Download L11 radiation

Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 2 for Electrical Engineering
Lecturers: Daniel Rohrlich , Ron Folman
Teaching Assistants: Ben Yellin, Yoav Etzioni
Grader: Gad Afek
Week 11. Electromagnetic radiation –   E • Divergence theorem •
Maxwell’s equations in differential form • Current conservation •
Electromagnetic waves • Poynting vector
Source: Halliday, Resnick and Krane, 5th Edition, Chap. 38.
This set of four fundamental equations for E and B,
 E  dA   0
q
(Gauss’s law)
 B  dA  0

d
E  dr    B
dt

d
B  dr   0 I   0 0  E ,
dt
are Maxwell’s equations in integral form.
(Faraday’s law)
(Ampère’s law
as modified by
Maxwell)
In 1929 – 65 years after Maxwell’s prediction – M. R. Van
Cauwenberghe was the first to measure “displacement current”,
on a round parallel-plate capacitor of capacitance C = 100 pF
and effective radius R = 40.0 cm. The potential across the
capacitor, which alternated with frequency f = 50.0 Hz, reached
V0 = 174 kV. We calculated that the maximum field B
measureable at the edge of the capacitor was 2.73×10–8 T.
You ask: What was so difficult about this experiment, that it
took 65 years to do? After all, the two terms I and ε0 dΦE/dt in
the Ampère-Maxwell law yield the same B, and it was not
difficult to verify Ampère’s law!
In 1929 – 65 years after Maxwell’s prediction – M. R. Van
Cauwenberghe was the first to measure “displacement current”,
on a round parallel-plate capacitor of capacitance C = 100 pF
and effective radius R = 40.0 cm. The potential across the
capacitor, which alternated with frequency f = 50.0 Hz, reached
V0 = 174 kV. We calculated that the maximum field B
measureable at the edge of the capacitor was 2.73×10–8 T.
Two physicists who did the measurement in 1985 (thinking
they were the first) answered this question in their paper:
From the integral form of Maxwell’s equations, we anticipate
electromagnetic radiation: Intuitively, we see that a changing
electric flux will generate a transient magnetic field around it,
which will generate a transient electric field, etc. etc. while the
wave spreads through space.
Electromagnetic radiation is a phenomenon that spans more
than 18 orders of magnitude (in wavelength and frequency)!
To calculate all the features of electromagnetic waves, we need
Maxwell’s equations in their differential form.
With the aid of Stokes’s theorem, we converted two of
Maxwell’s equations to differential form:
B
E  
t
E
  B   0 J   0 0
.
t
To convert the other two of Maxwell’s equations, we need the
“divergence theorem”.
 E
The divergence of a vector field E(x,y,z), written   E, is the
following:



E 
Ex  E y  Ez
x
y
z
.
Note that   E is a scalar field.
Note that the notation   E is reasonable because   E is
    
(formally) the scalar product of the vectors    , , 
 x y z 
and E = (Ex, Ey, Ez), just as   E is (formally) the vector
product of the same two vectors.
 E
Example 1: What is   V ?
 E
Example 1: What is   V ?
  V    V
Answer:   V 

  
x  x  y  y
   V 
  

 z  z 
2
2 
 2


V .



 x 2 y 2 z 2 


This sum of second derivatives is called the “Laplacian” and
written as “del-squared”:
2
2 
 2



      2 .


 x 2 y 2 z 2 


 E
Example 2: What is   r ?
 E
Example 2: What is   r ?
Answer:



  r  x    y   z   3 .
x
y
z
 E
Example 3: E(r) is the electric field of a uniformly charged
ball of radius R with total charge q. What is   E(r) ?
 E
Example 3: E(r) is the electric field of a uniformly charged
ball of radius R with total charge q. What is   E(r) ?
Answer: As we know, the electric field E(r) of any
electrostatic configuration is minus the gradient of the electric
potential V(r), i.e. E(r)  V (r). Hence   E  2V ,
 q
 4 r ,

0
V

where
 (3R 2  r 2 )q

,
 8 0 R 3
  E  
2
q
8 0 R
(3R  x  y  z ) 
2
3

r  R

 . For r ≤ R we have
r  R

2
2
2
3q
4 0 R
3
.
 E
Example 3: E(r) is the electric field of a uniformly charged
ball of radius R with total charge q. What is   E(r) ?
Answer: As we know, the electric field E(r) of any
electrostatic configuration is minus the gradient of the electric
potential V(r), i.e. E(r)  V (r). Hence   E  2V ,
 q

