CHAPTER 5

LIGHT IS NOT A WAVE!

• Introduction

• Photoelectric effect• Photons

• Compton Effect

• Gravitational effects

Introduction

In the 17th century there was a difference in opinion whether light consisted of a beam of particles or waves. The chief protagonists were Isaac Newton, who used a particle theory to explain reflection and refraction, although he was forced to assume that light traveled faster in glass or water, and Robert Hooke and Christian Huygens, who explained refraction using a wave theory and assuming that light traveled more slowly in glass water than in air. Although there was no clear “winner”, the reputation of Newton was such that the particle theory was accepted through the 18th century.

However, Thomas Young’s double-slit experiment (1801) forced the particle theory aside as the weight of evidence now appeared to favor the wave hypothesis.

In addition, the studies of Augustin Fresnel, involving many experiments on interference and diffraction, put the wave theory on a rigorous mathematical basis.

The classical wave theory of light was further established by James Clerk Maxwell in 1860 with his theory of electromagnetism. He predicted the speed of

electromagnetic waves at 3!108 m/s, the same as the

speed of light, which suggested that light is an electromagnetic wave. Thus, there was clear, irrefutable evidence that light has the properties of waves.

However, experiments at the end of the 19th and beginning of the 20th centuries suggested that maybe light consisted of particles with discrete amounts of energy.

The photoelectric effect

The quantization of light was introduced by Einstein in 1905 to explain the photoelectric effect (for which he later received the Nobel Prize). Various experimenters

had observed that light, with a frequency greater than a certain threshold frequency, causes electrons - called photoelectrons - to be ejected from a metal surface (C). Measurements of the kinetic energy of the electrons gave the surprising result that the maximum kinetic energy

Kmax( ) did not depend on of the intensity of the incident

light. Subsequent experiments did show, however, that the maximum kinetic energy of the photoelectrons did depend on the wavelength (frequency) of the incident light.

A

C

e! V

IIncident light

Evacuated vessel

Kmax is determined by applying a reverse potential

difference, called the stopping potential Vs( ), so the

emitted electrons are repelled, and the photocurrent I = 0. Then Kmax = qeVs, where qe is the charge on an electron.

Einstein explained the observations by suggesting that light consists of a stream of particles - later called photons - each with a discrete amount of energy that depends on the wavelength (frequency) of the light. The energy of each photon is

E = hf = hc! ,

where f is the frequency and h is a constant now known as Planck’s constant.

Vs

Photocurrent

V Kmax = qeVs

Through interactions within the metal, the photons transfer their energy to the electrons. If the energy of the electrons within the metal is above a certain minimum (!), they are

ejected through the surface of the metal.

Einstein’s photoelectric equation says that the maximum kinetic energy of the photoelectrons is

Kmax =

12

mv2"

# $

%

& '

max= hf ( !,

where ! is called the work function and is characteristic of

the metal surface. It represents the minimum amount of energy required to eject an electron from the surface.

hf

hf! Kmax = 1

2mv2"

# $ % & '

Energy

electrons in the metal

maximum energy ofelectrons inside the

metal

surface region

So, according to Einstein, a plot of Kmax against the

frequency (f) of the light, should be linear, with slope h.Shown alongside aredata obtained by Millikan (1916), from a lithium surface. It provided the first comprehensive

verification of Einstein’s photoelectric equation and it afforded a measurement of Planck’s constant. From the slope = h( ), we find

h = 4.136 !10"15eV # s = 6.626 !10"34J # s,

and the cut-off when Kmax = 0 gives

$ = hft = (4.136 !10"15eV # s) ! (42.5 !1013Hz)

= 1.76eV.Work functions ($) are typically a few electron-volts. Note: if hf < $, no electrons can be emitted.

Kmax (eV)

0

1

2

3

40 60 80 100 120! 1013Frequency (Hz)

ft = 42.5! 1013Hz

Question 1: Electrons are ejected from a surface with speeds ranging up to 770km/s when radiation of wavelength 392nm illuminates the surface. (a) What is the work function of the material? (b) What is the threshold frequency for the surface? (c) What is the stopping potential for the surface?

