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Using time to measure distance
Question 20W: Warm-up Exercise
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Read the information below and answer the questions that follow.
A Swedish instrument to measure distance
It was in pursuit of the value of the speed of light that a Swedish physicist named Erik Bergstrand
began experimenting with a system that bounced…light off of a reflector at a precisely measured
distance. Since the time between the outgoing and return of the signal was so infinitesimally short
that it could not be measured directly, …an oscilloscope was used to measure light travel time.
Having used distances to determine the speed of light to his satisfaction, Bergstrand then
proposed to reverse the process. In the late 1940s, he asked the company Svenska Aktiebolaget
Gasaccumulator (AGA) to develop the first commercial geodetic distance meter, and thus
development of the Geodimeter began.’
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Open the JPEG file
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Top: An AGA 2A made in 1959, showing the front control pane. Bottom: The AGA 2A from the side
showing the separate measuring and optical units.
First produced in Sweden in 1953, the Geodimeter was based on the principle that a beam of
light could be used to measure the distance from one point to another. The Geodimeter could be
used to measure base lines with greater precision than previously possible and accomplish in
hours what had previously taken weeks. The instrument worked by sending out a beam of light to
a reflector and then measuring the time required for the beam to travel to the reflector and back.
Using the known constant speed of light, the exact distance between the measurement points
could be determined. The Geodimeter was the first electronic distance measurement instrument
to use visible light to measure distance.
Source: http://celebrating200years.noaa.gov/distance_tools/0411.html (NOAA = National Oceanic
and Atmospheric Administration)
Answer these questions
Geodimeters were used for surveying distances from a few tens of metres to several kilometres. The
speed of light is 3.00  108 m s–1.
1. How long would it take light to travel 1.00 km?
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2. The images show the geodimeter itself. What additional piece of apparatus must be used to make
a distance measurement when surveying?
3. What time interval would have to be measured if a geodimeter was used to survey a distance of
1.0 km?
The timebase on an oscilloscope is set to 2.0 s cm–1 and the oscilloscope is used with the
geodimeter above to measure a distance of 1.0 km. The operator sees two pulses on the screen, one
representing the emitted pulse and the other representing the returning pulse, as shown in the
diagram below.
direction of movement of oscilloscope beam
4. Given these data, how far apart (in cm) must the two peaks be?
5. For peaks to be clearly distinguished they must be at least 0.20 cm apart on the screen. What is
the shortest distance that can be surveyed with this system?
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A major part of any electronic distance measurement instrument (EDMI) is the reflector that
returns a beam of light back to an instrument which measures the time required for the light to be
emitted and bounce back. Over the evolution of EDMI, the types of reflector systems used have
also evolved. Initially, Swedish physicist Erik Bergstrand, inventor of the Geodimeter, used a
plane mirror as a reflector, then a spherical mirror, and then a prism system.
Source: http://celebrating200years.noaa.gov/distance_tools/0411.html
Open the JPEG file
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These images show a laser signal and prism reflector system in action.
6. What is the advantage of using a spherical mirror over using a plane mirror?
7. What are the advantages of lasers over conventional light sources for this method of distance
measurement?
8. Modern electronic distance measurers (EDMs) can achieve a precision of 1.0 mm. What time
difference between emitted and reflected pulses does this correspond to?
A similar technique has been used to measure the distance from the Earth to the Moon to a precision
of a few cm:
The first laser ranging retroreflector was positioned on the Moon in 1969 by the Apollo 11
astronauts. By beaming laser pulses at the reflector from Earth, scientists have been able to
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determine the round-trip travel time that gives the distance between the two bodies at any time to
an accuracy of about 3 centimeters. The laser reflector consists of 100 fused silica half-cubes,
called corner cubes, mounted in a 46-centimeter square aluminium panel. Each corner cube is
3.8 centimeters in diameter. Corner cubes reflect a beam of light directly back toward the point of
origin.
Source: http://sunearth.gsfc.nasa.gov/eclipse/SEhelp/ApolloLaser.html
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9. The average time for light to make a round trip to the Moon is 2.567 s. What is the average
distance to the Moon?
Units for distance measurement
Question 30W: Warm-up Exercise
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These questions are about the various units used for measuring distance. Two units used in
astronomy are:
the astronomical unit (AU) which is the mean distance of Earth from Sun – 1 AU = 1.5 1011 m
the parsec (pc) – 1 pc = 2.1  105 AU
Take c = 3.00  108 m s–1.
You will have to research or estimate some of the data needed for the questions.
Distances in light-seconds
Convert each of the following to distances in light-seconds
1. The distance from London to Manchester (about 300 km).
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2. The distance to the Moon (about 385 000 km).
3. The distance to the Sun (1 AU = 150 000 000 km).
4. The distance to Jupiter when it is 4.2 AU from Earth.
5. The distance to the nearest star, Alpha Centauri (1.3 pc).
Choice of unit for distance
What units or units would be appropriate for measuring each of the following distances?
6. The diameter of the Moon.
7. The distance from Earth to the Moon.
8. The distance from the Sun to Mars.
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9. The diameter of the Milky Way.
Time needed for space travel
It took the Apollo astronauts about 4 days to reach the Moon (about 385 000 km).
10. What was their average speed in m s–1?
11. How long would it take a spacecraft travelling at the same speed as Apollo to reach the nearest
star? Comment on your answer.
Cosmological expansion
Question 40W: Warm-up Exercise
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Look around you and pick out two objects, one (let’s call it A) at about 1 m from you and the other (B)
at about 2 m from you. Now imagine that the room has been uniformly enlarged by a factor of 2 so
that all the distances have doubled.
1. What is the new distance to object A?
2. What is the new distance to object B?
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Now assume that the expansion took place steadily over a period of 5 s.
3. How would both A and B have appeared to move during this time?
4. What was the average speed of A relative to you during the expansion?
5. What was the average speed of B relative to you during the expansion?
6. Why does B have a greater average speed?
7. Sketch a graph of recession (moving away) speed against object distance (use the original
positions) and put on points for A and B.
8. Add points for objects that were originally at 0.5 m and 1.5 m from you.
9. Complete the graph by drawing a best-fit straight line.
10. Describe the mathematical relationship between object distance and recession speed that is
shown by your graph.
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11. In the 1920s. Vesto Slipher and Edwin Hubble gathered data that showed that distant galaxies
are all moving away from us and that the speed at which they are moving away is directly
proportional to their distance. How do cosmologists account for these observations?
When does the speed of light matter?
Question 50W: Warm-up Exercise
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The speed of light is a defined constant: c = 299 792 458 m s–1 (about 3.0  108 m s–1). This is much
faster than any speed you will meet in everyday life, so the effects of special relativity are usually too
small to be detected. This question will give you a better idea of how fast light travels and of when you
might have to take account of relativistic effects.
1. The distance from the Earth to the Moon is about 400 000 km. How long would it take light to
travel this distance?
2. Express the distance to from the Earth to the Moon in ‘light-seconds’.
3. The speed of the Apollo 11 spacecraft on its journey to the Moon was about 36 000 km h–1. This
is over 100 times greater than the maximum speed of a formula 1 car. Express this speed as a
fraction of the speed of light.
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4. Roughly how long did it take for the Apollo spacecraft to travel from the Earth to the Moon?
5. Radio waves were used to communicate with the astronauts after they had landed on the surface
of the Moon. What is the minimum delay between sending a signal from Earth and receiving a
reply from the astronauts on the Moon?
6. NASA is considering sending a manned mission to Mars. When Mars is closest to the Earth its
distance is still 2.0  1011 m. What is the minimum time delay between sending a message from
Earth to an astronaut on the surface of Mars and receiving a reply? Is there any way we can
reduce this delay?
The relativistic time-dilation factor 
Question 60W: Warm-up Exercise
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The size of relativistic effects depends on the value of the ‘gamma-factor’:

1
1
v2
c2
For example, if you compare time on a your wristwatch with the time on a wristwatch worn by a friend
in a rocket moving past you at high speed (e.g. by exchanging text messages via mobile phone) then
her watch will run slow compared to yours. If gamma equals 2 then her watch ticks at half the rate of
yours. For an hour passing on your watch only half an hour passes on hers.
1. Complete the table below to show how the rate of her watch (as seen by you) varies with her
speed. The first line has been done for you.
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Speed
Ratio v / c
‘gamma-factor’
Rate of time
passing on moving
watch compared to
your watch
Time passed on rocket
compared to 1 h of
your time
0.20c
0.20
1.02
0.98
59 min
0.40c
0.60c
0.80c
0.90c
0.95c
0.99c
2. Plot a graph to show how the ‘gamma-factor’ changes with speed.
3. What is the value of the ‘gamma-factor’ for v = c? How much time would pass in the rocket during
1 h of your time if the rocket could travel at the speed of light?
4. Do physicists think we will ever be able to build a rocket that can travel at the speed of light?
Measuring distances within the solar system and beyond
Question 20S: Short Answer
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These questions encourage you to think about the scale of the galaxy and beyond. They do this by
showing how astronomers use different distance units as the scale of the distance requires, going
from metres to astronomical units (linking with parallax measurement) to parsecs and finally to
light-years.
Distances in metres
From the Sun to the Earth is 149.6 million kilometres, or 149.6 thousand million metres. Pluto is 5907
million kilometres from the Sun, a number well beyond the scale of ordinary human experience.
The astronomical unit
The mean radius of the Earth’s orbit is called an astronomical unit or AU.
1. If Jupiter is 778.3 million kilometres from the Sun, how many times further away than the Earth is
it? How far is this in AU?
2. Estimate the distance from the Sun to Pluto in AU
The parsec
A unit of distance in common use amongst astronomers is the parsec. As the Earth moves in its orbit
around the Sun, the position of nearby stars against the background of very distant stars seems to
change.
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distant stars
nearby star

d
Earth
R
Sun
The parallax of a star is the angle labelled  on the diagram. These angles are very small because
even nearby stars are very distant compared with the Earth’s orbital diameter. Small angles can be
expressed in minutes and seconds; one minute of arc is 1 / 60 of a degree; one second of arc is 1 /
60 of a minute. If the radius of the Earth’s orbit is R then the distance to a star with parallax  is:
d
R
.
tan
A star at a distance of 1 parsec (1 pc) from us has a stellar parallax of 1 second.
3. How far is 1 parsec in metres?
4. How far is this in astronomical units?
The light-year
The nearest star to us is Proxima Centauri and light takes 4.2 years to reach us from it. On the Earth,
we are used to light apparently travelling instantaneously, but even at the ultimate speed in the
Universe (approximately 3.0  108 m s–1) light still takes 3.3 microseconds to travel a kilometre.
Distances even in the solar system are vast and the radio signals that travelled from the first men on
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the Moon to Earth took long enough to cause considerable delay to communications.
5. The Moon is 3.84  105 km from Earth. What would the minimum time delay be between a
question asked on Earth and its reply arriving from the Moon?
6. How long does light take to reach us from the Sun?
7. How far does light travel in one year? How far is Proxima Centauri from us in metres? In AU? In
parsecs?
These last two answers should have convinced you that using light travel time is the only sensible
distance unit to use for the stars and beyond. Our Milky Way galaxy is 100 000 light-years in
diameter; the nearest galaxy to ours is the great Andromeda galaxy, M31, a distance of 2.2 million
light-years from us. In fact, this is said to be the most distant object visible to the unaided eye – on a
very clear night.
8. How far is it to Andromeda in AU? In parsecs?
Comparing intensities for lamps
Question 40S: Short Answer
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Comparing brightnesses
A low-power lamp at close range can often appear as bright as a high-power lamp much further
away. Similar reasoning is used to calculate the brightness of stars.
Two filament lamps
Here you compare two incandescent filament lamps, one marked 2.5 V, 0.3 A and a second 100 W,
240 V. Assume that approximately 20% of the power dissipated by the lamps results in visible
photons being emitted uniformly, and that both are being run at normal brightness.
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1. Estimate the brightness of the 100 W lamp from 3 metres away, that is the luminous flux density
in W m–2.
2. Calculate how close you must be to the 2.5 V lamp to see it as being the same flux density.
Two floodlights
This question is about a 15 W bicycle headlight and a 1500 W exterior security floodlight. Both the
floodlight and the headlight have approximately the same shaped beam, a cone that covers
approximately 1 / 10 of a sphere.
3. Estimate the flux density of the bicycle lamp from 20 metres away.
4. Your pupil has a diameter of approximately 0.01 m.
How much light enters your eyeball per second?
5. The floodlight looks just as bright as the bicycle lamp.
How far away is it?
Jupiter and Saturn close together in the sky
Question 45S: Short Answer
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Is Jupiter really brighter than Saturn?
The data given in these questions let you compare how bright Jupiter and Saturn really are, knowing
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how bright they look and how far away they are.
Jupiter and Saturn at opposition together
In the winter of the years 2000 to 2001 Jupiter and Saturn appeared very close together in the night
sky. They were both in the south at midnight, which means that the Earth was directly between them
and the Sun. This is called an ‘opposition’. It also means that the two planets were both at the same
time at their closest to the Earth.
At this time Jupiter was the brightest object in the night sky apart from the Moon. Saturn was very
bright too, though not as bright as Canis Major (Sirius or the ‘Dog Star’) to be seen below and to the
left of the constellation Orion.
Measurements of the light received at the Earth from Jupiter and Saturn showed that Jupiter
appeared to be about 10 times brighter than Saturn.
The radii of the approximately circular orbits around the Sun of Earth, Jupiter and Saturn are:
Planet
Approximate radius of orbit in millions of km
Earth
150
Jupiter
778
Saturn
1430
1. Remembering that at this time the Earth lay directly between the Sun and both Jupiter and
Saturn, calculate the distances from Earth to Jupiter and from Earth to Saturn.
2. Show that at that time Saturn was very nearly twice as far from the Earth as was Jupiter.
3. Suppose for a moment that Jupiter and Saturn were actually equally bright; that is, reflected the
same amount of light from the Sun. If that were true, how many times brighter than Saturn would
Jupiter have looked, bearing in mind their different distances from the Earth?
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4. Actually, Jupiter appeared 10 times brighter than Saturn, as seen from the Earth.
Calculate the ratio of the actual brightnesses (amount of light sent out) of Jupiter and Saturn.
5. Suggest one reason why Saturn sends out less light than Jupiter.
Brighter stars aren’t always nearer
Question 46S: Short Answer
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Can the brightness of stars indicate their distance?
If all stars shone equally brightly, the ones that are further away would look fainter. But if stars vary a
lot in actual brightness, how bright they look is a poor guide to how far away they are. Here you
calculate actual brightnesses from how bright stars look and how far away they are.
A selection of well-known stars
If the Sun were a typical fairly nearby star in the sky, it would be rather faint, in fact only just visible to
the naked eye. In the table below, the brightness of the Sun as it would look on Earth if it were 10
parsecs (32.6 light years) away, is arbitrarily taken as 1 unit. Other apparent brightnesses are given
as multiples of this standard brightness. The stars listed are all amongst the brightest-seeming stars
in the night sky of the northern hemisphere. They all therefore give out more light than the Sun.
Star
type
Distance in light
years
Apparent
brightness as
multiple of
brightness of
Sun if it were at
a distance of
32.6 light years
(10 parsec)
Absolute
brightness:
actual power
emitted as
multiple of
power from
Sun
Sun
Ordinary yellow star
If moved to a distance of
1
1
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Star
type
Distance in light
years
Apparent
brightness as
multiple of
brightness of
Sun if it were at
a distance of
32.6 light years
(10 parsec)
Absolute
brightness:
actual power
emitted as
multiple of
power from
Sun
Sun
Ordinary yellow star
If moved to a distance of
32.6 light years (10
parsec)
1
1
The ‘Summer Triangle’ of three bright stars high in the summer sky
Altair
Quite nearby white star
16
42
10 approximately
Vega
Quite nearby bright
blue-white star
25
83
49
Deneb
Distant blue-white
supergiant. One of the most
luminous naked eye stars
known
1630
26
65 000
The two brightest stars in Orion the Hunter, and Sirius the 'Dog Star' near Orion
Betelgeuse
Very large unstable red
giant, diameter 300 – 400 ×
Sun
1040
54
55 000
Rigel
Blue-white supergiant
864
76
53 000
Sirius
Nearby bright white star
8.64
330
23
360
13
1600
The Pole Star
Polaris
Yellow supergiant Cepheid
variable
Three red giants. Antares is the red ‘eye’ of the scorpion in Scorpio. Aldebaran is the red eye of the
bull in Taurus. Arcturus is the 4th brightest star in the sky, and may be like the Sun will be at a later
stage of its evolution
Antares
Red supergiant, diameter
300 × Sun
619
35
13 000
Arcturus
red giant, mass similar to
Sun, diameter 27 × Sun
33
88
90
Aldebaran
Red giant
62
39
141
Data adapted from J B Kaler ‘Stars’ Scientific American Library 1992
You are going to show that these stars vary very greatly in how much light they actually emit.
Print out the table to answer the following questions.
1. List the stars in order of their distance from us, from nearest to furthest.
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2. List the stars in order of how bright they appear, from brightest to faintest.
3. Remember that in the comparisons the Sun’s apparent brightness at a ‘standard distance’ of 32.6
light years (10 parsec) is taken as 1. Notice that Altair is almost exactly half this distance away
(16 light years). By what factor would Altair’s apparent brightness decrease if it were twice as far
away, at the ’standard distance’?
4. Show that Altair emits about 10 times more light than the Sun.
5. Deneb is the most distant of these stars, at 1630 light years. If it were at the ‘standard distance’ of
32.6 light years, how many times brighter would it appear than it now does?
6. Show that Deneb emits between 50 000 and 100 000 times more light than the Sun.
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7. Sirius is one of the brightest looking stars in the sky. The Pole Star Polaris is not remarkably
bright. Show from the data that Polaris emits more than 50 times as much light as Sirius.
8. After Deneb, which two stars are the next most powerful emitters of light?
9. Aldebaran is about twice as far away as Arcturus, and appears roughly half as bright. What
inference can you draw about the ratio of the amounts of light they emit?
10. List these stars in order of their actual brightness (the amount of light they emit).
Trip times tell distances
Question 50S: Short Answer
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Constant speed, different times
Electromagnetic waves travel at constant speed. Choose a wavelength that is not absorbed too
much, find a good reflector, and then you can use a pulse of waves to see how far away the reflector
is. Take the speed of light, c = 3.00  108 m s–1.
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Moon, Sun
1. A radar measurement of the distance to the Moon gives a round-trip time of 2.57 s.
Calculate how far away the Moon is.
2. The Sun is said to be 8 light-minutes away.
How far is this in km?
Planets
3. Earth has a radius of orbit of 1.496  1011 m. Venus has a radius of orbit of 1.082  1011 m.
Based on this information alone, what might you expect to be the minimum and maximum
round-trip times for radar pulses reflected from Venus?
4. Pluto has a radius of orbit of 59.13  1011 m, and a radius of 1151 km.
Outline the problems that might arise in trying to radar range Pluto. Provide some calculated
estimates to support your assertions where possible.
Doppler detection
Question 52S: Short Answer
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These questions are about uses of the Doppler effect by the police and by the military.
Doppler speed traps
The police use ‘speed cameras’ to catch motorists who break the speed limit. A speed camera works
using the Doppler effect, in which the frequency of a transmitted radio wave reflected off a vehicle
changes when there is relative motion between the camera and the motor vehicle. If both objects are
standing still there is no relative motion between the camera and the vehicle and the reflected signal
has the same frequency as the transmitted signal. The change in frequency f is given by the Doppler
shift relationship
f v