,
r

R
 4 r



0
V

where
 (3R 2  r 2 )q
 . For r ≥ R we have

,
r  R
 8 0 R 3

2  1    x 
1
x2
       3 , so   E  0.


x 2  r  x  r 3 
r3
r5
 E
Example 4: What is   (  A) ? (A is any vector field.)
 E
Example 4: What is   (  A) ? (A is any vector field.)
Answer: We write out   A:
 






  A   Az  Ay , Ax 
Az ,
Ay 
Ax  ,
z
z
x
x
y
 y

so
2
2
2
    A  
Az 
Ay 
Ax
xy
xz
yz
2
2
2

Az 
Ay 
Ax
yx
zx
zy
 0.
Divergence theorem
The divergence theorem is Stokes’s theorem in 3D, but without
curvature.
We extend our notation: If V denotes any volume, then ∂V
denotes the boundary of the volume (the surface bounding V ).
∂V
V
Divergence theorem
To prove the divergence theorem, we assume that we can
reduce any volume V to infinitesimal cubes:
Double pyramid
reduced to cubes
Heart
made of
cubes
Divergence theorem
The divergence theorem tells us that
 E  dA over any closed
V
surface ∂V is equal to the sum
 nE  dA over all the cubes
n
in the volume V that has ∂V as its boundary.
Divergence theorem
To see why the divergence theorem is true, all we have to do is

look at two neighboring cubes and at E  dA for each one:
Divergence theorem
To see why the divergence theorem is true, all we have to do is

look at two neighboring cubes and at E  dA for each one:
The contributions of their common side cancel!
Divergence theorem
When we put all the cubes together, the only sides that
contribute (that are not cancelled by facing sides) are the faces
along ∂V !
The contributions of their common side cancel!
Divergence theorem
When we put all the cubes together, the only sides that
contribute (that are not cancelled by facing sides) are the faces
along ∂V !
The divergence theorem thus tells us
 E  dA  n nE  dA,
V
i.e. that the integral over ∂V equals the sum of the contribution
of each cube.
We will now compute the contribution of an infinitesimal cube.
In the limit of infinitely many cubes, the sum over cubes will
become an integral over the volume V.
Divergence theorem

We compute the flux E  dA for an infinitesimal cube. The
cube is centered at a point r = (x,y,z) with the area elements dA
pointing along these axes. There are six contributions to the
total flux:
Ey(x,y+Δy/2,z)ΔxΔz
y
z
Ez(x,y,z+Δz/2)ΔxΔy
x
Ex(x+Δx/2,y,z)ΔyΔz
–Ex(x–Δx/2,y,z)ΔyΔz
–Ey(x,y–Δy/2,z)ΔxΔz
–Ez(x,y,z–Δz/2)ΔxΔy
Divergence theorem

We compute the flux E  dA for an infinitesimal cube. The
cube is centered at a point r = (x,y,z) with the area elements dA
pointing along these axes. There are six contributions to the
total flux, and they add up to

E x ( x  x/ 2, y, z )  E x ( x  x/ 2, y, z )
E  dA 
xyz
x

E y ( x, y  y/ 2, z )  E y ( x, y  y/ 2, z )
y
xyz
E z ( x, y, z  z/ 2)  E z ( x, y, z  z/ 2)

xyz
z
Divergence theorem

We compute the flux E  dA for an infinitesimal cube. The
cube is centered at a point r = (x,y,z) with the area elements dA
pointing along these axes. There are six contributions to the
total flux, and they add up to

 E x E y E z 
E  dA  


 xyz    E xyz .
y
z 
 x

The sum of E  dA over all infinitesimal cubes then goes over
to a volume integral, and we obtain…
Divergence theorem
…the divergence theorem:


E  dA  (  E) dxdydz ,
V
V
i.e. the flux of E through the boundary ∂V of the volume V
equals the integral of   E over the volume.
Maxwell’s equations in differential form
Gauss’s law says
 E  dA   0   0   (r)dxdydz ,
q
1
V
V
where ρ(r) is charge density. But the divergence theorem says
 E  dA   (  E) dxdydz
V
so
V
 (r )