(a) The photoelectric equation is Kmax = hf ! " =

hc#

! ",

i.e., " =

hc#

! Kmax =hc#

!12

me vmax( )2

=

6.63 $10!34 $ 3$108

392 $10!9 !12$ 9.11$10!31 $ 7.70 $105( )2

= 2.373 $10!19J %

2.373 $10!19

1.602 $10!19 = 1.48eV.

(b) The threshold frequency is given by hft = ",

&ft =

"h=

1.48eV4.136 $10!15eV ' s

= 3.58 $1014Hz.

(c) The stopping potential Vs is related to Kmax by

qeVs = Kmax , i.e., the potential difference required to stop

the photoelectrons.

Hence qeVs =

12

me vmax( )2 % Vs =me2qe

vmax( )2

=

9.11$10!31 $ (7.70 $105)2

2 $1.602 $10!19 = 1.69V.

Class discussion problem: If the photoelectric effect is observed when blue light is incident on a certain metal surface, will it be observed for red light?

Class discussion problem: (a) If the photoelectric effect is observed with cesium, can you guarantee that it will be observed in sodium? (b) If the photoelectric effect is observed with tungsten, can you guarantee that it will be observed in potassium? Assume the following values for the work functions:

Cesium 1.9eV Potassium 2.1eV Sodium 2.4eV Tungsten 4.58eV.

Question 2: The wavelength of red light emitted by a 3mW He-Ne laser is 633nm. If the diameter of the laser beam is 1mm, what is the density of photons in the beam? Assume it is constant across the beam.

Question 2: solution

The density of photons is ! = N

V.

The total power is P = NE

"t , i.e., N = P"t

E.

The volume of the beam containing the photons is

V = A # (c"t ).

But E = hc

$

%! =

P$

hAc2

=3 &10'3 & 633&10'9

6.63 &10'34 & (4& (1&10'3)2)

* +

,

- . & (3&108 )2

= 4.05 &1013m'3.

Class discussion problem: Suppose the photoelectric effect occurs in a gas (rather than a solid), will photoelectrons be produced at all energies of the incident photons?

Class discussion problem: An electron and a photon both have energies of 10eV. Which one, if either, has the greater momentum?

A: The electron.B: The photon.C: The same.

Compton Effect

A second experiment that confirmed the idea that photons carry energy and momentum was established in 1923 when A.H. Compton discovered that x-rays scattered from free (isolated) electrons suffer a change in wavelength that varied with scattering angle.

According to classical theory, if an electromagnetic

wave (frequency f1) is incident on free charges, the

electric field of the e-m wave will cause the charges to

oscillate with frequency f1 and to re-radiate

electromagnetic waves with the same frequency ( f1).

Incoming x-ray

Scattered x-ray

m

!"

However, Compton pointed out that if the process was like a “classical” collision between a photon and a free electron, the electron would recoil and absorb energy. So, the scattered photon would have reduced energy, i.e., a lower frequency (a longer wavelength) compared with the incident photon.

In chapter 4, we found the following relation between energy and momentum for a particle of mass m:

E = pc( )2 + (mc2 )2 .

Since a photon has zero mass, its energy and momentum are, respectively:

E = pc = hf =

hc!

and p =

h!

.

Incoming x-ray

Scattered x-ray

m

"#

Conservation of momentum gives:

! p 1 =! p 2 + ! p e.

From the momentum diagram we find

pe2 = p1

2 + p22 ! 2p1p2 cos".

The energy of the scattered electron is

Ee = pec( )2 + (mec2)2 ,

where me is the mass of the electron. Therefore,

conservation of energy gives

p1c + mec2 = p2c + pec( )2 + (mec2)2 .

Substituting for pe2, we find

! p e

! p 1

! p 2

"

Incoming x-ray

p1 = h

#1

Scattered x-ray

p2 = h

#2

m

"$

pe

p1c+ mec2 = p2c+ (p1c)2 + (p2c)2 ! 2p1p2c2 cos" + (mec2)2 .

