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LECTURE 3
Electromagnetic Radiation
Electromagnetic Radiation
This course is about electromagnetic energy sensors – other types
of remote sensing such as geophysical will be disregarded.
• For proper analysis and interpretation of remote sensing data, it
is important to understand
• the source of electromagnetic radiation,
• its nature,
• its propagation, and
• its interaction with atmosphere and other matter.
Some basic terminology
Wavelength is formally defined as the mean distance between maximums (or minimums)
of a periodic pattern and is normally measured in micrometers (µm) or nanometers (nm).
Frequency is the number of peaks that pass a point per unit time.
Amplitude is the height of crest from the mid-point
A wave that sends one peak every second (completing one cycle) is said to have a frequency
of one cycle per second or one hertz, abbreviated 1 Hz.
Units
1 Micrometer (μm) = 10-6 m 1 Nanometer (ηm) = 10-9 m
1Angstrom (Å) = 10-10 m
1 Hertz (Hz) = 1 cycle per second
1 Kilo Hertz (KHz) = 103 Hz
1 Mega Hertz (MHz) = 106 Hz
1 Giga Hertz (Ghz) = 109 Hz
1eV = 1.602×10−19 J
• Human hair – 10-100 micron
• The smallest object that the human
eye can see (unaided) is 5 micron
Electromagnetic (EM) radiation
EM Energy
source- the SUN
Electromagnetic energy source : The Sun
Surface regions of the Sun the photosphere, the
chromosphere, and the corona.
The photosphere corresponds to
the bright region normally
visible to the naked eye.
(`Temperature ~ 6000 K)
The chromosphere lies ~5,000
km above the photosphere.
Short-lived, projections may
extend for several thousands of
kms from the chromosphere.
(Temperature ~ 20,000 K)
The corona is the outermost
layer of the Sun; this region
extends into the region of the
planets.
Distance from the Earth - 1.5x108 km
Nature of electromagnetic radiation:
the electromagetic spectrum
The Sun produces a
continuous spectrum of
energy from gamma
rays to radio waves
Various parts of the EM
spectrum may be
differentiated using
wavelength (measured
in micrometers or
nanometers, i.e., μm or
nm) or electron volts
(eV).
2
Visible portion – 0.4 to
0.7 μm (~10-7 m range)
Optical Regions
Visible (V)
Near Infrared (NIR)
Shortwave Infrared (SWIR)
0.4-0.7 μm
0.7-1.0 μm
1.0-3 μm
Thermal Regions
Thermal Infrared (TIR)
3-15 μm
Mid-Far Infrared Regions
Mid Infrared (MIR)
Far Infrared (FIR)
15 – 100 μm
> 100 μm
Microwave Regions
P band
L band
S band
C band
X band
Ku band
K band
Ka band
0.3-1 GHz (30-100 cm)
1-2 GHz (15-30 cm)
2-4 GHz (7.5-15 cm)
4-8 GHz (3.8-7.5 cm)
8-12.5 GHz (2.4-3.8 cm)
12.5-18 GHz (1.7-2.4 cm)
18-26.5 GHz (1.1-1.7 cm)
26.5-40 GHz (0.75-1.1 cm)
Spherical polar coordinates (θ , φ, Ω)
θ = zenith angle,
usually relative to local
vertical, i.e.,
θ = 0o is overhead, θ
= 90o is the horizon.
φ = azimuth angle,
e.g., the direction with
respect to the
geographic north
Ω = distance from
origin
Direction plays an important role in any discussion of radiation
Angle and Solid angle
• Angle: Transverse distance at a distance
(radians)
Angle in radians 
• Solid angle: Transverse area at a distance
(steradians)
Traverse length (Arc length)
(Distance or Radius ) 2
Traverse Area
Solid angle (in steradians ) 
(Distance or Radius ) 2
Solid angle ( Ω): the steradian (sr)
How much of the visual field of
view is occupied by an object?
Ω
1 steradian
• Three-dimensional analog of the planar radian (ratio of arc length : radius)
• Defined as  = area of surface A / r2 (Solid angle is dimensionless)
• Sphere subtends a solid angle of ??? sr
• The entire sky above the horizon subtends ??? sr
Solid angle: the steradian
sin θ accounts for the convergence of ‘longitude’ lines at the ‘pole’
Solid angle over the entire hemisphere
can be found by integrating infinitesimal
solid angles over the entire sphere
2 