E
dxdydz


 dxdydz .
V
V
0

.
This is true for any V, hence   E 
0
,
Maxwell’s equations in differential form

Gauss’s law for B says B  dA  0 .
V


But the divergence theorem says B  dA  (  B) dxdydz ,
V
therefore
V
   B dxdydz  0 . This is true for any V, hence
V
B  0 .
Maxwell’s equations in differential form
Using Stokes’s theorem and the divergence theorem, we have
converted all four Maxwell equations to differential form:

E 
0
B  0
B
E  
t
E
  B   0 J   0 0
.
t
Current conservation
What happens when we apply the div operator   to Ampère’s
law, with and without Maxwell’s modification?
E
yields
Applying   to   B   0 J   0 0
t
   E 
    B    0  J   0 0
.
t
But we proved   (  A)  0 for any A, and Maxwell’s
equations say   E = ρ/ε0, so we conclude

0  J 
t
What does this equation mean?
.
Current conservation

The equation 0    J 
is called the “continuity equation”.
t
What does it mean?
Let’s integrate both sides of this equation over the volume V of

dq
,
an infinitesimal box. We obtain 0    J dxdydz 
dt
V
where q is the amount of charge in the box. Next, using the
divergence theorem, we can rewrite this equation as

dq
J  dA  
.
dt
V
Current conservation

dq
The left side of the equation J  dA  
is the net flux of
dt
V
current out of the box. The right side is the rate of change of
the total charge in the box.
Jy(x,y+Δy/2,z)ΔxΔz
y
z
Jz(x,y,z+Δz/2)ΔxΔy
x
Jx(x+Δx/2,y,z)ΔyΔz
–Jx(x–Δx/2,y,z)ΔyΔz
–Jy(x,y–Δy/2,z)ΔxΔz
–Jz(x,y,z–Δz/2)ΔxΔy
Current conservation

dq
The left side of the equation J  dA  
is the net flux of
dt
V
current out of the box. The right side is the rate of change of
the total charge in the box.
Ampere’s law, without Maxwell’s modification, implies that
the net flux of current out of any box vanishes:
 J  dA  0.
V
No charge can build up anywhere. That is certainly not true for
a circuit with a capacitor inside.
Current conservation

dq
The left side of the equation J  dA  
is the net flux of
dt
V
current out of the box. The right side is the rate of change of
the total charge in the box.
Ampere’s law with Maxwell’s modification simply states that

dq
. Anywhere that the net
charge is conserved: J  dA  
dt
V
flux of current out of a box is not zero, charge must build up in
that box, or leave that box.
Electromagnetic waves
Let’s learn more tricks with the curl   and div   operators,
and with Maxwell’s equations.
Example 1: What is   (  E) ?
Electromagnetic waves
Let’s learn more tricks with the curl   and div   operators,
and with Maxwell’s equations.
Example 1: What is   (  E) ? Answer: We have
E x E z
E z E y
(  E) x 

, (  E) y 

and
y
z
z
x
E y
E x
(  E) z 

, from which we can compute the
x
y
x-component of   (  E). It is
  (  E)x
  E y E x    E x E z 
 
 


 .

y  x
y  z  z
x 
Electromagnetic waves
We have


  (  E)x    Ez    E y
y
z
  E y E x    E x E z 
 
 




y  x
y  z  z
x 
  E x E y E z    2
2
2

 





2
2
x  x
y
z   x
y
z 2

   E    2 E x ,
x
where we added and subtracted ∂2Ex/∂x2 on the right side.

E x


Electromagnetic waves
Since   (  E)x

   E   2 E x , it is easy to guess
x
how the other components will turn out; we obtain
  (  E)    E   2E .
Both terms on the right side are vectors, as they must be since
the left side is a vector.
Electromagnetic waves
Now let’s consider Maxwell’s equations in the vacuum (no
charge ρ or current density J):
E  0
B  0
B
E  
t
E
  B   0 0
.
t
Electromagnetic waves
B
Let’s apply   to the Maxwell equation   E  
and
t
E
.
insert the vacuum Maxwell equation   B   0 0
t
B