f
c
where v is the speed of the vehicle.
1. A common type of speed camera is called the GATSO; this transmits a radar beam at a
frequency of 24.11 GHz. A car comes towards a fixed GATSO speed camera at 50 mph (22.4 m
s–1) in a 30 mph zone. What is the change in frequency of the reflected radar signal?
Military target identification
Doppler radar can be used to identify a target such as a helicopter, as well as to measure its speed
relative to the radar set.
The main body of the helicopter reflects a Doppler shifted signal.
However, the rotating helicopter blades also reflect the radar signal. This spreads out the returned
signal over a range of frequencies, as shown in the graph below. The pattern of the spectrum of
returned frequencies can be used to identify the type of target.
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motion of rotor blade
motion of rotor blade
m otion of aircraft body
range at Doppler shifted signal (due to blade rotation)
Doppler shift
of main body
transmitted
frequency
Doppler shift of
rotor blades
relative to the
main body
echo from
main body
echo from
rotor blades
0
frequency
2. Explain how the helicopter blades produce the wide spectrum of frequencies shown.
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Doppler shift
of main body
transmitted
frequency
echo from
main body
0
frequency
Doppler shifts in astronomy
Question 55S: Short Answer
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Calculating speeds and changes of wavelength
You are given data for a number of astronomical objects, relating their speed of approach or
recession to shifts in the wavelength of light they emit.
Doppler shifts
As long as the relative radial velocity v of an object is small compared to the speed of light c, then
because of the Doppler effect, light of wavelength from the object will be shifted in wavelength by an
amount , where
 v


c.
The wavelength increases if the object is receding and decreases if it is approaching. The speed of
light c = 3.00 108 m s–1.
The Crab nebula is the remains of a supernova which exploded in 1054 AD, being seen as a new
bright star by Chinese astronomers. Now it is a cloud of rapidly expanding gas, with a neutron star
(pulsar) at the centre. The gas is expanding with a radial velocity of 1300 km s–1.
1. Calculate the shift in wavelength of the red hydrogen line at 656 nm.
2. Calculate the shift in wavelength of the blue hydrogen line at 486 nm.
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3. What is the percentage change in wavelength in the two cases? Why is it the same for both?
4. Estimate the present width of the gas cloud in light years. What probably false assumption do you
have to make?
An undistinguished-looking star unglamorously called P-Cygni, in the constellation of Cygnus (The
Swan), turns out to be one of the brightest stars ever seen, brighter than two million Suns. So violent
is it that it is throwing out matter at the rate of 10–4 Suns per year. The speed with which gas is thrown
out from the star can reach 3% of the speed of light. The spectrum of P-Cygni shows lines from
ionised gas, with a dark absorption line on the blue-side of each line. The absorption comes from the
light being partly absorbed by the fast moving gas thrown out of the star. The absorption line is shifted
to the blue because of the Doppler effect, being absorbed by gas moving towards us.
5. The emission line from ionised helium at 438.8 nm has an absorption line next to it at 425.6 nm.
Calculate the change in wavelength due to the Doppler effect.
6. Express the change in wavelength as a percentage.
7. Calculate the velocity of the absorbing gas.
8. There is another emission line from ionised helium at 400.9 nm. At what wavelength will its
absorption line appear, if the same mechanism applies?
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Atoms and molecules in hot gases move randomly at speeds of the order of hundreds of metres per
second. A particle emitting a photon may be travelling in any direction, towards or away from the
observer. As a result, narrow spectral lines become broader, due to Doppler shifts.
9. Explain why the spectral lines are broadened.
10. Make an estimate of the width of the line as a percentage of its wavelength, due to this thermal
broadening.
11. Suggest how observing widths of spectral lines could be used to estimate the temperature of the
gas in a distant star.
Binary stars
Question 60S: Short Answer
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Orbiting binary stars:
A type of variable star
This type of variable star consists of two stars orbiting around each other. When the dimmer star is in
front of the brighter one, the observed intensity is at a minimum, and when the two stars are side by
side, the observed intensity is at a maximum.
Relative velocities in an orbiting binary star system
The two stars in this question are similar in mass, but star B is cooler and dimmer than star A. They
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orbit about their common centre of gravity X, as shown in the diagram.
C
v
A
X
B
v
In one such pair of stars, the time for one revolution is 73 days.
The stars are a distance of 1.0 × 1011 m apart.
1 day = 8.64 × 104 s
1. What are the velocities of stars A and B, relative to the point X?
2. What is the velocity of star A, relative to star B?
The point X is not still with respect to Earth. It is moving at a velocity of 64 km s–1 in the direction
away from Earth.
3. What are the velocities of stars A and B, relative to Earth?
4. What is the velocity of either star, relative to Earth, when it is at the position C?
The Doppler shift in wavelength for a star moving at a velocity v relative to the observer is
 v


c
if the speed v is small compared with the speed of light, c. The speed of light, c, is 3.00 × 108 m s–1.
5. Find the wavelength shift  that an astronomer (on Earth) would observe in a spectral line of
wavelength 589.0 nm from star A in the position shown in the diagram. Is this a red shift or a blue
shift?
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6. Find the wavelength shift  that an astronomer (on Earth) would observe in a spectral line of
wavelength 589.0 nm from star B in the position shown in the diagram. Is this a red shift or a blue
shift?
7. Find the wavelength shift  that an astronomer (on Earth) would observe in a spectral line of
wavelength 589.0 nm from either star when it is in the position C on the diagram. Is this a red shift
or a blue shift?
8. In many binary stars, the two stars are not perfectly lined up when seen from Earth. This means
that there will not be any dimming or brightening of the light, because the dimmer star will not
block out the light from the brighter one. How might an astronomer tell, from the spectrum, that
there are in fact two stars moving about their common centre of mass as described in this
question?
Practice with the relativistic Doppler shift equation
Question 70S: Short Answer
Teaching Notes | Key Terms | Answers
Quick Help
These questions give you practice with the relativistic Doppler shift equation, which says that the ratio
of wavelength observed from a moving source to the wavelength emitted by the source is
k
 observed
1 v / c

 emitted
1 v / c
v / c = 0.5
Imagine that a source of light waves is moving away from you at half the speed of light.
1. For v / c = 0.5:
write down the value of the term 1 + v / c
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write down the value of the term 1 – v / c
write down their ratio
calculate the value of the Doppler shift factor k
compare the value of k with the value of the non-relativistic shift 1 + v / c. Which is the larger?
v / c = 3 / 5 = 0.6
2. Repeat the calculations in question 1 for a larger relative velocity, v = 3 / 5 c.
v / c = 0.9
3. Repeat the calculations in question 1 for the very large relative velocity, v = 0.9 c.
v / c = 0.1
4. Repeat the calculations in question 1 for the smaller relative velocity, v = 0.1 c.
Now that you have seen some examples of how the relativistic Doppler shift varies with the ratio v / c
you should be able to estimate some values:
5. Estimate roughly the value of v / c at which the relativistic Doppler shift is twice the non-relativistic
value.
6. Estimate to three decimal places the relativistic Doppler shift for v / c = 0.01.
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Calculating v / c from the Doppler shift factor k
From the equation
k
 observed
1 v / c

 emitted
1 v / c
you can show that:
v k2 1

c k2 1
7. The distant quasar 3C48 has a Doppler shift factor k = 1.368. If this were attributable to a speed
of recession, what would be the ratio v / c?
8. The factor k2 is also the ratio of the frequency of a speed trap radar signal reflected back from a
car coming towards the speed trap, to the frequency of the signal transmitted towards the car. If
the speed trap equipment can detect a change in frequency of 1 part in 108, what is the smallest
car speed it can detect?
Relativistic Doppler effect assuming time dilation
Question 75S: Short Answer
Teaching Notes | Key Terms | Answers
Quick Help
In the Advancing Physics A2 student’s book the relativistic Doppler shift is derived from arguments
about radar measurements of distance and speed. The time dilation factor

1
1
29
v2
c2
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is shown to follow as a consequence. However, there are independent arguments to predict that time
dilation exists, notable the light clock argument. So here you can show that if you assume that time
dilation exists, the relativistic Doppler equation follows.
Imagine a distant star that is emitting radiation of wavelength 0 (in the reference frame of the star).
The star is in a galaxy that is moving away from us at a velocity v . The motion of the star causes a
time dilation effect that increases the time period T0 of the radiation by a factor

1
1
v2
c2
Non-relativistic argument
1. Forgetting time dilation for the moment, write down an expression for the distance the star moves
in time T0.
2. Write down an expression for the extra time the light takes, to travel this extra distance.
3. Use the answer to question 2 to show that the time between wave peaks arriving at Earth will be
T = ( 1 + v / c )T0.
Relativistic argument
4. Now introduce the effect of time dilation. As seen from Earth, the time T0 is stretched by the factor
. Modify the answer to question 3 to take this into account.
5. Use the result of question 4 to show that the wavelength  of radiation arriving at Earth will be
larger than the wavelength 0 emitted by the star, in the ratio:
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
  1  v / c 
0
6. Show that this reduces to the classical expression

 1 v / c
0
when v << c.
7. How serious is the error if you use the non-relativistic formula when v = 0.10 c?
8. How serious is the error if you use the non-relativistic formula when v = 0.90 c?
Practice with the relativistic time dilation equation
Question 80S: Short Answer
Teaching Notes | Key Terms | Answers
Quick Help
These questions give you practice with the relativistic time dilation equation, which says that time
intervals ticked out by a clock moving with you (wristwatch time) are related to the time intervals t
ticked out by a clock moving relative to you by the relation t = where

31
1
1 v 2 / c 2
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v / c = 0.5
Imagine that a source of light waves is moving away from you at half the speed of light.
1. For v / c = 0.5:
write down the value of the term v 2 / c 2
write down the value of the term 1 – v 2 / c 2
calculate the reciprocal 1 / ( 1 – v 2 / c 2 )
calculate the square root, to get the factor

1
1 v 2 / c 2
if there were no time dilation would be exactly 1. How important is time dilation in this case,
when v / c = 0.5?
v / c = 3 / 5 = 0.6
2. Repeat the calculations in question 1 for a larger relative velocity, v = 3 / 5 c.
v / c = 0.9
3. Repeat the calculations in question 1 for the very large relative velocity, v = 0.9 c.
v / c = 0.1
4. Repeat the calculations in question 1 for the smaller relative velocity, v = 0.1 c.
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Estimating values
Now that you have seen some examples of how the time dilation varies with the ratio v / c you should
be able to estimate some values:
5. Estimate roughly the value of v / c at which the time dilation  = 2.
6. The answer to question 4 (v / c = 0.1) suggests that for small v / c the value of  is approximately
1 + (½) v 2 / c 2. Using this rule of thumb, estimate the time dilation factor for v / c = 0.01.
Estimating v / c from the time dilation factor 
7. Particles in an accelerator can easily reach = 1000. For  to be 1000:
what must be the value of 1 / = (1 – v 2 / c 2 )½?
what then must be the value of 1 – v 2 / c 2 ?
by how much must v 2 / c 2 differ from 1?
approximately by how much must v / c differ from 1? (Use the fact that (1 – x)½ = 1 – (½) x if x
is small.)
Light clocks and time dilation
Question 85S: Short Answer
33 Advancing Physics
Teaching Notes | Key Terms | Answers
Quick Help
This question will lead you through a derivation of Einstein’s famous time dilation factor:

1
1 v 2 / c 2
The derivation is based on thinking about a light clock – this is a clock that ‘ticks’ as light bounces
back and forth between a pair of parallel mirrors. In the light clock the mirrors are placed a distance d
apart. You are going to compare a light clock that moves with you (i.e. is at rest relative to you)
against one that moves sideways past you at some speed v.
si tting b esid e the cloc k
d = c
m irrors d = c apart
time out an d ba ck
(1 tic k ) = 2
c lock reco rds
w ristw atc h tim e 
The clock ticks each time a light pulse leaves the lower mirror. The spacing between the mirrors is
such that the interval between ticks is 2.
Now think about another identical clock moving sideways past you at speed v. You will work out the
new time between ticks of this clock moving past you.
clock travelling pa st you at speed v
ct
c
x = vt
ct
x = vt
you see the light ta ke a longe r path
but the speed is still c
so the time t is l onger
the moving clock ticks more slowly
1. How does the motion of the light clock affect the speed of light?
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2. How does the motion of the light clock affect the length of light path between ticks (as seen by
you)?
3. Write down an expression in terms of v and t for the horizontal distance x moved by the lower
mirror as a light pulse goes from the lower mirror to the upper one.
In this time t the light must move a distance ct because the speed of light is unchanged by relative
motion. The distance between the mirrors is unchanged, so is still equal to c, where  is time as
recorded on the clock when it travels with you.
4. Use Pythagoras’s theorem to write down a relation between the distance c between the mirrors
and the distances ct and x. Rewrite as a relation between , t, v and c.
5. Use the answer to question 4 to show that 2 = t 2(1 – v 2 / c 2 ).
6. Now use this to show that t = , where

1
1 v 2 / c 2
7. Calculate t when  = 1 ns and v = 0.75c. Explain why this means that time for the ‘moving’ clock
runs slow.
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8. There is also a digital wristwatch travelling with the moving light clock. Which of the two light
clocks will it keep time with?
9. How would the rates of the two mirror clocks compare for an observer travelling past you
alongside the moving clock?
Red-shifts of quasars
Question 95S: Short Answer
Teaching Notes | Key Terms | Hints | Answers
Quick Help
Red shifts from the expansion of the Universe
The questions below are about the red-shift of light from quasars (‘quasi-stellar radio source’
abbreviated to QSR). You will need to know that the change in the wavelength of the light is due to
the expansion of the Universe. Thus the ratio:
 received
 sent
is equal to the factor by which the Universe has expanded during the time of travel of the light. The
red shift z is usually expressed as:
z
change in wavelength 