Rearranging, and squaring both sides of the equation, we find

p1c ! p2c( ) + mec2[ ]2 = p1c( )2 + p2c( )2

!2p1p2c2 cos" + (mec2)2,

#(p1c)2 + (p2c)2 + (mec2)2 ! 2p1p2c2 + 2p1mec3 ! 2p2mec3

= (p1c)2 + (p2c)2 ! 2p1p2c2 cos" + (mec2)2.

Collecting like terms and dividing through by c2, we get

p1p2 + p2mec ! p1mec = p1p2 cos".

Dividing through by p1p2, we get

1+ mec

p1! mec

p2= cos" ,

i.e.,

1p2

! 1p1

= 1mec

1! cos"( ).

Or, in terms of wavelengths of the incoming and scattered photons:

$2 ! $1 = h

mec1! cos"( ).

These are the Compton scattering equations.

!2 " !1 = #! =

hmec

1" cos$( ).

The quantity,

hmec

, has the dimensions of length and is

called the Compton wavelength

!C =

hmec

=6.626 %10"34J & s

9.11%10"31kg( ) % 3 %108 m/s( ) = 2.42 %10"12 m = 2.42pm.

Note that the increase in wavelength, #! , does not

depend on !1 and is a maximum for $ = 180!, when

#! = 2 % 2.42 = 4.84pm.

Note also, that in deriving the Compton scattering equation we used the relativistic form of the energy-momentum relation for the electron. The reason is that, in general, with high energy photons (x-rays), the energy acquired by the electron is a significant fraction of its rest energy (511keV).

Question 3: Compton [A. H. Compton, Phys. Rev. 21, 483; 22, 409 (1923)] carried out his experiments by

scattering Mo K! x-rays, with a wavelength of

"1 = 0.0711nm, from a sample of graphite What values

of "2 did he observe for # = 45!, 90! and 135!?

It costs only a few eV to remove a free electron from graphite. However, the energy of the x-rays is

E1 =

hc!

=1240eV " nm

0.0711= 17440eV,

so the amount of energy required to “free” the electron is negligible compared with E1.

The Compton equation is !2 = !1 +

hmec

(1# cos$).

For $ = 45!:

hmec

(1# cos$) = 2.42 %10#12(1# 0.7071)

= 7.09 %10#13m = 0.709pm.

&!2 = (71.1+ 0.709)pm = 71.809pm,

i.e., '!

! = 1%.

For $ = 90!:

hmec

(1# cos$) = 2.42 %10#12 m = 2.42pm.

&!2 = (71.1+ 2.42)pm = 73.52pm,

i.e., '!

! = 3.4%.

For ! = 135!:

hmec

(1" cos!) = 2.42 #10"12(1+ 0.7071)

= 4.13 #10"12 m = 4.13pm.

$%2 = (71.1+ 4.13)pm = 75.23pm,

i.e., &%

% = 5.8%.

Although small, Compton was able to determine these changes by measuring the changes in the angle of diffraction from a calcite crystal for the incoming and scattered x-rays from a graphite sample. He used a graphite target because atoms with a low atomic number

ZC = 12( ) have a larger percentage of loosely bound, i.e.,

nearly free, electrons.

The x-rays were scattered from graphite and observed at different

angles with a spectrometer. The spectrometer consists of a rotating

framework with a calcite crystal to diffract the x-rays and an

ionization chamber for detection of the x-rays. The spacing of the

crystal planes in calcite is known, so the angle of diffraction gives

an accurate measurement of the wavelength.

6.5! 7.0! 7.5!

6.5! 7.0! 7.5!

6.5! 7.0! 7.5!

!2 = 0.0718nm

!2 = 0.0735nm

!2 = 0.0752nm

6.5! 7.0! 7.5! !1 = 0.0711nm

" = 0

"

" = 45!

" = 90!

" = 135!

x-ray tube

calcitecrystal

detector

graphitesample

Class discussion problem: Why do both the incoming and scattered wavelengths appear in the data obtained by Compton?

