Introductory Physics Part Three Modern...

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Introductory PhysicsPart Three

Modern Physics

Saint Mary’s CollegeChris RayFall 2019

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Contents

1 Oscillations and Waves1 Oscillation as a Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Complex Representation of an Oscillation . . . . . . . . . . . . .63 Complex Representation of an Oscillator . . . . . . . . . . . . 104 Complex Representation of a Periodic Wave . . . . . . . . . 115 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 Special Relativity1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 Galilean Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .194 Lorentz Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 Intervals in Special Relativity . . . . . . . . . . . . . . . . . . . . . . . 226 Causality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 Velocity Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .258 Energy and Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 Energy and Momentum Transformation . . . . . . . . . . . . . 30

10 Invariant Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3111 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3412 More Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3 Nuclear Physics1 Nuclear Activity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392 An Individual Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403 Decay Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414 Attenuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .42

4 Light - Waves as particles1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 Review of Some Properties of Light . . . . . . . . . . . . . . . . . .453 Blackbody Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .464 Photoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 Photon Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516 Light in terms of it’s electric fields . . . . . . . . . . . . . . . . . . .537 Taylor’s Double Slit Experiment . . . . . . . . . . . . . . . . . . . . .538 The Meaning of the Wave Function . . . . . . . . . . . . . . . . . .579 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

5 Electrons - Particles as waves1 Line Spectra of Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632 The development of a model of the atom . . . . . . . . . . . . 633 Electron Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

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4 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706 Quantum Theory

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 712 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .713 Inner product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734 Expectation Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .745 Quantum Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766 Photon polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

7 Quantum time evolution1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 832 The Hamiltonian operator . . . . . . . . . . . . . . . . . . . . . . . . . . .833 Time Evolution of the Wave Function . . . . . . . . . . . . . . . 834 Stationary States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .865 General time dependent states . . . . . . . . . . . . . . . . . . . . . . .906 Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .907 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 938 More Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

8 Quantum Applications1 Unbound States - Barrier penetration . . . . . . . . . . . . . . . 952 Angular Momentum and Spin . . . . . . . . . . . . . . . . . . . . . . . 983 Multi-particle States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1014 Chapter Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

9 Classical Thermodynamics1 Kinetic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1072 Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1123 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1124 Latent Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1145 Work and Internal Energy . . . . . . . . . . . . . . . . . . . . . . . . . .1166 Thermodynamic Processes with an Ideal Gas . . . . . . . 1197 Heat Engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1238 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1269 Methods of Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . 126

10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13210 Statistical Mechanics

1 Microstates and Macrostates . . . . . . . . . . . . . . . . . . . . . . . 1332 Maxwell-Boltzmann Distribution . . . . . . . . . . . . . . . . . . . 1363 Molecular speeds in an ideal gas . . . . . . . . . . . . . . . . . . . .1404 Average Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1425 Separable Contributions to the Energy . . . . . . . . . . . . . 1456 Temperature Dependence of Specific Heat . . . . . . . . . . 1457 Specific Heat and the Partition Function . . . . . . . . . . . 149

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8 Indistinguishable Particles . . . . . . . . . . . . . . . . . . . . . . . . . .1509 Derivation of the Fermi-Dirac Distribution . . . . . . . . . 154

10 Bose-Einstein Condensation . . . . . . . . . . . . . . . . . . . . . . . . 15611 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15812 Summary: Statistical Mechanics . . . . . . . . . . . . . . . . . . . .159

A HintsB Derivations

1 Compton Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .166C Useful Mathematics

1 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1682 Fundamental Derivatives and Integrals . . . . . . . . . . . . . 1683 More Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1684 Power Series Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

D Physical Constants and Data1 Fundamental Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

1.1 Oscillation as a Rotation 5

1 Oscillations and Waves§ 1.1 Oscillation as a Rotation

Suppose that you have a mass oscillating on the end of a spring.We know that the position of the mass is given as a function of time as

x(t) = a cosωt

where a is the amplitude of the oscillation and ω is the frequency inunits of radians per second, ω = 2πT , where T is the period of oscillation.

Now consider an object moving clockwise in a circle of radius awith an angular velocity of dθdt = −ω. The position of this object isgiven by

x(t) = a cos(−ωt) = a cos(ωt)y(t) = a sin(−ωt)

Thus we find that the horizontal component of a circular motion is thesame as the motion of an oscillator. We will find that complex numbersgive us a very simple way to represent circular motion and thus a simpleway to represent oscillations.

. Problem 1.1Below is shown an oscillator at four different points in it’s cycle. Drawthe corresponding location of a clockwise circular motion for each ofthe four points, and estimate the angle, φ = −ωt, for each. This angleis called the phase of the oscillator at that point in time.

x0 a-av

v

v

v

(a)

(b)

(c)

(d)

The representation x(t) = a cos(−ωt) and y(t) = a sin(−ωt) onlyworks if the phase at t = 0 is zero, that is if the motion starts atthe most positive value for x. We can however account for any initial

6 Oscillations and Waves 1.2

condition by adding the phase at t = 0, that is the following expressioncan represent the general oscillator.

x(t) = a cos(φ0 − ωt)y(t) = a sin(φ0 − ωt)

. Problem 1.2

Suppose that an oscillator is started with an initial position x0 withvelocity v0 at time t0. Find the amplitude a and phase φ0 that willallow the expression x(t) = a cos(φ0 − ωt) to represent the oscillation.

§ 1.2 Complex Representation of an Oscillation

It ends up that periodic systems (electrical circuits, mechanicaloscillators, quantum waves . . . ) are most easily represented by usingcomplex numbers, and doing so is common practice in most fields. Forthis reason we are now going to do a short review of complex numbersin order to remind you (or demonstrate to you for the first time) theproperties of complex numbers that will be useful to our understandingof quantum mechanics.

Definition of a Complex Number

Define a new number i that has the following property.

i2 = −1This number is called an imaginary number since the square of anyreal number is positive. A complex number z is defined to be a linearcombination of imaginary and real parts

z = x+ iy

The real numbers x and y in the expression above are called the realand imaginary parts of z

Re[z] = x

Im[z] = y

Complex Numbers as Vectors

Since a complex number has two components it is often useful tothink of a complex number as a vector in two dimensions, with the realand imaginary parts being the x and y components of the vector.

1.2 Complex Representation of an Oscillation 7

Re

Im

✓x

y

z

r

It is often useful to visualize algebraic operations in this way. Forexample, suppose a complex number is the sum of two other complexnumbers z = z1+z2 then algebraicly we can add the numbers as follows

z = (x1 + iy1) + (x2 + iy2)

= (x1 + x2) + i(y1 + y2)

So we find that complex numbers add like vectors, in that one simplyadds component by component. Graphically the addition is as follows.

Im

Re

z

z1

z2

Multiplication and Polar Coordinates

Let z = z1z2 then

z = (x1 + iy1)(x2 + iy2)

= (x1x2 − y1y2) + i(x1y1 + x2y1)so if x and y are the components of the product z then

x = x1x2 − y1y2y = x1y1 + x2y1

This is not very intuitive. But if you use polar coordinates to representthe complex numbers then the polar coordinates of the product arerelated to the polar coordinates of the elements of the product in asimple way

r = r1r2

θ = θ1 + θ2

as shown in the diagram below.


Re

Imz

✓2

✓1

r 1

r2

z1

z2✓2

✓ = ✓1 + ✓2

r=

r 1r 2

As a specific example suppose z = z1z2 with

z1 = 2 + 3i −→

{r1 =

√22 + 32 = 3.606

θ1 = arctan(3/2) = 0.983rad

z2 = 5 + i −→

{r2 =

√52 + 12 = 5.099

θ2 = arctan(1/5) = 0.197rad

z = (2 + 3i)(5 + i) = 10 + 2i+ 15i+ 3i2 = 10− 3 + 17i = 7 + 17i

Sor =

√72 + 172 = 18.385

θ = arctan(17/7) = 1.180rad

and we see that r = r1r2 = (3.606)(5.099) and θ = θ1 + θ2 = 0.983 +0.197. That the product of complex numbers works in this way can beproved using Euler’s Formula

Theorem: Euler’s Formula

eiθ = cos θ + i sin θ

The relationship between rectangular and polar coordinates is

x = r cos θ

y = r sin θ

1.2 Complex Representation of an Oscillation 9

So that

z = x+ iy = r cos θ + ir sin θ = r(cos θ + i sin θ) = reiθ

Theorem: Polar Form of a complex numberIf r and θ are the polar coordinates of a complex number z then

z = reiθ

Euler’s formula allows us to write out a complex number in polarcoordinates in a very compact notation. It is because of the simplicityof the algebraic form of a complex number in polar coordinates thatwe use complex numbers to represent periodic systems.

. Problem 1.3Use the polar representation of complex numbers prove that if z = z1z2then r = r1r2 and θ = θ1 + θ2.