0 0
0
  d    sin  d d  2  sin  d  4
4
Z
Area = dA = d λ x dβ
dω = dA/r2 = (d λ x dβ)/r2
h
dφ
r‫׳‬
z
r
dθ
θ
z
y
φ
Y
h
y
X
(Say pointing North)
r‫ = ׳‬r.sinθ
Angle (in radians) = Arc length / radius
Arc length (d λ) = Angle (dθ) x radius (r) = r.d
Arc length (d β) = Angle (dφ) x radius (r‫ = )׳‬r.sinθ.dΦ
dA = r.dθ x r.sinθ dφ = r2.sinθ.dθ.dφ
dω = dA/r2 = r2.sinθ.dθ. dφ /r2 = sinθ.dθ. dφ
Solid angle problem
• Mean distance of the moon from Earth = 3.84 x 105 km
• Radius of the moon = 1.74 x 103 km
• Mean distance of the Sun from Earth = 1.496 x 108 km
• Radius of the Sun = 6.96 x 105 km
• What is the angular diameter subtended by the Sun and moon?
• What is the solid angle subtended by the Sun and moon?
• Which appears larger from the Earth?
Some terminology
Radiant energy (Q) (Joules, N.m, W.s)
• The energy carried by an EM wave.
• A measure of its capacity to do physical work (heat, change of state, movement etc.)
when interacting with the matter.
• Refers to the amount of energy propagating into or propagating through or emerging
from a surface of a given area in a given period of time.
• All wavelengths contained in the radiation are included.
• Called Spectral Radiant Energy (Qλ), if we consider the radiant energy at a particular
wave length:
dQ
QA 
d
Radiant Flux (Φ) (J.s-1, N.m.s-1, W)
•The rate at which radiant energy is emitted, transferred, or received in the form of EM
radiation from a point or a surface to another surface.
• Since it represents energy per unit time, it corresponds to Power
dQ

dt
•Called Spectral Radiant Flux (Φλ), if we consider the radiant flux per unit wavelength at a
particular wave length (λ) (W.μm-1):
d
 
d
Irradiance (E) and Exitance (M) (Wm-2)
•Radiant flux per unit area OR radiant flux density:
d
E
da
• Radiant flux density arriving at a surface is called Irradiance; while the radiant flux
density leaving (or being emitted by) a surface is called Exitance.
• The exitance is inversely proportional to the square of the distance from the source.
What is solar extance, given that the solar flux is ~3.96 x 1026 W? (Solar
radius ~ 7x105 km)
6.42 x 107 W m-2
What is the solar irradiance at the earth?
(sun-earth distance ~1.5 x 108 km)
1378 W m-2
Radiant Intensity (W.sr-1)
•Radiant flux leaving a point source per unit solid angle:
d
I
d
• For the source radiating in an isotropic medium:

I
4
• A source can be considered a point source if its dimension is much smaller than the
observation distance (<<1/10 of the observation distance)
RADIANCE (L) (W.m-2.sr-1)
Z
•Radiant flux leaving an extended source in a given
direction per unit projected area in that direction per
unit solid angle :
d
L ( x, y ,  ,  ) 
d.da. cos 
Radiating element
Area projected in
the given
direction
= da.cosϴ
Element
solid angle
= dω
Clearly,
E   dL( ,  )
Radiating element
area = da
Y

X
(x, y)
Relationship between Irradiance and
radiance
L
•irradiance/radiant exitance from a surface is
found by integrating I over radiance over solid
angles
• Since radiance is defined as the flux per unit
area per unit solid angle normal to the beam,
the contributions to the flux must be
weighted by a factor of cos θ
Relationship between radiance and irradiance
The upward-directed irradiance (or radiant exitance) from a surface is given by:
2
2  / 2
0
0 0
E    L cos  d  

E 

L
 ( , ) cos sin  d d
2  / 2

L
 ( ,  ) cos sin  d d

0 0
/ 2
 2L  cos  sin  d
0
/ 2
 2L 
0
1
sin 2 d
2
 L
Irradiance    Radiance
Inverse square law
Flux density at surface of sphere
E
E
E
E
• Irradiance (or flux density) decreases as the square of distance from the source
• What about radiance? Radiance is invariant with distance
• Solar ‘constant’ (irradiance at top of Earth’s atmosphere) is ~1370 W m-2
Global insolation
• How much total solar radiation Φ is incident on Earth’s atmosphere?
• Consider the amount of radiation intercepted by the Earth’s disk
1370 W m-2
  S 0 R
2
E
 1.74 10 W
17

• Applies for mean Sun-Earth distance of 1.496 x 108 km
• But Earth’s orbit is elliptical, so the solar flux (S) actually varies from 1330 W m-2
in July to 1420 W m-2 in January
In the beginning there were four Equations,
two fluxes and two curls,
obeying charge conservation.
And Maxwell said, Let there be light:
and there was light.
WAVE MODEL: Maxwell’s Equations
 