6.5! 7.0! 7.5!

!2 = 0.0752nm

!1 = 0.0711nm

" = 135!

Question 4: An x-ray photon with wavelength 0.01nm makes a collision with a stationary electron. If the

photon is scattered at an angle ! = 90!, find (a) the

wavelength of the scattered photon, (b) the scattering angle of the electron, (c) the velocity of the electron.

" " p 1

" p 2

" p e

90!

(a) !1 = 0.01nm = 10pm. The wavelength of the

scattered photon is

!2 = !1 +

hmec

(1" cos#)

= 10 $10"12 m + 2.42 $10"12 m

= 1.242 $10"11m (12.42pm).

(b) The photon momenta are: p1 =

h!1

and p2 =

h!2

.

%& = tan"1 p2

p1

'

( )

*

+ , = tan"1 !1

!2

'

( )

*

+ , = tan"1 10.0

12.42' ( )

* + ,

= 38.8!.

& " p 1

" p 2

" p e

90!

" p e

" p 2

" p 1

&

(c) Conservation of energy in the Compton process gives:

E1 + mc2( ) = E2 + Ke + mc2( ).

!Ke = E1 " E2 = hc

1#1

"1#2

$

% &

'

( ) = 3.88 *10"15J

+

3.88 *10"15J

1.602 *10"19J/eV= 24.2keV.

Classically: v =

2Keme

= 9.22 *107 m/s + 0.31c.

Since this is a significant fraction of the speed of light, it was appropriate to use the relativistic form for the electron

energy. Also, note that Ke (24.2 keV) is almost 5% of the

electron rest mass (511keV). Using the relativistic form for the kinetic energy,

Ke = (, "1)mec2,

i.e., , =

Ke

mec2+ 1 = 1.047,

we find v = 0.30c.

Question 5: In a Compton collision with an electron, a photon of violet light ( ! = 400nm) is backscattered

through an angle of 180!. (a) How much energy (eV) is

transferred to the electron? (b) What energy would an electron acquire in a photoelectric process with the same photon?

(a) In a Compton scattering event, the photon wavelength changes by

!" =

hmc(1# cos$) =

2hmc

.

The energy of the photon is E =

hc"

, so the energy lost by

the photon (and transferred to the electron as Ke) is

!E = !

hc"

% & '

( ) * = #

hc"2

% & '

( ) * !" = #

2h2

m"2 .

Therefore, the energy transferred to the electron is

2h2

m"2 =2 + 6.626 +10#34J , s( )2

9.11+10#31kg( ) + 400 +10#9 m( )2

= 6.02 +10#24J - 3.77 +10#5eV.

(b) In the photoelectric process, the energy gained by the electron is

E =

hc"

=6.626 +10#34J , s( ) + 3 +108 m/s( )

400 +10#9 m

= 4.97 +10#19J - 3.11eV.

The energy gained in the photoelectric process ( 3.11eV) is

~ 105 greater than the energy of the electron scattered in

the Compton process ( 3.77 !10"5eV). So, violet light

cannot eject electrons from a metal by the Compton effect since the energy transferred to the electron is much less than the work function (typically a few eV). Actually, the photo-emitted electron will have less energy than 3.11eV, because of the work function.

A rough rule of thumb is for Compton (elastic) scattering with a free electron, the wavelength of the incoming

photon must be comparable with #C = hmc ~ 10"12 m( ).

Now, we face a dilemma, interference and diffraction effects seem to confirm that light is wave-like but the photoelectric effect and Compton scattering provide strong evidence that when light interacts with matter it behaves as if it were composed of particles (with energy

hf and momentum h! ). But we are left with some very

difficult questions. For example, what is the meaning of “frequency” and “wavelength” when we are dealing with a particle? In fact, we will see that neither classical mechanics nor (modern) quantum theory can be used exclusively to describe electromagnetic waves like light; a complete understanding is obtained only if the two models are combined in a complementary manner. Thus, we are left with an uneasy compromise between wave and particle concepts.