. Problem 1.4Prove the follow two useful identities.

cos θ =eiθ + e−iθ

2

sin θ =eiθ − e−iθ

2i

Complex Conjugate

Definition: Complex ConjugateThe complex conjugate of z = x+ iy = reiθ is defined to be

z∗ = x− iy = re−iθ

Theorem: Conjugate Properties

Re[z] =z + z∗

2

Im[z] =z − z∗

2i

r2 = z∗z

. Problem 1.5Prove the above theorem.


. Problem 1.6

Let a = 1 + 5i, and b = 4− 3i.(a) Compute c = ab, the product of a and b.

(b) Write a, b, and c in polar form.

(c) Check to see that rc = rarb and θc = θa + θb

. Problem 1.7

Let z = 2 + 2i. Draw z as a vector in the complex plane. Now considerthe the complex number w = eiθz with θ = π/2.

(a) Draw w in the complex plane.

(b) How does it compare with z?

(c) Repeat for θ = π/4, θ = −π/4 and θ = π.(d) Now generalize. What does multiplying an arbitrary z by eiθ do?Prove your generalization.

. Problem 1.8

Use Euler’s formula to show the following identities

cos(a+b) = cos(a) cos(b)− sin(a) sin(b)

sin(a+b) = sin(a) cos(b) + cos(a) sin(b)

§ 1.3 Complex Representation of an Oscillator

The position of a general harmonic oscillator can be written as

x(t) = a cos(φ− ωt)Where the constant φ gives the phase of the oscillator at t = 0. If welet ξ(t) = aei(φ−ωt) then we can write x in terms of ξ.

x(t) = Re[ξ(t)]

since

Re[ξ(t)] = Re[aei(φ−ωt)

]= Re[a cos(φ− ωt) + ia sin(φ− ωt)]= a cos(φ− ωt)

The function ξ(t) is the complex representation of the function x(t).Note that the complex representation is simply exploiting our observa-tion that an oscillation is a component of a constant velocity rotation.It is good to keep this in mind.

. Problem 1.9

Below is shown four different oscillators at t = 0. Write out the complexrepresentation of each oscillator.

1.4 Complex Representation of a Periodic Wave 11

x0 a-a

v

v

v

(a)

(b)

(c)

(d)

§ 1.4 Complex Representation of a Periodic Wave

We have seen that it is possible to represent an oscillator as aconstant velocity rotation in the complex plane. Now we would liketo do the same for traveling waves. We know this will work, becauseat each point in space, the value of the field of the wave oscillates.For example, if you are standing in the water with your eyes closed,you would feel the water rising and falling as the waves in the waterpass by you. The depth of the water would oscillate. This is also trueat any other location in the water. Thus at each location, there isan oscillation. But the different locations do not all oscillate in sync.Each place will reach its maximum depth at its own time. Thus fora traveling wave, we find that the phase is different at each location.In addition the amplitude is sometimes different as well. So if we letΨ(x, t) be the displacement of the wave at the location x at time t then

Ψ(x, t) = a(x)ei(φ(x)−ωt)

• a(x) is the amplitude of the wave at the position x.• φ(x) is the phase of the wave at the position x.Specifying the amplitude and phase as a function of position completelydescribes the wave.


The following examplegives the phase for a planewave. In the figure to theright a traveling sine wave atfive different times is shown.All together they span an in-terval of time equal to onethird of a period of the wave.The “clocks” below each graphgive the phase of the wave atthat location. From this di-agram we can deduce φ(x).Looking at the first row ofclocks we see that as youmove to the right the phaseincreases linearly, φ = kx,at a rate of one quarter ofa turn (π2 ) every one quarterof a wavelength, or 2π everywavelength. Thus

φ(x) = kx =2π

λx

and so

Ψ(x, t) = a(x)ei(kx−ωt)

v

x

ωt=0π/6

ωt=1π/6

ωt=2π/6

ωt=3π/6

ωt=4π/6

Ψ

Ψ

Ψ

Ψ

Ψ

Definition: Wave NumberThe constant 2πλ is called the wave number.

k =2π

λ

Note that since 2πλ =2πv/f =

2πfv =

ωv that we may also define the

wave number ask =

ω

v

Theorem: Plane Wave

Ψ(x, t) = aei(kx−ωt)

1.4 Complex Representation of a Periodic Wave 13

By examining the phaseof the left moving wave pic-tured to the right, we can seethat

φ(x) = −2πλx

so that

k = −2πλ

=ω

v.

So we see that if the velocityof the wave is negative thenthe wave number is negativealso. The second form forthe wave number, k = ωv ,is good because it remindsus that the wave number hasthe same sign as the velocity.

v

ωt=0π/6

ωt=1π/6

ωt=2π/6

ωt=3π/6

ωt=4π/6

xΨ

Ψ

Ψ

Ψ

Ψ

Theorem: Addition of two wavesSuppose that ψ = ψ1 + ψ2 with ψ1 = a1e

i(φ1−ωt) and ψ2 =a2e

i(φ2−ωt). Then ψ can be written as ψ = aei(φ−ωt) with

a2 = a21 + a22 + 2a1a2 cos(φ2 − φ1)

. Problem 1.10

Prove the above theorem.

. Problem 1.11

A line of point scatterers are spaced a distance a apart. A beam of lightwith wavelength λ strikes the scatterers at an angle θi from the normalto this line of scatterers. A detector is placed at an angle θr from thesame normal. The width of the beam is very very large compared withthe spacing a. The width of the beam is very very small compared withthe distance form the scatterers to the detector or light source.


line of scatterers

✓i

✓r

source detector

(a) Show that the light scattered from all of the different scatterers isin phase at the detector if θr = θi.(b) Show that the light scattered is also in phase if aλ (sin θi − sin θr) isan integer.

. Problem 1.12A double slit is set up but one slit is narrower than the other. Lightwith a wavelength of 600nm is incident on the slits. The detector isat a location that is 10.000mm from slit A and 10.001mm from slit B.When just slit A is open, 900 particles of light are counted in a detectorper time interval at the observation screen. When just slit B is open,400 particles of light are counted in a detector per time interval at theobservation screen. We will see that the number of particles detectedis proportional to ψ∗ψ, assume that the number of particles detected isN = ψ∗ψ and use the “addition of two waves” theorem to answer thefollowing.(a) How many photons do you expect to count with both slits open?(b) Now suppose the wavelength of the light can be adjusted between200nm and 1000nm. Make a graph of the count versus the wavelength.

1.5 Chapter Summary 15

§ 1.5 Chapter SummaryComplex Variables:

i =√−1

z = x+ iy = reiθ

z∗ = x− iy = re−iθ

r2 = z∗z

Re[z] =z + z∗

2

Im[z] =z − z∗

2i

eiθ = cos θ + i sin θ

A sinusoidal traveling wave can be written as

ψ(x, t) = aei(kx−ωt)

with k = 2πλ and ω =2πT = 2πf .

If ψ = ψ1 +ψ2 with ψ1 = a1ei(φ1−ωt) and ψ2 = a2e

i(φ2−ωt) then ψcan be written as ψ = aei(φ−ωt) with

a2 = a21 + a22 + 2a1a2 cos(φ2 − φ1)

16 Special Relativity 2.5

2.1 Introduction 17

2 Special Relativity§ 2.1 Introduction

The laws of physics allow us to see order in the observations wemake of the physical world. For example, we can use Newton’s lawsto understand the trajectory x(t) of a stone. Measurements of thetrajectory are made in a coordinate system, x and t. This coordinatesystem is called a frame of reference or simply a frame. We have somefreedom in choosing our coordinate system. For example, suppose thatwe are measuring the trajectory of an apple as it falls from a tree, wecan choose the location of the origin of our coordinate system to beat the ground, or at the branch where the apple is hanging, or at thecenter of the earth, or at pluto. In some sense the choice of the origindoes not make any difference. Of course the trajectory function x(t)will be different, but other things, such as the time it takes to reach theground, will not be different. The laws of physics themselves (F = mafor example) will also remain the same regardless of our choice of origin.

You may recall that we are also free to choose a frame of referencethat is moving at a constant speed (an inertial frame) and the laws ofphysics will be the same in this inertial frame. So if you do experimentson a train that is traveling at a constant speed, the laws of physics youfind on the train will be the same as those on the ground.

By the beginning of the twentieth century it had been determined,through careful experimental work, that the speed of light is the samein any inertial frame. This was very surprising since light was knownto be an electromagnetic wave and all waves have a fixed speed relativeto the medium the wave travels through. As an example of what isexpected of waves traveling in a medium, imagine that you are in aboat being carried along by current in a smoothly flowing river. Alsoimagine that you are making waves. From your point of view thewaves move out from you in circles: the waves have the same speed inall directions. In contrast to this a person on the shore sees that thewaves moving upstream are moving more slowly than the waves thatare moving downstream. People worked very hard to measure a similarshift in the speed of light waves, but no variation in the speed was everfound. Apparently one of the laws of physics is that the speed of lightis the same in all inertial frames.


The constancy of the speed of light lead to a contradiction in thelaws of physics. The law for the addition of velocities in classical me-chanics does not allow for something to have the same speed in twoframes that are moving relative to each other, see the problem belowfor an example. Something had to give. Einstein developed specialrelativity in order to adapt our ideas of space and time to be consistentwith the observed fact that the speed of light is the same in all inertialframes.

