Today’s agenda:
Electromagnetic Waves.
Energy Carried by Electromagnetic Waves.
Momentum and Radiation Pressure of an Electromagnetic
Wave.
Energy Carried by Electromagnetic Waves
Electromagnetic waves carry energy, and as they propagate
through space they can transfer energy to objects in their path.
The rate of flow of energy in an electromagnetic wave is
described by a vector S, called the Poynting vector.*
S=
1
E B
0
This is derived from
Maxwell’s equations.
The magnitude S represents the rate at which energy flows
through a unit surface area perpendicular to the direction of
wave propagation (energy per time per area).
Thus, S represents power per unit area. The direction of S is
along the direction of wave propagation. The units of S are
J/(s·m2) =W/m2.
*J. H. Poynting, 1884.
y
E
B
z
For an EM wave E  B = EB
EB
so S =
.
0
1
S = E B
0
S
c
x
Because B = E/c we can write
E2
cB 2
S=
=
.
0 c 0
These equations for S apply at any instant of time and
represent the instantaneous rate at which energy is passing
through a unit area.
EB E2
cB 2
S=
=
=
0 0 c 0
EM waves are sinusoidal. Ey =Emax sin kx - t 
Bz =Bmax sin kx - t 
EM wave propagating
along x-direction
The average of S over one or more cycles is called the wave
intensity I.
The time average of sin2(kx - t) is ½, so
2
2
EmaxBmax Emax
cBmax
I = Saverage = S =
=
=
20
20 c
20
Notice the 2’s in
this equation.
This equation is the same as 32-29 in your text, using c = 1/(00)½.
The magnitude of S is the rate at which energy is
transported by a wave across a unit area at any instant:
 energy

 power 
time


S=
=



area
area

instantaneous


instantaneous
Thus,
 energy

 power 
time


I= S =
=



area
area

average

average
Note: Saverage and <S> mean the same thing!
Energy Density
The energy densities (energy per unit volume) associated
with electric and magnetic fields are:
1 B2
uB =
2 0
1
uE = 0E2
2
Using B = E/c and c = 1/(00)½ we can write
 
E
1 B2 1 c
uB =
=
2 0 2 0
2
1 0 0E2 1
=
= 0E2
2 0
2
1
1 B2
2
uB = uE = 0E =
2
2 0
remember: E and B are
sinusoidal functions of time
1
1 B2
2
uB = uE = 0E =
2
2 0
For an electromagnetic wave, the instantaneous energy density
associated with the magnetic field equals the instantaneous
energy density associated with the electric field.
Hence, in a given volume the energy is equally shared by the
two fields. The total energy density is equal to the sum of the
energy densities associated with the electric and magnetic
fields:
B2
2
u = uB +uE = 0E =
0
2
B
u = uB +uE = 0E2 =
0
instantaneous energy densities
(E and B vary with time)
When we average this instantaneous energy density over one
or more cycles of an electromagnetic wave, we again get a
factor of ½ from the time average of sin2(kx - t).
1
2
uE = 0Emax
,
4
2
1 B max
uB =
, and
4 0
2
B
1
1
2
max
u = 0Emax
=
2
2 0
2
2
1 Emax
1 cBmax
Recall S average = S =
=
so we see that S = c u .
2 0 c 2 0
The intensity of an electromagnetic wave equals the average
energy density multiplied by the speed of light.
Help!
“These factors of ¼, ½, and 1 are making my brain hurt!”
It’s really not that bad. These are the energy densities
associated with E(t) and B(t) at some time t:
1
1 B2
2
uB = uE = 0E =
2
2 0
Add uB and uE to get the total energy density u(t) at time t:
2
B
u = uB +uE = 0E2 =
0
Help!
Again, these are the energy densities associated with E(t)
and B(t) at some time t:
2
1
1
B
uB = uE = 0E2 =
2
2 0
If you average uB and uE over one or more cycles, you get an
additional factor of ½ from the time average of sin2(kx-t).
1
2
uE = 0Emax
4
2
1 B max
uB =
,
4 0
The Emax and Bmax come from writing E = Emax sin(kx-t) and
B = Bmax sin(kx-t), and canceling the sine factors.
Help!
These are the average energy densities associated with E(t)
and B(t) over one or more complete cycles.
1
2
uE = 0Emax
4
2
1 B max
uB =
,
4 0
Add uE and uB to get the total average energy density over
one or more cycles:
2
2
B
B
1
1
1
1
2
2
max
max
u = uE + uB = 0Emax
+
= 0Emax
=
4
4 0
2
2 0
Help!
Summary:
At time t:
Average:
At time t:
Average:
2
1
1
B
(t)
2
uB (t) = uE (t) = 0E (t) =
2
2 0
1
2
uE = 0Emax
4
2
1 B max
uB =
,
4 0
2
B
(t)
u(t) = 0E2 (t) =
0
2
B
1
1
2
max
u = 0Emax
=
2
2 0
If you use a starting equation that is not valid for the problem scenario, you will get incorrect results!