William Henry Bragg (co-discoverer with his son of “Bragg’s Law” in x-ray diffraction) once said

“Light behaves like waves on Mondays, Wednesdays and Fridays, like particles on Tuesdays, Thursdays and Saturdays, and like nothing at all on Sundays”

Also, if light is truly composed of particles, is it affected by a gravitational field?

Gravitational effects (“Newtonian” approach)

Although photons have zero mass, we can define an

apparent inertial mass, mi, because they have

momentum, viz: p = h

! = mic.

"mi = h

c! = hfc2

= Ec2

# $ % &

' ( .

The inertial mass determines how the photon responds in a collision, like Compton scattering. If we equate the inertial mass with the gravitational mass, then if a photon “falls” through a height H, i.e., travels from

A ) B, its gravitational potential energy decreases

by migH. Since the total

energy remains constant, its energy E = hf( ) will increase. Therefore, the

photon frequency will increase also, i.e., fo > fs.

A

B

H

Source of photons( fs)

Detector( fo)

Applying conservation of energy we have

EB + UB = EA + UA,

i.e., hfo = hfs + !U,where

!U = UA " UB

= migH =

hfc2 gH.

#fo = fs 1 +

gHc2

$ % &

' ( ) *

!ff

=fo " fs

f=

gHc2 ,

i.e., fo > fs, as expected.

In the 1960’s, R.V. Pound, G.A. Rebka and J.L. Snider verified this so-called gravitational blue-shift (since

fo > fs) using + -ray photons. With H = 22.6m,

!ff

=9.81, 22.6

3 ,108( )2 = 2.46 ,10"15,

requiring incredibly precise measurements (which they were able to achieve to a few percent).

A

B

H

Source of photons( fs)

Detector( fo)

Gravitational Redshift

If a photon moving from a position of higher gravitational potential energy to a position of lower gravitational potential energy gains energy, then a photon moving from a position of lower gravitational potential energy to a position of higher gravitational potential energy loses energy. Consider light emitted from the surface of a star

of mass Ms and radius Rs.

Using a similar approach to before we have:

E + U[ ]r =! = E + U[ ]Rs

.

If hfs is the photon energy at the source and hf! is the

photon energy where the gravitational field is negligible, i.e., ( r " !),

hf! = hfs # G

MsmiRs

= hfs #GMsRs

hfs

c2

$

% &

'

( ) .

f!

r "!

fs

Rs

Ms

!f" = fs 1#

GMs

Rsc2

$

% & &

'

( ) ) ,

so the change in frequency is

*ffs

=f" # fs

fs= #

GMs

Rsc2

.

Since f" < fs, the frequency (and energy) of the photon

is reduced when it escapes the gravitational field of a massive object; this is called the gravitational red-shift.

• Note, if

GMsRsc

2 + 1, i.e., Rs ,

GMs

c2, then f" , 0,

so the light is shifted to zero frequency and is not observed! An object for which this condition is met is

called a black-hole. Here, Rs is not the Schwarzschild

radius rS( ); using a more rigorous approach (General

Relativity) the Schwarzschild radius is actually,

rS =

2GM

c2.

Question 6: Using the Newtonian approach, what are

the critical values of Rs for which f! = 0, for objects

with the mass of (a) the Earth and (b) the Sun.

(a) For the Earth we have ME = 5.98 !1024kg and

G = 6.67 !10"11N #m2/kg2. So, for f$ = 0,

Rs =

GME

c2=

6.67 !10"11 ! 5.98 !1024

(3 !108 )2

= 4.43 !10"3m% 4.43mm.

(b) For the Sun we have MSun = 1.99 !1030 kg and

G = 6.67 !10"11N #m2/kg2. So, for f$ = 0,

Rs =

GMSun

c2=

6.67 !10"11 !1.99 !1030

(3 !108 )2

= 1.47 !103m%1.47km.

As mentioned earlier, these radii are not the actual Schwarzschild radii associated with black-holes. Our calculation is a non-relativistic determination, but it does give a result within a factor of 2.