. Problem 2.1Suppose that you throw a ball while riding a bicycle. You throw theball at a speed of 9ms relative to the bike. The bike is moving with aspeed of 7ms .(a) What is the speed of the ball relative to the ground?(b) What is the speed of the ball relative to the ground, if the ball isthrown backwards?(c) If you instead shine a flashlight (throw light at a speed of c) fromthe bike, what is the speed of the light relative to the ground?

§ 2.2 EventsSpecial relativity is most easily understood in terms of events. An

event, in special relativity, is anything that happens at a single pointin space at a single point in time. For example the explosion of afirecracker. The combination of a point in time (e.g. t = 5.34s) and apoint in space (e.g. x = 1.43m) is called a space-time point. We will

write this space-time point as

[5.34s1.43m

], which is a column vector in

the language of matrix algebra. We can now state the definition of anevent in special relativity.

Definition: EventAn event is something that occurs at a single space-time point.

If an event occurs at a space-time point p in one frame of reference,then that same event will occur at a different space-time point p′ in adifferent frame. There is a rule that tells you how to compute p′ fromp. This rule is the part of classical mechanics that Einstein had tochange in order to make physics consistent. In classical mechanics therule for this conversion is called the Galilean transformation. Beforegoing on to the relativisticly correct transformation we will review theGalilean transformation, so that the similarities and differences will bemore apparent.

2.3 Galilean Transformation 19

. Problem 2.2Is a concert an event? Is the stroke of midnight on new years eve anevent? Give an example of an event.

§ 2.3 Galilean TransformationIn order to understand the Galilean transformation, we will look

at a single event as seen from two different inertial frames. Let usconsider the specific event of a drop of water landing on the windshieldof a moving car, four seconds after the car passes the county line. Thecar is driving at a speed of 8ms .

To begin with we must define our two frames of reference. The firstinertial frame, S, is a frame of reference moving with the car, with theorigin of S at the rear bumper of the car. The second inertial frame,S′, is the frame of reference of the road. Let us pick the origin of theroad’s frame to be the county line. In both frames we choose the zeroof time to be when the rear bumper of the car crosses the county line.In this way the origins coincide at t = 0. We will always choose framesso that the origin’s coincide at t = 0.

Let us assume that the windshield of the car is 3 meters from thebumper. So we know that in the car’s frame of reference the eventoccurs at the position 3 meters, and at the time 4 seconds. Thus the

event occurs at the space-time point,

[tx

]=

[4s3m

]. We can now find

the location of this event in the road’s frame of reference, by using ourknowledge about the displacement of an object moving at a constantspeed, that is ∆x = v∆t. So we know that the bumper of the carwill have moved a distance vt = (8ms )(4s) from the county line in fourseconds. Thus the rain drop hits the windshield when the bumperis (8ms )(4s) meters from the county line, and so the rain drop hits thewindshield (being 3 meters further on than the bumper) is at a distanceof (8ms )(4s)+3m from the county line. So in the road’s reference frame[

t′

x′

]=

[4s

(8ms )(4s) + 3m

]=

[4s

35m

]In general this transformation can be stated as follows.

Definition: Galilean TransformationIf a frame S is moving with a velocity v relative to a frame S′ then

an event that occurs at the point

[tx

]in S will occur at the point[

t′

x′

]in S′ with

[ct′

x′

]=

[1 0v/c 1

] [ctx

], where c is some reference

velocity. The use of c is needed to make the matrix unitless.


For the raindrop example with c = 2ms , the computation goes as follows:[ct′

x′

]=

[1 0

(8ms )/(2ms ) 1

] [(2ms )(4s)

3m

]=

[1 04 1

] [8m3m

]=

[(1)(8m) + (0)(3m)(4)(8m) + (1)(3m)

]=

[8m35m

].

So that x′ = 35m and ct′ = 8m −→ t′ = (8m)/c = 4s.The matrix representation to do this problem might seem like

overkill (and it is) but bear with the process as it will help soon. Usethe matrix method to solve the following problems.

. Problem 2.3An event happens at the space-time point [2s , 5m] in the frame S.Frame S is moving at a speed of 7ms in the positive direction withrespect to S′ At what space-time point does the event happen in S′?

. Problem 2.4An event happens at the space-time point [2s , 5m] in the frame S.Frame S is moving at a speed of 7ms in the negative direction withrespect to S′ At what space-time point does the event happen in S′?

. Problem 2.5An object is moving with a constant velocity u in the frame S. Theobject is at position x1 at time t1 and is at position x2 at time t2 Theframe S is moving with a velocity v with respect to another frame S′.(a) Use the Galilean transformation to find the differences in S′, ∆x′ =x′2 − x′1 and ∆t′ = t′2 − t′1, in terms of the differences in S, ∆x and ∆t.(b) Show that velocity of the object u′ in frame S′ is given by u′ = u+v.

§ 2.4 Lorentz TransformationThe Galilean transformation works fine for objects that are mov-

ing with speeds much smaller than the speed of light. The Galileantransformation does not work for light itself. We saw in the last sec-tion that the velocity of something moving in one frame is not the samein another frame (u′ = u+ v). This is an unavoidable property of theGalilean transformation.

Hendrik Lorentz

Light has never been observed to have such achange in its speeds. So the Galilean trans-formation fails for light, since it would pre-dict the wrong velocity for light in a movingframe of reference. It also fails for any ob-ject moving at an appreciable fraction of thespeed of light. Hendrik Lorentz was able tocome up with a transformation that worked

2.4 Lorentz Transformation 21

and starting from the assumption that thespeed of light is the same in all inertial frames

Einstein derived that the correct transformation is the Lorentz trans-formation.

Definition: Lorentz Transformation[ct′

x′

]=

[γ γβγβ γ

] [ctx

]where β =

v

cand γ =

1√1− β2

where c = 3.00× 108 ms is the speed of light. Keep in mind that vis the velocity of S as measured by S′.

Notice that c in the Lorentz transformation is not arbitrary butmust be the speed of light. One can think of the quantity ct as thetime expressed in the units of meters.

. Problem 2.6

Compute β and γ for v = 103 ms , 106 m

s , 0.1 c, 0.5c, 0.9c and 0.999c.Compute β and γ algebraically for v = 35c,

√3

2 c,45c.

. Problem 2.7

An event occurs at the point x = 5.0m, ct = 3.0m in frame S. FrameS is moving with a speed of 45c in the positive direction relative to S

′.At what space-time point does the event occur in S′.

. Problem 2.8

An event occurs at the point x = 5.0m, ct = 3.0m in frame S. FrameS is moving with a speed of 45c in the negative direction relative to S

′.At what space-time point does the event occur in S′.

. Problem 2.9

Work the raindrop example again but this time use the Lorentz trans-formation. Does the Lorentz transformation give a different result thanthe Galilean transformation did? Is this what is to be expected forspeeds that are small compared with the speed of light?

. Problem 2.10

Some space kids are out riding their space bikes. They pass a spaceschool with at velocity of v = 0.6c relative to the school. One of thebikes has a leaky reactor that squirts out one drop of radioactive coolantevery 409 ns as observed by someone on the bike. The drops squirt outsideways and strike the side of the school, leaving a line of glowingspots on the wall. When the window washer cleans the school how farapart will he find the spots on the wall?


. Problem 2.11A baby and her father are traveling on a space ship. The followingstory is written from the point of view of someone on the ship. Thefather is looking for his keys at the front of the ship. The baby is atthe back of the ship (45 m away) and has dropped the keys into thetoilet. The ship passes (at a speed of 0.8c) an asteroid where the baby’smother is working. The baby grins at the mom as they pass each otherand 60ns later flushes the toilet. At the same time (on the ship) thatthe baby grins at the mom, the dad opens a drawer to look inside.(a) Fill-in the following chart.

Ship Ship Mom MomEvent ct x ct′ x′

grinflushopen

(b) Retell the story, chronologically, from the mom’s point of view.(c) In the mom’s frame, does the dad open the drawer before or afterthe baby flushes?

§ 2.5 Intervals in Special RelativityWe have learned from the homework problem with the flushed keys

that something very strange is happening with time. Even the order ofevents is relative. It is very disconcerting to lose the absolute orderingof time, the loss seems to undermine the simplest of observations. Forexample, how can there be any significance to even a simple observa-tion like “A happens after B” if this statement is only true for someobservers. In this section we will try to understand this complicationmore fully. In the end we will see that there are some things that arestill absolute.

We will start by making the following definition.

Definition: Space Time IntervalIf we have two events, e1 and e2, the pair of numbers ∆x = x2−x1and ∆t = t2−t1 is called the interval between the events. Intervalstransform via the Lorentz transformation.

. Problem 2.12Prove that space-time intervals transform via the Lorentz transforma-tion.

In order to begin this investigation, do the following problem.