original wavelength
 .
Since
 received   sent   ,
the ratio
 received

 
 sent
 1 z
 sent
 sent
Thus for a red shift z the corresponding expansion of the Universe is z + 1.
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The first quasar identified
In 1962, Australian radio astronomers got an accurate position for one of the sources in the Third
Cambridge Catalogue of radio sources. Its catalogue number was 3C273. They did it by watching
3C273 being eclipsed by the edge of the Moon. Optical telescopes picked it up as a very faint
star-like object with a faint ‘jet’ coming out of it. Opinion was generally that it was a peculiar nearby
star. But when Maarten Schmidt took its spectrum, he recognised the spectrum as part of the
hydrogen spectrum, shifted to longer wavelengths by 15.8%.
1. One of the hydrogen lines Schmidt observed normally has wavelength 486 nm. What wavelength
did Schmidt observe it to have?
2. Another hydrogen line has wavelength 434 nm. What wavelength does it have, red shifted by the
same fraction?
3. By what factor has the Universe expanded since light now reaching us left 3C273?
Another quasar, already seen
Earlier, another source, 3C48 had been identified with an even fainter ‘star’. Its spectrum was
reported as showing 'a combination of lines unlike that of any other star known'. Given Schmidt’s
identification of red shifted lines in the 3C373 spectrum, others quickly identified lines in the 3C48
spectrum that had an even bigger red-shift, of 36.8%.
4. If a line in the 3C48 spectrum appeared in the red at 650 nm, what would its original wavelength
have been? Is this in the visible region?
5. If the scale of the Universe is now about 14 billion light years, what was its scale when the light
left 3C48?
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Red-shifts make invisible radiation visible
Hydrogen atoms emit strongly in the ultraviolet. One such line is at wavelength 122 nm. This line was
found appearing in even more distant quasars at the far blue end of the spectrum, wavelength about
360 nm.
6. What is the percentage red shift in this case?
7. What was the scale of the Universe when this light was emitted?
Extreme red-shifts
For a time it was thought that there were no quasars to be found beyond 200% red shift. However,
since then red shifts discovered have reached as high as a factor 10.
8. In what part of the spectrum would a photon now seen in the visible have started out, if the red
shift is 5?
9. By what factor has the photon energy decreased in this case?
10. By what factor has the Universe expanded in this case?
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Cosmic microwave background radiation
Question 100S: Short Answer
Teaching Notes | Key Terms | Hints | Answers | Key
Skills
Quick Help
The Universe as a perfect (black-body) radiator at 2.7 K
This question is about the temperature of the Universe and the radiation that has filled it since neutral
atoms were first produced.
Plasma absorbs photons
It is thought that the early Universe was too hot for electrons and protons to combine together to form
neutral atoms, but they existed as a mixture of positive ions and electrons (a plasma). In these
conditions photons cannot travel freely, as electromagnetic photons continually interact with the ions,
so dense plasma is not transparent.
1. 'Hot atoms' are atoms with large quantities of random kinetic energy. Explain why such atoms
would break down into electrons and positive ions.
core  107 K
radiative zone  107 K
convective zone
photosphere 5800 K
2. The radiative zone of a star like our Sun consists of dense plasma. Explain why the photons
generated in the Sun’s core take millions of years to travel across the radiative zone.
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The photons generated in the core are gamma photons of frequency  4 × 1017 Hz. The radiative
zone gets cooler and cooler from the core to the convective zone. By continual absorption and
emission in the radiative zone, photons of lower energy are produced. Eventually, the photon energy
is transferred by convection to the photosphere (the Sun’s surface), where photons of frequency  6 ×
1014 Hz are emitted.
3. Show that, for each gamma photon produced in the core, about 600 photons are emitted from the
Sun’s surface.
The Universe becomes transparent
Electromagnetic radiation could not travel through the early Universe, filled with charged particles like
the radiative zone of a star. As the Universe expanded, it cooled.
About 105 years after the Big Bang, the Universe had cooled to 3000 K, when the ions in the plasma
combined to form stable neutral atoms. The photons produced in the combination were now free, in
the absence of the plasma, to travel throughout the Universe.
Wilhelm Wien investigated the radiation emitted by a ‘black-body’ (a perfect emitter) at a temperature
T (in kelvin) and deduced that the peak of the distribution of energies it emits occurs at a wavelength
 given by
T = 2.9 × 10–3 m K
Emission curve of black-body at 8000 K
0
200
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600
800
wavelength / nm
1000
1200
This is Wien’s displacement law, for which he received the Nobel Prize in 1911. You can apply this to
the entire Universe, which acts as a perfect emitter, giving out the radiation with which it fills itself.
4. Use Wien’s law to find the peak wavelength emitted by the Universe at 3000 K.
5. In what region of the electromagnetic spectrum would you expect to detect this radiation?
6. 3000 K is about the temperature of the tungsten wire in a light bulb. Why do tungsten light bulbs
produce so much heat? Why does the glass envelope get so hot?
The Universe expands to its current size
As the Universe has expanded, the photons that fill it have been stretched in wavelength.
7. Theory suggests that the Universe has now cooled to a temperature of 2.7 K. Use Wien’s law to
find the peak wavelength emitted by the Universe at 2.7 K. Why is this radiation referred to as
cosmic microwave background radiation?
8. Show that the diameter of the Universe is currently about 1000 times larger than it was when it
became transparent.
9. Arno Penzias and Robert Wilson first detected the cosmic microwave background radiation from
an observatory on Earth's surface. Suggest reasons why accurate measurements of this
radiation, which confirm the temperature of the Universe as 2.735 K, needed to be done from a
satellite.
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Time dilation for muons
Question 120S: Short Answer
Teaching Notes | Key Terms | Answers
Quick Help
Muons are unstable subatomic particles created near the top of the Earth’s atmosphere (about 60 km
above the surface) by cosmic rays. They travel down toward the surface of the Earth at a velocity
close to the speed of light. However, muons have a half-life of about 2 s so some will decay before
they reach the surface.
1. What is the maximum distance a muon can travel during one half-life?
2. As seen from the Earth, how long would it take a muon travelling at the speed of light to reach the
surface of Earth, 60 km below? Express your answer in muon half-lives.
3. Use your result from question 2 to estimate the fraction of muons created at 60 km that reach the
surface. Comment on your result.
4. In fact about one in eight muons created at 60 km do reach the surface. How is this possible?
5. If one in eight muons reach the surface, how many half-lives must have passed, so far as the
muons are concerned? Compare this with the answer to question 2 to find , the time dilation
factor, for these muons.
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6. How fast are the muons travelling? Use
  1 1 v 2 / c 2
7. If you were travelling with the muon you would pass through the Earth’s atmosphere in about 6 s
and yet you would agree with a terrestrial observer that your relative speed is about 0.9995 c.
How is this possible?
Time dilation and length contraction for particles
Question 150S: Short Answer
Teaching Notes | Key Terms | Answers
Quick Help
Questions 1 to 4 are about time dilation: c = 3.00 × 108 m s–1.
Cosmic radiation constantly bombards the Earth from outer space. The majority of these cosmic rays
are protons which, when they hit the Earth’s upper atmosphere, create sub-atomic particles called
pions, which then quickly decay into muons.
1. Suppose a cosmic ray proton hits a molecule of nitrogen in the upper atmosphere at a distance of
50 km above the Earth’s surface. If the pion produced has a velocity of 0.9999c and has an
average lifetime in its own wristwatch time of  = 2.60 × 10–8 s, how far would the pion travel if
there were no time dilation?
2. If the muon produced has a downward velocity of 0.999943c and an average lifetime in its own
wristwatch time of  = 2.20 × 10–6 s, how far would the muon travel before it decayed if there were
no time dilation?
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However, special relativity says that the speed of light is the same for all observers no matter how
they move, and that this has the consequence that ‘moving clocks run slow’. This means that pion
and the muon appear to take a longer time t to decay when observed from someone on the Earth
relative to whom they are moving, as compared with the particles’ own wristwatch time .
The time dilation is given by t =  with

1
1 v 2 / c 2
which is always greater than 1.
Now consider the effect of time dilation.
3. How far would the pion travel before decaying taking into the effects of time dilation?
4. How far would the muon travel before decaying taking into the effects of time dilation?
You should notice the effect that time dilation has on the passage of the pions and muons. The large
time dilation allows the muons to reach the surface of the Earth and be detected by ground-based
instruments. The fact that the pion decay time is 100 times smaller means that the pions almost never
make it to the ground even though their time to decay is also longer due to the effects of time dilation.
Observations of particles with a variety of velocities have shown that time dilation is real. In fact the
only reason cosmic ray muons ever reach the surface of the Earth before decaying and can be
detected at sea level is because of the time dilation effect.
Questions 5 to 9 are about time dilation and length contraction: c = 3.00 × 108 m s–1.
At Stanford, USA, a particle accelerator called SLAC (Stanford Linear ACcelerator) routinely
generates subatomic particles called taus. The lifetime of a tau (in its own wristwatch time ) is about 
= 3.05 × 10–13 s.
5. Suppose a tau is accelerated by SLAC to a speed of 0.9987 c. How far would it travel before it
decayed if you use the familiar equation distance = speed x time?
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However, particle physicists at SLAC measure the tau particle to have travelled further than your
answer to question 5. This is the result of time dilation. The decay of the tau is ‘slowed down’
relative to an observer with respect to whom it is moving. The time dilation given by t =  with
1

1 v 2 / c 2
so the longer decay time due to time dilation
3.05  10 13 s
t
1  0.9987c  c
2
 5.98  10 12 s
2
6. Considering the effect of time dilation how far has the tau travelled?
Now look at the situation from the tau’s point of view, that is, with the tau ‘at rest’ and the accelerator
rushing past it at high speed. The tau’s half-life in its own wristwatch time does not change. ‘Common
sense’ suggests that the tau would see the distance the accelerator moves to be the same as your
answer in question 5. However, special relativity says that the fact that the speed of light is the same
in all reference frames means that the length the tau perceives of things that travel past it is shorter or
contracted. The length contraction is given by
L
L0

with

1
1 v 2 / c 2
that is
L  L0 1  v 2 / c 2
Note that
1 v 2 / c 2
is always less than 1.
7. What is the distance that the tau sees the accelerator travel past before the tau decays?
8. According to the tau, how long did it take the accelerator to travel this distance? Compare with
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the tau lifetime in its own wristwatch time.
9. Now apply the same analysis to the decay of the muon in question 4 looking at the situation from
the muon’s rest frame. Through what distance does the muon see the Earth travelling towards it?
Apparent star brightnesses and logarithmic scales
Question 30C: Comprehension
Teaching Notes | Key Terms | Answers
Quick Help
What to do
Read through the text which describes how the magnitude scale for measuring apparent star
brightness developed from an empirical one to a mathematical one. Ensure that you understand the
calculations given as examples, then try questions 1–3. If your answers are correct, go on to use a
graphical method to show how a logarithmic scale relates stars' magnitudes to apparent
brightnesses. Note the mathematical validation of this in the answers.
Apparent star brightnesses and logarithmic scales
Astronomers of 2000 years ago, of whom Hipparchos (150 BC) was one of the greatest, catalogued
the brightness of stars according to their apparent brightness, i.e. how bright stars appear to the
naked eye. Stars were divided into six brightness classes ranging from the first magnitude, m1, for the
20 brightest stars in the sky to the sixth magnitude, m6, for stars at the limit of naked eye visibility. As
the invention of telescopes brought fainter stars into view, these were assigned to the seventh, eighth
and subsequent magnitudes.
By the mid-nineteenth century a visual photometer was available which could measure the brightness
of stars very accurately. This allowed the English astronomer, Norman Pogson, to give the scale a
firm mathematical basis. He defined a star of magnitude 1 to be 100 times brighter than one of
magnitude 6. So a difference of five magnitudes corresponds to a factor of 100 in terms of brightness.
This mathematical scale closely fits with the observed apparent brightnesses of stars. However, a
strict mathematical scale meant that apparent brightnesses had to be re-catalogued, leading to stars
of fractional magnitude and negative magnitude. On this scale the star Capella in Auriga has
magnitude 0.1, Proxima Centauri is of magnitude 11.1, and the very bright Sirius is of magnitude –
1.5. The magnitude of the Sun on the same scale is – 26.7.
Mathematically then, a difference in magnitude of 1 is equivalent to a brightness ratio of 1001/5  2.5.
The difference in magnitude of Sirius and Capella is 1.6 magnitudes; therefore Sirius, having the
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lower magnitude, is (2.5)1.6 = 4.4 times brighter than Capella. Remember that these are apparent
brightnesses which do not take into account that the two stars are different distances away from
where the observations are made.
Consider the catalogued stars below and answer the following questions about them:
Star
Magnitude,
m
A
–1.5
B
0.1
C
3.7
D
4.7
E
5.7
F
8.2
1. What is the difference in magnitude between stars B and C, and between C and D?
2. How many more times brighter is star B than star C, and star C than star D? These are the
brightness ratios for these pairs of stars.
3. Calculate the brightness ratios for all the pairs AB, AC, AD, AE and AF.
How is the brightness ratio related mathematically to the magnitude difference? A clue comes from
the fact that the magnitudes were originally assigned by naked-eye observations of star brightness.
The manner in which the eye responds to brightness is known to be logarithmic. Could it be that the
relationship between brightness ratio and magnitude difference is a logarithmic one?
4. Using your answers to question 3, construct a new table which shows the magnitude difference
between each star and star A, the brightness ratio of star A to each of the other stars, and the
logarithm to base 10 of these brightness ratios.
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5. Plot a graph with the magnitude differences on the vertical axis, and the base 10 logarithm of the
brightness ratios on the horizontal axis. Does the graph show proportionality? What is the
gradient of the graph?
6. Write down a relationship between magnitude difference m2 – m1 and brightness ratio S1 / S2
which would allow you to calculate the difference in magnitude between two stars whose
brightness ratio has been measured.
Thinking relatively
Question 50C: Comprehension
Teaching Notes | Key Terms | Answers
Quick Help
Leaning about relativity requires a special way of thinking: ‘thinking relatively’. You have to learn to
describe events seen from different points of view, and work out what changes and what stays the
same.
These questions provide simple exercises in thinking in this way. They do not require knowledge of
the results of the theory of relativity.
Travel by plane
You are flying to Madrid in Spain to go on holiday. You have just been served a drink, which is sitting
on the tray in front of you. You glance out of the window at the ground slipping past beneath you.
1. Describe the motion of the cup containing your drink, relative to you. Describe it again, relative to
a person on the ground below.
2. Describe your own motion, as seen from Air Traffic Control in Madrid. Describe the motion of
Madrid as seen from inside the aeroplane. Name a quantity that is the same in both descriptions.
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3. Your plane is due in Madrid in half an hour. When does Madrid reach your plane?
Watching the sun set
You are looking out to sea, as the Sun sets on the horizon.
4. Describe the motion of the Sun relative to you, taking yourself to be standing still and upright.
Now describe the relative motion again, as seen by an observer on the Moon watching the Earth
spin, with you standing on it. Name a quantity that is the same in both descriptions.
Going skating
Suppose that you are just learning ice-skating. You have learned how to get moving, but not how to
slow down or stop. You are heading straight at the barrier round the ice rink.
5. Describe the motion of the barrier as seen by you. Now describe the relative motion as seen by a
friend standing by the barrier. Name a quantity that is the same in both descriptions.
Falling off the back of a lorry
A sack of sand falls off the back of a lorry travelling along a motorway.
6. Describe the motion of the sack relative to the lorry. Now describe the motion relative to the
surface of the motorway. Ignore air resistance in both cases. Name a quantity that is the same in
both descriptions.
Being a snooker ball
A snooker player cues the white cue ball so that, relative to the table, the white ball hits the black
head on and stops, after which the black runs forward into a pocket.
7. Describe this event from the point of view of the cue ball, thinking of it as at rest throughout. Now
describe it from the point of view of the black ball, again thinking of it as at rest throughout. Name
a quantity that is the same in both descriptions.
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The Michelson–Morley experiment
Question 100C: Comprehension
Teaching Notes | Key Terms | Answers
Quick Help
By the latter part of the 1800s physicists knew that light was an electromagnetic wave. They also
knew how to derive a value for the speed of light from Maxwell’s equations of the electric and
magnetic fields. This gave a speed of 3.0  108 m s–1, which was in agreement with the (rather poor)
experimental measurements that existed at that time. However, they were not sure how to interpret
this speed because they did not know what to measure it relative to.
It is much simpler with mechanical waves. The speed of sound is measured relative to still air. The
speed of water waves is measured relative to still water. But against what could the speed of light be
measured? The ‘obvious’ solution to this problem was to assume that light itself was a disturbance in
some unseen medium (just as sound is a disturbance in the air) and that the speed of light is
measured relative to this all-pervading ‘ether’. This made quite a lot of sense. Electric and magnetic
fields could then be understood as disturbances (stresses and strains) in the ether (rather like
stresses and strains in an invisible solid material). Of course, this would also mean that the ether filled
all of space and that we, on Earth, would be moving through the ether as we orbit the Sun.
The idea that we are moving through the ether gave physicists a chance to test the ether hypothesis
because our movement should affect the speed of light relative to our laboratories. Several
experiments were carried out to test this, the most famous of which was the Michelson–Morley
experiment. This experiment was set up to ‘race’ two light beams over similar paths at right angles to
one another, with one path running parallel to the Earth’s motion and the other running at 90 degrees
to it. The easiest way to think about this is to imagine the effect of the Earth’s motion as creating an
‘ether wind’ through the laboratory. Light travelling parallel to the Earth’s motion is slowed relative to
the laboratory as it goes into the headwind and is speeded up as it moves with the wind behind it.
Light going at 90 degrees to the Earth’s motion is delayed in both directions in the same way that a
boat is delayed when it sails across a fast-flowing river (its path is effectively lengthened because it
has to sail at an angle to counter the effect of the current).
The arrangement of apparatus is shown in the diagram. Simple geometry shows that if light has a
constant speed relative to the ether then the time to travel OBO is greater than the time to travel OAO
so when light recombines at the half-silvered mirror it will be out of step. This produces an
interference pattern at the detector. If the apparatus is then rotated through 90° the fringe pattern
should move because the two racing beams have exchanged places. Michelson had calculated that
an ether wind equal to the velocity of the Earth’s orbital motion should result in a shift of 4 / 10 of a
fringe spacing. However, Michelson and Morley found that ‘if there is any displacement (of the
interference fringe pattern) due to the relative motion of the Earth and the luminiferous ether, this
cannot be much greater than 0.01 of the distance between the fringes’. Their experiment must count
as one of the most important ‘null results’ of all time.
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m ir ro r
A
ether
h a lf -s ilve re d
m irro r
O
li g h t
s o u rce
B
m irro r
d e te c to r
(in te rf e re n ce
frin g e s)
O A = OB = L
This experiment (or versions of it) has been repeated many times by different experimenters. Here
are some of the results.
Name
Year
Arm length / m
Fringe shift
expected
Fringe shift
measured
Michelson and Morley
1887
11.0
0.4
< 0.01
Morley and Miller
1902–4
32.2
1.13
0.015
Michelson et al
1929
25.9
0.9
0.01
Joos
1930
21.0
0.75
0.002
1. Why did Michelson think that the speed of light would differ in different directions on Earth?
2. Assuming that light has a constant velocity relative to the ether, give an expression for the relative
velocity of light along arm OB of the apparatus (moving toward B).
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3. Assuming that light has a constant velocity relative to the ether give an expression for the velocity
of light relative to the apparatus along arm BO of the apparatus (moving toward O).
4. Write an expression for the time taken for light to travel from O to B.
5. Write an expression for the time taken for light to travel from B to O.
6. Combine the expressions from questions 5 and 6 to find the total time for path OBO and show
that it can be written as
t OBO 
2cL
(c  v 2 )
2
7. Show that the time taken to travel from O to A and back is given by the expression
2L
t OAO 
c v2
2
HINT: find the relative velocity parallel to OA by using Pythagoras’s theorem.
8. Show that the ratio of times for these two routes is given by the ‘gamma-factor’