Question 7: A certain white dwarf has a mass equal to

the Sun ( MS = 1.99 !1030kg) and a radius equal to that

of Earth ( RE = 6.37 !106m). (a) Show that the

gravitational field of the Earth is negligible compared with that of the white dwarf. (b) What would be the change in the observed wavelength on Earth of the 434nm blue line in the hydrogen spectrum?

Source of photons( fs)

Detector( f )

(a) At the Earth’s surface, gE ! 9.81m/s2, and at the star’s

surface, gs =

GMSRE

2

=

6.67 "10#11 "1.99 "1030

(6.37 "106)2= 3.27 "106 m/s2.

$gs >> gE.

(b) Since the gravitational field at the Earth’s surface is negligible compared with that at the star’s surface,

%ff

= #GMSREc2 = #

6.67 "10#11 "1.99 "1030

6.37 "106 " (3 "108 )2

= #2.32 "10#4.

But f =

c&

, i.e., %f = #

c

&2%&

$

%ff

= #%&&

' %& = #&%ff

.

$%& = # 434 " (#2.32 "10#4 )( ) = +0.101nm.

An increase in wavelength corresponds to a red-shift.

The expression f! = f 1"

GM

Rc2

#

$ %

&

' ( prompts the question ...

since the frequency of a source and the periodic time of the wave are related does the gravitational field influence the periodic time?

By definition, the periodic time of a wave is ) = f "1, so,

the observed periodic time of wave at a position where

the gravitational field can be ignored ()!) is related to

the periodic time of the source in the gravitational field ()). From above we find:

)! = ) 1"

GM

Rc2

#

$ %

&

' ( "1

.

*)! > ).

Note, if

GM

Rc2<< 1 then

)! + ) 1 +

GM

Rc2

#

$ %

&

' ( ,

i.e.,

,))

=)! " )

)=

GMRc2 .

So, if we associate the periodic time of the source with the ticking of a clock, i.e., we replace ! by t, then we see that a gravitational field affects time, i.e.,

t" # t 1 +

GM

Rc2

$

% &

'

( ) if

GMRc2 << 1.

*

+tt

=t" , t

t=

GMRc2 .

Therefore, time is slowed down in a gravitational field

since t" > t . This time shift is different from time-

dilation in special relativity; it is a gravitational effect.

Question 8: In question 7 we calculated the shift in the wavelength of the 434nm blue line in the hydrogen emitted from spectrum white dwarf, whose mass was equal to the Sun and whose radius was equal to that of Earth. What would be the gravitational time shift of a clock on the white dwarf as observed from Earth? Assume the gravitational field of the Earth is negligible compared with that of the white dwarf.

White dwarf( t)

Earth( t!)

From the previous discussion, we have

t! = t 1"

GMRc2

# $ %

& ' ( "1

, i.e., )t! = )t 1"

GMRc2

# $ %

& ' ( "1

where )t is a time interval on the “source” clock (on the white dwarf ), where there is a large gravitational field, and )t! is the observed time interval where the

gravitational field is negligible (on Earth).

For the white dwarf,

GMRc2 =

6.67 *10"11 *1.99 *1030

6.37 *106 * 3*108( )2 = 2.32 *10"4,

which is << 1.

+)t! = )t 1"

GMRc2

# $ %

& ' ( "1

, )t 1+GMRc2

# $ %

& ' ( .

If )t = 1.00s, then

)t! = 1.00 1+ 2.32 *10"4( ) = 1.000232s,

i.e., the observed time interval is greater by 232µs.

Class discussion question: If a clock at the edge of a black hole “ticks” once every second, what would be the time interval observed on Earth?

What time-shift is produced by the gravitational field of the Earth at a height H above the Earth’s surface?

EA + UA = EB + UB, i.e., EB = EA ! UA ! UB( )

i.e., hfB = hfA ! GMEm

1RE

!1

RE + H( )"

# $

%

& ' ,

where m =

hfAc2 =

hc2(A

.

