chapter 09 - electrodynamics · p. j. grandinetti chapter 09: electrodynamics maxwell’s discovery...

Chapter 09Electrodynamics

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 09: Electrodynamics

James Clerk Maxwell1831-1879

Since Maxwell’s time, physical reality has beenthought of as represented by continuous fields, andnot capable of any mechanical interpretation. This

change in the conception of reality is the mostprofound and the most fruitful that physics has

experienced since the time of Newton– Albert Einstein


Imagine a two-dimensional velocity vector field: v(r)

13.5°N

17.5°N

21.5°N

25.5°N

29.5°N

33.5°N

102.5°W 94.5°W 86.5°W 78.5°W

latit

ude

longitude


Imagine a three-dimensional velocity vector field: v(r)


Vector Calculus


Divergence of a vector field is a scalar field

DefinitionDivergence of a vector field, A(r), is a scalar field

∇ ⋅ A(r) =𝜕Ax(r)𝜕x

+𝜕Ay(r)𝜕y

+𝜕Az(r)𝜕z

, Recall ∇ = ex𝜕𝜕x

+ ey𝜕𝜕y

+ ez𝜕𝜕z

+

Source

Positive divergence Zero divergence

–

Sink

Negative divergence

Positive divergence means field region is a source. Negative divergence means field region is a sink.


Curl of a vector field is a vector field

DefinitionCurl of vector field is a vector field and given by determinant

∇ × A(r) =

||||||||||||

ex ey ez

𝜕𝜕x

𝜕𝜕y

𝜕𝜕z

Ax(r) Ay(r) Az(r)

||||||||||||Curl gives circulation per unit area.

Curl right hand fingers in direction of circulation then thumb points in direction of curl vector.


Wind Velocity Vector Field of Summer 2016

latit

ude

4°S

11°N

26°N

20 40 60 80 100 120

Wind velocity at 10 m / (km/h)

96°W 81°W 66°W 51°W 36°W 21°Wlongitude

Where are regions of high and low curl in velocity vector field?P. J. Grandinetti Chapter 09: Electrodynamics

Fluid velocity vector field above the bathtub drain: v(r)

In this region...is the divergence of the water velocity vector field positive or negative, ∇ ⋅ v(r)?which direction is the curl of the water velocity vector field, ∇ × v(r)?


Maxwell’s Equations

In the 1860s James Clerk Maxwell unified fields of electricity, magnetism, and optics with setof equation summarizing relationships between electric vector field, (r, t), and magneticvector field, B(r, t) with given charge density, 𝜌(r, t), and current density, J(r, t):

∇ ⋅ (r, t) = 𝜌(r, t)∕𝜖0, (A) ∇ × (r, t) = −𝜕B(r, t)𝜕t

, (B)

∇ ⋅ B(r, t) = 0, (C) ∇ × B(r, t) = 𝜇0J(r, t) + 𝜇0𝜖0𝜕(r, t)

𝜕t. (D)

𝜇0 is magnetic constant or permeability of free space,𝜖0 is electric constant or permittivity of free space.


Maxwell’s Equations∇ ⋅ (r, t) = 𝜌(r, t)∕𝜖0 is Gauss’ law. You can derive it from Coulomb’s law.

▶ positive divergence from positive charge, negative divergence from negative charge.

∇ × (r, t) = −𝜕B(r, t)𝜕t

is the differential form of Faraday’s law of induction.▶ electric field curls around changing magnetic field vector.

∇ ⋅ B(r, t) = 0 is the differential form of Ampere’s law.▶ no magnetic monopoles – no magnetic sources or sinks.

∇ × B(r, t) = 𝜇0J(r, t)⏟⏟⏟

Ampere’s law

+ 𝜇0𝜖0𝜕(r, t)

𝜕t⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟

displacement current

is Ampere-Maxwell’s law.

▶ Magnetic field curls around flowing charge direction.▶ Displacement current term discovered by Maxwell in attempts to make equations consistent.▶ Maxwell’s discovery of displacement current term was essential in leading him to wave

solutions which he identified as light waves.P. J. Grandinetti Chapter 09: Electrodynamics

Electromagnetic Vector Wave Equation


Maxwell’s Equations in empty space

In empty space where no charges or currents are present: 𝜌(r, t) = 0 and J(r, t) = 0.

Maxwell’s equations become

∇ ⋅ (r, t) = 0, (A) ∇ × (r, t) = −𝜕B(r, t)𝜕t

, (B)

∇ ⋅ B(r, t) = 0, (C) ∇ × B(r, t) = 𝜇0𝜖0𝜕(r, t)

𝜕t. (D)

This set of coupled first-order differential equations can be decoupled.