1
1
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9. Use your result from question 8 to confirm that the route via B takes longer than the route via A.
10. The table shows how the results from several of these experiments compared with the theoretical
prediction based on the ether theory. What conclusions can be drawn from these results?
Evidence for a hot early Universe
Question 110C: Comprehension
Teaching Notes | Key Terms | Answers
Quick Help
Discovery and importance of the cosmic microwave background
Read the extracts below about the discovery and importance of the cosmic microwave background
radiation. The questions which follow are about the bold words or phrases in these extracts.
When the universe was not see-through
Here Matts Roos explains what the universe is like when it is so hot that all its atoms are ionised,
consisting of a gas of electrons and charged nuclei, emitting and absorbing photons:
Since the photons continue to meet electrons, they do not propagate in straight paths for very
long distances. In other words, the universe is opaque to electromagnetic radiation, including
light. It would have been impossible to do astronomy if this situation had persisted until
now.
When the electrons become slow enough, they are captured into atomic orbits by the protons,
forming stable hydrogen atoms and other light atoms….When this combination process is
completed, at a temperature of about 3000 K, the photons find no more free electrons to scatter
against…Thus the photons decouple, continuing in straight lines from the point of their last
scattering….Hence the Universe becomes transparent to radiation.
[There is] an analogy with the surface of the Sun. Photons inside the Sun are continually
scattered so that it takes millions of years for some photons to reach the surface. But once they
emerge they scatter no more, but continue in straight lines towards us. Therefore we can see the
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surface of the Sun, which is the last scattering surface of the solar photons, but we cannot see
the solar interior.
From Matts Roos 1997 Introduction to Cosmology 2nd edition (Wiley).
Accidental discovery of the cosmic microwave background
Here Michael Rowan-Robinson describes the events surrounding the discovery of the microwave
background:
On May 13th 1965 a modest and laconic letter entitled ‘A measurement of excess antenna
temperature at 4080 MHz’ reached the Astrophysical Journal. The paper, only 600 words long,
describes in a brief, almost brusque, style the first project undertaken by two young radio
astronomers, Arno Penzias and Robert Wilson, since their move to Bell Telephone Laboratories.
They had been working with the 20 feet horn reflector built by Bell Laboratories engineers for the
Echo satellite project…Using this antenna, they had measured the brightness of the sky when the
antenna was pointing in the zenith (vertical) direction….Any measured intensity or brightness can
be translated into an equivalent black body temperature which would give the same brightness at
that wavelength. Penzias and Wilson found a temperature of 6.7 K, of which 2.3 K was due to
atmospheric absorption and 0.9 K was due to ohmic losses in the antenna. This left 3.5 K
unaccounted for…
They noted that ‘this excess temperature is, within the limits of our observations isotropic,
unpolarised, and free from seasonal variation’.
…Penzias and Wilson did not themselves immediately recognise the significance of their
discovery. However, they had the good fortune to be near Princeton, where the group led by Bob
Dicke and Jim Peebles was only too aware of the significance of a 3 K microwave background.
This was just what would be expected in a hot Big Bang universe which is initially dominated by
radiation.
…By 1965 Dicke, in collaboration with Jim Peebles, had reinvented or revived a version of the
Hot Big Bang model of George Gamow, and was predicting the existence of cosmic background
radiation, and had two young researchers, P G Roll and David Wilkinson, building a system to
[detect] this background. The plans and dreams of the Princeton group must have been rudely
shattered by the news of Penzias and Wilson’s discovery. Peebles, Dicke, Roll and Wilkinson
hastily wrote an elegant paper explaining the significance of the radiation as a relic of the
‘primeval fireball’ phase of the hot Big Bang.
From M Rowan-Robinson 1993 Ripples in the Cosmos (Freeman).
What happens to the radiation as the universe expands
Here Jim Peebles explains how the wavelength of the photons is stretched by the expansion:
The number of photons in radiation at temperature T depends on the ratio of the energy E = hf of
a photon to the quantity kT which represents roughly the energy per particle at temperature T. k
is the Boltzmann constant. In the expansion of the Universe after the photons decouple, the
number of photons remains constant, since they do not interact with anything. Thus the ratio hf /
kT stays constant. As the temperature T falls, the frequency f decreases in proportion.
This means that the wavelength increases with proportional to 1 / T. Since the expansion
of the universe between two times stretches the wavelength in proportion to the ratio R2 / R1 of
the scale of the universe at the two times, the temperature of the radiation must fall by the same
factor. That is:
T1/R
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Adapted from J Peebles 1993 Principles of Physical Cosmology (Princeton University Press).
Estimating the red-shift of the microwave radiation
Michael Rowan-Robinson shows how to calculate the red shift of the microwave background:
The effect of the expansion of the universe on the radiation is to preserve its spectrum, with the
radiation temperature continuing to fall according to the equation T  1/R. This provides the most
natural explanation of the 2.7 K microwave background radiation…The discovery of this radiation
provided the most spectacular confirmation to date of the hot Big Bang picture of the universe.
The photons last scattered from matter when the temperature had fallen to about 3000 K, when
electrons and protons combined to form neutral atoms.
…Now from the equation T  1/R we can identify the epoch of decoupling as being when:
now/decoupling = Rnow/Rdecoupling = 3000 K / 2.7 K  1000
The microwave background provides a look at the Universe when it was 1000 times
smaller than it is now.
Adapted from M Rowan-Robinson 1996 Cosmology 3rd edition (Oxford University Press).
Questions
The questions are about the highlighted words or phrases copied (in order) from the passages above.
Use information in the passages and you knowledge of physics to answer the questions about them.
1. atoms are ionised. What does it mean for an atom to be ionised? Give an example of another
case where atoms are ionised.
2. emitting and absorbing photons Write a sentence or two explaining what is meant by the term
‘photon’.
3. It would have been impossible to do astronomy if this situation had persisted until now.
Give some reasons why it would have been impossible to do astronomy.
4. the Universe becomes transparent to radiation. Relate this statement to your answer to
question 3.
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5. we cannot see the solar interior. Why not?
6. 4080 MHz What wavelength were Penzias and Wilson using? Give an example of another device
that operates at this kind of wavelength.
7. Using this antenna, they had measured the brightness of the sky What does ‘brightness’
mean here? Is it how much visible light is coming from the sky?
8. isotropic, unpolarised, and free from seasonal variation. What does ‘isotropic’ mean? What
does ‘unpolarised’ mean? Why do these facts matter for the interpretation of the radiation as
cosmic background?
9.
‘primeval fireball’ phase of the hot Big Bang. Read the first extract again, and explain more
carefully what ‘cosmic fireball’ means here.
10. energy E = hf of a photon. State the meaning of this phrase in your own words.
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11. the ratio hf / kT stays constant. As the temperature T falls, the frequency f decreases in
proportion. Why is f proportional to T if the ratio hf / kT is constant?
12. This means that the wavelength  increases with  proportional to 1/T. Why does this follow
from the statement in question 11?
13. T  1 / R Justify this equation from the information in the extract.
14. The microwave background provides a look at the Universe when it was 1000 times
smaller than it is now. Justify this statement from the equation now/decoupling = Rnow/Rdecoupling =
3000 K / 2.7 K  1000 given just before it in the extract.
The brighter stars in the night sky
Question 52D: Data Handling
57 Advancing Physics
Question 52D: Data Handling
Quick Help
Teaching Notes | Key Terms | Answers
Brightness and distance data
The data in the spreadsheet provided give distances and brightness of a number of the better known
stars in the night sky. They include the Pole Star, and the very bright star Sirius, for example. Notice
how many of the names are of Arabic origin.
You will use these data to plot a number of graphs to answer the question: 'Are the brighter stars also
the nearer stars?'
Brightness and distance of some of the brighter naked-eye stars
star
distance in
parsec*
Sirius
distance in
light years
apparent
brightness as
multiple of
apparent
brightness of
Pole Star
absolute
brightness
as multiple
of amount of
light emitted
by Sun
2.65
9
24.66
23
Canopus
70
228
12.47
8132
Arcturus
10.3
34
6.67
94
7.5
24
6.25
47
Capella
12.5
41
5.97
124
Rigel
265
864
5.75
53764
Procyon
3.4
11
4.53
7
Betelgeuse
320
1043
4.06
55246
5
16
3.16
11
19
62
2.94
141
190
619
2.65
12750
Spica
67
218
2.61
1557
Pollux
10.6
35
2.25
34
Fomalhaut
6.7
22
2.21
13
Deneb
500
1630
2.03
67599
Regulus
22
72
1.85
119
Castor
15
49
1.50
45
Mirzam
53
173
1.04
388
110
359
1.00
1610
10
33
0.08
1
Vega
Altair
Aldebaran
Antares
Pole Star
Sun if seen from
a distance of 10
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star
distance in
parsec*
distance in
light years
apparent
brightness as
multiple of
apparent
brightness of
Pole Star
absolute
brightness
as multiple
of amount of
light emitted
by Sun
parsec
*1 parsec = 3.26 light year
There are two values of the ‘brightness’ of each star. One is how bright it looks – its apparent
brightness. This depends on how far away it is – nearer stars will look brighter than more distant
ones, if they give out the same amount of light. The apparent brightness is given as a multiple of the
apparent brightness of the Pole Star.
But the stars do differ in how much light they give out – their absolute brightness. This is given as a
multiple of the amount of light given out by the Sun. You can see that all the stars here give out more
light than the Sun.
Open the Excel Worksheet
Apparent brightness and distance
Open up the spreadsheet.
1. Make a scatterplot of the apparent brightness of these stars against their distances.
2. Is the graph improved by plotting the values logarithmically? Why?
3. What relationship, if any, is there between apparent brightness and distance?
Bias in choice of stars
4. Make a scatterplot of the absolute brightness against distance.
5. Is the graph improved by plotting the values logarithmically? Why?
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6. Explain why the graph shows evidence of bias in the selection of stars. How would this naturally
come about?
Checking the inverse square law
If there is no absorbing matter in space, you would expect the apparent brightness of a star to be
proportional to its absolute brightness divided by the square of its distance.
7.
Use the spreadsheet to construct a new column proportional to the absolute brightness
divided by the square of the distance. Use it to plot a graph to test whether the inverse square law
applies to these data.
Using a log-log plot to test the inverse square law
If the apparent brightness is a, the absolute brightness is b, and the distance is r , then
b
ak 2
r
where k is a constant.
8. Show that the ratio
a
k

b r2
9. Construct a new spreadsheet column of the ratio apparent brightness / absolute brightness. For
convenience, multiply each value of the ratio by 106. Plot a graph of the ratio apparent brightness
/ absolute brightness against distance. Does this look like an inverse square relation?
10. Convert the scales to both be plotted logarithmically, in powers of ten. How does this graph show
that the inverse square law applies?
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Using orbital data to calculate masses
Question 70D: Data Handling
Teaching Notes | Key Terms | Hints | Answers
Quick Help
Calculating Jupiter’s mass from observations of its moons
This question is about the calculation of the mass of a planet or star by observing the period and
radius of an orbiting satellite.
The four largest moons of Jupiter were discovered by Galileo in 1610, and he thought they could be
used as an accurate ‘clock’ to aid mariners in measurement of longitude. This proved difficult in
practice, and it was complicated by the time taken for light to travel from Jupiter to the Earth, which
varies by 16 minutes depending on the relative positions of the two planets in their orbits. However,
the accurate observation of orbital periods and radii allowed calculation of the mass of Jupiter, once
the gravitational constant G had been determined.
This method has been extended to other planets, and to planets orbiting distant stars.
Orbital period and centripetal force
This section relates the periodic time to the orbital speed and thus to the central force keeping it
orbiting.
1. Show that the speed v of an object moving in a circle of radius R in a time T is given by v = 2  R
/ T.
2. Use the centripetal force equation F = m v2 / R to show that the force per kilogram making this
object circulate in its orbit of radius R in a periodic time T is
F 4 2R

