Maxwell's Equations & Electromagnetic Waves: Luttermoser's PHYS-2020 Lecture Notes, Study notes of Physics

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PHYS-2020: General Physics II

Course Lecture Notes

Section IX

Dr. Donald G. Luttermoser East Tennessee State University

Edition 3.

Abstract

These class notes are designed for use of the instructor and students of the course PHYS-2020: General Physics II taught by Dr. Donald Luttermoser at East Tennessee State University. These notes make reference to the College Physics, Enhanced 7th Edition (2006) textbook by Serway, Faughn, and Vuille.

IX–2 PHYS-2020: General Physics II

equation for the divergence of the B-field:

∇ ·^ ~ B~ = 0 ,

the ‘zero’ simply means that there are no magnetic monopoles.

c) A varying B-field induces an emf and hence electric (E) field =⇒ Faraday’s law re-expressed in Maxwell’s form:

∇ ×^ ~ E~ = −∂ B~

∂t

The ‘∇×~ ’ operation is called the curl in higher mathe- matics. Whereas the divergence produces a scalar, the curl produces a vector.

d) Magnetic fields are generated by moving charges (or cur- rents) =⇒ Ampere’s law re-expressed in Maxwell’s form:

∇ ×^ ~ B~ = μ◦ J ,~

where J(∝ ∂q/∂t) is the current density.

2. Maxwell’s last 2 laws allow electromagnetic waves (i.e., radiation) to self-propagate at a velocity of

c = (^) √^1 μ◦◦ = 2. 99792 × 108 m/s. (IX-1)

a) “c” is called the speed of light, since visible light is a form of electromagnetic radiation.

b) An oscillating electric charge produces an E-field that varies in time, which produces a B-field that varies in time, which produces a new E-field that varies in time,

Donald G. Luttermoser, ETSU IX–

and so on.

y

directionof wave propagation

E (^) field z wavecrest

Bfield λ

x

c) The ratio of the maximum magnitude E-field to the max- imum magnitude of the B-field of an E/M wave is

Emax Bmax^ =^ c^.^ (IX-2)

d) We will see in the Optics portion of this course that the above value of c is the value of the speed of light in a vacuum. Light slows when it enters a medium (i.e., air [though not by much], glass, etc.).

e) Even though the solution to Maxwell’s equations clearly showed that light (or any E/M radiation) can self-propagate, it took another 50 years before Einstein demonstrated this with the Special Theory of Relativity in 1905. Prior to this, light was assumed to propagate on a medium in space called the Ether.

f) In the late-1800s, Michelson and Morley tried to measure the Earth’s motion through the Ether and was unable to detect it — either the Ether moved with the Earth, or the Ether didn’t exist. Einstein chose the latter and used this fact as one of the underlying principles to relativity.

Donald G. Luttermoser, ETSU IX–

vi) This is not just a technical barrier, such as the sound barrier, it is impossible for anything with mass, even an electron, to be accelerated to v = c.

vii) Also note that if v > c, then we get a negative number in the square-root =⇒ an imaginary num- ber, which is a meaningless quantity for velocity in our physical Universe.

3. E/M waves also behave like particles! a) Sometimes joking called a wavicle!

b) As such, Planck called a packet of electromagnetic radia- tion a photon in the early 1900s and this term has been used ever since to describe E/M radiation.

c) The photoelectric effect demonstrates this “particle” picture nicely: If a photons hits a certain type of metal, it can “knock” an electron off of an atom in the metal and produce an electric current. (Einstein won his Nobel Prize for figuring this out. He never won a Nobel Prize for either the Special or General Theories of Relativity since at the time of their development, they could not be fully tested due to the lack of technology.)

d) At this point we need to distinguish between two types of radiation. The word radiation simply means something flowing from one point to another. This “something,” however, can either be particles of energy or particles of matter: i) The flow of photons (energy particles) is called ra- diation.

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ii) Also the motion of atomic and subatomic (mat- ter) particles given off during radioactivity events is called radiation. This was discovered by the Curies in the early 1900s. The Curies discovered three types of particle radiation:

• Alpha particles have positive charge and were later determined to be helium (He) nuclei. As such, He nuclei are often referred to as alpha par- ticles.
• Beta particles have negative charge and were later found to be the simple electron.
• Gamma particles have no charge and were later realized to be high-energy photons. These parti- cles were renamed gamma rays and represent the highest energy form of E/M radiation.

4. E/M waves are transverse waves, since the electric and magnetic fields are ⊥ to the direction of propagation of the wave and to each other.
5. E/M waves carry both energy and momentum (despite have no mass). a) The energy is proportional to frequency and inversely pro- portional to wavelength.