Maxwell’s discovery that light is an electromagnetic waveElectric field vector wave equation1st, take curl of Faraday’s law and use vector calculus identity to get

∇ × (∇ × ) = ∇(∇ ⋅ ) − ∇2 = ∇ ×(−𝜕B𝜕t

)

∇ × (∇ × ) = ��:0∇(∇ ⋅ ) − ∇2 = ∇ ×

(−𝜕B𝜕t

)2nd, use Gauss’ law to eliminate 1st term in middle expression and get

−∇2 = ∇ ×(−𝜕B𝜕t

)3rd, use Ampere-Maxwell law to get a vector wave equation for (r, t):

∇2(r, t) = 𝜇0𝜖0𝜕2(r, t)

𝜕t2


Maxwell’s discovery that light is an electromagnetic wave

∇2(r, t) = 𝜇0𝜖0𝜕2(r, t)

𝜕t2

Similarly one can derive a vector wave equation for B(r, t):

∇2B(r, t) = 𝜇0𝜖0𝜕2B(r, t)

𝜕t2

We now have 2 decoupled second-order partial differential vector equations (i.e. 6 wave equations).

Each vector component of and B satisfy classical wave equation with 1∕v2p = 𝜇0𝜖0 or

vp = 1√𝜇0𝜖0

= c0,

c0 is speed of light in vacuum.


Maxwell’s discovery that light is an electromagnetic waveFor Maxwell’s equations to be satisfied at all r and t we need k and 𝜔 to be the same forall components: x, y , and z, and Bx, By, and Bz.Consider plane wave (harmonic) solutions

(r, t) = 0ei(k⋅r−𝜔t), and B(r, t) = B0ei(k⋅r−𝜔t)

Both and B depend on k ⋅ r. This defines a plane of constant phase that isperpendicular to k, direction of propagation.Plugging these solutions into wave equations lead to

𝜔2 =|k|2𝜇0𝜖0

or 𝜔 = c0|k|Maxwell’s equations further restrict and B to be perpendicular to k.To prove this substitute (r, t) and B(r, t) above into ∇ ⋅ (r, t) = 0 and ∇ ⋅ B(r, t) = 0.


Electromagnetic wavesAlso ∇ × (r, t) = −𝜕B(r, t)∕𝜕t leads to k × (r, t) = 𝜔B(r, t) so must be ⟂ to B.Cross product also tells us |k||| = k = 𝜔|B| = 𝜔B

Taken together our E&M harmonic wave is

x

y

z

= c0B for harmonic plane wave

(r, t) = 0 cos(kz − 𝜔t) ex

B(r, t) =0c0

cos(kz − 𝜔t) ey


Discovery of Radio WavesBetween 1886 and 1889 Heinrich Hertz experimentally proved that electromagnetic wavesmove at the speed of light, and in the process he discovered radio- and micro-waves and howto generate and detect them.


Classification of electromagnetic spectrum.Classification Frequency Range/Hz Wavelength Range/m

Extremely low frequency 3 - 30 Hz 100 - 10 MmSuper low frequency 30 - 300 Hz 10 - 1 MmUltra low frequency 300 - 3000 Hz 1 - 0.1 MmVery low frequency 3 - 30 kHz 100 - 10 km

Radio Low frequency 30 - 300 kHz 10 - 1 kmWaves Medium frequency 300 - 3000 kHz 1 - 0.1 km

High frequency 3 - 30 MHz 100 - 10 mand Very high frequency 30 - 3 MHz 10 - 1 m

Ultra high frequency 300 - 3000 MHz 1 - 0.1 mMicro- Super high frequency 3 - 30 GHz 100 - 10 mmWaves Extremely high frequency 30 - 300 GHz 10 - 1 mm

Far Infrared 300 - 3000 GHz 1 - 0.1 mmMid Infrared 3 - 30 THz 100 - 10 𝜇m

Visible Near Infrared 30 - 300 THz 10 - 1 𝜇mNear ultraviolet 300 - 3000 THz 1 - 0.1 𝜇m

Extreme ultraviolet 3 - 30 PHz 100 - 10 nmIonizing Soft X-rays 30 - 300 PHz 10 - 1 nm

Radiation Soft X-rays 300 - 3000 PHz 1 - 0.1 nmHard X-rays 3 - 30 EHz 100 - 10 pmGamma Rays 30 - 300 EHz 10 - 1 pm


Polarization


PolarizationDefinitionOrientation of transverse vector wave displacement relative to direction of propagation iscalled wave polarization.

x

y

z

Example of linear polarization wheretransverse oscillations remain in singleplane.

(r, t) = 0 cos(kz − 𝜔t) ex

B(r, t) =0c0

cos(kz − 𝜔t) ey


Circular PolarizationCircular polarization of light is where field vector moves counter clockwise in x-y plane as itpropagates forward along z.