2.5 Intervals in Special Relativity 23

. Problem 2.13Suppose there are two events in a frame S. Define ∆t = t2 − t1 and∆x = x2−x1. Frame S is moving with a velocity v relative to anotherframe S′.(a) Show that if |∆x| < c|∆t| then:

(1) There is a frame S′ in which the relative location of the eventsis reversed (that is, the sign of ∆x′ is the opposite of the sign of ∆x).

(2) The order of events is the same in any frame S′.(b) Show that if |∆x| > c|∆t| then:

(1) There is a frame S′ in which the order of the events is reversed.(2) The relative location of the events is the same in any frame S′.

We see from the previous problem that even in relativity somethings are still absolute. These absolutes are significant enough thatthe following adjectives have been defined.

Definition: TimelikeIf two events are such that |∆x| < c|∆t| then the interval betweenthe events is said be timelike. The time ordering of timelike dis-placements is absolute, that is, the order of events is the same inall frames of reference.

Definition: SpacelikeIf two events are such that |∆x| > c|∆t| then the interval betweenthe events is said be spacelike. The space ordering of spacelikedisplacements is absolute.

Definition: LightlikeIf two events are such that |∆x| = c|∆t| then the interval betweenthe events is said be lightlike.

. Problem 2.14Consider the four events pictured in the space-time diagram below.There are six intervals between the six pairs of events. Classify eachinterval as spacelike or timelike.

ct (m)

x (m)

0 1 2 3 4 5 6 70

1

2

3

4 A

B

C

D


. Problem 2.15

Based on the results of the previous problem, describe the region inspace-time around a point that is in the absolute future relative to thepoint. Do the same for the absolute past, absolute right and absoluteleft.

§ 2.6 Causality

We now make an important observation. If the order of events isnot absolute then that ordering cannot have any physical significance.Thus if event B is not absolutely after event A then there can be nophysical significance to the fact that B is after A in some particularframe. As we will see in a moment, this observation is important be-cause of what it tells us about causally related events.

Part of the reason it is disconcerting to discover the relativity oftime ordering, is that this lack of ordering disrupts our understandingof causality. In order for event A to cause event B, it is essential thatthe cause (A) must precede the effect (B). Without this time orderingof cause we would live in one of those science-fiction nightmares, wherepeople change the present by causing a change in the past. Fortunately,the solution to this apparent difficulty, is close at hand. We can reasonas follows: If A is the cause of B then A must precede B. But whetherA causes B or not, should not depend on the frame of reference. Sowe can strengthen this requirement to, if A is the cause of B then Amust precede B in all frames of reference. This is the main result ofthis section.

Theorem: CausalityIf two events are causally related then the displacement betweenthe events must be timelike.

. Problem 2.16

Argue that there can be no force that acts instantaneously on a distantobject. Note that the model of gravitational and electric force thatyou were taught in your introductory physics classes have just such aninstantaneous action at a distance.

. Problem 2.17

Show that in order for a signal to travel between two events that areseparated by a spacelike displacement that the signal must travel at aspeed greater than the speed of light.

2.7 Velocity Transformation 25

The result of this problem gives us a way of understanding why acause must be in the absolute past: we see that this is equivalent tosaying that a cause must travel at a speed no greater than the speedof light. This is in accord with the notion that forces are transmittedvia fields (e.g. electric and magnetic fields) and that these fields havea finite speed of propagation.

It is interesting to note that the displacement of a pair of eventscannot be both spacelike and timelike. Thus we find that for a displace-ment either the time ordering or the space ordering (but not both) hasa physical significance. We will find a similar “complimentarity” whenwe study quantum physics.

§ 2.7 Velocity Transformation

Remember that the way we knew something was wrong with theGalilean transformation, was that it did not correctly transform thevelocity of light. Let us see how velocities are transformed by theLorentz transformation.

. Problem 2.18

Suppose that an apple is moving with a velocity u as measured in aframe S, and that S is moving with a velocity v relative to anotherframe S′. Consider two events along the trajectory of the apple. Sincewe know that the apple has a velocity u in the frame S we can say thatfor the displacement between the events on the trajectory ∆x = u∆t.Show that the velocity u′ of the apple as measured in S′ is given by

u′ =u+ v

1 + uvc2

This result is called the velocity transformation. It can also be writtenas

βu′ =βu + βv1 + βuβv

Theorem: Velocity TransformationIf and object is moving with velocity u in frame S and S is movingwith velocity v in frame S′ then the velocity u′ of the object in S′

is given by

u′ =v + u

1 + vuc2


. Problem 2.19Use the velocity transformation to show that if an object has a speedu = c in a particular frame then it has a speed u′ = c in any otherframe. Thus showing that light has the same speed in all frames! Verystrange.

. Problem 2.20A ship is traveling toward the earth at a speed of 0.5c. A supply packagehas been sent to the ship from the earth at a speed of 0.4c. What isthe speed of the package as measured by the the ship?

. Problem 2.21A ship is traveling away from the earth at a speed of 0.8c. The shiplaunches a scout vehicle ahead at a speed of 0.9c relative to the ship.What is the speed of the scout vehicle as measured by the the earth?

. Problem 2.22A ship is traveling away from the earth at a speed of 0.8c. The shipsends a package back towards earth at a speed of 0.9c relative to theship. What is the speed of the package as measured by the the earth?

§ 2.8 Energy and MomentumOne of the consequence of the development of special relativity

was the prediction of the existence of nuclear energy. The often quotedrelationship E = mc2 was a direct result of the effort to understandwhat special relativity tells us about energy. In this section we will seehow this result arises.

Recall that in classical mechanics, if two (or any number) of par-ticles interact with each other and nothing else, then their total energyand total momentum are the same before and after the interaction.This constancy is called the conservation of energy and momentum.Energy and momentum are conserved in all inertial frames.

It is natural to consider this same situation again but employingwhat we know about special relativity. Let us look at what happensto the conservation of momentum in a particular example. Supposethat we have two particles m2 = 2m and m1 = m headed toward eachother as shown below. The “after” picture below is predicted from the“before” picture assuming that the classical energy and momentum areconserved.

u2 = 0.4 c u1= -0.8c

After in S

Before in S

uf2 = -0.4 cm2

uf1 = 0.8 cm1

m2 m1

2.8 Energy and Momentum 27

Now let us look at this collision from a frame of reference, S′,moving with the particle that was initially moving to the left. In S′, theoriginal frame S will have a velocity v = 0.8c. By using the relativisticvelocity transformation on all four velocities we find the following result

u’2 = 0.909 cm2

u’1 = 0m1

After in S’

Before in S’

u’f2 = 0.588 cm2

u’f1 = 0.976 cm1

Now let us compute the total momentum before and after the collisionas measured in S′.

p′i = m1u′1 +m2u

′2 = 0 + 2m(0.909)c = 1.818mc

while

p′f = m1u′f1 +m2u

′f2 = m(0.976)c+ 2m(0.588)c = 2.152mc

So we see that in the frame S′ that the classical momentum is not con-served. It ends up that the predictions we made for the “after” picturein S are not correct either. But be careful what you learn from thisexample. The failure in the above example arises because the failure ofthe classical formulas for momentum, mu, and energy, 12mu

2, when thevelocities are large. Momentum and energy are still conserved,we just need to use the correct formulas. This is similar to the situationwith the Galilean transformation, it works fine so long as the veloci-ties are small compared with the speed of light, but if the velocitiesare large the Lorentz transformation must be used. With the follow-ing definitions of energy and momentum, if energy and momentum areconserved in one frame then they are conserved in any frame. Therehas never been an observed violation of either conservation law.

Definition: Energy and Momentum

E =mc2√

1− (u/c)2= γumc

2

p =mu√

1− (u/c)2= γumu

The momentum does not look so different, it is just the classicalmomentum multiplied by γ and since γ ≈ 1 for “every day” velocitiesit is no wonder that the classical definition works fine for “every day”situations. The energy on the other hand looks very different from


what we have been thinking of as the energy of a particle. Let us lookmore closely to see what is going on.

Notice first that even when the velocity is zero (and thus γ = 1)that the energy is not zero, but instead is mc2. So that the energy doesnot reduce to the kinetic energy when the velocity is small.

Definition: Rest EnergyWe call the energy the particle has when it is at rest, mc2, the restenergy.

Definition: Kinetic EnergyWe call the remainder of the energy the kinetic energy:

K = E −mc2 = (γ − 1)mc2

. Problem 2.23Compute the Taylor series expansion of γ to show that

γ = 1 +1

2β2 +

3

8β4 + · · ·

With the result of the previous problem we can expand the kineticenergy in powers of β.

K = (γu − 1)mc2 =(

1

2β2u +

3

8β4u + · · ·

)mc2

=1

2β2umc

2

(1 +

3

4β2u + · · ·

)=

1

2

(uc

)2mc2

(1 +

3

4

(uc

)2+ · · ·

)=

1

2mu2

(1 +

3

4

(uc

)2+ · · ·

)We see then that in the limit u� c that the relativistic kinetic energydoes reduce to the classical kinetic energy.