m
T2
Centripetal force and gravitational field strength
The centripetal force per kilogram on any orbiting object is the gravitational field strength g = G M /
R2, where M is the mass of the source of the gravitational field and G is the gravitational constant,
6.67 × 10–11 N m2 kg–2.
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3. Combine these two answers to show that
T2 
4 2 3
R
GM
(note that T2  R3 is Kepler’s third law).
The table shows the mean radius of orbit and orbital period of the four ‘Galilean moons’ of Jupiter
discovered by Galileo.
Satellite
R/m
T/s
Io
4.22 × 108
1.53 × 105
Europa
6.71 × 108
3.07 × 105
Ganymede
1.07 × 109
6.18 × 105
Callisto
1.88 × 109
1.44 × 106
4. Use the equation in question 3 and the data in the table to find 4 ² / G M and from this M, the
mass of the planet Jupiter. This can be done by plotting a suitable straight-line graph, possibly
with a spreadsheet which calculates a regression equation, or by repeated calculation and
averaging. Which method is better?
Pluto and Charon
The outermost planet Pluto has a moon, Charon, which is thought to be nearly a quarter the size of
Pluto itself. The orbital period can be measured reasonably accurately, as can the separation
between Pluto and Charon.
5. The mean distance between Pluto and Charon is 1.91 × 107 m, and the orbital period is 5.52 ×
105 s. Use the equation
T2 
4 2 3
R
GM
to calculate the mass of Pluto.
6. The orbiting moon Charon is large compared with the planet Pluto.
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Explain why this is an inaccurate method of calculating the mass of Pluto.
Extra-solar planets
Much astronomical research is now focused on the search for planets orbiting stars other than our
Sun. The presence of such a planet causes the position of certain stars to oscillate in the sky. Recent
spectroscopic measurements made on such a star have shown the existence of a planet, because
the planet and star differ in colour. Interest is centred on the possibility of finding planets at the right
distance from their star for life to develop, as it has on Earth.
The mass of the star and the orbital period need to be known before you can determine the radius
from the equation T 2 = (4 2 / G M) R 3.
7. How would the orbital period be determined?
8. How might the mass of the star be determined?
Planets observed so far in this way have been large, at least as big as Jupiter.
9. What difficulties would there be in making observations of a planet like Earth orbiting a distant
star?
Astronomical distances
Question 80D: Data Handling
Teaching Notes | Key Terms | Answers
Quick Help
Finding the distances of galaxies using Cepheid variable stars
The first measurements of astronomical distances beyond the limits set by geometrical arguments
depending on parallax were based on observations on a type of variable star first identified in the
constellation Cepheus. The period of the star’s variation in brightness was simply related to its
absolute brightness, and this was used to show that there were galaxies outside of our own. From
this base Hubble built up evidence for the expansion of the Universe by relating the apparent
brightness of galaxies to their red shift.
If a galaxy isn’t too far away astronomers can detect a Cepheid variable. The brightness of these
stars varies in a regular way:
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1.2
1.0
0.8
0.6
0.4
0.2
0
0
10
20
30
time / days
40
50
60
Henrietta Swan Leavitt studied examples of these stars and measured the periods and real
brightnesses of a number of stars all in the Small Magellanic cloud, and so at the same distance from
Earth. She found that the brighter the star the longer was its period of variation in brightness, from
maximum to maximum, say. Of course she had to allow for the distances of the stars to calculate their
real brightness from their apparent brightness. The further away a star is the dimmer it looks. The
apparent brightness of a star gets less with distance according to the inverse square law.
Here are some of her data:
0
–1
–2
–3
–4
–5
–6
–7
0
10
20
30
period / days
40
50
So to find the distance of a galaxy astronomers look for a Cepheid variable. By measuring the time
period they know exactly (give or take some measurement errors) its real brightness. Then they use
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the inverse square law to work out its distance. Stars vary so much in brightness that astronomers
use a logarithmic scale to measure real and apparent brightness in a unit called magnitude. The
brightest stars we can see have magnitude 1 or less, the faintest we can see with the eye alone have
a magnitude 6. The scale is logarithmic and works ‘backwards’ – the fainter the star the greater its
magnitude number. Jupiter is often brighter than the brightest star, and so has to be given a negative
magnitude, e.g. –2.
Real brightness has the symbol M and apparent brightness the symbol m.
This graph links apparent brightness (apparent magnitude) m to distance for a set of real
brightnesses (real magnitudes) M.
distance chart for apparent and absolute magnitudes
10000000
M = –6
M = –5
M = –4
M = –3
M = –2
1000000
100000
10 000
1000
100
10
4 5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20
apparent magnitude, m
You have to...
...use the Cepheid variable star method to estimate the distances of stars and galaxies, by:
1. Using the observed time period of a Cepheid variable to estimate its absolute magnitude (real
brightness) M by reading the value of absolute magnitude against period from the graph.
2. Reading along the plotted line of absolute magnitude (or interpolate for intermediate values) on
the graph of distance against apparent magnitude, looking for an intersection with the apparent
magnitude to estimate its distance.
The distances are plotted on a logarithmic axis and in the astronomers’ distance unit of the parsec.
One parsec is about 3.3 light-years.
Example
The star Delta Cepheus has a variation period of 5.4 days and an apparent magnitude m of 4.0.
From graph 2, the period of 5.4 days means it has a real (absolute) magnitude M of –3.5. This means
that if it were a standard distance away it would appear to us to be a very bright star indeed.
Now find the point on the graph of distance against apparent magnitude which you estimate to link an
m of 4 with the M value –3.5. It is half way between the red line for M = –3.5 and the purple line for M
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= –4. The distance is about 300 parsecs or 1000 light-years.
Questions
The North Star (Polaris) is a Cepheid variable of period 4.0 days. Its apparent magnitude m is 2.0.
What is its distance:
1. In parsecs?
2. In light-years?
3. A Cepheid variable is observed in the Andromeda galaxy (M31). Its period is 27 days and it has
an apparent magnitude of 18.7. Estimate the distance of the Andromeda galaxy.
4. Find the distances of the following Cepheid variable stars:
Star
Apparent magnitude / m
Period / days
P
12
8
Q
12
40
R
8
7
Calculating the age of the Universe
Question 90D: Data Handling
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Teaching Notes | Key Terms | Hints | Answers
Quick Help
Hubble’s constant
The Hubble constant can be used to estimate the age of the Universe. Over time, estimates of this
constant have changed and so estimates of the age of the Universe have changed. This question
asks you to work out the changing estimates of the age of the Universe from the Hubble constant.
What does the constant mean?
In the Advancing Physics A2 student’s text, you can read about Hubble’s work. He realised that the
light from distant galaxies is red-shifted and that the more distant they are, the greater the red shift.
Today, this is interpreted to mean that the Universe is expanding: that is, the space between clusters
of galaxies is expanding and therefore the distance between galaxies is increasing. Hubble’s
interpretation was slightly different: he took the results to mean that galaxies were receding from each
other through space. Following Hubble’s explanation, we can write:
v  Hd
where v is the recession speed of a distant galaxy, d its distance from us and H is the Hubble
constant. Now, if you imagine the Universe running backwards, how long would it take a distant
galaxy to reach you? Answering this question tells you how long ago it is since all the galaxies were
together in the same place, i.e. how long ago the Big Bang occurred. The time taken for a galaxy
travelling at speed v to travel a distance d is:
d
1
 .
v H
Therefore, the value of 1 / H gives us an estimate of the age of the Universe. This is only as an
estimate because as the galaxies in the time-reversed Universe fall towards one another, they would
be expected to speed up.
Hfuture
Hnow
Hpast
past
now
future
distance
A more precise description of 1 / H is the ‘expansion timescale’, or ‘Hubble time’ rather than the ‘age
of the Universe’.
A graph of measured values of Hubble’s constant
The following graph shows how measured values of the Hubble constant have changed since the
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1920s, when Hubble made his original measurement.
How estimates of the Hubble time have changed
+ Hubble 1929
600
+
+ +
+
2
400
+
+
200
100
0
1920
1940
+
age of Earth
4.6 Gyr
over 300 measurements in 5
+
disputed range 50–100
++
++ +
10
+
+
20
de
Vaucouleurs
1960
Sandage
1980
2000
current best value
14 Gyr
The units of H on the graph are km s–1 Mpc–1 (km per second per million parsecs). This is a useful
unit for astronomers as it tells them how fast, in km s–1, a galaxy is receding if its distance is a certain
number of parsecs from us. The two different units are useful to astronomers for measuring the two
different quantities, speed and distance, but if you want to find the age of the Universe from 1 / H,
then you must have H in units of (time)–1. The first thing you have to do is change the units of H into
s–1.
1. What is 1 Mpc in km?
2. By writing 1 km s–1 Mpc–1 in terms of km s–1 km–1, write down a conversion factor that changes
km s–1 Mpc–1 into s–1.
3. The earliest point on the graph gives a Hubble constant value of approximately 600 km s–1 Mpc–1.
What is this in s–1?
4. For this early value of the Hubble constant, calculate an estimate for the age of the Universe.
What are the units of your answer? Convert the answer to years.
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5. Read from the graph the upper and lower limits of the recent disputed values of the Hubble
constant. Calculate the two different 'ages of the Universe' that they imply.
Logarithmic scales
Question 10X: Explanation–Exposition
Quick Help
Teaching Notes | Key Terms | Answers
Instructions
Read through the information on stellar distances and galactic dimensions and make sure you
understand the development of the logarithmic scale. Apply the method to distances in the solar
system.
The solar system
The solar system contains nine planets, one of which is the Earth, and these rotate about the star that
we call the Sun. You may have some ideas about the relative sizes of planets and their distances
from the Sun from your GCSE studies, and will appreciate the difficulties in getting a sense of these
measures because of the scale of the solar system. These difficulties increase when we start to
measure distances outside the solar system. For instance, light takes 8.31 minutes to travel from the
Sun (the brightest star) to the Earth. From the next brightest star, Sirius, it takes 8.6 years. We say
Sirius is 8.6 light-years (written as 8.6 ly) away. By comparison our galaxy, the Milky Way, which
contains hundreds of billions of stars including the Sun and Sirius, is a disc approximately 100 000 ly
across, and 12 000 ly thick. The scale of the Galaxy is massive, and getting a sense of stellar
distances relative to the size of the Galaxy is difficult.
Suppose we try to construct a scale which allows us to place the distances away from us of the Sun
and Sirius in relation to the size of the Galaxy. We draw a line 10 cm across to represent the diameter
of the Galaxy’s disc:
0
20 000
40 000
60 000
80 000
100 000
distance / ly
Try marking the distances to the Sun and Sirius on this scale.
1. How far along the scale would the Sun be placed?
2. How far along the scale would Sirius be placed?
It just is not possible. Let us consider the second brightest star in the night sky, Canopus. It is far
more distant than Sirius, at a distance of about 310 ly. On the scale above Canopus’ distance would
be 0.31 mm away from the zero, and still too small to mark accurately. This scale is a linear scale.
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Because the Galaxy size is tens of thousands of times bigger than the distances to these stars, we
cannot use such a scale to make sensible comparisons. Linear scales are useful if the largest value
to be represented is in the range of 10 to 100 times the smallest value, so even plotting distances of
planets from the Sun has disadvantages using a linear scale.
The solution to this problem is to write the distances to Sirius, Canopus and the Galaxy size in
standard form, and construct another type of scale called a logarithmic scale.
0
Distance to Sirius
8.6  10 ly
Distance to Canopus
3.1  10 ly
Size of Galaxy
1.0  10 ly
2
5
If we construct a scale which is linear in powers of ten, as shown below, then all of these distances
can be represented on that scale:
Sirius
100
0
101
1
Beta Centauri
102
2
103
3
Galaxy
104
4
105 distance / ly
5
log (distance / ly)
This is called a logarithmic scale because if we take logarithms (to base 10) of the powers of 10, then
we simply get a linear scale running from 0 to 5. The distances to the two stars relative to the size of
the Galaxy can then be compared on the scale. Whereas, on a linear scale, the length along the
scale represents the quantity, on a logarithmic scale it represents the ratio of the quantities. We are
saying that Canopus is about 40 times more distant than Sirius, and the length between these on the
scale shows this ratio.
Making your own scale
In what follows you will compare the use of linear and logarithmic scales to represent the distances
between the planets and the Sun.
Draw a horizontal line 10 cm long to represent the distance from the Sun to beyond Pluto at 10 000 
109 m, and using the data provided below, mark on as many of the planets as possible above the line
using a linear scale. Then, by first expressing the distances in standard form, and then taking
logarithms to base 10, construct a logarithmic scale marking the planets below the line this time. Your
line and the markings on the linear and logarithmic scales should look something like this:
linear
0
10
5000
11
12
logarithmic
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10000
13
distance / 109 m
log (distance)
You should note that using the linear scale the planets, and in particular the inner ones, are closely
bunched together on the axis. The logarithmic scale has the advantage of spacing them out along the
scale, but if shown fully, it would have the disadvantage of spreading out the first 10  109 m too
much!
Data
Planet
Distance from Sun
to planet / 109 m
Mercury
57.9
Venus
108.2
Earth
149.6
Mars
227.9
Jupiter
778.3
Saturn
1427
Uranus
2870
Neptune
4497
Pluto
5907
Books about relativity
Reading 100L: Reading List
Teaching Notes | Key Terms
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There are very many books explaining relativity theory. Here are a few that can be specially
recommended.
1. Hermann Bondi 1980 Relativity and Common Sense (Dover) ISBN 0-486-24021-5
Bondi’s book is a classic, still in print. It is understandable by any A-level candidate. Our
approach in Advancing Physics follows its line of thought, starting with distance and speed
measurement via the Doppler effect. This is sometimes called the k-calculus.
2. N David Mermin 2005 It’s About Time – Understanding Einstein’s Relativity (Princeton University
Press) ISBN 978-0-691-12201-4
This is a very well written introduction to relativity, understandable by A-level candidates. It is very
careful and thorough, discussing the fundamental arguments in depth, from a variety of points of
view. A good resource for the critical student to tackle.
3. Edwin F Taylor and John Archibald Wheeler 1992 Spacetime Physics – Introduction to Relativity
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2nd edition (W H Freeman) ISBN 0-7167-2327-1
An excellent textbook, very comprehensive and thorough. It has in-depth discussions of the
meaning of time dilation, the ‘twin paradox’ and length contraction. The meaning of mass in
relativity is discussed with great care. There are many witty and wise responses to students’
difficulties and confusions. It is a particularly good resource for teachers to have to hand.
The ladder of astronomical distances
Reading 20T: Text to Read
Teaching Notes | Key Terms
Quick Help
Few measurements in physics have proved as difficult to make reliably, or have caused as much
argument and dispute, as the seemingly simple question, ‘How far is it to the stars and galaxies’, or,
in modern terms, ‘How big is the Universe?’. Here you can find out a little about some of those
arguments, and what they were about.
Challenge:
It’s half as big as you say
The prestigious meeting of the International Astronomical Union in 1976 was startled to be told that
the Universe is only half as big as the astronomers present all thought, and therefore only half as old.
The challenger was the French-American astronomer Gerard de Vaucouleurs; the leader of the
challenged orthodoxy was Allan Sandage, who had inherited the mantle of the American astronomer
Edwin Hubble, who had in the 1920s first assembled evidence for the expansion of the Universe from
red-shifts of galaxies. Sandage’s main collaborator was Gustav Tammann.
De Vaucouleurs mounted a sarcastic attack on Sandage and Tammann’s arguments for the
distances of galaxies, accusing him and his collaborators of confusion, circular argument and
observational errors. Both sides admitted uncertainties in their estimates, but the ranges of estimates
of uncertainty did not even overlap. The dispute lasted for 20 years: today – with some irony – the
generally agreed values seem to have settled down half way between the two. But for those 20 years
astronomers had either to decide for one value or the other, or in despair to split the difference.
How could such a strong difference of opinion about the facts come about? It arose because
establishing distances to stars and galaxies does involve a long chain of reasoning, and an error at
one step propagates through all the others. There was also a matter of belief involved. The size and
age of the Universe which Sandage and Tammann preferred was large and long enough for models
of the evolution of the Universe in which he believed to be right; the smaller size and shorter age
which de Vaucouleurs advocated made serious trouble for these models. The Universe according to
de Vaucouleurs came out dangerously close to being younger than some of the oldest objects, such
as globular star clusters, to be found in it.
This was by no means the first time that changes to astronomical distance scales had been
suggested. Successive revisions from the 1920s had in fact lengthened the scale, with Sandage
himself having at least quadrupled the scale over that time. You may wonder how there can be such
uncertainty in a measurement. The reason is that distances have to be worked out by making links in
a long chain, each link introducing uncertainties and doubts of its own.
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It all hangs by a hair
There are two ways to estimate the distances to the nearby stars, one of which is simple but bad and
the other of which is difficult but good. The simple bad method is to assume that stars are all much of
a muchness, and that all are about as bright as the Sun. Then comparing how bright they look with
how bright the Sun looks gives you the ratio of the two distances, using the inverse square law. This
turns out to be a bad method because stars vary in brightness by a factor of a million, though the
astronomer Herschel used it to guess at the scale of the Milky Way, for lack of anything better. But
bad as it is, this method illustrates several important features of astronomical distance measurement:
If you have a class of objects of the same or known brightness, you can get distances just from
the inverse square law of light intensity. Astronomers call such objects ‘standard candles’. It just
turns out that stars are bad standard candles.
The result you get from this kind of method is a ratio of distances, not the distance itself. One of
the distances has to be linked back somehow to a known distance, often through a series of other
estimated distance ratios.
Questionable assumptions have to be made: for example that the light is not dimmed by
absorption along the way.
There is a built-in bias (called Malmquist bias) because at great distances one only sees the
brightest stars and compares them with average stars nearby. Thus large distances consistently
get underestimated. It is one reason why the astronomical distance scale has usually tended to
enlarge when corrected.
The ultimate distance, found directly by radar measurement, is the scale of the solar system, and in
particular the diameter of the Earth’s orbit.
Parallax measurement of distance of star
width W
angle  =
width W
distance d
angle 
Earth’s orbit
star
distance d
The difficult but good method of finding the distances of nearby stars is to look for a shift in their
direction, against the background of much more distant stars, as the Earth goes round its orbit. The
effect is called parallax. You use it unconsciously yourself to judge distances of nearby things, using
the baseline between your two eyes, which see things in slightly different directions. The trouble is
that even the nearest star, Alpha Centauri, is more than 200 000 times further away than the diameter
of the Earth’s orbit. The shift in angle is less than 1 second of arc, less than the thickness of a hair
seen across a large room. It was not until the mid nineteenth century that the first such parallax shift
was measured. The distance at which a star would shift by 1 second of arc to and fro as the Earth
moves round its orbit is called 1 parsec, equal to roughly 3 light-years.
Difficult as measurements of parallax are, they are not the source of substantial uncertainties in the
astronomical distance scale. They are the place where, in the end, other estimates of ratios of
distances link to the known diameter of the Earth’s orbit. In recent years, good parallax
measurements, previously limited to a few thousand stars, have been extended to greater distances
by the scientific satellite Hipparcos. Crucially, Hipparcos provided parallax measurements of a