(r, t) = 0 cos(kz − 𝜔t) ex + 0 sin(kz − 𝜔t) ey

Similar behavior for B field vector, as long as it remains ⟂ to at all times.P. J. Grandinetti Chapter 09: Electrodynamics

Light PolarizationMost light sources—sun, burning candle, incandescent and fluorescent light bulbs—produceunpolarized light consisting of vector waves with random polarizations. These sources alsoproduce incoherent light waves with a variety of frequencies and phases.

Light waves can become polarized orpartially polarized by interacting withmatter.Sunlight that reaches earth becomespartially polarized due to scattering frommolecules in atmosphere. See lightintensity in the sky change when tiltinghead while wearing polaroid sun glasses.


Electromagnetic waves in dispersive media


Electromagnetic waves in dispersive mediaFor light waves are moving in dielectric instead of vacuum replace 𝜖0 and 𝜇0 with 𝜖 and 𝜇𝜖 and 𝜇 are electric permittivity and magnetic permeability of medium substance.Unlike in vacuum, 𝜖 and 𝜇 depend on wavelength in a dielectric.

vp(k) =√𝜇(k)𝜖(k)

In dielectrics speed of light is reduced to

c(k) =c0

n(k)where n(k) =

√𝜇(k)𝜖(k)𝜇0𝜖0

This also means light disperses in a dielectric. Dispersion relationship is not linear.

𝜔(k) =c0kn(k)

Spectroscopy is devoted to understanding and exploiting this relationship.P. J. Grandinetti Chapter 09: Electrodynamics

Electromagnetic Field Energy


Electromagnetic Field EnergyDefinitionEnergy density of electromagnetic field at any instant and point in space is

u(r, t) = 12𝜖02(r, t) + 1

21𝜇0

B2(r, t)

2(r, t) = |(r, t)|2 and B2(r, t) = |B(r, t)|2For monochromatic plane wave, where E = c0B, we find

u(r, t) = 𝜖02(r, t)

Substituting (r, t) = 0 cos(k ⋅ r − 𝜔t) gives

u(r, t) = 𝜖020 cos2(k ⋅ r − 𝜔t)

Averaging over one complex cycle k ⋅ r − 𝜔t = 0 → 2𝜋 gives the average energy density of amonochromatic plane wave ⟨u⟩ = 1

2𝜖02

0


Poynting vectorDefinitionEnergy flux of electromagnetic wave is along k and given by :

Poynting vector S(r, t) = 1𝜇0

(r, t) × B(r, t)

S is energy per unit area per unit time, or power per unit area.

Averaged over a complete cycle for a monochromatic plane wave we find

⟨S⟩ = 12

c0𝜖020 ek = c0⟨u⟩ek

Light (or optical) intensity, I, for a monochromatic plane wave is defined as the magnitude ofthe Poynting vector averaged over a cycle,

I = 12

c0𝜖020 = c0⟨u⟩


Radiation from Oscillating Dipoles


Radiation from Oscillating DipolesHow are light waves generated? One example: Oscillating electric dipole

𝜇(t) = 𝜇d cos(𝜔t)ez

produces (r, t) and B(r, t) at large distances away, r ≫ c0∕𝜔, given by

(r, t) = −(𝜇0

4𝜋

)𝜇d𝜔

2( sin 𝜃

r

)cos

[𝜔(t − r∕c0)

]e𝜃

B(r, t) = −(𝜇0

4𝜋

) 𝜇d𝜔2

c0

( sin 𝜃r

)cos

[𝜔(t − r∕c0)

]e𝜙

z

y

x

Spherical coordinate unit vectorsMonochromatic spherical waves radiating from oscillating dipole.Amplitude decreases as 1∕r.When spherical waves are far from dipole source we can approximate them as plane waves.


Radiation from Oscillating DipolesWe can get radiated energy from Poynting vector

S(r, t) = 1𝜇0

((r, t) × B(r, t))=

𝜇0c0

{𝜇d𝜔2

4𝜋

(sin 𝜃r

)cos

[𝜔(t − r∕c0)

]}2

er

If we averaged S(r, t) over one dipole oscillator cycle we get

⟨S(r)⟩ = 𝜇04𝜋

(𝜇2

d𝜔4

8𝜋c0

)sin2 𝜃

r2 er

Note: no radiation along the long axis of the dipole, where sin 𝜃 = 0. Intensity profile lookslike a donut.Total power radiated is found by integrating ⟨S(r)⟩ over sphere of radius r.

⟨P⟩ = 𝜇04𝜋

(𝜇2

d𝜔4

3c0

)Electric Dipole Power Radiated


Radiation from Oscillating Dipoles

Source: Student Guide to Waves, Fleisch, Kinnaman


Web App Demos

3D Waves


http://www.falstad.com/wavebox/fullscreen.html

chapter 09 - electrodynamics · p. j. grandinetti chapter 09: electrodynamics maxwell’s discovery...

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