 E  dA  4  dV
s
v
 
 B dA  0
S
 
d
c E  dl   dt s B  dA
 


d 


B

d
l



I


E

dA
0 
0
c


dt s


 
.E  4
 
.B  0
Gauss’ law for
electric field
Gauss’ law for
magnetic field

 
B
 E  
t
Faraday’s law of
Induction

 
E
 B  J 
t
Ampere-Maxwell
law
Maxwell’s Equations
• The flux of the electic field though a closed surface is due to the
charge density contained inside - electric charges produce electric
fields!
• The flux of the magnetic field through a closed surface is zero and
no matter how much we wish magnetic monopoles do not exist!
• The curl of electric time rate is equal to the negative of the time
rate of change of magnetic field – that a magnetic field that is
changing in time produces an electric field!
• The curl of the magnetic field is due to the current flow and a
changing electric field
The wave model of electromagnetic radiation
The wave model of electromagnetic radiation
c   v
v
c


v

c
Frequency, v, is inversely
proportional to wavelength
λ, The longer the
wavelength, the lower the
frequency, and vice-versa.
c is the velocity of light.
The wave model of electromagnetic radiation
• Speed of light: 3 x 108 m/second calculated as:
cvacuum 
1
 0 0
where ε0 (=8.85 x 10-12 Farad/m) is the permittivity of free space and μ0 (=4π x 10-7
Henry/m) is the permeability of free space.
• For quick calculations : speed of light = 1 ft per nanosecond (1 ft per10-9 s).
•The electromagnetic wave consists of two fluctuating fields—one electric and the
other magnetic. The two vectors are at right angles (orthogonal) to one another,
and both are perpendicular to the direction of travel.
EM Radiation Equations
• Heat energy is the kinetic energy of radiation of random motion of the particles of the matter
• The random motion results in excitations (electronic, vibrational or rotational) due to collisions
followed by random emission of EM during decay.
• Because of its randomness, this type of transformation leads to emission over wide spectral band.
• If an ideal source (called blackbody) transforms heat energy into radiant energy at the maximum
rate permitted by thermodynamic laws, then the spectral emission is given by:
S ( ) 
8hC
5

1
eCh / kT  1
(power per area per wavelength – spectral irrradiance)
This is Planck’s formula. In this, h is Planck’s constant (6.62x10-34 m2 Kg/sec), k is the Boltzmann
constant (1.38 × 10-23 m2 kg / s2 K1), C is the velocity of light, and T is the absolute temperature in
degrees Kelvin.
Applications of Planck’s formula for radiation
S ( ) 
8hC

5

1
eCh / kT  1
EM Radiation Equations
S ( ) 
8hC
5

1
e Ch / kT  1
Integrating the Planck’s formula over wavelength, we obtain the Stefan-Boltzmann Law
2 5 k 4 4
4
L
T


T
15h3C 2
Irradiance or power per area, σ is Stefan Boltzmann
constant = 5.6697x10-8 W m-2 K -4
Multiplying the irradiance by the area gives us the total emitted radiation:
2 5 k 4 4
4
P
T

A


AT
15h3C 2
Apply the first derivative to the wavelength form of Planck's law to determine the peak
wavelength as a function of temperature (Wien displacement law)
hc 1
2898
max ( m) 

kx T
T
Blackbody radiation curves
6000 K (SUN)
3000 K
(TUNGSTON FILAMENT)
800 K
300 K (EARTH)
195 K (DRY ICE)
79 K (LIQUID AIR)
•The area under each curve may
be summed to compute the total
radiant energy exiting each
object.
•The Sun produces more radiant
exitance than the Earth because
its temperature is greater.
•As the temperature of an object
increases, its dominant
wavelength (λmax ) shifts toward
the shorter wavelengths of the
spectrum.
Radiation curves of the earth and Sun
Spectral Bands of Landsat and ASTER
The particle model of electromagnetic radiation
Niels Bohr (1885–1962) and Max Planck proposed the quantum
theory of electromagnetic radiation.
This theory states that energy is transferred in discrete packets
called quanta or photons.
The relationship between the frequency of radiation expressed by
wave theory and the energy of a quantum is:
Q  h v
where Q is the energy of a quantum measured in joules, h is
the Planck constant (6.62610-34 Js), and v is the frequency of
the radiation.
Quantum energy - wavelength relation
Q  h v
Or
v = c
λ
Q

Q
v 
h
(Particle model)
(Wave model)
hc

Thus, the energy of a quantum is inversely proportional to its wavelength,
i.e., the longer the wavelength involved, the lower its energy content.
Naturally emitted long wavelength radiation (e.g., MW emissions) is more
difficult to sense remotely than shorter wavelength emissions (e.g., TIR).
Another implication is that long wavelength sensors must view larger areas
of the earth at any give time in order to obtain a detectable signal
Generation of electromagnetic radiation
Nuclear processes
(From : Elachi)
Energy transfer mechanisms
Energy may be transferred three ways: conduction, convection, and radiation.
a) Energy may be conducted directly from one object to another as when a pan is in
direct physical contact with a hot burner.
b) The Sun heats the Earth’s surface with radiant energy causing the air near the ground
to increase in temperature. The less dense air rises, creating convectional currents in the
atmosphere.
c) Electromagnetic energy in the form of electromagnetic waves are radiated through
the vacuum of space from the Sun to the Earth.