Download ODU booklet 2 Teachers booklet Sept 2014 (7.5MB Word)

Higher Physics
Our Dynamic Universe
Notes
Teachers Booklet 2
Previous Knowledge
This section builds on the knowledge from the following key areas from
Dynamics and Space Booklet 2 - Forces
 Newton’s Laws
Learning Outcomes
At the end of this section you should be able to :
o State Newton’s Universal Law of Gravitation
o Use Newton’s Universal Law of Gravitation to calculate the force
between two masses separated by a distance, r
o Compare the gravitational force between objects on different
scales
o Everyday objects
o Planets
o On a subatomic level
o Explain how the slingshot effect is used to travel in space
o Describe lunar and planetary orbits.
2
Gravitational Field
A gravitational field is a place where there is a force on a mass.
weight
The gravitational field strength ‘g’ is defined as the force per unit mass.
We often call the gravitational force acting on an object its weight.
W = mg
W = weight (Newton – N)
m = mass (kilogramme – kg)
g = gravitational field strength (Nkg-1)
We can calculate that the value for ‘g’ is different in different places.
Place
Earth
Moon
Mars
Jupiter
Gravitational Field Strength (Nkg-1)
9.8
1.6
4
25
The force of gravity pulls objects towards the centre of the planet. It
has an effect over a large distance. The Moon orbits Earth due to the
force of gravity acting on it.
Newton formulated a law of universal gravitation. His theory was that the
force of gravitational attraction is dependent on the masses of both
objects and is inversely proportional to the square of the distance that
separates them.
We experience the force due to gravity because of the mass of the
planet.
Gravitation
3
Newton’s Law of Universal Gravitation
F= Force of attraction (N)
G= Gravitational constant (Nm2kg-2)
m1= mass of object 1 (kg)
m2= mass of object 2 (kg)
r = distance between objects (m)
F = G m1m2
r2
Gravitational constant = 6.67 x 10-11 Nm2kg-2 (in data sheet)
1. Force between everyday objects
Example 1
0.2 m
nge
Calculate the force between an apple of mass 0.1kg and an orange of
mass 0.15kg, sitting on a worktop, separated by 0.2m.
F = G m1m2 = 6.67 x 10-11 x 0.1 x 0.15pic = 2.5 x 10-11 N
r2
0.22
Calculate the force due to the Earth on the apple
Mass of Earth – 5.97 x 1024 kg
Radius of Earth – 6.38 x 106 m
F = G m1m2 = 6.67 x 10-11 x 0.1 x 5.97 x 1024
r2
(6.38 x 106) 2
= 0.978 N
W = mg = 0.1 x 9.8 = 0.98 N
Gravitation
4
Newton’s Law of Universal Gravitation
2. Subatomic Scale
Example 2
Calculate the force of attraction due to gravitation between a proton
and a neutron.
Mass of a proton = 1.67 x 10-27 kg
Mass of a neutron = 1.67 x 10-27 kg
Radius of proton : radius of neutron = 0.84 x 10-15m
F = G m1m2 = 6.67 x 10-11 x 1.67 x 10-27 x 1.67 x 10-27
r2
(0.84 x 10-15) 2
= 2.64 x 10-34 N
3. Force of attraction on a planetary scale
Example 3
A satellite orbits a planet at a distance of 5.0 x 107 m from the centre
of the planet.
The mass of the satellite is 3.5 x 104 kg.
The mass of the planet is 6 x 1024 kg.
a)
Calculate the gravitational force acting on the satellite due to
the planet is
(2013 SQA Revised Higher Q4 adapted)
F = G m1m2 = 6.67 x 10-11 x 3.5 x 104 x 6 x 1024
r2
(5.0 x 107) 2
= 5.6 x 103 N
b)
The radius is now doubled. What effect does this have on the
force observed on the planet?
It is quartered because the force is proportional to 1/r 2
Gravitation
5
Newton’s Law of Universal Gravitation
Example 4
Two small asteroids are 12 m apart.
