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7/29/2019 emw_engphys

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Introduction to Light, Atoms, the E/M Spectrum(all uncreditied figures courtesy of Thomas Arny)

The Electromagnetic Spectrum/Electromagnetic Waves

E/M waves are disturbances that travel through space E/M waves are transverse, and harmonic E/M waves have particle and wave properties E/M waves travel at 3 108 m/s in free space, c E/M waves transport energy (electromagnetic energy) fc =

chhfE ==

decreasing wavelengths increasing energy longer wavelengths lower frequency E/M waves reflect, refract, scatter, diffract, interfere, and may be polarized


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Origins of E/M Radiation

Visible, ultraviolet and infrared light originates from electronic transitions in atoms. Gamma

rays originate from similar events in the nuclei of atoms. X rays may form in any of severalways but most commonly from the rapid acceleration of atoms. At the other end of the

spectrum, microwaves may originate from molecular transitions, radio waves result from the

oscillations of large numbers of charged particles.

E/M Radiation from Space and the Transparence of Earths Atmosphere


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E/M Radiation and Temperature

All objects emit thermal radiation. An analysis of thermal radiation shows that it actually consists of a broad spectrum

of e/m waves that have a peak intensity related to the temperature of the object.The temperature of this page, for example, is about 20oC and its emission spectra

peaks in the infrared (long wavelength) region of the spectrum at around 0.7m. The human eye cannot detect infrared radiation, so the light that one sees coming

from this page is reflected light from other sources.

Certain objects are both very good absorbers and very good emitters of allradiation in the form of e/m waves. Such objects are called blackbodies.

A blackbody is any object that is 100% efficient at absorbing all of theelectromagnetic radiation that falls on it. Because such an object reflects no light

it appears to be black as long as it is cool.

When blackbodies get hot, either through absorbing radiation or other externalmeans they are also very efficient radiators of energy.

Most blackbodies are high-density materials such as solids. There are very few perfect blackbodies, but many objects are close enough that we

may assume that they are essentially blackbody radiators. Wiens Lawrelates temperature to color for any blackbody or near blackbody When an object is a selective absorber and emitter of electromagnetic radiation it

obeys Kirchoffs laws. Wiens Displacement Law relates the temperature of a blackbody to the wavelength

of maximum emitted intensity: (max)T = 0.2898 10-2 mK , where max is the

wavelength at which maximum emission occurs. When all objects are heated to any

temperature above absolute zerothey emit a broad spectrum of

wavelengths but emit most strongly

in a narrower region of wavelengths

closely associated with a particulartemperature.

In blackbodies the relationshipbetween intensity, temperature and

color obeys Wiens law. Hotterobjects emit more strongly in the

shorter wavelength region.

Why do very hot objects emit whitelight, instead of blue or ultraviolet?


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For the Sun max = 0.2898 10-2 mK/6000K, which is roughly equal to 0.520m or

520nm. This indicates that the suns radiant energy peaks in the yellow-greenportion of the visible spectrum.

The earths surface temperature averages 300K (810F). In this case max = 0.2898 10-2 mK/300K, which is roughly equal to 10m, 100nm, or 10,000 angstroms. This

corresponds to emission in the infrared part of the spectrum. An alternative mathematical

expression of Wiens law is:

max

6103

=KT

where max is in nanometers (10-9).

For our sun:K

nmTK 5800

520

1036

=

=

Luminosity

Luminosity is energy radiated per second, i.e., power. Expressed in Joules/sec or watts.

The suns luminosity is about 4 1026 watts

The Inverse Square Relationship

24 r

L

A

PI

==


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The Stefan-Boltzman Law

Stefan-Boltzman relates luminosity to temperature.

4TE =, where E= L/A, = 5.67 10

-8

W m

-2

K

-4

(the Stefan-Boltzmanconstant).

More conveniently written:244 RTL = for a spherical isotropic radiator.

Not universally applicable works with blackbody radiation Note that a small increase in temperature results in a large increase in E.

Applying Stefan-Boltzmann Law to the Sun:E = (5800K)4

= () 1.132 x 1015 K4

= (5.67 x 10-8W m-2 K-4)(1.1 x 1015K4)= 6.4 107 W m-2

This is a whopping amount of energy when the entire surface area of the sun is taken into

account (about 4 1026 watts).

Example If the sun radiates 4 1026 watts, and we are 150 million kilometers awaywhat is the intensity of the radiation from the sun at the top of the earths atmosphere?

