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Transcript
24.4 Heinrich Rudolf Hertz


1857 – 1894
The first person
generated and received
the EM waves


1887
His experiment shows
that the EM waves follow
the wave phenomena
1
Hertz’s Experiment




An induction coil is connected to a
transmitter
The transmitter consists of two
spherical electrodes separated by a
narrow gap to form a capacitor
The oscillations of the charges on
the transmitter produce the EM
waves.
A second circuit with a receiver,
which also consists of two
electrodes, is a single loop in
several meters away from the
transmitter.
2
Hertz’s Experiment, cont




The coil provides short voltage surges to the
electrodes
As the air in the gap is ionized, it becomes a
better conductor
The discharge between the electrodes
exhibits an oscillatory behavior at a very high
frequency
From a circuit viewpoint, this is equivalent to
an LC circuit
3
Hertz’s Experiment, final



Hertz found that when the frequency of the
receiver was adjusted to match that of the
transmitter, the energy was being sent from
the transmitter to the receiver
Hertz’s experiment is analogous to the
resonance phenomenon between a tuning fork
and another one.
Hertz also showed that the radiation generated
by this equipment exhibited wave properties


Interference, diffraction, reflection, refraction and
polarization
He also measured the speed of the radiation
4
LC circuit
5
24.5 Energy Density in EM
Waves




The energy density, u, is the energy per
unit volume
For the electric field, uE= ½ eoE2
For the magnetic field, uB = B2 / 2mo
Since B = E/c and c  1 m o e o
2
1
B
2
uB  uE  e o E 
2
2 mo
6
Energy Density, cont

The instantaneous energy density
associated with the magnetic field of an
EM wave equals the instantaneous
energy density associated with the
electric field

In a given volume, the energy is shared
equally by the two fields
7
Energy Density, final

The total instantaneous energy density
in an EM wave is the sum of the energy
densities associated with each field


u =uE + uB = eoE2 = B2 / mo
When this is averaged over one or more
cycles, the total average becomes

uav = eo (Eavg)2 = ½ eoE2max = B2max / 2mo
8
Energy carried by EM Waves



Electromagnetic waves carry energy
As they propagate through space, they
can transfer energy to objects in their
path
The rate of flow of energy in an EM
wave is described by a vector called the
Poynting vector
9
Poynting Vector

The Poynting Vector is
defined as
S


1
mo
EB
Its direction is the direction
of propagation
This is time dependent


Its magnitude varies in time
Its magnitude reaches a
maximum at the same instant
as the fields
10
Poynting Vector, final

The magnitude of the vector represents
the rate at which energy flows through a
unit surface area perpendicular to the
direction of the wave propagation


This is the power per unit area
The SI units of the Poynting vector are
J/s.m2 = W/m2
11
Intensity


The wave intensity, I, is the time average of S
(the Poynting vector) over one or more cycles
When the average is taken, the time average
of cos2(kx-wt) equals half
I  Savg

2
2
Emax Bmax
Emax
c Bmax



2 mo
2 mo c
2 mo
I = Savg = c uavg

The intensity of an EM wave equals the average
energy density multiplied by the speed of light
12
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15
24.6 Momentum and Radiation
Pressure of EM Waves



EM waves transport momentum as well as
energy
As this momentum is absorbed by some
surface, pressure is exerted on the surface
Assuming the EM wave transports a total
energy U to the surface in a time interval Dt,
the total momentum is p = U / c for complete
absorption
16
Measuring
Radiation Pressure



This is an apparatus for
measuring radiation pressure
In practice, the system is
contained in a high vacuum
The pressure is determined by
the angle through which the
horizontal connecting rod
rotates


For complete absorption
An absorbing surface for which all
the incident energy is absorbed is
called a black body
17
Pressure and Momentum

Pressure, P, is defined as the force per
unit area
F 1 dp 1 dU dt
P 

A A dt c A

But the magnitude of the Poynting vector
is (dU/dt)/A and so P = S / c
18
Pressure and
Momentum, cont



For a perfectly reflecting surface,
p = 2 U / c and P = 2 S / c
For a surface with a reflectivity somewhere
between a perfect reflector and a perfect
absorber, the momentum delivered to the
surface will be somewhere in between U/c
and 2U/c
For direct sunlight, the radiation pressure is
about 5 x 10-6 N/m2
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