The masses of the asteroids are 2.0 x 103kg and 0.050 x 10
gravitational force acting between the asteroids is
F = G m1m2 = 6.67 x 10-11 x 2.0 x 103 x 0.050 x 10
r2
12 2
= 4.6 x 10-8 N
3
kg. The
3
SQP Q4 adapted
Example 5
What is the gravitational pull of a 5 kg mass on a 3 kg mass when their
centres are 0.15 m apart? What is the force of the 3 kg mass on the
5 kg mass?
F = G m1m2 = 6.67 x 10-11 x 3 x 5
r2
0.152
= 4.4 x 10-8 N
The force of the 3 kg mass on the 5 kg mass is the same magnitude
(4.4 x 10-8 N) but in the opposite direction.
Example 6
One of the boys in a physics class turns to the girl sitting next to him
and says ‘I feel a force of attraction towards you’. Explain, in terms of
physics, whether or not he is correct, calculating a value for the force.
Estimate the mass of the pupils to be 50 and 65kg, with a distance of
30cm between them.
F = G m1m2 = 6.67 x 10-11 x 50 x 65
r2
0.302
= 2.4 x 10-6 N
Although it is a force of attraction it would be too small to feel the
effect of. Possibly biology rather than physics?
Gravitation
6
ODU Problem booklet 2
P2
Q1-6
Special Relativity
Learning Outcomes
o Describe the motion of an object in terms of an observer’s frame
of reference.
o State that the speed of light in a vacuum is the same for all
observers in all reference frames.
o Explain how the constancy of the speed of light led Einstein to
derive his theory of Special Relativity.
o Use the time dilation and length contraction equations to analyse
real and observed times and lengths.
o Explain that relativistic effects are only observed when objects
are moving with velocities close to the speed of light.
Describe examples which illustrate Einsteins postulates
Explain an example of time dilation
Use the equation for time dilation to calculate the new time
Explain an example of length contraction
Use the equation for length contraction to calculate the new length
Use relativistic effects to explain the observation that more
muons are found at sea level than expected
o Carry out calculations to calculate the half life of muons when
travelling close to the speed of light
o Carry out calculations to find the length contraction for muons
o Explain situations using special relativity
o Twin’s Paradox
o Muons
o Ladder
o Light
o
o
o
o
o
o
Special Relativity
7
Frames of Reference
How you describe the motion of an object depends on how you observe it.
1. Two buses travelling at 60 miles per hour in opposite directions.
If you were sitting next to the window of a bus how would the following
observers describe your motion?
Observer
1
Location
Observation
Sitting next to you on
the bus
You are stationary
2
A person standing on the
pavement as the bus
passes.
You are travelling to the
right at 60 miles per hour.
3
A person sitting beside
the window on the bus
travelling to the left at
60 miles per hour
You are travelling to the
right at 120 miles per hour.
All of these are possible.
2.Two spacecraft travelling at 2 x 108 ms-1 in opposite directions.
Observer
1
Location
Observation
Sitting next to you on
the spacecraft
You are stationary
2
A person standing on the
pavement as the
spacecraft passes.
You are travelling to the
3
A person sitting beside
the window on the
spacecraft travelling to
the left at 2 x 108 ms-1
You are travelling to the
right at 2 x 108 ms-1
right at 4 x 108 ms-1.
Observations 1 and 2 are possible, but observation 3 is not because you
cannot exceed the speed of light.
Special Relativity
8
Einstein’s Postulates
1. When two observers are moving at constant speeds relative to one
another, they will observe the same laws of physics
2. The speed of light (in a vacuum) is the same for all observers,
regardless of their motion relative to the light source.
Speed = distance
time
Since speed is constant, then either distance or time must alter.
Special relativity describes the behaviour of objects travelling at
very high speeds. It is ‘special’ relativity because it is the ‘special’
case where the motion between observers is uniform
Special Relativity
9
ODU Problem booklet 2
P3
Q 1 - 11
Time Dilation
A ball is fired from the
floor of a railway carriage,
hits the roof, then falls
back to the ground again.