The rest energy of an object is all the energy contained in thatobject. For example, suppose you take a frozen banana and weigh it.You now warm up the banana. The thermal energy of the bananaincreases and so the internal energy increases and so does the restenergy. But the rest energy is simply mc2, so m must increase. Thus ifyou weigh the warm banana it will be heavier than the frozen banana.

In the end we find that mass is nothing but balled-up energy. Ifyou bring an electron and a positron together, essentially all that willemerge is energy in the form of light.

2.8 Energy and Momentum 29

. Problem 2.24How much does a kilogram of ice increase in mass when it is thawed?

. Problem 2.25Compute the rest energies of an electron and a proton. Express theenergies in electron-volts.

. Problem 2.26Estimate the rest energy of the pen or pencil that you are writing with.How long would this amount of energy run a 100 watt lightbulb?

. Problem 2.27What percentage of the mass of your car would need to be “used up”in order to raise the car 3000 meters (e.g. driving up a tall mountain).Consider only the gravitational energy.

. Problem 2.28The spring of your car has a spring constant of k = 1000N/cm. If thespring is compressed 5cm by how much does the mass of the springincrease?

. Problem 2.29Two particles are moving toward each other and collide. During thecollision the masses fuse into a single mass.

E1 p1m1

E2p2m2

E3 p3m3

After: In Lab Frame

Before: In Lab Frame

E1 = 8�, p1 = 7�/c, E2 = 2�, p2 = −1�/cwhere � is a small amount of energy. Write all your answers in termsof � and c.(a) Find the energy and momentum of m3 (write the answer in termsof � and �/c).(b) Find the velocities u1, u2 and u3 of the three particles (write theanswer in terms of c). Hint: There is a very easy way to find thesevelocities directly from the known values of the energy and momentum,look at the definitions of energy and momentum and figure out how toget the velocity.(c) Find the masses of m1,m2 and m3 (write the answer in terms of�/c2).(d) Is it true that m3 = m1 +m2?(e) Find the velocities u′1, u

′2 and u

′3 of the three particles in a frame of

reference S′ that is moving to the right with a speed of 0.6c.


(f) Find the three energies and momentums in S′ and show that energyand momentum is conserved in this frame also.

§ 2.9 Energy and Momentum TransformationIn the last section we used the conservation of energy and momen-

tum to transform the energy and momentum of an object in one frameinto the energy and momentum of that object in a different frame. Themethod was somewhat involved. One of the stranger things in relativityis that there is a much easier way to do this transformation.

Theorem: Transformation of Energy and MomentumIf the energy and momentum of and object are E and p in theframe S and they are E′ and p′ in the frame S′, then[

E′

p′c

]=

[γ γβγβ γ

] [Epc

]were v is the velocity of frame S in frame S′.

The energy-momentum vector transforms via the Lorentz trans-formation, just like a space-time vector. This amazing simplicity stemsfrom the structure of space-time.

. Problem 2.30Use the Lorentz transformation to transform the energy and momentumof particle 1 in the previous problem. Do you get the same results?

. Problem 2.31Use the Lorentz transformation to show that the velocity of the framein which the total momentum is zero is

β =v

c=pc

Ewhere β is from the transformation and E is the total energy and p isthe total momentum. This frame is called the center of mass frame.

E1p1 E2

p2

Show that the total energy in the center of mass frame is

E′ = E/γv =√E2 − p2c2

This energy E′ =√E2 − p2c2 is called the rest energy since it is the

energy in the frame where the total momentum is zero.

The CM frame is the frame where the energy is the smallest, butall of the energy in the CM frame is “available” for the creation ofanother particle.

2.10 Invariant Quantities 31

. Problem 2.32Suppose you have a positron with a speed of 0.9c that strikes a station-ary electron. How much energy is available in the CM frame? What isthe most massive particle that could be created in this collision?

. Problem 2.33Suppose that an electron and positron have a head on collision, eachhas a speed of 0.45c before the collision. What is the most massiveparticle that could be created in this collision?

§ 2.10 Invariant QuantitiesWe have found in our study of relativity that while many things

are relative, there are some things that are universal. For exampleenergy is conserved in any frame of reference. This does not meanthat energy is the same in all frames of reference, it only means thatwithin any single frame the total energy does not change over time.Surprisingly there are some quantities that are the same in all framesof reference.

Definition: InvariantIf something is the same in all reference frames it is said to beinvariant.

If a quantity is the same for all observers it has a more fundamen-tal physical reality: remember in our discussion of causality, the im-portance of the “absolute past”. Invariant quantities are the buildingblocks of any fundamental model of the world. For example, QuantumChromodynamics is written down entirely in invariants. In this sectionwe will see how to construct invariant quantities.

Theorem: Invariants (forming)

Suppose that you have two vectors

[ab

]and

[fg

]that transform

via the Lorentz transformation when you move from one frame toanother. The quantity af−bg is the same in all frames of reference

We have used a, b, f and g for the components because the vectors canbe either space-time vectors or energy-momentum vectors.

Here is a proof of the above theorem. Since they transform via theLorentz transformation we know that in S′ the vectors are[

a′

b′

]=

[γ γβγβ γ

] [ab

]and

[f ′

g′

]=

[γ γβγβ γ

] [fg

]


What we need to show is that

a′f ′ − b′g′ = af − bgSo let us start with the LHS and then work toward the RHS.

a′f ′ − b′g′ = [a′ b′][

1 00 −1

] [f ′

g′

]= [a b]

[γ γβγβ γ

] [1 00 −1

] [γ γβγβ γ

] [fg

]= [a b]

[γ γβγβ γ

] [γ γβ−γβ −γ

] [fg

]= [a b]

[γ2(1− β2) 0

0 −γ2(1− β2)

] [fg

]= [a b]

[1 00 −1

] [fg

]= af − bg

So we have proved that the quantity af − bg is invariant.

One invariant that is often useful is found when you let

[ab

]=[

fg

]=

[Epc

]. Then one finds that the quantity E2 − p2c2 is an invari-

ant. Similarly if you start with

[ab

]=

[fg

]=

[ctx

]Then one finds

that the quantity c2t2 − x2 is an invariant.

Theorem: Invariants (energy-momentum and space-time)The quantities E2 − p2c2 and c2t2 − x2 are invariant.

Notice that if the energy-momentum vector is for a single parti-cle, that in the frame of reference where that particle is at rest themomentum is zero and the energy is simple mc2 so that

E2 − p2c2 = (mc2)2 − 0 = m2c4

So we see that the quantity√E2 − p2c2 is the rest energy of the particle

in all frames. We can also think of the relationship E2 − p2c2 = m2c4as being like the Pythagorean theorem since it can be rewritten as

E2 = (pc)2 + (mc2)2

2.10 Invariant Quantities 33

Emc2

pc

So that the same particle viewed in different frames will have differentenergies and momentums but in such a way that the triangle con-structed will always have the same height, mc2.

Emc2

pc

mc2

pc

E mc2

pc

E

This graphical representation may help you to remember the relation-ship. The relationship is sometimes helpful to find the energy, momen-tum or mass. If you know two of the three you can find the other withthis “Pythagorean Thoerem”.

. Problem 2.34A particle has a mass of 3.0MeV/c2 and has a total energy of 5.0MeV.What is the particles momentum? Express the answer in units ofMeV/c.

Notice that in order to solve this problem directly from the defini-tions you would need to first find γ from the energy and mass.

E = γumc2 −→ γu =

E

mc2

Next you would find the speed of the electron from the calculated valueof γ.

γu =1√

1− u2/c2−→ u = c

√1− 1

γ2u

Finally you would compute the momentum from the definition.

p = γumu

. Problem 2.35We saw that

√E2 − p2c2 is the rest mass.

(a) For a timelike interval what is the significance of the quantity√c2∆t2 −∆x2?

(b) For a spacelike interval what is the significance of the quantity√∆x2 − c2∆t2?


§ 2.11 SummaryLorentz Transformation:[

ct′

x′

]=

[γ γβγβ γ

] [ctx

]where v is velocity of S as measured by S′ and

β =v

cand γ =

1√1− β2

Velocity Transformation:

βu′ =βu + βv1 + βuβv

Displacements:Timelike: |∆x| < c|∆t|Spacelike: |∆x| > c|∆t|Lightlike: |∆x| = c|∆t|Causally related events must have a timelike displacement.

Energy and Momentum:Energy, momentum and kinetic energy are defined as follows.

Eu = γumc2

pu = γumu

Ku = Eu −mc2

The quantity mc2 is called the rest energy.The energy-momentum vector transforms via the Lorentz transforma-tion. [

E′

p′c

]=

[γ γβγβ γ

] [Epc

]

Invariant Quantities:

Suppose that you have two vectors

[ab

]and

[fg

]that transform via

the Lorentz transformation when you move from one frame to another.The quantity af−bg is the same in all frames of reference. For example,

E2 − p2c2 = m2c4 and c2t2 − x2

2.12 More Homework 35

§ 2.12 More Homework

. Problem 2.36You are in a ship that is moving past the earth at a velocity v. Supposethat you measure a time interval ∆t on your ship by using a stopwatch.Show that in the earths frame of reference it takes a time ∆t′ = γ∆tbetween the starting and stopping of the stopwatch. We see then thatthe watch on the ship run’s slower by a factor γ. This stretching oftime is called time dilation. It is important to note that in order forthe formula ∆t′ = γ∆t to be correct the two events in S must occur atthe same location.