Also,

1RE

!1

RE + H( ) =H

RE RE + H( ).

)fB = fA 1! G

HMEc2RE RE + H( )

"

# $ $

%

& ' ' ,

i.e., fB < fA (a red-shift) as expected.

H

EA = hfA = h

(A:

UA = !G

MEmRE

EB = hfB = h

(B:

UB = !G

MEmRE + H( )

Since f = !"1, we obtain

1!B

=1!A

1" GHME

c2RE RE + H( )#

$ % %

&

' ( ( ,

i.e., !B = !A 1" G

HMEc2RE RE + H( )

#

$ % %

&

' ( (

"1

.

Generalizing, by replacing ! with t, we find

tB = tA 1" G

HMEc2RE RE + H( )

#

$ % %

&

' ( (

"1

.

If G

HMEc2RE RE + H( )

<< 1, then

tB = tA 1 + G

HMEc2RE RE + H( )

#

$ % %

&

' ( ( ,

and the fractional time shift is

)ttA

=tB " tA

tA= G

HMEc2RE RE + H( )

,

i.e., tB > tA, so time passes more quickly at altitude.

Although this is a very small shift, in October 1971 the effect was verified by J.C. Hafele and R.E. Keeting (Science 177, 166 (1972)) who flew around the world several times on commercial flights carrying cesium atomic clocks!

Time difference (ns)Eastward Westward

Gravitational 144± 14 179± 18Kinetic (time dilation) !184 ± 18 96± 10Predicted net !40 ± 23 275± 21Observed values !59 ± 10 273± 21

"t

t

1# 10!10

5# 10!11

0 1# 106 8# 105 6# 105 4# 105 2# 105 H (m)

ISS

"tt= G

HMEc2RE RE + H( )

Question 9: The International Space State (ISS) orbits the Earth at an average height of 369km. (a) Do clocks on the ISS tick faster or slower compared with identical clocks on Earth, due to the gravitational effect? (b) What is the time difference between identical clocks on the ISS and on Earth over a period of one day (24 hours)? Use the Newtonian approximation.

(The radius of the Earth is 6.37 !103km and the mass of

the Earth is 5.98 !1024 kg.)

(a) A clock in a smaller gravitational field ticks faster than a clock in a larger gravitational field. Since the gravitational field is smaller at a height of 369km

( 3.69 !105m) compared the value at the surface of the

Earth, the clock on the ISS ticks faster than an identical clock on Earth.

(b)

If time intervals on the ISS and Earth are tB and tA, respectively, we have

tB = tA 1" G

HMEc2RE RE + H( )

#

$ % %

&

' ( (

"1

.

But G

HMEc2RE RE + H( )

= 3.81!10"11, which is << 1.

EA = hfA = h

)A:

UA = "G

MEmRE

EB = hfB = h

)B:

UB = "G

MEmRE + H( )

H

! tB = tA 1 + G

HMEc2RE RE + h( )

"

# $ $

%

& ' ' ( tA 1 + 3.81)10*11( )

i.e., +t = tB * tA = 3.81)10*11tA.

With tA = 24 hours = 8.64 )104s, we find clock B (the ISS

clock) gains 3.29µs a day.

Note: if H << RE, i.e., close to the Earth’s surface, then

G

HMEc2RE RE + H( )

, GHMEc2RE

2 =gHc2 .

! tB = tA 1 +

gHc2

" # $

% & '

i.e.,

+ttA

=gHc2 = (1.09 )10*16)H

with H in meters.

So, what does a photon “look” like if it has to have both wave-like and particle-like properties?

Photons are sometimes visualized as wave-packets. In chapter 1 we saw that a wave-packet can be formed by the superposition of waves spanning a central wavevector. The electromagnetic wave has frequency and wavelength (like a wave), yet it is also discrete and localized in space and time (like a particle).

! E (x)

x

Wave packet

!

v = c = f!

Class discussion question: Can you think of a reason why such a picture for a photon is not an entirely appropriate description?

! E (x)

x

Wave packet

!

v = c = f!