An observer standing on the platform films the train pass, then plays
back the film frame by frame. The position of the ball is shown in the
frames below.
Y
X
1
2
1
3
Z
4
1
5
1
6
1
7
1
The observer on the platform sees the ball travel much further than an
observer inside the train – but this all takes place over the same time
interval.
Thought Experiment 1
The experiment is repeated on a spaceship travelling at velocity v. A
beam of light detected by a photodiode replaces the ball. The path taken
by the light can be represented by the following diagram.
Y
Y
½ ct’
2d
½ ct’
X
X, Z
Special Relativity
Z
½ vt
½ vt
10
Time Dilation
On the left is the path seen by the person on the train. On the right is
the path seen by the observer on the platform watching the train pass.
t =
t’ =
v =
c =
time measured on the platform by the observer on the
platform
time measured on the train by the observer on the train
speed the object is moving at (ms-1)
speed of light (ms-1)
t = 2d (for the observer on the train)
c
The horizontal distance travelled by the train = vt’.
By forming a right angled triangle the following relationship can be
calculated.




(½ ct’)2 = (½ ct)2 + (½ vt’)2
(ct’)2 = (ct)2 + (vt’)2
c2t’2 = c2t2 + v2t’2
(c2 –v2)t’2 = c2t2
(1 – v2/c2)t’2 = t2 (dividing both sides by c2)

t’ =
𝑡
√1−(𝑣)
2
𝑐
Example 7
A rocket is travelling at a constant 2.7 x 108 ms-1 compared to an
observer on Earth. The pilot measures the journey as taking 240
minutes. How long did the journey take when measured from Earth?
t = 240 minutes
v = 2.7 x 108 ms-1
c = 3 x 108 ms-1
t’ = ?
t’ =
𝑡
𝑣
1−( )2
𝑐
=
240
2.7 𝑥 108 2
)
3 𝑥 108
1−(
t’ = 550 minutes
Special Relativity
11
ODU Problem booklet 2
P5
Q1-9
Length Contraction
Thought Experiment 2
1
2
To the observer on the
platform the ball travels to
the end of the carriage, but it
has moved forwards, so the
total distance travelled is less.
For the observer on
the train the ball goes
from one end of the
train to the other and
back
Length contraction is only observed when objects travel at a
speed close to the speed of light and in a direction parallel to the
direction in which the observed body is travelling.
l
l’
v
c
=
=
=
=
length measured on the platform by the observer
length measured on the train by the observer
speed the object is moving at (ms-1)
speed of light (ms-1)
l’ = l
𝑣
1 − ( )2
𝑐
Example 8
A rocket has a length of 10m when at rest on the Earth. An observer
on Earth watches the rocket passing at a constant speed of
1.5 x 108ms-1. Calculate the length of the rocket as measured by the
observer.
l = 10m
v = 1.5 x 108 ms-1 = 0.5c
c = 3 x 108 ms-1
t’ = ?
Special Relativity
l’ = l
𝑣
1 − ( )2 = 10 1 − ( 3 𝑥 108 )2
1.5 𝑥 108
𝑐
l’ = 8.7m
12
ODU Problem booklet 2
P7
Q1-9
Lorentz Factor (γ)
The Lorentz factor takes into account the speed of the object.
γ=
1
√1−(𝑣)
2
𝑐
This allows us to re-write both the time dilation equation and the length
contraction equation as shown below.
t’ = t γ
l’ =
l
γ
11
10
9
Lorentz Factor
8
7
6
5
4
3
2
1
0
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1
Speed (multiples of c)
For small speeds (less than 0.1c) the Lorentz factor is approximately 1
and relativistic effects are negligibly small.
Thought Experiment 3 – The Twins Paradox
One twin stays at home, the other goes on a journey through space
travelling close to the speed of light. What happens?
Time runs more slowly for the twin who goes into space, so when they
return they should be younger than their twin who stayed at home.