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number of the bright variable stars called Cepheid variables, which play an important part in the story
of the astronomical distance scale.
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Ladder of astronomical distances
Red-shift
Assume that the speed of recession
as measured by wavelength shift is
proportional to distance
1010
Brightest galaxy
Assume that brightest
galaxies in clusters are
all equally bright
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Coma cluster
Supernovae
Type Ia supernovae all have
the same absolute brightness
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Virgo cluster of galaxies
107
M31 Andromeda
106
Magellanic clouds
Tully-Fisher
Faster rotating galaxies
have greater mass and
are brighter
Blue supergiants
Assume that the brightest star
in a galaxy is as bright as the
brightest in another
105
104
Cepheid variables
These very bright pulsating stars can
be seen at great distances. The bigger
they are the brighter they shine and
the slower they pulsate
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102
Colour-luminosity
The hotter a star the bluer its light, and the brighter
it shines. If the type of star can be identified there
is a known relationship between colour and
brightness. Distance then found by comparing
actual with apparent brightness
10
1
Parallax
Shift in apparent position as Earth moves in
orbit round Sun. Recently improved by using
satellite Hipparcos: now overlaps Cepheid scale
Baseline
all distances based on measurement of solar system, previously using parallax,
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using radar
Colour and brightness
The first step beyond parallax in building a distance scale is a development of the simple but bad
method of judging the distance of stars by their brightness. As stars were surveyed, it became clear
that a large number of ‘normal’ stars ran on a scale from being very massive, very hot and very bright
to being less massive than the Sun, less hot than the Sun and not so bright. The temperature of a star
can be measured from its spectrum, intense in the blue if very hot and intense towards the red if
relatively cool. So although it was a mistake to suppose that all stars were more or less equally bright,
how bright many of them truly were could be found by looking at the spectrum of their light. Knowing
the ratios of brightness of two stars, and again using the inverse square law, the ratio of their
distances can be found. Again, by linking to the Sun, a link to the measured size of the Earth’s orbit
could be made.
Brightness and hotness of stars:
Hertzsprung-Russell diagram
supergiants
hot bright
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cool big bright
Cepheid
variables
‘main
sequence’
of ‘normal’
stars,
including
Sun
10
1
red giants
Sun
white dwarfs
10–1
cool dim
hot small dim
100 000
30 000
10 000
3000
1000
surface temperature / K
Henrietta Leavitt measures Cepheid variables
In 1912 women of talent found it hard to get university positions to do science. But they were often
welcome doing work such as making calculations for their male colleagues. The American Henrietta
Leavitt found herself such a role at Harvard. In 1912, photography was beginning to be used not just
to make pictures through telescopes, but to make careful measurements. Over time, Miss Leavitt
became a world authority on the new science of measuring the optical density of images of stars, so
as to measure how bright they appear to be. She made a study of a special kind of variable star which
are so bright that they could be detected outside the Milky Way Galaxy, in a ‘satellite’ star cluster
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called the Large Magellanic Cloud, visible to the naked eye in the Southern Hemisphere. These
Cepheid variable stars vary in brightness over periods from 2 to 200 days. Detecting them calls for
painstaking examination and measurement of repeated images of thousands of stars, to detect those
which are varying, and how rapidly.
Henrietta Leavitt found that there is a simple relationship between the brightness of the Cepheid
variables and their period of oscillation: the slower the brighter. She could assume that differences in
the apparent brightness of the Cepheid stars in the Magellanic Cloud were due to real differences in
brightness, since they were all more or less at the same distance. She could not know how bright the
stars were, or how far away they were. But she provided a way for astronomers to compare true
brightnesses just by clocking the time of variation, and so to get ratios of distances from ratios of
apparent brightness.
Since then, Cepheid variables have been an essential link in the astronomical distance scale. With
parallax measurements to nearby Cepheids from the Hipparcos satellite, and with the Hubble
telescope able to detect Cepheid variables as far away as the Virgo cluster (tens of millions of
light-years), it turns out that Henrietta Leavitt identified one of the strongest and longest links in the
chain of constructing an astronomical distance scale. The value of her work was recognised also,
back in the 1920s, when Edwin Hubble observed Cepheid variables in the nearest large galaxy, M31
in Andromeda, establishing it as an ‘island universe’ outside and on the same scale as the Milky Way.
So Henrietta Leavitt contributed also to the first solid evidence that the building bricks of the Universe
are the galaxies, each with perhaps 1011 stars. But, just to illustrate the difficulties, you should know
that Hubble’s 1920s estimate of the distance of M31 was half as big as the current estimate, because
of the indirect way he had to use to estimate the distance of the Magellanic Clouds.
More brightness guesswork
Beyond the range of the Cepheid variables, only the very brightest of stars could be picked out in
distant galaxies. These are the blue supergiant stars, very big, bright and hot. The distance scale
could be pushed out further if one assumed that there is some limit to how bright and big a star can
get. If so, the very brightest stars in a galaxy would all be equally bright, and could be used as
standard candles as far out as they could be seen. You will see for yourself that this is a rather
dangerous assumption to make. Could there not be other perhaps unknown kinds of even brighter
object? (and in fact there are). But for a time it was the only way available to get distances of distant
galaxies from the brightness of stars in them.
A lucky break:
Brighter galaxies spin faster
Beyond the distance where individual stars could be seen in galaxies, the only hope was to make
further dangerous assumptions, for example that galaxies of the same type are equally bright, or
equal in size. Neither method is helped by the fact that galaxies are seen at all angles to the line of
sight, head-on, sideways, and in-between.
In 1972 two young American graduate students, Brent Tully and Richard Fisher, had the idea of using
radio astronomy to measure how fast distant galaxies were rotating. The idea was that the radio
emission at 21 cm from hydrogen in a galaxy (then recently used to map the arms of the Milky Way)
would show through Doppler shifts in its frequency the difference in speeds of the gas towards or way
from us as the galaxy rotates. The point of finding out how fast a galaxy rotates is that more massive
galaxies should rotate faster, and more massive galaxies should be brighter. Tully and Fisher were
after a clock which could measure galaxy brightness, much as Henrietta Leavitt had found for
Cepheid variables.
People were not slow to point out the dangerous assumptions involved. Most serious was that most
of the brightness of a galaxy comes from its hot blue stars, while most of its mass is in the bulk of
cooler redder ones and indeed in dust and gas clouds. So the empirical relation they could show for
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galaxies whose distances had been estimated by other means, between brightness and rotation
speed, was a relationship between not really connected things. Luckily for Tully and Fisher, in more
recent years infrared telescopes on satellites have confirmed that the infrared brightness is also
related to rotation speed. The Tully–Fisher step on the distance ladder has proved very useful.
Standard outbursts
One very special and very rare kind of stellar object can be seen out to nearly the largest distances
telescopes can reach. They are the stellar explosions called supernovae, in which a cataclysm in a
dying star briefly makes it shine as brightly as a whole galaxy. There are two types, Type 1 and Type
2. Type 2 explosions are of giant stars, and vary in brightness depending on the mass of the star. But
Type 1 are remarkable standard candles, each as bright as another. This is because they come from
a collapsed dwarf star which collects matter from a neighbouring star. It happens that such a
collapsed star cannot ever be more than 1.4 times the mass of the Sun. Beyond that mass, it cannot
support its own gravitational attraction, and collapses yet further. These second collapses all start
from the same point, the point when the dwarf has just collected the maximum mass possible, and all
work in the same way. This makes every such explosion equally big. But they are also rare, perhaps
20 or 30 a year in a given galaxy. So to use them for distance measurement means looking at
thousands of distant galaxies in the hope of detecting a few such outbursts.
To think about
1. Suggest as many ways as possible to compare the distances of approaching cars at night,
including ways depending only on having photographs of the car headlights.
2. How far away would the brightest main sequence stars in the Hertzsprung–Russell diagram have
to be to look as bright as the Sun?
3. If the outer regions of a galaxy rotate at 300 km s–1, what shift in the wavelength of the 21 cm
hydrogen line would the Tully–Fisher method have to detect?
4. How far away would a Type 1 supernova shining with the brightness of 1011 Suns have to be to
look as bright as the Sun would look at a distance of 10 light-years?
5. Variable stars and rare supernovae are sometimes hunted down by comparing pairs of
photographs of the same objects, flashing the two images alternately on the same screen in the
same place. What would the astronomer have to look for?
6. Why will it generally take a longer exposure to get a spectrum of a galaxy than to take a picture of
the galaxy?
Einstein’s 1905 relativity paper
Reading 40T: Text to Read
Teaching Notes | Key Terms
Quick Help
The following extract is taken from the 1905 paper on special relativity by Einstein. The introduction
has been deleted. The first two sections are very fundamental to developing an understanding of
relativity.
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The first section deals with time from a radical new view point. It forces you to abandon Newton’s idea
of a universal time shared by all and puts in its place local times defined for given frames of
reference. Judgements about the simultaneity of an event and a local clock reading must be made to
establish the time of an event.
The second section shows us that intuitive actions such as measuring the length of a body or a time
interval between two events must be carefully considered in the light of the changes introduced in the
first section of the paper. Strange results are to be expected.
On the Electrodynamics of Moving Bodies
By A Einstein
30 June 1905
I. Kinematical part
§1. Definition of Simultaneity
Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In
order to render our presentation more precise and to distinguish this system of co-ordinates verbally
from others which will be introduced hereafter, we call it the ‘stationary system’.
If a material point is at rest relative to this system of co-ordinates, its position can be defined relatively
thereto by the employment of rigid standards of measurement and the methods of Euclidean
geometry, and can be expressed in Cartesian co-ordinates.
If we wish to describe the motion of a material point, we give the values of its co-ordinates as
functions of the time. Now we must bear carefully in mind that a mathematical description of this kind
has no physical meaning unless we are quite clear as to what we understand by ‘time’. We have to
take into account that all our judgments in which time plays a part are always judgments of
simultaneous events. If, for instance, I say, ‘That train arrives here at 7 o'clock’, I mean something
like this: ‘The pointing of the small hand of my watch to 7 and the arrival of the train are simultaneous
events’.
It might appear possible to overcome all the difficulties attending the definition of ‘time’ by substituting
‘the position of the small hand of my watch’ for ‘time’. And in fact such a definition is satisfactory when
we are concerned with defining a time exclusively for the place where the watch is located; but it is no
longer satisfactory when we have to connect in time series of events occurring at different places, or
– what comes to the same thing – to evaluate the times of events occurring at places remote from the
watch.
We might, of course, content ourselves with time values determined by an observer stationed
together with the watch at the origin of the co-ordinates, and co-ordinating the corresponding
positions of the hands with light signals, given out by every event to be timed, and reaching him
through empty space. But this co-ordination has the disadvantage that it is not independent of the
standpoint of the observer with the watch or clock, as we know from experience. We arrive at a much
more practical determination along the following line of thought.
If at the point A of space there is a clock, an observer at A can determine the time values of events in
the immediate proximity of A by finding the positions of the hands which are simultaneous with these
events. If there is at the point B of space another clock in all respects resembling the one at A, it is
possible for an observer at B to determine the time values of events in the immediate neighbourhood
of B. But it is not possible without further assumption to compare, in respect of time, an event at A
with an event at B. We have so far defined only an ‘A time’ and a ‘B time’. We have not defined a
common ‘time’ for A and B, for the latter cannot be defined at all unless we establish by definition
that the ‘time’ required by light to travel from A to B equals the ‘time’ it requires to travel from B to A.
Let a ray of light start at the ‘A time’ t A from A towards B, let it at the ‘B time’ t B be reflected at B in
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the direction of A, and arrive again at A at the ‘A time’ tA .
In accordance with definition the two clocks synchronize if
tB  t A  t A  tB
We assume that this definition of synchronism is free from contradictions, and possible for any
number of points; and that the following relations are universally valid:
1. If the clock at B synchronizes with the clock at A, the clock at A synchronizes with the clock at B.
2. If the clock at A synchronizes with the clock at B and also with the clock at C, the clocks at B and
C also synchronize with each other.
Thus with the help of certain imaginary physical experiments we have settled what is to be
understood by synchronous stationary clocks located at different places, and have evidently obtained
a definition of ‘simultaneous’, or ‘synchronous’, and of ‘time’. The ‘time’ of an event is that which is
given simultaneously with the event by a stationary clock located at the place of the event, this clock
being synchronous, and indeed synchronous for all time determinations, with a specified stationary
clock.
In agreement with experience we further assume the quantity
2AB
c
t A  t A
to be a universal constant – the speed of light in empty space.
It is essential to have time defined by means of stationary clocks in the stationary system, and the
time now defined being appropriate to the stationary system we call it ‘the time of the stationary
system’.
§2. On the Relativity of Lengths and Times
The following reflexions are based on the principle of relativity and on the principle of the constancy of
the velocity of light. These two principles we define as follows:
1. The laws by which the states of physical systems undergo change are not affected, whether
these changes of state be referred to the one or the other of two systems of co-ordinates in
uniform translatory motion.
2. Any ray of light moves in the ‘stationary’ system of co-ordinates with the determined velocity c,
whether the ray be emitted by a stationary or by a moving body. Hence
velocity 
light path
time interval
where time interval is to be taken in the sense of the definition in §1.
Let there be given a stationary rigid rod; and let its length be l as measured by a measuring-rod which
is also stationary. We now imagine the axis of the rod lying along the axis of x of the stationary
system of co-ordinates, and that a uniform motion of parallel translation with velocity v along the axis
of x in the direction of increasing x is then imparted to the rod. We now inquire as to the length of the
moving rod, and imagine its length to be ascertained by the following two operations:
(a) The observer moves together with the given measuring-rod and the rod to be measured, and
measures the length of the rod directly by superposing the measuring-rod, in just the same way
as if all three were at rest.
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(b) By means of stationary clocks set up in the stationary system and synchronizing in accordance
with §1 the observer ascertains at what points of the stationary system the two ends of the rod to
be measured are located at a definite time. The distance between these two points, measured by
the measuring-rod already employed, which in this case is at rest, is also a length which may be
designated ‘the length of the rod’.
In accordance with the principle of relativity the length to be discovered by the operation (a) – we will
call it ‘the length of the rod in the moving system’ – must be equal to the length l of the stationary rod.
The length to be discovered by the operation (b) we will call ‘the length of the (moving) rod in the
stationary system’. This we shall determine on the basis of our two principles, and we shall find that it
differs from l .
Current kinematics tacitly assumes that the lengths determined by these two operations are precisely
equal, or in other words, that a moving rigid body at the epoch t may in geometrical respects be
perfectly represented by the same body at rest in a definite position.
We imagine further that at the two ends A and B of the rod, clocks are placed which synchronize with
the clocks of the stationary system, that is to say that their indications correspond at any instant to the
‘time of the stationary system’ at the places where they happen to be. These clocks are therefore
‘synchronous in the stationary system’.
We imagine further that with each clock there is a moving observer, and that these observers apply to
both clocks the criterion established in §1 for the synchronization of two clocks. Let a ray of light
depart from A at the time t A , let it be reflected at B at the time t B, and reach A again at the time t A .
Taking into consideration the principle of the constancy of the velocity of light we find that
tB  t A 
r AB
c v
and
t A  tB 
r AB
c v
where r AB denotes the length of the moving rod – measured in the stationary system. Observers
moving with the moving rod would thus find that the two clocks were not synchronous, while
observers in the stationary system would declare the clocks to be synchronous.
So we see that we cannot attach any absolute signification to the concept of simultaneity, but that
two events which, viewed from a system of co-ordinates, are simultaneous, can no longer be looked
upon as simultaneous events when envisaged from a system which is in motion relatively to that
system.
End of extract from Einstein’s paper.
Questions to discuss
Read the following text from the first paragraph again, carefully:
‘The laws by which the states of physical systems undergo change are not affected, whether these
changes of state be referred to the one or the other of two systems of co-ordinates in uniform
translatory motion.
Any ray of light moves in the ‘stationary’ system of co-ordinates with the determined velocity c,
whether the ray be emitted by a stationary or by a moving body. Hence
velocity 
light path
time interval
where time interval is to be taken in the sense of the definition in §1.’
1. What is the first statement saying about the laws of physics as referred to different frames of
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reference?
2. Why is the second statement so strange for people who have not studied relativity in detail?
There are two equations towards the end of the first section.
tB  t A  t A  tB
This equation could be described as a radar equation.
Study the text below taken from the paper:
‘Let a ray of light start at the ‘A time’ t A from A towards B, let it at the ‘B time’ t B be reflected at B in
the direction of A, and arrive again at A at the ‘A time’ tA.’
3. Show that the above equation can be arranged as:
t B  (t A  t A ) / 2
A
B
t´ A
t
tA
x
4. The space time diagram shown summarises the argument above. The red lines show the world
lines for light rays. Make a copy of the diagram and mark on it t B the reflection time. Show that
the equation above gives the t B you would assign to this event assuming that light travels
identically from A to B and back from B to A. Explain how this diagram shows light being used like
radio waves in a radar measurement.
The second part of the paper needs to be read quite carefully.
Einstein derives these equations:
tB  t A 
r AB
c v
and
t A  tB 
r AB
c v
then says because t B – t A does not equal tA – t B then the clocks are not synchronised.
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5. The diagram below represents the situation discussed in the second section of the paper. The rod
AB is moving and this is shown in the space time diagram by the sloping lines. The light rays still
have the same slope as light waves always move at c relative to any moving body.
A
B
t´ A
t
tB
tA
r AB
x
x
Use the space-time diagram above to find x. Then simplify your answer to obtain the first of the
equations above. The second light path can be similarly analysed to find the second equation.
Remember that the diagram represents a rod moving along a straight line. In the space-time diagram
we show for example the position of A at certain times as measured in the lab frame. Lab frame time
is being used. There is no one special time: this is a key point to grasp. Lab frame clocks are