The inverse square law for a spherical isotropic radiator:( )29

26

2101504

104

4 m

watts

r

L

A

PI

===

,

yielding a little over 1400 watts per square meter. Most of this is visible light. About 1000

watts per square meter reaches the surface of the earth in equatorial latitudes.


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Spectroscopy

Spectral analysis of light emitting objects yields information about elemental composition,

relative abundance of elements, temperature, relative velocity and other information by

splitting the light given off by a glowing object into its component colors. Each of these colors

is indicative of a specific atomic transition. All elements have a unique spectral fingerprint

There are three types of spectra: bright lineor emission line spectra, continuous spectra,

dark line or absorption line spectra. Emission spectra are produced by low-density

gasses that radiate energy at specific

wavelengths characteristic of the element or

elements that make up the gas. The spectrumconsists of a number of bright lines against a

dark background.

Continuous spectra are produced by solids,liquids or dense gases. The spectrum appearsas a smooth transition of all colors in the

visible spectrum from the shortest or the longest wavelength without any gaps

between the colors. Blackbodies emit continuous spectra. Absorption spectra are produced when a cooler gas absorbs specific wavelengths of

light passing through it. The wavelengths absorbed are determined by the elements

that compose the gas. Since no two elements absorb the exact same wavelengths, it is

possible to determine the elemental composition of the gas by examining the spectra.

A dark line or absorption spectrum appears as a continuous spectrum of all colors witha number of dark lines through it. The K and H lines in the solar spectrum, for

instance, are due to ionized calcium in the outer layers of the sun's atmosphere. If the

dark lines are closely spaced in some parts the clumps of dark lines are known as bands.


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Spectrometers may use either prisms or diffraction gratings to separate light into its

component colors.


8/17


9/17

Doppler Shift

Any relative motion between an observer and a source of light or any form of e/m radiation

results in a Doppler Shift, i.e., a shifting of spectral lines toward either shorter or longerwavelengths. Objects moving towards an observer undergo a blue shift and objects moving

away from an observer undergo a red shift.


10/17

E0

B0

k

Electromagnetic Waves and Maxwells Equations

Maxwells Equations describe all classical electromagnetic phenomena in much the same

manner as Newtons Laws describe all classical mechanical phenomena

0

encqAdE =vr

Gauss Law

0= AdBvv

Gauss Law for Magnetism

dt

dsdE m

=

v

r

Faradays Law

dt

dIsdB Eenc

+= 000

v

r

Amperes Law

BqvEqFvvv

+= The Lorentz Force Equation

For harmonic transverse waves traveling to the right along the x-axis of an arbitrary

coordinate system (assuming that y= 0 when x= 0 and t= 0) the displacement of the

medium with respect to time may be expressed as: ( )tkxAy = sin

An equivalent form of these equations is ( )tkx= sin0 where a + sign indicates thewave is moving to the left. In general the displacement of the medium is represented by a

general coordinate,, so that: ( )tkx= sin0 where:

2=k

r

, f2= ,T

f1

= ,

kfv ==

for e/m waves is the amplitude of the Eor Bfield E, Band kare all perpendicular

( )

)sin(

sin

0

0

tkxBB

tkxEE

=

=rr

rr

where E0 and B0 are maximum amplitudes

cBE = in free space 00 vBE = in general


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The Energy in E/M waves

The energy transported in an e/m wave is stored in the Eand Bfields.

The respective energy densities:2

02

1

EuE = ,2

02

1

BuB =

( ) EBEB uuuEc

Eu ===

= 2

0

00

2

0 22

1

, hence the energy of an e/m wave is

equally divided between the electric and magnetic fields

Total energy 2

0

2

0

122 BEuuuuu BEBE

====+=

Power carried by E/M waves The Poynting Vector

Consider a cube with sides of areaA

that encloses some flux of e/m energy in free space. Theenergy in this volume of space is:

E= energy density volume = ( )AcdtE20

Note that power is energy per unit time:( ) ( )AcE

dt

AcdtEP 2

0

2

0

==

Define: BEcBEccEcBcEArea

PowerS

rrrrv

===== 202

00

2

0

More commonly The Poynting Vector is written BESvrr

0

1

.Note that the Poynting

vector is a measure of the intensityof an e/m wave.

cdt


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E0

B0

S

The Eand B,vectors are perpendicular to each other an the Poynting Vector points in thedirection of the wave travel.