However – space travel requires acceleration. This can only be dealt with
using General Relativity, which is beyond this course.
Special Relativity
13
Muons
Muons are created in the upper atmosphere. The half life of the muon is
1.56 x 10-6 s and they travel at a speed of 0.98c. If a million muons start
off at a height of 10km we should find very few muons at sea level,
however substantial numbers are detected – what is the reason?
1,000,000 muons
Non-relativistic calculation
t = distance =
10,000
speed (0.98 x 3 x 108)
=3.5 x 10-5 s
?
muons
This is equivalent to 21.8 half
lives, meaning we would expect
only 0.3 muons at ground level.
Time dilation
Time from muon’s perspective
1.56 x 10-6 s
Length Contraction
Length from muon’s perspective
Time from our perspective
t’ =
𝑡
𝑣 2
𝑐
1−( )
=
1.56 𝑥 10−6
0.98
1 − (1.00)2
= 7.84 x 10-6 s
This is equivalent to approx. 4
half lives, which predicts that
roughly 50,000 muons would be
observed.
𝑣
l’ = l
1 − ( )2
= 10
1−(
𝑐
0.99 2
)
1.00
= 2 km
Length from our perspective
10 km
t = distance =
2,000
speed (0.98 x 3 x 108)
=6.8 x 10-6 s
Again this is equivalent to
approx. 4 half lives, predicting
50,000 muons at ground level.
We see many more muons than we would expect because, travelling at
0.98c, they experience relativistic effects. From the muon’s
perspective the 10km is 2km because of length contraction.
Alternatively the half life of 1.56 μs is increased to 7.84 μs due to
time dilation.
Either of these situations has the effect of allowing more muons to
reach sea level.
Special Relativity
14
Muons
Example 9
Muons are sub-atomic particles produced when cosmic rays enter the
atmosphere about 15km above the surface of the Earth. Muons have
a mean lifetime of 2.2 x 10-6 s in their frame of reference. Muons are
travelling at 0.999c relative to an observer on Earth.
a) Show that the mean distance travelled by the muons in their
frame of reference is 660m
b) Calculate the mean lifetime of the muons as measured by the
observer on Earth
c) Explain why a greater number of muons are detected on the
surface of the Earth than would be expected if relativistic
effects were not taken into account.
a)
b)
c)
d = vt = (3 x 108 x 0.999) x 2.2 x 10-6 = 660m
−6
𝑡
2.2 𝑥 10
t’ =
=
2
2
√1−(𝑣)
√1−(0.999)
𝑐
1
= 4.9 x 10-5s
For an observer on Earth’s frame of reference the mean
life of the muon is much greater
OR
The distance in the muon frame of reference is shorter.
SQA SQP Q4 (adapted)
Summary
Property
Length – L
Time – T
Speed
Measured by
observer on
spaceship
Rest
L
Close to c
L
Rest
T
Close to c
T
Special Relativity
15
Measured by
observer on
planet
L
Less than L
T
Greater than T
ODU Problem booklet 2
P9
Q1-9
The Expanding Universe
This section builds on the knowledge from the following key areas from
Dynamics and Space Book 3 – Space
 Cosmology
Learning Outcomes
At the end of the section you should be able to
o Describe the Doppler Effect in terms of the changing frequencies
of sound and light for moving objects
o Use the Doppler Effect equation for calculations involving the
sound emitted by moving objects
o Understand that light from distant galaxies is red-shifted because
they are moving away from the Earth
o Carry out calculations involving the red-shift and the recession
velocity of a distant galaxy
o State that the Doppler Effect equations used for sound cannot be
used with light from fast moving galaxies because relativistic
effects need to be taken into account
o State that, for slow moving galaxies, redshift is the ratio of the
velocity of the galaxy to the velocity of light.
o Explain Hubble’s Law as the relationship between the recession
velocity of a galaxy and its distance from the observer
o Use Hubble’s Law in calculations involving recession velocity and
distance from observer
o Understand how Hubble’s Law can be used to estimate the age of
the universe.
o Describe evidence for the expansion of the universe
o Describe that the orbital speed of the Sun and other stars gives a
way of determining the mass of our galaxy
o Define the term ‘dark matter’.
o Explain the evidence for dark matter in terms of measurements of
the mass of our galaxy and others.
o Define the term ‘dark energy’.
o Explain the evidence for dark energy in terms of measurements of
the expansion rate of the Universe and the concept of something
that overcomes the force of gravity.