synchronised in the lab frame and measuring rulers used to determine x are at rest in this frame.
The ‘breakthrough of the year’
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This is a reading about a discovery hailed as the ‘Breakthrough of the Year’ for 1998 by the journal
Science. Read it carefully and decide how certain you think the discovery is. If you have access to the
Internet, a web search will give you the latest information about the status of this discovery.
What follows is the text as released by the Royal Astronomical Society.
Supernova cosmology project a winner
By observing distant, ancient exploding stars, astronomers at the Institute of Astronomy, Cambridge
with colleagues at the Lawrence Berkeley Laboratory, California, and elsewhere, have determined
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that the Universe is expanding at an accelerating rate - an observation that implies the existence of a
mysterious, self-repelling property of space first proposed by Albert Einstein, which he called the
cosmological constant. This extraordinary finding has been named by the journal Science as
‘Breakthrough of the Year’ for 1998.
The Supernova Cosmology Project, led at Cambridge by Professor Richard Ellis, Dr Richard
McMahon and Dr Mike Irwin, and at Berkeley by Dr Saul Perlmutter, shares the citation with the
High-z Supernova Search Team, another international collaboration involving astronomers in
Australia, Germany and the USA.
The surprising discovery that the expansion of the Universe is accelerating – and hence likely to go
on expanding forever – is based on observations of stellar explosions known as type Ia supernovae.
These supernovae all have the same intrinsic brightness. This means that their apparent brightness
as observed from Earth reveals their distance.
By comparing the distance of these exploding stars with the red-shifts of their host galaxies,
researchers can calculate how fast the Universe was expanding at different times in its history. Good
results depend upon observing many type Ia supernovae, both near and far. The Supernova
Cosmology Project has fully analysed the first 42 out of more than 80 supernovae it has discovered,
and more analysis is in progress.
Type Ia supernovae are rare. In a typical galaxy they may occur only two or three times in a thousand
years. And to be useful they must be detected within a week or two of the explosion, while the
supernova is still increasing in brightness. Prior to implementation of search techniques developed by
the Supernova Cosmology Project during the first five years of its existence, finding supernovae –
even those in relatively nearby galaxies – was a haphazard proposition which made it difficult to
secure telescope time to observe them.
Origin of the Supernova Cosmology Project
‘I first attempted this project in 1986 with a small telescope in Chile’ explains Richard Ellis. ‘After
months of effort I found the first distant supernova and this demonstrated the feasibility of the current
programme. Today’s award from Science comes after over a decade of hard work by a large team of
astronomers at Cambridge and Berkeley.’
‘The most important technical breakthrough was the agreement between Cambridge and Berkeley to
employ a new panoramic camera, produced by Perlmutter’s group, on UK telescopes in the Canary
Islands’, adds Cambridge astronomer Richard McMahon. ‘Our team was then able to develop a new
strategy that assured discovery of numerous supernovae “on demand”.’
How the technique works
Project member Mike Irwin explains how the strategy works. ‘Just after a new Moon, when the sky is
dark, we make images of 50 to 100 patches of sky. Each contains roughly a thousand distant
galaxies. Three weeks later the same patches are imaged again. Supernovae occurring anywhere in
these fields show up as bright points of light.’ In these three weeks, the supernovae typically have not
yet reached their brightest moment.
‘This guarantees that we will have supernovae to study during the best nights for observation, right
before the new Moon’, says project member Richard McMahon. He adds, ‘Type Ia supernovae are so
similar, whether nearby or far away, that the time at which an explosion started can be determined by
simply looking at its spectrum. Type Ia supernovae which exploded when the Universe was half its
present age behave the same as they do today.’
Early success
By 1994 the Supernova Cosmology Project had proved repeatedly that, with this search technique, a
few nights on the world’s best telescopes dependably resulted in many new supernova discoveries.
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‘While some of us are surveying distant galaxies from the Cerro Tololo Interamerican Observatory
(CTIO) in the Chilean Andes, or at the Isaac Newton Group telescopes in the Canary Islands, others
at home are retrieving the data over the Internet and analysing it to find supernovae’, says team
member Richard Ellis. ‘Then, with the powerful Keck Telescope in Hawaii we confirm spectra and
measure red-shifts. We call the Hubble Space Telescope into action to study the most distant
supernovae, as these require much more accurate measurements than we can get from the ground.’
Among the supernovae discovered by the Supernova Cosmology Project are the most distant, and
therefore the most ancient, ever seen. In 1997, the Cambridge–Berkeley team announced that a
supernova with a red shift of 0.83, equivalent to an age of seven billion years, had been found using
the CTIO and Keck telescopes and subsequently observed by the Hubble Space Telescope.
As early as 1994, as their early supernova discoveries began to accumulate, members of the
Supernova Cosmology Project developed key analytic techniques that could be used to interpret
supernova measurements and thereby determine the cause of the expansion rate of the Universe. At
the time almost everybody assumed that the Universe was slowing down, due to gravity acting on the
matter in the Universe. The question was how quickly was it slowing? What is the mass density of the
Universe? And, finally, is there enough mass density to eventually reverse the expansion, leading to a
‘big crunch’ finale for the Universe.
Evidence for Einstein's cosmological constant
There was also the possibility, unlikely as it seemed, that some intrinsic property of empty space was
in play, something called the cosmological constant – a term originally proposed by Einstein in 1917,
in an attempt to balance the equations of General Relativity and preserve a picture of a stable
Universe that would neither expand nor collapse on itself. A dozen years after Einstein introduced the
cosmological constant, astronomer Edwin Hubble found that the Universe is indeed expanding;
Einstein dismissed his cosmological constant idea as ‘the biggest blunder of my life’.
But observations of distant type Ia supernovae placed them significantly farther away than expected
from their red-shifts, suggesting that Einstein spoke too soon. Something is pushing everything
farther apart faster than it did in the early Universe. The cosmological constant is the best candidate.
Thus instead of slowing down, as everybody had expected, the expansion of the Universe is in fact
speeding up. In early January 1998 the Supernova Cosmology Project presented the first compelling
evidence that the expansion of the Universe is accelerating and that this acceleration is due to the
cosmological constant, known by the Greek letter lambda, which may represent as much as 70 per
cent of the total mass–energy density of the Universe. Subsequently, the High-z Supernova Search
Team announced that they had found the same result in their data.
Barring change in the value of lambda – whose exact nature remains a mystery – the Universe will
expand forever.
Continuing research
‘It is important that the High-z team has joined in the quest to learn the nature and fate of the cosmos
by studying supernovae’, says Richard Ellis. ‘In major scientific programmes such as this it is
important to have two competing teams; it leads to healthy scepticism and a high standard of
scientific rigour. The most important conclusion so far is, astonishingly, that both teams agree the
Universe is accelerating!’
‘We are now searching for more supernovae with high red-shifts in order to get more information
about the early Universe’, says team member Richard McMahon. ‘But, we are also looking for
supernovae with low red-shifts – nearby supernovae – to make sure that young and old type Ia
supernovae are essentially the same, and make for dependable standard candles. We want to be
sure we aren’t being fooled by interstellar dust dimming the supernovae, or stellar explosions that are
somehow weaker in the distant past. So far we haven’t found anything to shake our confidence, but
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this is such an unexpected discovery that we’ll keep looking for any loopholes.’
Using the world’s best telescopes, including the Keck Telescope, the Hubble Space Telescope and
the UK’s Isaac Newton Group’s telescopes on the Canary Islands, the Supernova Cosmology Project
continues to pursue studies aimed at confirming these astonishing results.
The sky is dark at night:
A reason to think that the Universe has evolved
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Thinkers as far back as Edmond Halley (1656–1742) realised that there is important cosmological
information in the simple fact that the sky is dark at night. You can read here what some modern
cosmologists have said about the cosmological implications of the dark night sky.
Olbers’ paradox
‘Why is the sky dark at night?’ Heinrich Olbers (1758–1840), German astronomer and comet hunter,
raised this question in 1826, though not for the first time. Olbers pointed out that if the Universe is
infinite and filled uniformly with stars, there must be a star in every direction you can look. So the sky
should be as bright as the surface of a star. That’s what ‘infinite’ means.
In an infinite Universe, uniformly populated with stars, there
will be a star somewhere along every possible line of sight
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But the sky is rather black at night. Most lines of sight do not seem to end on a star. This was called
‘Olbers’ paradox’. Olbers himself suggested that dust in space intercepts and absorbs the missing
starlight. An objection to this is that the light would heat up the dust until it glowed as bright as a star,
so that the sky would be bright again. If the Universe were infinitely old, this must have happened
already! Back to the paradox.
‘Our evolving Universe’
Here is what Cambridge astronomer Malcolm Longair has written about Olbers’ paradox:
In its simplest form, Olbers’ paradox arises because, if the Universe were infinite, static and
uniformly filled with stars, the sky would be as bright as the surface of the stars, clearly in
contradiction with our experience. At least one of these assumptions about the nature of our
Universe has to be wrong….The fact that the sky is not as bright as the surface of the Sun
provides us with some very general information about the Universe. Probably the most general
way of expressing the significance of this observation is that the Universe must, in some sense,
be far from equilibrium, though the way in which it is in disequilibrium cannot be deduced from
this very simple observation. The fact that the Universe is expanding and has a finite age are two
contributions to the resolution of the paradox.
Malcolm S Longair (1996) Our Evolving Universe (Cambridge University Press).
‘Getting rid of the debris’
Professor Jim Peebles, American cosmologist, looks at the paradox in a different way. He asks,
‘Where has all the starlight gone?’ He points out that if the radiation in the Universe just grew and
grew, it would put out the stars:
If energy is conserved, the stars cannot be permanent, for they must eventually exhaust their fuel
supply. If energy conservation as usually understood were violated within stars, so they could
shine forever, what would become of the starlight? If the material Universe were a finite island in
infinite empty space, the starlight could stream away. In a homogeneous static Universe,
however, the mean energy per unit volume of starlight would have to grow with time. The energy
density in starlight presumably would continue to grow until the radiation temperature reached the
internal temperature of the material within stars, making it impossible for the material to release
any more energy unless the stars got hotter, and of course making space quite uninhabitable for
us. This is the effect known as Olbers’ paradox. The effect has been more or less understood
since at least the seventeenth century, though people have not always been eager to face up to
it, as we see from the fact that Einstein did not take note of this problem with his world model.
…The assumption of a homogeneous unchanging Universe leads to a double bind: we have to
postulate an eternal source for new stars and galaxies, and we need some provision for getting
rid of the debris. In the absence of the latter the energy density of starlight would build up until it
became intense enough to suppress energy production within stars.
P J E Peebles 1993 Principles of Physical Cosmology (Princeton University Press).
‘The microwave background is old starlight’
The microwave background radiation fills the Universe with photons emitted long ago. It does not
dazzle us as the original photons would have, because it is red shifted by a factor of about a
thousand. Its temperature is only a few degrees kelvin, so it is cold and the sky looks dark. This is
how Professor Michael Rowan-Robinson of Imperial College looks at Olbers’ paradox:
…the phenomenon of the microwave background radiation, interpreted as the relic of the fireball
phase of the Hot Big Bang, on which so much of our present cosmological story is pinned. This
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can be experienced gratuitously by looking at the random dots on your television screen after
broadcasting has ceased. Part of this ‘noise’ is due to the microwave background radiation. This
radiation is also in a sense related to the darkness of the night sky, which in a supposedly infinite
Universe of stars perplexed astronomers and philosophers for centuries (the so-called Olbers’
paradox) until Edgar Allan Poe explained the darkness as due to the finite age of the Universe.
The light from the most distant stars has not had time to reach us yet. But as I like to point out,
the sky is pretty bright in the microwave band, being about as bright in energy terms as the Milky
Way is in the visible band.
Michael Rowan-Robinson 1993 Ripples in the Cosmos (W H Freeman).
To think about…
1. Can so much really be decided from so little?
2. What might make people want to believe that the Universe is infinite and eternal?
3. ‘The light from a star at distance r is dimmed by a factor 1 / r 2. So distant stars make little
contribution, and Olbers was wrong.’ What is wrong with this proposal? If you have no idea, look
at the next question.
4. ‘The number of stars at distance r increases as r 2, because they fill a shell of surface area 4 r 2.
So the more distant the stars the bigger the contribution they make. Olbers was wrong.’ What is
wrong with this proposal? If you have no idea, look at the previous question.
5. List as much evidence as you can think of which suggests that the Universe has not always been
the same.
Life CV:
Edwin Powell Hubble
Reading 70T: Text to Read
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Introduction
What follows is a brief description of Hubble’s life. It outlines his major scientific work and describes
some interesting details about him as a person.
Early life
Born: November 1889 in Marshfield, Missouri.
Father: John Powell Hubble; Mother: Virginia Lee James.
Grandparents: Martin and Mary Jane Hubble (Judge) and Dr William H and Lucy Ann James. The
James family were related distantly to Jesse and Frank James, the famous bank robbers.
Education
Educated at central school, Wheaton, Chicago 1901–6. University of Chicago 1906–10, where he
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excelled in athletics and used all his 6 ft 2 in of height in the basketball team. He studied maths and
science with, amongst others, Albert Michelson and was a laboratory assistant to Robert Millikan. He
was awarded a Rhodes scholarship in 1910 and entered Queen’s College Oxford until 1913. He
studied law as his father desired and started an English degree, which he dropped for Spanish. Whilst
there he is supposed to have sparred with Georges Charpentier, the French heavyweight champion
of the world. On a tour round Europe he leant to duel with the Schläger (rapiers) in Germany and
fought a pistol duel with a German officer after rescuing his wife from the Kiel canal!
Career
Back in America after his father’s death, he taught Spanish and Physics at the New Albany High
School, Indiana. He then studied for a PhD in astronomy under Edwin Frost, director of Yerkes
Observatory. In 1917 he was offered a position at the new Mount Wilson Observatory by Adams, but
he joined up before he could take up the offer. In the First World War Hubble reached the rank of
major with the 86th Division and was wounded near Metz.
Discoveries
In 1919 he joined the staff of the new 100 inch Hooker Telescope at Mount Wilson, where his
Anglophile idiosyncrasies reached new heights. He spent most of his time photographing ‘nebulae’,
which he always refused to call galaxies even when appropriate! This important work led to his
famous discovery of the ‘cosmological red-shifts’ and the expansion of the Universe. His co-worker in
this was one Milton Humason, a brilliant self-taught observer, who joined the Mount Wilson staff early
on as a teenager in charge of the mules, which brought equipment up the mountain. Many people
visited the famous telescope, including the Hollywood stars of the times.
War service and later life
During the Second World War Hubble did research in ballistics, running the Aberdeen Proving Range
in Chesapeake Bay. Afterwards he worked on the Hale Telescope at Mount Palomar, the new 200
inch telescope which came into operation in 1948. On 27 September 1953 he died of a heart attack,
having suffered a previous attack in 1949. He missed being awarded a Nobel prize; they are not
awarded posthumously.
Suggestions for additions
1. From the Internet, or other sources, find a photograph of Hubble and of the telescopes he used to
add to this biography.
2. Construct similar biographies for:
Vesto Slipher who measured red shifts
Allan Sandage who developed distance scales
Fritz Zwicky who proposed missing mass
Henrietta Leavitt who discovered the period–luminosity relationship for Cepheid variables
Vera Rubin who discovered missing mass in galaxies
Hermann Bondi who, jointly with Fred Hoyle, proposed the steady-state theory
Fred Hoyle who, jointly with Hermann Bondi, proposed the steady-state theory
Walter Baade who corrected distance scales
Gerard de Vaucouleurs who challenged the distance scale
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Subrahmanyan Chandrasekhar who found the reason stars collapse
Why we believe in special relativity: Experimental support for
Einstein’s theory
Reading 100T: Text to Read
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These notes are based on those provided by kind permission of John S Reid, Department of Physics,
University of Aberdeen.
Einstein’s theory of special relativity is sometimes presented as if it were a piece of philosophy or
mathematics that arose purely from abstract thinking about space and time. This is not the case.
Einstein based his theory firmly on experimental results known to him. Subsequently, the implications
of his theory have been widely tested over the past century. No repeatable and generally accepted
experiment has been found in disagreement with special relativity.
Introduction
One of the influences that motivated Einstein throughout his life was a search for ‘the truth’. It has
been, and is, a motivation shared by many scientists and you can be sure that neither his
contemporaries nor his successors would be prepared to sit back and accept such a radically new
way of looking at nature as special relativity without asking ‘is this really true?’. It is one of the
strengths of science that we don’t ask questions like this of people but we ask nature herself. By 1905
there were a range of relevant experimental results On the Electrodynamics of Moving Bodies (the
title of Einstein’s key paper), mainly optical results, that Einstein knew supported his theory.
Three pre-1905 results
The astronomer James Bradley discovered in 1727 that a small correction had to be made to the
direction of a telescope aimed towards a star because of the motion of the telescope caused by the
Earth orbiting the Sun. This phenomenon is known as ‘stellar aberration’ because if you don’t make
the correction the location of a star appears follows a small ellipse over a year. The ellipse can have a
diameter as great as 41 arc seconds; not much but measurable with care, even in 1727. Bradley’s
result is just what special relativity predicts. It is a more important result than you might suspect
because it rules out one of the proposed alternative theories to special relativity, namely that the
special frame of reference referred to as the ether was always fixed with respect to the Earth.
Bradley’s result is not a definitive test but it is one relativity has to pass. Airy in 1871 tested whether
the aberration remained the same if the telescope was filled with water, in which light travels more
slowly. The result does and this is consistent with special relativity.
In 1851 the French physicist Fizeau was the first to measure the speed of light by a laboratory
technique. In fact he was able to measure the speed of light through a moving medium (remember
the title of Einstein's paper). In Fizeau’s experiment the speed of light travelling against the motion of
moving water was compared with its speed travelling with the motion of the water. His result was
predicted earlier by Fresnel and is just what Maxwell’s electromagnetic equations and special
relativity later predicted.
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Much more recently, there was some controversy over what happens when the medium is moving at
right angles to the direction of travel of the light. The result was settled by an experiment carried out in
Aberdeen in 1971, with Professor R V Jones building the equipment and Professor Mike Player
covering the theory. The result vindicated the predictions of Maxwell’s equations and the predictions
of special relativity.
Einstein has been criticised for not putting any references into his famous 1905 paper. Nevertheless,
