside, the wires have no resistance).
Between the plates:
𝑙𝑜𝑜𝑝
𝐵 ∙ 𝑑𝑙 = 𝜇0 𝜖0
𝑑𝐸
∙ 𝑑𝐴
𝑑𝑡
𝑠𝑢𝑟𝑓
𝐵 𝑟 = 𝑎 ∙ 2𝜋𝑎 = 𝜇0 𝜖0
𝑑𝐸 2
1
𝑑𝐸
𝜋𝑎 𝐵 𝑎 = 𝜇0 𝜖0 𝑎
𝑑𝑡
2
𝑑𝑡
The energy flow from outside through the cylindrical surface of radius a and height h:
1
1
1
𝜕𝐸
𝑑𝐸 𝑑𝑈𝑒
2
𝑆 ∙ 2𝜋𝑎ℎ = 𝐸𝐵 ∙ 2𝜋𝑎ℎ = 𝐸 ∙ 𝜇0 𝜖0 𝑎
∙ 2𝜋𝑎ℎ = 𝜋𝑎 ∙ ℎ 𝜖0 𝐸
=
𝜇0
𝜇0
2
𝜕𝑡
𝑑𝑡
𝑑𝑡
This energy isn’t coming down the wires, it comes from the 𝐸 and 𝐵 fields
16
surrounding the capacitor!
Radiation Pressure
1m
Pressure: force per unit area.
For particles bombarding a unit area:
v
v1s
n–
particle
density
The net force:
𝐹𝑛𝑒𝑡
Pressure:
𝑑𝑝
𝐹𝑖 =
𝑑𝑡
𝑑𝑝
𝑑𝑝
2
= 𝑛 𝑣𝑜𝑙𝑢𝑚𝑒/1𝑠
= 𝑛 1𝑚 ∙ 𝑣
𝑑𝑡
𝑑𝑡
𝑑𝑝
𝑃 = 𝑛𝑣
𝑑𝑡
In the e.-m. wave, the particles are photons (massless particles moving with 𝑣 = 𝑐)
Photon energy:
𝑃𝐸𝐵
𝑃𝐸𝐵
𝑆
=
𝑐
𝑇𝑝ℎ = 𝑐𝑝𝑝ℎ
𝑑𝑝𝑝ℎ 𝑑 𝑛𝑇𝑝ℎ
1 𝑑 𝑛𝑐𝑇𝑝ℎ
𝑆
= 𝑛𝑐
=
=
=
𝑑𝑡
𝑑𝑡
𝑐
𝑑𝑡
𝑐
- assuming that the energy of all photons bombarding
the wall is absorbed. For reflection, the pressure is
twice as large.
Radiation Pressure
Find the pressure of sunlight at the earth’s surface, using intensity 𝑆 = 1,000𝑊/𝑚2
and assuming that the radiation is absorbed.
𝑃𝐸𝐵
𝑆 1,000𝑊/𝑚2
= =
≈ 3 × 10−6 𝑃𝑎
8
𝑐 3 × 10 𝑚/𝑠
Radiation pressure has had a major effect on the development of the cosmos,
from the birth of the universe to ongoing formation of stars and shaping of
clouds of dust and gasses on a wide range of scales.
Next time: Lecture 20. Inductance. Magnetic Field Energy.
§§ 30.1 - 3
19
Appendix I: Alternating Currents (AC circuits)
Consider two cases (see the figures): (a) the ac transmission line is loaded by a resistor;
(a) the ac transmission line is loaded by a capacitor. Use Poynting formalism to describe
the energy flow in these two cases.
(a)
𝑰
𝑺
Energy
always flows
towards the
resistor.
𝑰
𝑩
𝑬
(b)
𝑰
𝑰
𝑺
𝑩
𝑬
Energy flows
back and forth,
the energy flux
on average is
0.
Appendix II: Energy Conservation in Electromagnetism
1
𝑆=
𝐸×𝐵
𝜇0
𝐸 and 𝐵
fields
One can show (by using vector analysis)
that indeed S is related to losses of
electromagnetic energy in a volume:
𝑆
𝑠𝑢𝑟𝑓
𝜕
𝑆 ∙ 𝑑𝐴 = −
𝜕𝑡
net flux of
energy out of
the volume
𝑣𝑜𝑙𝑢𝑚𝑒
𝜀0 𝐸 2 𝐵2
+
𝑑𝜏 +
2
2𝜇0
net loss of the e.-m.
energy in the volume
𝐸 ∙ 𝑗 𝑑𝜏
𝑣𝑜𝑙𝑢𝑚𝑒
Joule heat in
the volume
21
Appendix III: The Total Energy Flow
General statement: whenever there is a flow of energy (it might be field energy or any
other kind of energy), the energy flowing through a unit area per unit time is equal to
the momentum density  c2.
Non-zero-mass particles
1m
v
Massless particles (photons)
particle
density
n
1m
c
v1s
Total energy of a particle: 𝑇 =
c1s
𝑚𝑐 2
1−
Momentum of a particle: 𝑝 =
𝑚𝑣
Total energy of a photon: 𝑇𝑝ℎ = 𝑐𝑝𝑝ℎ
𝑣2
𝑐2
Momentum of a photon:
𝑣2
1− 2
𝑐
Energy flux:
𝑛𝑇
𝑛 1𝑚2 ∙ 𝑣 ∙ 1𝑠
𝑆=
=
1𝑚2 ∙ 1𝑠
1𝑚2 ∙ 1𝑠
photon
density
n
𝑝𝑝ℎ
Energy flux:
𝑚𝑐 2
1−
𝑣2
𝑐2
=
𝑛𝑝𝑐 2
𝑛 1𝑚2 ∙ 𝑐 ∙ 1𝑠
𝑆=
𝑐𝑝𝑝ℎ = 𝑛𝑝𝑝ℎ 𝑐 2
2
1𝑚 ∙ 1𝑠
The Total Energy Flow
General statement: whenever there is a flow of energy (it might be field energy or any
other kind of energy), the energy flowing through a unit area per unit time is equal to
the momentum density  c2.
Massless particles (photons)
c
1m photon
density
n
c1s
Momentum density
𝑝𝐸𝐵 ≡ 𝑛𝑝𝑝ℎ :
Radiation pressure 𝑃𝐸𝐵
(the momentum transferred
to 1 m2 per 1s)
Energy of a photon:
𝑇𝑝ℎ = 𝑐𝑝𝑝ℎ
𝑝𝑝ℎ
Momentum of a photon:
Energy flux:
𝑝𝐸𝐵 =
𝑛 1𝑚2 ∙ 𝑐 ∙ 1𝑠
𝑆=
𝑐𝑝𝑝ℎ = 𝑛𝑝𝑝ℎ 𝑐 2
2
1𝑚 ∙ 1𝑠
1
𝑆 = 𝜖0 𝜇0 𝑆
𝑐2
𝑃𝐸𝐵 = 𝑝𝐸𝐵 1𝑚2 ∙ 𝑐 ∙ 1𝑠 /1𝑠 =
𝑆
𝑐
𝑃𝐸𝐵 =
𝑆
𝑐
(this is for absorption, the pressure is twice as
much for reflection)