( )tkxcEcES == 22002

0 sinr

Soscillates at twice the frequency of the wave and is always positive

Define irradianceEe the time average valueof the absolute value of the Poynting vector.

( )tkxcEsEe ==22

00 sinr

To find the time average value of tf 2sin= over one full cycle of the function:

2

1sin

2

0

2

0

2

==

=

=

=

=

t

t

t

t

dt

tdt

f

So: 200

2

0

00

0

2

0

3

0000

2

002

11

2

1

2

1

2

1

2

1B

cBcBcccBcBcEEe

=====

t

S


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The Speed of E/M waves

In free space the speed of an e/m wave is00

1

== fc ,were 0 is the permittivity of free

space (electricity) and 0 is the permeability of free space (magnetism).

In matter

1

== fv .

Derivation of the speed of light in free space from Maxwells equations

Consider an e/m wavefront moving through space as shown above. As the wavefront moves

through space the Evector sweeps out area svdt, wheresis length of the Evector

E

B

vdt

s E

vdt


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This area swept out by the Efield experiences a change in magnetic flux. Note that there is a

simultaneous Bfield present with field lines that penetrate this changing area. By applyingFaradays law to this we may determine the relationship between Eand B.

Summing up the contributions of Esfrom each segment of the loop, ccl:1. Es2. 0

3. 04. 0

So from Faradays Law: BvEsvBdt

dsE m

vrrr

==

=

By symmetry the Bfield sweeps out an area in the same time that is penetrated by electric

flux from the Efield. We may therefore apply Amperes law to determine the relationshipbetween Eand Bhere.

dt

dIsdB Eenc

+=

000

v

r

Amperes Law

B

vdt

E

1

2

3

4

1 3

2

4

dt

dsdE m

=

v

r

Faradays Law


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Once again, ccl around the loop:

1. Bs

2. 0

3. 0

4. 0

So from Amperes Law:

Evsdt

dsB E 0000 =

=

v

r

vEBrr

00=

0000

11

v

BEB

vE

r

vvr

==

If Maxwells equations are valid, then both Amperes and Faradays laws must be valid andboth expressions must be true.

Hence:

( )( )

18

2721212

00

2

00

10998.2

1041085.8

11

=

==

==

sm

ANmNC

vv

v

BBvE

r

vr

The quantity c0 is known as the impedance of free space, has the SI unit of ohms and the

value:

== 3770

0

0

c


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Radiation Pressure

E/M waves transport momentum as well as energy. This means that when an e/m waveimpinges on a surface it exerts pressure on that surface.

Ifp, U, and crepresent linear momentum, energy and wave speed respectively:

c

Up= for an absorbing surface

c

Up

2= for a reflecting surface

Note that the first expression applies strictly only to blackbodies. In terms of the Poyntingvector (Pis the radiation pressure):

c

SP = for an absorbing surface

c

SP

2= for a reflecting surface

Radiation pressures are small for sunlight on earth (about 26105 mN in direct sunlight).

Example The average intensity of light (power per unit area or the value of the Poynting

vector) from the sun on the earths surface worldwide is about 340 watts per square meter(the maximum amount is about 1000 watts per square meter near the equator). Calculate the

power delivered to a solar panel with dimensions of 8 15 meters (about the size of the roofof a house), assuming complete conversion.

( )( ) kWmmWSAP 840158340 22 .===

In practice most solar panels are about 15% efficient for a yield of about 6 kW. This is good

for about a 50 ampere service which is less than that the 100 200 amp service required by

most homes.

What is the radiation pressure on the roof? Assuming complete absorption:

26

18

2

1031103

340

=

== mN

sm

mW

c

SP .

A small fraction of the normal load any roof supports.


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Computing the distances to stars

There are basically two methods of computing distances to stars: theparallax methodand themethod of standard candles.

The parallax method uses trigonometry to compute the distance to a star based on ameasuring a parallax angle formed due to the apparent shift of a stars position againstthe background of stars as the earth orbits the sun. This method works with stars up

to 250 parsecs (800 ly) distant. The method of standard candles uses spectral data and Wiens law to determine a

stars surface temperature, Stefan-Boltzman to determine its actual luminosity, andthe inverse square law to determine the distance from its apparent luminosity. This

method works well for stars greater than 250 parsecs (800 ly) distant.

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