. Problem 2.37There are many types of elementary particles besides electrons, protonsand neutrons. Most of these particles spontaneously decay. Some typesof particles are more stable and on average live longer, while others havea shorter life time on average. For example the average life time of acharged pion is τπ = 26ns, while the average lifetime of a free neutronis 15 minutes. Suppose that a pion is created in a lab and has a speedof 0.9 c relative to the lab.(a) How far would you expect the pion to travel if you did not takerelativistic effects into account?(b) How far do you expect the pion to travel in the lab? Keep in mindthat the lifetime τπ is in the pion’s rest frame.(c) Repeat these two calculations for a 10% increase in the speed of thepion, v = 0.99c.(d) By what percentage does the distance traveled increase when thespeed increases by 10%?

. Problem 2.38There is a stick of length L at rest in a frame S. This stick is passingyou by at a speed v. You want to measure the length of the stick. Youset up a sheet of photo-sensitive paper behind the stick and a flash bulbin front the stick, so that as the stick passes you can record the shadowof the stick and thus measure it’s length without stopping the stick.Show that the shadow of the stick will be a length L′ = L/γ. Notethat the flash is set up so that the entire stick is illuminated at thesame instant in time (in your frame of reference). Hint: Consider thetwo events of the light striking each end of the stick: you know whenthese events happen in your frame and where these events happen inthe sticks frame.

This effect of an object being smaller in a frame of reference inwhich it is not at rest is called length contraction. Note that the formulaL′ = L/γ only works if the object is at rest in the frame S.


. Problem 2.39

Suppose that a stick of length L is at an angle θ to the x-axis in theframe S. The frame S is moving with a velocity v relative to anotherframe S′. Show that the angle of the stick to the x-axis in S′ is givenby

tan θ′ = γ tan θ

Assume that in the y direction (perpendicular to the velocity) theLorentz transformation is simply y′ = y.

. Problem 2.40

Andy has an airplane that is 10 meters long. Andy’s friend Bill hasa barn that is 8 meters long, with a door at each end. Clearly Bill’sbarn is not long enough for Andy’s plane to fit inside. UnfortunatelyAndy and Bill have heard a little bit about length contraction, and soa disagreement arises about what will happen if Andy flies his planethrough Bill’s barn. Bill figures that if Andy flies his plane through thebarn at a speed such that γ = 2 then, since Andy’s plane will be lengthcontracted by a factor of 2, his plane will be only 5 meters long andthus will easily fit into the barn. Andy on the other hand maintainsthat from his point of view in the plane, it is the barn that is movingand so the barn will be length contracted to only 4 meters and thuseven less of his plane will fit in the barn. Settle the argument betweenAndy and Bill.

. Problem 2.41

A space ship has a string of blinking christmas lights strung from thefront of the ship to the rear. All but two of the bulbs have burned out.The lights flash on irregularly, but every time the light near the frontof the ship flashes, the light closer to the rear flashes exactly 70ns later.The two lights are a distance L = 33 meters apart.

(a) Is it possible that the flashing of the front light causes the flashingof the rear light?

The space ship now passes a space station at a speed of 0.8c.

(b) In the frame of reference of the space station how much time passesbetween the front flash and the rear flash? Which flash is first?

(c) In the frame of reference of the space station how far apart are thetwo flashes? Which flash is first?

. Problem 2.42

A single particle, 1, is moving to the right at shown. Particle 1 spon-taneously disintegrates into two particles, 2 and 3. The constant � is aquantity of energy.

2.12 More Homework 37

1Before Disintegration

2After Disintegration 3

E1 = 20 εp1c = 12 ε

E3 = ?p3c = ?

E2 = 8 εp2c = 4 ε

(a) What is the energy E3 of particle 3?(b) What is the momentum p3 of particle 3?(c) What is the velocity u1 of particle 1?(d) What is the mass m1 of particle 1?(e) What are the energy and momentum of particle 2 in the frame ofreference where particle 1 is at rest?

. Problem 2.43Karl has taken off from Earth in his space ship at a speed of 0.6c. Heleft Earth at exactly midnight 00:00 AM. Unfortunately a classmatefrom his collegiate seminar class has rigged Karl’s wrist watch so thatwhen the watch strikes 07:00 AM it will release a demon that will driveKarl crazy by demanding citations from The Book of the City of Ladies.Fortunately this diabolical plot is discovered. Alex grabs her demonneutralizer, jumps into her ship, and flies at top speed (0.8c) in order tocatch up to Karl, and save him from certain mental destruction. Alexleaves Earth at exactly 02:00 AM (Earth time).

0.8 c

0.6 c

KarlAlex

(a) In the Earth’s frame of reference where and when will the demonbe released, if there is no intervention?(b) How many straight jackets should be ordered, (i.e. will Alex reachKarl before the demon is released)?(c) At what speed does Karl see Alex’s ship approaching?

38 Nuclear Physics 3.12

3.1 Nuclear Activity 39

3 Nuclear PhysicsIn most interactions between atoms the nuclei of the atoms are

unchanged. There are times however when the nucleus itself changes.For example, the nuclei of some isotopes of Uranium will spontaneouslybreak into smaller parts. This chapter will be a brief introduction tothe physics of changes to the nucleus of an atom.

Nuclei are formed of neutrons and protons. Neutrons and protonshave approximately the same mass, but protons have a +e charge whileneutrons have no net electric charge. The coulomb force tries to pushthe protons away from each other but there is another force that actsbetween nucleons when they are very close together (∼ 10−15 meters).At this close distance this force becomes very strong and overpowersthe coulomb repulsion. This close range nuclear force is called the stongforce. The strong force is what holds the nucleus.

§ 3.1 Nuclear ActivityThere five main ways in which a nucleus can change, alpha decay,

beta decay, gamma decay, fission, and fusion.

Alpha Decay

In alpha decay the nucleus ejects a helium nucleus. It ends up thata helium nucleus (two protons and two neutrons bound together) is avery stable configuration of nucleons.

Beta Decay

In beta decay the nucleus ejects an electron. In the process thenucleus ends up with one less neutron and one more proton. Youcan think of this as a process in which a neutron is converted into anelectron and proton and then the electron escapes the nucleus. Thereis another particle (an antineutrino) that is created at the same time.

Gamma Decay

In gamma decay the nucleus goes from an excited stated to a lowerenergy state and in the process emits a high energy photon. This oftenoccurs after or during alpha and beta decays.


Fission

In the fission process a nucleus breaks into parts. For example if a235U absorbs a fast moving neutron it can break apart into 92Kr and141Ba and three neutrons. This reaction can lead to further reactionsbecause the three freed neutrons can be absorbed by other 235U . Whichwill in turn produce three more neutrons. Thus a self sustaining chainreaction can occur in which more and more uranium atoms are split,released a large amount of energy in a short time. In a nuclear powerplant the chain reaction is controlled so that the number of reactionsdoes not increase but continues at a steady rate.

Fusion

In the fusion process two smaller nuclei join together to form asingle larger nucleus. For example, the energy that fuels the starscomes from the formation of helium from protons. First two protonsfuse and release a positron and neutrino. This first reaction produces anucleus (deuterium) with one proton and one electron. Second, anotherproton is absorbed by the deuterium so that there are two protons andone neutron in the nucleus now. This is called tritium. In the final steptwo tritium nuclei join together, ejecting two protons in the process andleaving two protons and two neutrons, a helium nucleus.

§ 3.2 An Individual DecayA single atom does not decay at any predictable time. All that

can be said is that if at time t = 0 the atom has not decayed then theprobability that it will decay during the time interval between t andt+ dt is given by

dP =1

τe−t/τdt

where τ is the mean lifetime of the particle which is a characteristic ofthe type of atom, and dt is very small. We refer to the function

ρ(t) =1

τe−t/τ

as the probability density for the decay.The probability that the atom decays between ta and tb is given

by

P (t ∈ [ta, tb]) =∫ tbta

ρ(t)dt =

∫ tbta

1

τe−t/τdt

. Problem 3.1Show that the probability that the atom has not decayed by the timeT is p = e−T/τ .

3.3 Decay Rates 41

. Problem 3.2Show that the probability that the atom has decayed by the time T isq = 1− e−T/τ .

. Problem 3.3Compute the probability that the particle decays after the mean life-time τ .

. Problem 3.4At what time T does the particle have a 50% chance of having decayed?

Definition: Expectation ValueThe expectation value of a function f(t) given a probability densityρ(t) is given by

〈f(t)〉 =∫f(t)ρ(t)dt

where ρ(t) is the probability density and the integral extends overthe domain of the function ρ(t).

Theorem: Expectation Value is LinearLet h(t) = af(t) + bg(t) where a and b are constants, then

〈h〉 = a 〈f〉+ b 〈g〉

Definition: Expectation ValueThe standard deviation σf of a function f is

σf =

√〈(f − 〈f〉)2

〉

Theorem: Standard deviation computationThe standard deviation of a function f can be compute from 〈f〉and

〈f2〉.