E = hν. (IX-3)

i) E is the energy (SI unit of J, cgs unit of erg) of the E/M wave.

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b) The momentum of a photon (E/M wave) is

p = h λ

(IX-6)

or p = hν c

. (IX-7)

c) Note that if we use Eq. (IX-3) in Eq. (IX-6), we get

p =

E

c or E = pc. (IX-8)

Example IX–1. An electromagnetic wave in vacuum has an electric field amplitude of 220 V/m. Calculate the amplitude of the corresponding magnetic field. Solution: We just need to use Eq. (IX-2) and solve for Bmax:

Bmax = Emax c

220 V/m

3. 00 × 108 m/s = 7. 33 × 10 −^7 T = 733 nT.

Example IX–2. Compare the wavelength, energy, and momen- tum of a radio wave at 108 Hz (e.g., a TV signal) to that of visible light at 500 nm (the yellow part of the spectrum). Solution: First use Eq. (IX-4) to get the wavelength of the radio wave:

νradio = νR = 10^8 Hz λR = c νR

3. 00 × 108 m/s 108 s−^1 = 3.00 m λV = 500 nm = 500 × 10 −^9 m = 5. 00 × 10 −^7 m λR λV

3 .00 m

5. 00 × 10 −^7 m

= 6. 00 × 106

Donald G. Luttermoser, ETSU IX–

=⇒ the radio wave is 6 million times longer than the visible light wave. Now use Eqs. (IX-3) and (IX-5) to compare the energies: ER = hνR = (6. 63 × 10 −^34 J·s)(10^8 s−^1 ) = 6. 63 × 10 −^26 J EV = hc λV^ =

(6. 63 × 10 −^34 J·s)(3. 00 × 108 m/s)

5. 00 × 10 −^7 m = 3.^98 ×^10

− 19 J

ER

EV

6. 63 × 10 −^26 J

3. 98 × 10 −^19 J

= 1. 67 × 10 −^7 ,

or EV = 6. 00 × 106 ER =⇒ the visible light photon is 6 million times more energetic than the radio photon. Finally, use Eq. (IX-6) to compare the momenta: pR pV

h/λR h/λV

λV λR

5. 00 × 10 −^7 m 3 .00 m

= 1. 67 × 10 −^7

or pV = 6. 00 × 106 pR!

B. The Electromagnetic Spectrum.

1. The E/M spectrum (and spectra in general) is defined to be the intensity, or flux, of an E/M wave as a function of wavelength, frequency, or energy. a) In spectroscopy, very small units of length are often used: i) 1 micrometer (μm or μ) = 10−^6 m.

ii) 1 nanometer (nm) = 10−^9 m.

iii) 1 ˚Angstrom (˚A) = 10−^10 m.

b) Visible light ranges from 4. 0 × 10 −^7 m to 7. 0 × 10 −^7 m, or 0.4 μm – 0.7 μm, or 400 nm – 700 nm, or 4000 ˚A – 7000 ˚A.

Donald G. Luttermoser, ETSU IX–

g) Radio waves: λ > 10 cm. These photons have the lowest energy, lowest frequency, and longest wavelengths and can be created by electrical circuits and from extremely cold gas (T < 0.03 K). The longest radio waves (λ > 1 km) are sometimes just referred to as long waves.

3. Some spectral regions are subdivided into smaller bands. In the case of visible light, we refer to those bands as colors since our eyes perceive them that way. The wavelength delineation (which are approximate) for each color from shortest to longest wave- lengths are: a) Violet (400 nm < λ < 450 nm).

b) Blue (450 nm < λ < 490 nm).

c) Green (490 nm < λ < 520 nm).

d) Yellow (520 nm < λ < 590 nm).

e) Orange (590 nm < λ < 630 nm).

f) Red (630 nm < λ < 700 nm).

4. The visible region is the spectrum is the narrowest. It corre- sponds to the wavelengths to which the human eye is sensitive. a) The Earth’s atmosphere is transparent (assuming clouds are not blocking your view) at visible wavelengths.

b) The human eye is most sensitive to the green-yellow part of the visible spectrum between 500–570 nm which is also the wavelengths where the Sun emits the peak of its in- tensity in the E/M spectrum (natural selection at work! ).

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Example IX–3. The eye is most sensitive to light of wavelength

5. 50 × 10 −^7 m, which is in the gree-yellow region of the visible elec- tromagnetic spectrum. What is the frequency of this light? Solution: We just need to use Eq. (IX-4):

ν = c λ

=^3.^00 ×^10

(^8) m/s

5. 50 × 10 −^7 m = 5. 45 × 1014 Hz.

