Electromagnetic Radiation

1. LECTURE 3 Electromagnetic Radiation

2. 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.

3. Some basic terminology Wavelengthis formally defined as the mean distance between maximums (or minimums) of a periodic pattern and is normally measured in micrometers (µm) or nanometers (nm). Frequencyis 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.

4. Units 1 Micrometer (μm) = 10-6m - 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

5. Electromagnetic (EM) radiation EM Energy source- the SUN

6. 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 chromospherelies ~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

7. 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). Visible portion – 0.4 to 0.7 μm (~10-7 m range) 2

9. Optical Regions Microwave Regions Visible (V) 0.4-0.7 μm Near Infrared (NIR) 0.7-1.0 μm Shortwave Infrared (SWIR) 1.0-3 μm Thermal Infrared (TIR) 3-15 μm Mid Infrared (MIR) 15 – 100 μm Far Infrared (FIR) > 100 μm P band 0.3-1 GHz (30-100 cm) L band 1-2 GHz (15-30 cm) S band 2-4 GHz (7.5-15 cm) C band 4-8 GHz (3.8-7.5 cm) X band 8-12.5 GHz (2.4-3.8 cm) Ku band 12.5-18 GHz (1.7-2.4 cm) K band 18-26.5 GHz (1.1-1.7 cm) Ka band 26.5-40 GHz (0.75-1.1 cm) Thermal Regions Mid-Far Infrared Regions

10. Spherical polar coordinates (θ , φ, Ω) θ = zenith angle, usually relative to local vertical, i.e., θ = 0ois overhead, θ = 90ois 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

11. Angle and Solid angle • Angle: Transverse distance at a distance (radians) • Solid angle: Transverse area at a distance (steradians)

12. 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

13. 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

14. Z Area = dA = d λ x dβ dω = dA/r2 = (d λ x dβ)/r2 h dφ dλ r׳ z r dβ z dθ y θ Y x h φ x 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

15. 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?

16. 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. • CalledSpectral Radiant Energy (Qλ), if we consider the radiant energy at a particular wave length:

17. 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 • CalledSpectral Radiant Flux (Φλ), if we consider the radiant flux per unit wavelength at a particular wave length (λ) (W.μm-1):

18. Irradiance (E) and Exitance (M) (Wm-2) • Radiant flux per unit area OR radiant flux density: • 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.

19. What is solar extance, given that the solar flux is ~3.96 x 1026 W? (Solar radius ~ 7x105 km) What is the solar irradiance at the earth? (sun-earth distance ~1.5 x 108 km) 6.42 x 107 W m-2 1378 W m-2

20. Radiant Intensity (W.sr-1) • Radiant flux leaving a point source per unit solid angle: • For the source radiating in an isotropic medium: • A source can be considered a point source if its dimension is much smaller than the observation distance (<<1/10 of the observation distance)

21. RADIANCE (L) (W.m-2.sr-1) • Radiant flux leaving an extended source in a given direction per unit projected area in that direction per unit solid angle : Z Radiating element Area projected in the given direction = da.cosϴ Element solid angle = dω ϴ Clearly, Radiating element area = da ϴ Y ϕ (x, y) X

22. 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

23. Relationship between radiance and irradiance The upward-directed irradiance (or radiant exitance) from a surface is given by:

24. 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? • Solar ‘constant’ (irradiance at top of Earth’s atmosphere) is ~1370 W m-2 Radiance is invariant with distance

25. 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 • 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

27. 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.

28. WAVE MODEL: Maxwell’s Equations Gauss’ law for electric field Gauss’ law for magnetic field Faraday’s law of Induction Ampere-Maxwell law

29. 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

30. The wave model of electromagnetic radiation

31. The wave model of electromagnetic radiation c v c v = l × = v l = l 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.

32. The wave model of electromagnetic radiation • Speed of light: 3 x 108 m/second calculated as: • 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.

33. 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: (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.

34. Applications of Planck’s formula for radiation

35. EM Radiation Equations Integrating the Planck’s formula over wavelength, we obtain the Stefan-Boltzmann Law 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: 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)

36. Blackbody radiation curves • 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. 6000 K (SUN) 3000 K (TUNGSTON FILAMENT) 800 K 300 K (EARTH) 195 K (DRY ICE) 79 K (LIQUID AIR)

37. Radiation curves of the earth and Sun

38. Spectral Bands of Landsat and ASTER

39. = h ´ v Q 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: 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.

40. = h ´ v Q h c = Q l Quantum energy - wavelength relation Q (Particle model) v = Or h c v = (Wave model) λ 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

41. Generation of electromagnetic radiation Nuclear processes (From : Elachi)

42. Energy transfer mechanisms • Energy may be transferred three ways: conduction, convection, and radiation. • Energy may be conducteddirectly from one object to another as when a pan is in direct physical contact with a hot burner. • 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 convectionalcurrents in the atmosphere. • Electromagnetic energy in the form of electromagnetic waves are radiatedthrough the vacuum of space from the Sun to the Earth.