he was aware of the most significant result that supported his ideas, namely the null outcome of the
Michelson–Morley experiment of 1887. The Michelson–Morley experiment is the third of the pre-1905
results. We will describe it in more detail later.
A table of results
In the century since Einstein’s 1905 paper, a raft of tests have been carried out to check whether the
background concepts upon which special relativity is based and its predictions are true. Only a few
are shown on this table, which you’re not expected to read in detail. Nature has replied through a
megaphone: ‘yes’.
Optical experiments of Arago,
Fizeau, Bradley, Airy with moving
bodies
Pre 1905 results
Tests of postulates
Test on time dilation etc
Relativistic kinematics
Round-trip tests of speed of light
Michelson–Morley experiment and
derivatives
One-way tests
Lasers, masers and Mossbauer
effect
Independence of speed of light
Alvager et al , Brecher (on motion
of source), Schaefer (on energy)
Limit on photon mass
Goldhaber and Nieto, Davis et al
Tests on Lorentz invariance
Trouton–Noble experiment and
others
Particle lifetimes
Muon lifetimes, Bailey et al
Doppler formulae
Ives and Stillwell, McGowan et al
Twin paradox
Hafele and Keating
Limiting velocity is c
Brown et al , Glashow and
Coleman, Stodolsky (neutrinos)
Variation of momentum with
velocity
Particle accelerators + special tests
E = mc2
Element transformations,
astrophysics
The tests certainly all support the bottom line that no repeatable and generally accepted experimental
result is in disagreement with special relativity.
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Time dilation
One of the most dramatic predictions of special relativity is time dilation. Time dilation implies that
clocks in a frame moving with respect to you appear to run slow. You and I are never likely to be
moving at near the speed of light but many elementary particles can hardly help themselves from
doing so.
The muon experiment
Mu-mesons, or muons as they are called these days, are elementary particles bearing some similarity
to massive electrons. They are created by collisions induced in particle accelerators and by the same
process in cosmic ray showers when particles called -mesons decay. Muons are among the particles
that travel from several kilometres up in the atmosphere to the surface of the Earth. The fact that they
reach the surface at all is very curious indeed. But reach the surface of the Earth they do, as you can
hear in a radiation detector that records the natural background radiation in which life on Earth
evolves. A significant fraction of the clicks you hear in the counter is caused by muons.
Muons have a short half-life of 2.197 s. This means 2.197 s before they have a 50% chance of
spontaneously decaying into an electron or positron (depending on their charge) and neutrinos. A
bunch of muons travelling at 0.99 times the speed of light (0.99c) will only go 650 m before half of
them have decayed. In the atmosphere, muons are created in a shower at a typical height of 10 km
and will need 15.3 half-lives of time to reach the ground, more if they are coming at an angle. Let’s
suppose there are 15 rows of an audience in a lecture theatre or small cinema and I were to give a
sum of money to the back row and ask each row to take out half the money and pass the rest
forward. How much would I have to give to the back row to ensure that I received at least 1 p at the
front? The answer is about £327. After 15 rows each taking out half, very little of the original is left. In
fact only one part in over 32,000. The same should happen to the muons travelling earthwards. It
doesn’t, as a Geiger counter testifies. (10 km is about the height that many commercial airlines fly and
if the muon count at that height was 32,000 times what we record on the ground then commercial
airlines would be grounded.)
Special relativity explains what is going on. For muons travelling at 0.99c, the time dilation factor is
about 7 ( = 7.09, to be exact). Their half-life observed in our ground frame of reference is longer by a
factor of 7.09 and hence according to relativity the time needed for muons to reach the ground is not
15.3 half lives but only 15.3 / 7.09 = 2.18 half-lives. If there are only two rows dividing the £327 before
I get it instead of 15, I’m not going to get just 1p but about £82, or 8000 times as much. To get more
real, the muon count at the height commercial airliners fly is only four times what it is at ground level,
which is quite acceptable.
The Rossi and Hall muon experiment 1941
The original experiment was done by Rossi and Hall in 1941, who measured muon fluxes not 10 km
high but at the top of Mount Washington in New England, about 2 km high, and at the base of the
mountain. The effect is less for a height difference of only 2 km but for their muon speeds of 0.994c,
relativistically the reduction should have been only a factor of 1.26 whereas without time dilation the
reduction would be a factor of 8.5. Rossi and Hall’s figures were consistent with the relativistic
prediction. The experiment has since been repeated by others with convincing results. This table
shows some figures for the Rossi and Hall experiment.
Relativity
Newtonian
Frame of muon
Frame of ground
Distance
0.219 km
2 km
2 km
Time
0.734 s
6.71 s
6.71 s
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Relativity
Newtonian
Half-life
0.337
0.337
3.08
Reduction
1.26
1.26
8.5
In 1979 Bailey et al at a CERN accelerator reported a similar experiment with CERN generated
muons of speeds 0.9994c, trapped in a particle accelerator, that were observed in the lab to have
29.3 times the muon rest lifetime, completely consistent with time dilation.
Tests of time dilation and transverse Doppler effect
The Doppler effect is the observed variation in frequency of a source when it is observed by a
detector that is moving relative to the source. This effect is most pronounced when the source is
moving directly toward or away from the detector, and in pre-relativity physics its value was zero for
transverse motion (motion perpendicular to the source–detector line). In special relativity there is a
non-zero Doppler effect for transverse motion, due to the relative time dilation of the source as seen
by the detector. Measurements of Doppler shifts for sources moving with velocities approaching c can
test the validity of special relativity’s prediction for such observations, which differs significantly from
classical predictions; the experiments support special relativity and are in complete disagreement with
non-relativistic predictions.
The Ives and Stilwell experiment
H E Ives and G R Stilwell 1938 An experimental study of the rate of a moving atomic clock J. Opt.
Soc. Am. 28 215–26 ; 1941 J. Opt. Soc. Am. 31 369–74
This classic experiment measured the transverse Doppler effect for moving atoms.
D Hasselkamp et al 1989 Direct observation of the transversal Doppler-shift Z. Phys. A 289 151
A measurement which is truly at 90° in the lab. Agreement with special relativity to an accuracy of a
few per cent.
Cosmic ray evidence for ultimate speed
One of the consequences of special relativity is that no bodies can travel at a speed faster than the
speed of light c. Nobel Prize winner Sheldon Glashow and collaborator Sidney Coleman showed in
1997 that the argument could be taken further. The mere existence of very high-energy cosmic ray
photons reaching the Earth is strong proof, without any extra experiment, of the existence of an upper
limit of the speed of light c for material bodies. Their argument is that photons decay by pair
production into electrons and positrons at a rate that can be calculated from particle physics. If the
upper limit to the speed of electrons differed from c by a small amount, then high-energy photons
(~20 TeV) would decay in nanoseconds and never travel any significant distance from their point of
creation. The detection of these particles on Earth sets a tight bound of an upper limit to the speed of
matter being within 1.5  10–15 of c.
The twin paradox
Another implication of special relativity is the famous twin paradox in which one twin who travels away
and returns finds the other twin who has remained behind has aged more than the travelling twin has.
The Hafele and Keating experiment of 1971 described how four caesium beam clocks, which are
highly accurate clocks stable to about 1 part in 1013, were sent on a global round trip on commercial
airliners. The time these clocks gained or lost was compared with a master clock that stayed at the
US Naval Observatory, which carried out the experiments. A difference in elapsed time measured by
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the moving clocks was expected both because of the time dilation of special relativity and because of
a gravitational effect of general relativity due to the difference in height of the surface clock and the
aircraft clock of about 9 km. The time differences were nanoseconds, but caesium beam clocks can
accurately measure such small differences.
Predicted
Time difference / ns
eastward
Time difference / ns
westward
Gravitational
144 ± 14
179 ± 18
Special relativity
–184 ± 18
96 ± 10
Net effect
–40 ± 23
275 ± 21
Observed
–59 ± 10
273 ± 21
The results are shown in the table for the difference in elapsed times for the aircraft clocks and the
surface clocks. They were in complete agreement with Einstein’s predictions. Subsequent repetitions
of this experiment in more recent times with better clocks making different trips have confirmed the
results.
The Global Positioning System (GPS) that is used all over the world for highly accurate positioning
recording was the creation of the US military. It is based upon highly accurate clocks orbiting the
world in satellites. Corrections have to be included for both special relativistic effects and for general
relativistic gravitational effects. Without these corrections the system would not produced the
accuracy it does, by a long way. Light travels 1 m in about 3 ns so to get 1 m accuracy, and the
military system can do better, the clocks and timing corrections need to be correct to within a few
nanoseconds. Would you bet your life on special relativity being true? Anyone who relies on GPS in
bad weather may be doing just that. Probably thousands of aircraft passengers and crew do so every
day.
Tests of the ‘twin paradox’
J C Hafele 1970 Relativistic behaviour of moving terrestrial clocks Nature 227 270–71 (1970)
J C Hafele and R E Keating 1972 Around-the-world atomic clocks: predicted relativistic time gains
Science 177 166–8
J C Hafele and R E Keating 1972 Around-the-world atomic clocks: observed relativistic time gains’
Science 177 168–70
R F C Vessot and M W Levine 1979 A test of the equivalence principle using a space-borne clock’
Gen. Rel. Grav. 10 181–204
R F C Vessot et al 1980 Test of relativistic gravitation with a space borne hydrogen maser’ Phys. Rev.
Lett. 45 2081–4
They flew a hydrogen maser in a Scout rocket up into space and back (not recovered). Gravitational
effects are important, as are the velocity effects of special relativity.
C Alley 1983 Proper time experiments in gravitational fields with atomic clocks, aircraft, and laser light
pulses’ in Quantum Optics, Experimental Gravity, and Measurement Theory ed P Meystre and M O
Scully (Plenum) pp 363–427
They flew atomic clocks in airplanes which remained localized over Chesapeake Bay, and also which
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flew to Greenland and back.
J Bailey et al 1977 Measurements of relativistic time dilatation for positive and negative muons in a
circular orbit’ Nature 268 301–5
J Bailey et al 1979 Final report on the CERN muon storage ring including the anomalous magnetic
moment and the electric dipole moment of the muon, and a direct test of relativistic time dilation’ Nucl.
Phys. B 150 1–75
They stored muons in a storage ring and measured their lifetime. When combined with
measurements of the muon lifetime at rest this becomes a highly-relativistic twin scenario (v ~ 0.9994
c), for which the stored muons are the travelling twin and return to a given point in the lab every few
microseconds.
For the muon lifetime at rest:
S L Meyer et al 1963 Precision lifetime measurements on positive and negative muons Phys Rev.
132 2693
M P Balandin et al 1974 Sov. Phys. JETP 40 811
G Bardin et al 1984 Phys. Lett. B 137 135 (Also a test of the clock hypotheses – see below.)
The clock hypothesis
The clock hypothesis states that the tick rate of a clock when measured in an inertial frame depends
only upon its velocity relative to that frame, and is independent of its acceleration or higher
derivatives. The experiment of Bailey et al referenced above stored muons in a magnetic storage ring
and measured their lifetime. While being stored in the ring they were subject to a proper acceleration
of approximately 1018 g (g = 9.8 m s–2). The observed agreement between the lifetime of the stored
muons with that of muons with the same energy moving inertially confirms the clock hypothesis for
accelerations of that magnitude.
C W Sherwin 1960 Some recent experimental tests of the ‘clock paradox’ Phys. Rev. 129 17
Discusses some Mössbauer experiments that show that the rate of a clock is independent of
acceleration (~1016 g) and depends only upon velocity.
The Michelson–Morley experiment
When James Clerk Maxwell predicted the existence of electromagnetic waves there was an
extraordinary feature of his prediction whose full implication wasn’t fully appreciated at the time.
Maxwell’s equations included a fixed constant c for the speed of the electromagnetic waves, the
speed of light. In everyday life, speeds aren’t fixed constants. They depend on how the observer
moves. Imagine that I am driving along the road at 60 mph, and that my speed is clocked by a police
radar gun sitting by the roadside. Now imagine that the radar gun is in a police car coming towards
me also at 60 mph. You would expect the radar gun now to read the closing speed between us of 120
mph. There isn’t one constant that describes my speed. It seemed that Maxwell’s speed of light must
therefore be the speed in one particular frame of reference, which contemporary physicists called the
ether. Michelson set himself the task of measuring the speed v of the ether relative to the Earth.
Einstein was aware of the result of the Michelson–Morley experiment, and was aware that this
experiment found that v was zero within experimental error. The Michelson–Morley experiment
underpins a central assumption in relativity. It is one of the classic experiments of physics.
The Michelson interferometer
Michelson hit upon an ingenious way of measuring the speed of the invisible ether. His problem was
that the speed of light is about 300,000 km s–1 and whatever our speed was relative to the invisible
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ether it should be at least 30 km s–1, due to the orbital speed of the Earth around the Sun, as Bradley
had demonstrated. To detect this Michelson needed to measure the speed of light to an accuracy of
about 10 km s–1. This accuracy was beyond the scope of technology in the 1880s, when he was
considering the issue. Michelson was the best optical experimenter of his generation. He had a stroke
of genius. He realised that he just needed to measure the difference in the speed of light in the
direction of the ether and at right angles to the ether. This could be done by a ‘round trip’ experiment
in a device with two light paths at right angles to each other. He invented such a device, now called a
Michelson interferometer. It has turned out to be an enormously versatile instrument with a huge
number of uses.
The action needed in the experiment is to rotate the equipment smoothly round and if the speed of
light is different in different directions the comparison times between light travelling along the two
arms will change. Comparison fringes seen in the equipment will shift with the rotation. Michelson
found no shift at all. The limit on Michelson’s sensitivity was about 15 km s–1.
Open the JPEG file
Source
The result
The result was that no velocity could be found. The experiment has been repeated on many
occasions with improved equipment, and several variations. It is not an easy experiment to do. The
strong consensus of results is that no ether can be detected. You'd be wrong to conclude that the
Earth is simply dragging the ether with it, for Bradley's experiment, the very first one mentioned in this
section, is evidence that this isn’t happening.
The failure of the Michelson–Morley experiment and the other early experiments to actually observe
the Earth’s motion through the ether became significant in promoting the acceptance of Einstein’s
theory of special relativity, as it was appreciated from early on that Einstein’s approach was more
elegant and parsimonious of assumptions than were other approaches.
The following table comes from R S Shankland et al (1955 Rev. Mod. Phys. 27 167–78), which
includes references to each experiment (resolution and the limit on Vether are from the original
sources). The expected fringe shift is what would be expected for a rigid ether at rest with respect to
the Sun and Earth’s orbital velocity (30 km s–1).
Name
Year
Arm length / m
Fringe
shift
expected
Fringe shift
measured
Michelson
1881
1.2
0.04
0.02
Michelson and Morley
1887
11.0
0.4
< 0.01
Morley and Morley
1902–4
32.2
1.13
0.015
Miller
1921
32.0
1.12
0.08
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Upper limit on
Vether / km s–1
8
Name
Year
Arm length / m
Fringe
shift
expected
Fringe shift
measured
Miller
1923–24
32.0
1.12
0.03
Miller (sunlight)
1924
32.0
1.12
0.014
Tomascheck (starlight)
1924
8.6
0.3
0.02
Miller
1925–26
32.0
1.12
0.088
Kennedy (Mt Wilson)
1926
2.0
0.07
0.002
Illingworth
1927
2.0
0.07
0.0002
Piccard and Stahel (Mt Rigi)
1927
2.8
0.13
0.006
Michelson et al
1929
25.9
0.9
0.01
Joos
1930
21.0
0.75
0.002
Upper limit on
Vether / km s–1
1
Conclusion
In addition to the experiments mentioned above, an earlier table in this reading highlighted tests of
‘relativistic kinematics’, in particular the famous relationship E = mc2. This relationship has to be taken
into account in the design of any high-energy particle accelerator. If things didn’t happen as Einstein
had deduced, then these machines simply would not work as designed. They do work, as smoothly
as an aeroplane, thanks to Einstein's fundamental insight.
Here finally is a table that made an impression on me when I first saw it many years ago in the
textbook by W Panofsky and M Phillips (Classical Electricity and Magnetism, Addison-Wesley, 1962).
It lists 13 key experiments that have a testing relevance to special relativity in the columns, and the
predictions of six alternative theories to special relativity in the rows. Some of the experiments are
additional to those discussed above. The blue boxes mark the places where the experimental results
disagree with the predictions of the theory. Only special relativity is in agreement with all testing
experiments.
Theory
Ether
theories
Emission
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Light propagation experiments
Experiments from other fields
1
2
3
4
5
6
7
8
9
10
11
12
Stationary ether,
no contraction
A
A
D
D
A
A
D
D
N
A
N
D
Stationary ether,
Lorentz
contraction
A
A
A
D
A
A
A
A
N
A
N
A
Ether attached
to ponderable
bodies
D
D
A
A
A
A
A
D
N
N
N
A
Original source
A
A
A
A
A
D
D
N
N
D
N
N
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Theory
Emission
theories
Light propagation experiments
Experiments from other fields
Original source
A
A
A
A
A
D
D
N
N
D
N
N
Ballistic
A
N
A
A
D
D
D
N
N
D
N
N
New source
A
N
A
A
D
D
A
N
N
D
N
N
A
A
A
A
A
A
A
A
A
A
A
A
Special
relativity
A: the theory agrees with experimental results. D: the theory disagrees with experimental results. N:
the theory is not applicable to the experiment.
1: Aberration. 2: Fizeau convection coefficient. 3: Michelson–Morley. 4: Kennedy–Thorndike. 5:
Moving sources and mirrors. 6: De Sitter spectroscopic binaries. 7: Michelson–Morley, using sunlight.
8: Variation of mass with velocity. 9: General mass–energy equivalence. 10: Radiation from moving
charges. 11: Muon decay at high velocity. 12: Troughton–Noble. 13: Unipolar induction, using a
moving magnet.
Revision Checklist
I can show my understanding of effects, ideas and relationships by
describing and explaining:
how radar-type measurements are used to measure distances in the solar system
how distance is measured in time units (e.g. light-seconds, light-years)
A–Z references: speed of light, invariance of speed of light
Summary diagrams: Velocities from radar ranging, Distances in light travel time
how relative velocities are measured using radar techniques
e.g. using a simple pulse technique
A–Z references: Doppler effect
Summary diagrams: Non-relativistic Doppler shift, Two-way radar speed measurement,
Two-way radar speed measurement 2, Doppler shift – two-way and one-way
Effect of relativistic time dilation
A–Z references: speed of light, invariance of speed of light, theory of special relativity
Summary diagrams: Relativistic effects, The light clock, Time dilation at v / c = 3 / 5
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the evidence supporting the Hot Big Bang model of the origin of the Universe:
cosmological red-shifts and the Hubble law
cosmological microwave background radiation
A–Z references: expansion of the Universe, microwave background radiation
Summary diagrams: Red-shifts of galactic spectra, Hubble's law and the age of the Universe,
The 'age' of the Universe, The cosmic microwave background radiation
I can use the following words and phrases accurately when
describing astronomical and cosmological effects and
observations:
microwave background, red-shift, galaxy
A–Z references: expansion of the Universe, microwave background radiation, galaxy
I can interpret:
charts, graphs and diagrams which use logarithmic ('times') scales to display data, such as
distance, size, mass, time, energy, power and brightness
Summary diagrams: Distances in light travel time, The ladder of astronomical distances, The
history of the Universe
I can calculate:
distances and ages of astronomical objects
Summary diagrams: Distances in light travel time, The history of the Universe, The ladder of
astronomical distances
distances and relative velocities using data from radar-type observations
e.g. using time-of-flight
A–Z references: Doppler effect
Summary diagrams: Velocities from radar ranging, Two-way radar speed measurement, Doppler
shift – two-way and one-way, Measuring black holes
Effect of relativistic time dilation with

1
1 v 2 / c 2
A–Z references: speed of light, invariance of speed of light, theory of special relativity
Summary diagrams: Relativistic effects, The light clock, Time dilation at v / c = 3 / 5
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