C. The Formation of Spectra and the Doppler Effect of Light.

1. In the 1860s, Kirchhoff realized that there are 3 types of spectra that objects emit which depend upon the state and orientation the object is in =⇒ Kirchhoff’s Laws: a) Law 1: A luminous opaque object (solid, liquid, or gas) emits light at all wavelengths (E/M spectrum), thus pro- ducing a continuous spectrum — a complete rainbow of colors without any spectral lines.

λ

I

Continuous Spectrum

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d) Kirchhoff’s laws can be summarized with the following cartoon:

Star (ContinuumSource)

CloudGas Absorption Spectrum Seen

Emission Spectrum Seen Continuous Spectrum

Seen

Kirchhoff’s Radiation Laws

2. The Doppler effect for E/M Radiation: a) The spectrum of an object will be blueshifted if it is approaching the observer.

b) The spectrum of an object will be redshifted if it is receding from the observer.

c) The wavelength shift in a spectral line is given by: ∆λ λ◦ = vr c

, (IX-9)

where ∆λ = λ − λ◦ (negative shift = blueshift), λ◦ = rest (lab) wavelength, vr = radial (i.e., line-of-sight) velocity of the object, and c = speed of light.

Example IX–4. We observe a hydrogen spectral line of Polaris (the North Star) with a wavelength of 6562.48 ˚A, which in the lab- oratory is measured to be at 6562.85 ˚A. What is the radial velocity of Polaris?

Donald G. Luttermoser, ETSU IX–

Solution: We are given λ = 6562.48 ˚A and λ◦ = 6562.85 ˚A, so ∆λ = 6562.48 ˚A − 6562 .85 ˚A = − 0 .37 ˚A.

v = ∆λ λ◦ c =

− 0 .37 ˚A

6562 .85 ˚A

3. 00 × 108 m/s = (− 5. 638 × 10 −^5 ) (3. 00 × 108 m/s) = − 1. 69 × 104 m/s = − 16 .9 km/s.

Polaris is moving towards us (negative sign and the line was blueshifted) at 16.9 km/s.

D. Blackbody Radiation.

1. Late in the 1800s and in the early part of the 20th century, Boltz- mann, Planck, and others investigated E/M radiation that was given off by hot objects. a) In general, matter can absorb some radiation (i.e., pho- tons converted to thermal energy), reflect some, and trans- mit some of the energy.

b) The color of cool objects, objects that don’t emit their own visible light, is dictated by the wavelengths of light they either reflect, absorb, or transmit. i) A blue sweater is “blue” because the material reflects blue light (from either room lights or the Sun) more effectively than the other colors of the rainbow.

ii) Coal is black because it absorbs visible light and reflects very little.

Donald G. Luttermoser, ETSU IX–

4. The total brightness, or luminosity (L), of a blackbody is just the flux integrated over all of the surface of the object. For a spherical object, the surface area is 4π R^2 , where R is the radius of the spherical blackbody, so the luminosity is

L = 4π R^2 F = 4π σ R^2 T 4. (IX-11)

Note that if we treat stars as blackbodies, we can eliminate the constants in the above equation by dividing both sides by solar values: L L = 4 π σ R

2 T 4

4 π σ R^2 T 4 L L

( R

R

) 2 ( T

T

) 4

. (IX-12)

5. In 1893, Wien discovered a simple relationship between T of a blackbody and the wavelength where the maximum amount of light is emitted =⇒ Wien’s displacement law (usually called Wien’s law for short):

λmax =

2. 897 × 10 −^3 m · K T ,^ (IX-13) or in the other wavelength units we have discussed:

λmax = 0 .2897 cm · K T = 2.^897 ×^10

(^6) nm · K T = 2.^897 ×^10

7 ˚A · K

T

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6. Blackbody radiators, being in thermal equilibrium, emit contin- uous spectra that are called Planck curves:

λ

Flux

λ

λ max

Planck Curve

Example IX–5. A star has a temperature of 10,000 K and a radius of 20 R , what is its energy flux and wavelength of maximum flux? What is its luminosity with respect to the Sun ( )? (Note that R = 6. 96 × 108 m and T = 5800 K.) Solution: For this problem, we will assume that stars are blackbody ra- diators. We then use the Stefan-Boltzmann law (Eq. IX-10) to determine the energy flux of this star:

F = (5. 67 × 10 −^8 W m−^2 K−^4 ) (10, 000 K)^4 = (5. 67 × 10 −^8 W m−^2 K−^4 ) (10^4 K)^4 = (5. 67 × 10 −^8 W m−^2 K−^4 ) (10^16 K^4 ) = 5. 67 × 108 W m−^2.

The wavelength of maximum brightness is given by Wien’s law