σ2f =〈f2〉− 〈f〉2

. Problem 3.5For probability density ρ(t) = 1τ e

−t/τ compute the following.(a) 〈t〉.(b) 〈t2〉.(c) σt.

. Problem 3.6Prove that σ2f =

〈f2〉− 〈f〉2.


§ 3.3 Decay Rates

Now consider the case that we have a large collection of atoms thathave not decayed. Let N0 be the number at the time t = 0. Each atomhas a probability of not having decayed of p = e−t/τ so on average thenumber that have not decayed at time t is

N(t) = N0p = N0e−t/τ

and we see that the number decreases exponentially in time.

The rate at which the number decreases is

dN

dt=N0e

−t/τ

−τ= −N

τ= −N0

τe−t/τ

Thus we see that the time rate of decay is −dNdt =Nτ =

N0τ e−t/τ which

is the number of decays per second.

The half-life is the time it takes for the number to decrease to halfof its original value. Let τ1/2 be the half-life, then we know that

12N0 = N(τ1/2) = N0e

−τ1/2/τ −→ τ1/2/τ = ln 2 −→ τ1/2 = τ ln 2

Theorem: Decay RateThe number of radioactive nuclei in a sample at time t is

N(t) = N0e−t/τ

The half-life isτ1/2 = τ ln 2.

The average number of decays per second is Nτ .

. Problem 3.7

You have a sample containing the titanium isotope 44Ti which has ahalf-life of 1.89× 109 seconds.(a) You count the number (500) of decays in six seconds. Estimate thenumber of 44Ti in the sample.

(b) How long will it take to decay to one percent of it’s original activ-ity?

§ 3.4 Attenuation

3.4 Attenuation 43

Theorem: AttenuationThe number of particles that penetrate a distance x into a materialis given by,

N = N0e−x/λ

The constant λ is called the attenuation length and it is the distanceover which the particle count has decreased by a factor of e. Thevalue of the attenuation length depends on both the type of particleand the type of material.

Because alpha and beta particles are charged, they interact stronglywith matter. Thus the attenuation length is very short. Alpha particlesfor example can usually be stopped by a piece of paper. Gamma radi-ation on the other hand can easily penetrate most materials, leaving apath of ionized particles in it’s wake.

. Problem 3.8Define the half-thickness to be the thickness of material that will de-crease the number of particles to half of the number that entered. Re-late the half thickness to the attenuation length.

44 Light - Waves as particles 4.4

4.2 Review of Some Properties of Light 45

4 Light - Waves as particles§ 4.1 Introduction

Around 1900 there were a few apparently strange phenomena thathad not been explained in terms of the classical model of physics. Itwas not anticipated that the resolution of the standing problems wouldlead to a significant change in our understanding of the world, but thisanticipation was in error. We have seen one example of this in Ein-stein’s special relativity, which was developed to deal with the failureof classical physics for objects that move very fast. Eistein’s theory ofspecial relativity requires a significant change in our understanding oftime. In this section we will learn about quantum physics, which isthe physics that was developed in order to understand the workings ofthe very small. We will see that quantum theory requires a significantchange in our understanding of what it means “to be”.

§ 4.2 Review of Some Properties of LightThere are a few properties of light you will need to recall. First

recall that light is an electromagnetic wave. We experience light ofdifferent wavelengths as different colors. The colors in a rainbow areordered by their wavelength, with red being the longest wavelength andviolet being the shortest.

Our eyes cannot perceive most wavelengths of electromagnetic ra-diation. We can only see light with a wavelength in the range 400-700nm. Radio waves, microwaves and infrared are longer wavelengthsthan we can see. Ulraviolet, x-ray, γ-rays are shorter wavelengths thanwe can see.

Example An infrared camera can see in the dark because the camera is sensitive

to light that our eye does not detect. From the fact that the cameracan see us in the dark we learn that even in the dark we are emittinglight, due to the motion of the atoms composing our skin. We will comeback to this soon.

The term light is sometimes used to refer to the range of electro-magnetic radiation that we can see, but there is no difference (otherthan wavelength) between the radiation we can see and that which wecan’t. Just as often the term “light” is used to refer to all electromag-netic waves regardless of wavelength: this is how I will use it in thisclass.


Spectra

Most sources of light do not emit simply one wavelength of light,but emit a bit of light of all different wavelengths. The amount of eachwavelength emitted by the source determines the “color” of the light.For example purple light is composed of red and blue light combined,and white light has an equal amount of each wavelength. This is some-thing like a recipe which tells how much of each ingredient is needed tomake the cake. The recipe for a source of light is called the it spectrumof the light.

Our eyes are not the best means of determining how much of eachwavelength is in a beam of light since our eyes can not always tell thedifference between spectra that are different. For example a light thatappears green to us could either be composed of some yellow and bluelight or it could simply be a single wavelength green light (about 500nm). The spectra are different but our eye sees them as the same. Wecan build detectors that measure precisely how much of each wavelengththere is in a particular source of light. The results of such a measure-ment is usually represented in a graph of intensity versus wavelength.For example the following graph shows a possible spectrum of light thatappears purple to us.

Intensity

Wavelength (nm)

Spectrum for a Purple Light

Red

Blue

Gre

en

Yello

w

Ora

nge

Viol

et

UV

IR

§ 4.3 Blackbody RadiationLight is emitted by all macroscopic objects. For example, a hot

piece of steel or a glowing coal in fire, will glow. This electromagneticradiation is due to the thermal motion of the charged particles com-posing the object. We do not usually notice this radiation because thelight emitted by room temperature objects is in the infrared and notdetected by our eyes.

This radiation due to the thermal motion of a material is calledblackbody radiation. It is called black in order to clarify that it is

4.3 Blackbody Radiation 47

not the light reflected from the surface of the object, but is the lightcreated by the material. Classical statistical mechanics predicted thatthe intensity of radiation from a blackbody radiator would continue toincrease as the wavelength got shorter.

Intensity

Wavelength (nm)

Spectrum for a Blackbody at 5500K

ExperimentalPrediction fromClassical Theory

Experimentally this was not the case, as the wavelength got shorterthe intensity initially increased, but enventualy the intensity peaked(at λmax) and then decreased to zero as the wavelength went to zero.Further is was observed that if you take the temperature of the objectand multiplied by the wavelength at the peak then you always get thesame number.

λmaxT = 0.002898 m ·K

There was no classical theory that could produce the experimentallyobserved spectrum.

. Problem 4.1At what wavelength is the radiation from a room temperature objectthe greatest? Why do we not see this light?

. Problem 4.2Imagine you measure the spectra of light coming the sun and find thatthe peak intensity is at about 500nm. What would you predict thesurface temperature of the sun to be?

Plank’s Quantum Hypothesis

Max Plank attacked the blackbody radiation problem back-wards. He found a function that fit the experimental spec-trum and then used the laws of physics to work backwardsfrom this function to find the properties of the system. Hefound that the energy of the motion of the charged parti-cles of the system must be restricted in a certain way. The


restriction was that energy of these particles could only be a multipleof hf with h = 6.63 × 10−34J · s and f is the frequency of the lightemitted. This is like saying that if you are swinging on a swing thenyou cannot swing with just any amplitude, but must swing at one ofsay 10 different amplitudes.

Plank did not expect that there was any physical significance tohis quantum hypothesis, but considered the quantizing of the energy ofthe particles to be a mathematical trick used to get the right answer.Five years later Einstein took Plank’s quantum hypothesis seriously,and used it to explain one of the other lingering problems in classi-cal physics, and in this way Einstein gave weight to the hypothesis.Einstein was awarded the Nobel Prize in Physics for this contribution,because it was a huge leap forward in our understand of the world, andwas a pivotal step in the development of quantum physics. Einstein’sanalysis is given in the next section.

§ 4.4 Photoelectric EffectWhen light strikes some metals, electrons are knocked loose from

the metal, this is called photo-emission.

Light

ee

ee

e e e e e e ee e

The emitted electrons have a range of kinetic energies. Experimental-ists illuminated the metal with light of a single frequency and observedthe kinetic energies of the emitted electrons. A number of propertieswere observed, that could not be explained with the classical model ofphoto-emission. These observations are listed below:(1) The kinetic energies of the emitted electrons has a maximum Kmax.(2) The value of Kmax does not depend on the intensity of the light.(3) The value of Kmax is a linear function of the frequency of the light.(4) For a source of light with a small enough frequency there are nophoto-electons emitted regardless of the intensity of the light. Thiscut-off frequency depends on the type of metal.(5) The number of photo electrons emitted is proportional to the in-tensity of the light.