 2E
We get   (  E)    
    B    0 0
,
2
t
t
t
and since   (  E)    E   2E   2E in the
vacuum, we arrive at a wave equation:  2 E   0 0
 2E
t
2
 0.
Electromagnetic waves
Let’s try a plane wave solution to this wave equation:
E(r,t) = E cos (k·r – ωt + δ) ,
where E is a constant vector that fixes the amplitude and the
polarization of the wave, k = 2π/λ is the wave number of the
wave, and ω is its angular frequency. Substituting this plane
wave into  E   0 0
2
 2E
 0, we obtain a solution if and
t
only if – k2 + μ0ε0 ω2 = 0, hence ω/k = 1 /  0 0  c.
2
Electromagnetic waves
We found that E(r,t) = E cos (k·r – ωt + δ) is a solution if
and only if ω/k = 1 /  0 0  c.
We can determine the speed of this wave by checking at what
speed a crest of the wave moves. At a crest of the wave, the
phase k·r – ωt + δ is unchanging over time. For the
component of r in the k-direction, we have
d
dr
dr ω
0  kr – ωt     k
 ω, hence  
dt
dt
dt k
1
0 0
c .
Around 1864, Maxwell calculated the speed of these waves.
Imagine how he felt when he obtained the speed of light!
Electromagnetic waves
The speed of electromagnetic waves is v = ω/k = 1 /  0 0  c.
Today’s data:
μ0 = 4π × 10–7 T·m/A, ε0 = 8.854187817 × 10–12 F/m,
c = 2.99792458 × 108 m/s.
“We can scarcely avoid the conclusion that light consists in the
transverse undulations of the same medium which is the cause
of electric and magnetic phenomena.” (J. C. Maxwell)
Electromagnetic waves
Heinrich Hertz was the first (in 1886) to verify Maxwell’s
prediction of electromagnetic waves travelling at the speed of
light.
Receiver
Spark Gap
Transmitter
Receiver
Spark Gap
Transmitter
Electromagnetic waves
Summary: Maxwell’s equations predict that electromagnetic
waves (including light waves) travel at speed c in vacuum. The
accepted value of c today is 299,792,458 m/s.
B
0  E 
t
Faraday’s law
B 

0      E 
t 

Vector identity


 (  B)
 (  E)   E 
t
Gauss’s law in a vacuum:
2
•E = 0
Ampère’s law, as
modified by Maxwell
2


E
1

E


2
2
  E    0 0
  2 2  E
t 
t  c t
Electromagnetic waves
What else can we find out about electromagnetic waves?
Applying the Gauss law 0    E to our wave solution E(r,t) =
E cos (k·r – ωt + δ), we obtain