4.4 Photoelectric Effect 49

f

Kmax

fminThis phenomena is known as the photoelectric effect. Within the

classical electromagnetic model of photo-emission there was no way tounderstand the above mentioned properties of the photoelectric effect.In the classical model, light is an electromagnetic wave. The frequencyof the wave is the frequency of the light and the square of the amplitudeof the wave is proportional to the intensity of the light. The oscillatingfields cause and oscillating force on the electrons which are thus shakenloose from the metal. Using the classical model one would expect thatthe kinetic energy of the emitted electron would depend strongly onthe intensity of the light, because the intensity is directly related tothe strength of the force applied to the electrons. But as stated beforeKmax is completely independent of the intensity: the classical theoryfails completely. The other properties listed are similarly inexplicable.

Einstein’s Explanation of Photo-Electric Effect

While relativity is the work for which Einstein is most widelyknown, it is his explanation of the photoelectric effect that earned himthe Nobel Prize in physics. In order to explain the photo-electric effectEinstein made three assumptions.(A) Light comes in packages with a fixed amount of energy E = hf .We call these particles (quanta) of light photons.(B) The intensity of a light source is proportional to the number ofphotons per time that are emitted by the source.(C) Each electron of the photo-emitting metal only interacts with onephoton.

The first assumption (A) is consistent with Plank’s quantum hy-pothesis, that the source of light can only have energies that are amultiple of hf , since then if the atom looses energy it must loose anamount hf in order to go down to the next lower energy level. Forexample if an atom goes from E = 47hf to E = 46hf then it loosesone hf of energy. Thus if an atom looses energy by emitting light, thelight must have an energy equal to the energy lost by the atom, hf .

With these three assumptions a simple explanation of the photo-electric effect follows directly. Since each electron only interacts withone photon and each photon has a fixed amount of energy hf , we knowthat there is only a fixed amount of energy available to the electron.Some of the energy received from the photon will be used up in doing


the work W required to overcome the attractive force that holds theelectron to the metal. Thus the actual energy that is available to theelectron after being emitted is hf −W , and the kinetic energy of theelectron cannot be more than this.

Kmax = hf −WWe see then that Einstein’s photon theory not only predicts the exis-tence of a maximum kinetic energy (1), but also predicts the remainingproperties (2-5) of the photoelectric effect.(2) Assumption (B) tells us that a more intense light has more photons,but by (C) this does not change the amount of energy available to asingle electron. So the maximum kinetic energy does not depend onthe intensity of the light.(3) The relationship Kmax = hf−W tells us that the maximum kineticenergy is a linear function of the frequency.(4) If f < W/h then no electrons will be emitted since a single photondoes not have enough energy to do the work W . Each type of metalwill require a different amount of work to remove the electron so thecut-off frequency will depend on the type of metal.(5) Assumption (B) tells us that the intense is proportional to the num-ber of photons, and the number of photons is in turn proportional tothe number of emitted electrons. thus the number of emitted electronsis proportional to the intensity of the light.

Einstein’s photon model is essential to our current understandingof the nature of light.

Definition: PhotonLight comes in packages with a fixed amount of energy.

E = hf

A particle of light is called a photon.

Most light sources provide not a single photon but a continuousstream of photons. The following definitions relate the energy of asource of light to the number of photons emitted from the source.

Definition: Photon-Flux and PowerThe power of light source, the energy per time, is directly relatedto the photon-flux Φ which is the number of photons per time.

P = hfΦ

4.5 Photon Momentum 51

Definition: Photon-Flux Density and IntensityThe intensity of light, the energy per area per time, is directlyrelated to the photon-flux density φ which is the number of photonsper area per time.

I = hfφ

. Problem 4.3Suppose you have a red (650nm) laser pointer with a power output of3.0 mW. How many photons per second are emitted by the laser.

. Problem 4.4A red (650nm) laser with a power output of 3.0 mW is directed at apiece of metal. The photo-electrons emitted from the metal becauseof this light are collected on an anode and passed back to the metalthrough a wire. The current in the wire is measured to be 5 nA.(a) What fraction of the photons striking the metal create photo-electrons?(b) What is the probability that a single photo will create a photo-electron?

. Problem 4.5Suppose that sunlight on a particular day has an intensity of 1000watts per square meter. What is the photon-flux density? (To makethis easier assume that the sunlight has a wavelength of 550 nm.)

§ 4.5 Photon MomentumIf you shine light on a cloud of free electrons then some of the light

is deflected from it’s original direction. This deflection of light is calledCompton Scattering. Careful observation of the scattered light showsthat the scattered light is not of the same wavelength as the incidentlight. In addition it was observed that the change in the wavelengthwas related to the change in the direction θ of the light, and the massme of the electron.

∆λ =h

mec(1− cos θ)

The discovery of Compton scattering was further support for thephoton theory of light. The shift in wavelength is in strong disagree-ment with the classical model of light, and in accord with the photonmodel. The classical model is that the light shakes the electron andthe shaken electron will then emit an EM wave. But in this model thefrequency of the ‘absorbed shake’ would be the same as the frequencyof the ‘emitted shake’. The photon model treats the scattering as a


collision between two particles, the photon and the electron. If it is as-sumed that energy and momentum are conserved in the collision thenthe predicted shift in wavelength matches the observed shift.

Before Collision After Collision

θ

In order to apply the conservation of momentum and energy toa photon description of Compton scattering we must first extend thephoton theory a bit more. We know that the energy of a photon isE = hf , but we do not know what the momentum of a photon is. It isknown from electromagnetic theory that the energy and momentum ofan electromagnetic wave are related by E = pc, this relationship holdsfor a single photon as well. Solving for the momentum we find that fora photon

p =E

c=hf

c=h

λ.

It ends up that the last expression will apply not only to light but toall particles.

Fact: Momentum and WavelengthThe momentum and wavelength of a particle are related via

p =h

λ

It is this relationship that is used to find the shift in wavelengthduring Compton scattering.

. Problem 4.6Show that for a photon E = hcλ and that hc = 1240eV · nm.

. Problem 4.7Compute the energy and momentum of a photon with a wavelength of620nm. Express the energy in electron-volts.

. Problem 4.8A laser with a power output of 5 watts is directed at a perfectly ab-sorbing brick. What is the force on the brick due to the pressure of thelight.

4.7 Taylor’s Double Slit Experiment 53

. Problem 4.9Compute the wavelength of the scattered photon in the case the scat-tered direction is 90◦ and 170◦ to the incident angle. The incidentwavelength is 40pm.

. Problem 4.10Suppose that a photon with a wavelength of 40pm strikes an electronand that the scattered photon is emitted at 90◦ to the incident angle.(a) What is the momentum of the scattered photon?(b) What is the energy of the scattered photon?(c) What is the kinetic energy of the recoiling electron?(d) What is the direction of the recoiling electron?

. Problem 4.11Show that if a photon has a wavelength λ in a frame S, and thatframe S is moving with a velocity v with respect to frame S′ then the

wavelength of the photon in S′ is given by λ′ = λ√

1−β1+β .

§ 4.6 Light in terms of it’s electric fieldsOne can show using the classical theory of electromagnetic fields

that the intensity of light is proportional to the square of the electricfield strength of the light.

Theorem: Intensity of light with field strength E

I = �0cE2

where �0 is the permittivity of free space.

. Problem 4.12Suppose that you set up a light source (λ = 400nm) and a photoncounter at some distance from the light source. The photon counterhas an active area of 1 square millimeter. The photon counter detects200 photons in one microsecond.(a) What is the electric field strength at the location of the photoncounter?(b) Show that in general the electric field strength can be computed as

E =

√hN

�0λ ∆A ∆twhere ∆A is the area of the detector, N is the number of photonsdetected over the integration time ∆t of the detector.(c) A linear polarizer is set up between the light source and the detector,and the count reduces to N = 100. Next a second polarizer at and angle


of 30◦ to the first polarizer is placed between the first polarizer and thedetector. How many photons are detected now?

§ 4.7 Taylor’s Double Slit ExperimentWe have learned that sometimes light exhibits particle like prop-

erties, and that the photon theory of light explains this particle likebehavior. But light still retains it’s wave character. So let us look atthe properties of light that display it’s wave character, and considerwhat this tells us about the photon theory. The wave character of lightis displayed most strongly in interference phenomena so let us considerthe double slit interference as a specific example. Keep in mind thatthe question we want to answer is, what can interference phenomenatell us about photons?

The way that double slit interference is explained in terms of elec-tromagnetic waves is as follows. The wave strikes the double slit maskand all of the wave is stopped except the parts that hit the holes inthe mask. These two parts of the wave pass through the mask and themixing of the these two parts at the observation screen causes the inter-ference pattern. We know now that a wave of light is actually composedof a collection of photons. We need to understand how the splittingand remixing of the wave can be understood in terms of photons.

It is tempting (but wrong) to assume that some of the photons gothrough one slit, while other photons go through the other slit. Thephotons from the two different slits then mix together at the obser-vation screen which causes the interference pattern. Note that thisexplanation would break the simplicity of Einstein’s supposition thatphotons interact with matter individually, since it requires that twophoton work together to create the interference.

In 1909 the following experiment was designedand performed by Geoffrey Ingram Taylor, inorder to see if interference is due to an interac-tion of photons with each other. In this exper-iment a simple double slit apparatus is set up,b

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