0    E   E x  E y  E z  cos(k  r  t   )
y
z 
 x
 k  E sin(k  r  t   ) .
Since sin (k·r – ωt + δ) does not vanish, it follows that that
k·E = 0: the electric field of an electromagnetic wave in a
vacuum is transverse, i.e. perpendicular to the wave vector k.
Electromagnetic waves
What else can we find out about electromagnetic waves?
B
Applying Faraday’s law   E  
to our wave solution
t
E(r,t) = E cos (k·r – ωt + δ), we obtain
B
 k  E sin(k  r  t   )    E  
;
t
if we try B(r,t) = B cos (k·r – ωt + δ), we obtain a solution
provided k × E = ωB! Thus B is perpendicular to E and to k
(i.e. B is transverse) and propagates in phase with E. Since we
have seen that ω = ck, we have E = cB.
Electromagnetic waves
B is perpendicular to E and to k (i.e. B is transverse) and
propagates in phase with E.
E
k
B
Electromagnetic waves
Example: Prove using one of Maxwell’s equations that B is
perpendicular to k (i.e. B is transverse).
E
k
B
Electromagnetic waves
Example: Prove using one of Maxwell’s equations that B is
perpendicular to k (i.e. B is transverse).
E
k
B
Answer: From 0    B we derive
0    B cos(k  r  t   )  k  B sin(k  r  t   ) .
Since the sine function does not vanish identically, it follows
that k·B = 0, i.e. B is transverse.
Poynting vector
EB
The Poynting vector S, defined as S 
, has units W/m2.
0
Poynting vector
EB
The Poynting vector S, defined as S 
, has units W/m2.
0
What is the energy density in an electromagnetic wave? We
have formulas for the energy densities uE and uB in constant
electric and magnetic fields E and B, respectively; the total
electromagnetic energy density in constant E and B is uE + uB =
ε0E2/2 + B2/2μ0. In a wave, the energy density is half as much,
because it is the time average of the oscillating fields. Now
using E = cB and c2 = 1/ε0μ0, we can write the energy density in
an electromagnetic wave as EB/2μ0c.
Poynting vector
EB
The Poynting vector S, defined as S 
, has units W/m2.
0
What is the current of energy density in an electromagnetic
wave? It is the energy density times the speed of light, c,
namely EB/2μ0. The direction of the energy flow is the same as
the direction of S, which is parallel to k.
E B
Thus the time-averaged Poynting vector Sav=
, also
20
called the intensity, represents the current of energy density in
an electromagnetic wave.
Poynting vector
Let electromagnetic radiation fall upon a totally absorbing
surface of area A. If we multiply the time-averaged Poynting
E B
vector S av 
by the area A, we obtain the rate of energy
2 0
absorption by the surface. It can be shown that SavA/c is the
force on the surface. Pressure is force per unit area, thus
(SavA/c)/A = Sav/c is the electromagnetic radiation pressure on
the surface.
If the surface totally reflects the radiation instead of absorbing
it, the pressure is doubled, i.e. to 2Sav/c, since the change in
momentum per unit surface is doubled.
Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 13:
The electromagnetic plane wave E(r,t) = E sin (kx – ωt) is
polarized along the y-axis and its wavelength is λ = 3.18 m.
The amplitude of the wave is E = 288 V/m. (a) What is the
frequency f of the wave? (b) What are the magnitude and the
direction of the magnetic part of the wave? (c) What are k
and ω? (d) What is the intensity? (e) If the wave falls upon a
perfectly absorbing sheet of area 1.85 m2, at what rate is
momentum delivered to the sheet and what is the radiation
pressure?
Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 13:
The electromagnetic plane wave E(r,t) = E sin (kx – ωt) is
polarized along the y-axis and its wavelength is λ = 3.18 m.
The amplitude of the wave is E = 288 V/m. (a) What is the
frequency f of the wave? (b) What are the magnitude and the
direction of the magnetic part of the wave? (c) What are k
and ω? (d) What is the intensity? (e) If the wave falls upon a
perfectly absorbing sheet of area 1.85 m2, at what rate is
momentum delivered to the sheet and what is the radiation
pressure?
Answer: (a) f = c/λ = (3.00 × 108 m/s)/(3.18 m) = 94.3 MHz.
(b) B = E/c = (288 V/m)/(3.00 × 108 m/s) = 9.60 × 10–7 T and B
points along the z-axis. (c) k = 2π/λ = 1.98/m and ω = ck =
(3.00 × 108 m/s) (1.98/m) = 5.93 × 108 Hz.
Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 13:
The electromagnetic plane wave E(r,t) = E sin (kx – ωt) is
polarized along the y-axis and its wavelength is λ = 3.18 m.
The amplitude of the wave is E = 288 V/m. (a) What is the
frequency f of the wave? (b) What are the magnitude and the
direction of the magnetic part of the wave? (c) What are k
and ω? (d) What is the intensity? (e) If the wave falls upon a
perfectly absorbing sheet of area 1.85 m2, at what rate is
momentum delivered to the sheet and what is the radiation
pressure?
Answer: (d) The time-averaged Poynting vector magnitude is
Sav = EB/2μ0 = (288 V/m)(9.60 × 10–7 T)/2(4π × 10–7 T·m/A) =
110 W/m2, and this is the intensity.
Halliday, Resnick and Krane, 5th Edition, Chap. 38, Prob. 13:
The electromagnetic plane wave E(r,t) = E sin (kx – ωt) is
polarized along the y-axis and its wavelength is λ = 3.18 m.
The amplitude of the wave is E = 288 V/m. (a) What is the
frequency f of the wave? (b) What are the magnitude and the
direction of the magnetic part of the wave? (c) What are k
and ω? (d) What is the intensity? (e) If the wave falls upon a
perfectly absorbing sheet of area 1.85 m2, at what rate is
momentum delivered to the sheet and what is the radiation
pressure?
Answer: (e) The radiation pressure is EB/2μ0c = B2/2μ0 =
(9.60 × 10–7 T)2 / 2(4π × 10–7 T·m/A) = 3.67 × 10–7 N/m2, and
the rate of transfer of momentum is the pressure times the area,
namely (3.67 × 10–7 N/m2) (1.85 m2) = 6.78 × 10–7 N.