The Expanding Universe
16
The Expanding Universe
o Explain that the temperature of a stellar object is related to the
distribution of wavelengths of emitted radiation
o Describe that the peak wavelength emitted is related to the
temperature of the object – the greater the temperature, the
shorter the wavelength.
o Understand the qualitative relationship between the temperature
of a star and the intensity of radiation – the greater the
temperature, the greater the intensity of radiation.
o Understand what is meant by the cosmic microwave background and
how this relates to the peak wavelength of the radiation
o Provide evidence to justify the model of the Big Bang for the
beginning and evolution of the universe.
The Expanding Universe
17
Doppler Effect
The Doppler Effect is the change in pitch observed as something making
a sound moves towards you, passes you and then travels away from you.
Stationary Source
A stationary source of sound producing waves at a
constant frequency will have wavefronts travelling in
all directions from the source .
All observers will hear the same sound, which is
equal to the actual frequency of the source, f.
Moving Source
If the source is moving to the right at a speed, v, wavefronts are
produced at the same rate as before, but each time a new wavefront is
produced the source has moved some distance to the right. This causes
the wavefronts on the left to be created further apart and the
wavefronts on the right to be created closer together.
Longer wavelength
Lower frequency
Shorter wavelength
Higher frequency
NOTE – The frequency of the source remains constant – it is the
observed frequency that changes.
This is why sounds seem to be higher
before something reaches you, then
lower when it has passed. You only hear
the true frequency when it is opposite
you.
The Expanding Universe
18
Doppler Effect
This is used to check the speed of motorists and to measure blood
pressure by detecting the speed of blood flow in the body.
Moving towards Observer
fo = fs (
Moving away from Observer
𝑣
fo = fs (
)
𝑣−𝑣
𝑣+𝑣𝑠
𝑠
Frequency increases
Where
𝑣
Frequency decreases
fo = frequency heard by the observer (Hz)
fs = frequency of the source of the sound (Hz)
v = velocity of the sound waves (ms-1)
vs = velocity of the source (ms-1)
fo = fs (
𝑣
𝑣−𝑣𝑠
= 10
fo = fs (
)
𝑣
𝑣+𝑣𝑠
340
= 50
(340−40)
= 11.3
= 11 Hz
The Expanding Universe
)
340
(340+10)
= 48.6
= 49 Hz
19
)
Doppler Effect
Example 10
A source of sound waves of
frequency 10 Hz is travelling
towards an observer at 40 ms-1.
Calculate the frequency heard
by the observer.
Example 11
A source of sound waves of
frequency 50 Hz is travelling
away from an observer at 10 ms-1.
Calculate the frequency heard by
the observer.
Example 12
The siren on a fire engine is emitting sound with a constant frequency
of 1000Hz. The fire engine is travelling at a constant speed of 20 ms-1
as it approaches and passes a stationary observer. The speed of sound
in air is 340 ms-1.
Which row in the table shows the frequency of the sound heard by the
observer as the ambulance approaches and as it moves away from the
SQA SQP Q6 adapted
observer?
A
B
C
D
E
Frequency as fire engine
approaches (Hz)
1000
1063
944
1063
944
The Expanding Universe
Frequency as fire engine
moves away (Hz)
1000
944
1000
1000
1063
20
B
ODU Problem booklet 2
P11
Q 1 - 18
Red Shift
The Doppler Effect affects light as well as sound. Astronomers have
observed that the characteristic line spectra for different elements
appeared to expand and are shifted towards the longer wavelengths at
the red end of the spectrum. This is known as Red Shift. It also suggests
that the universe is expanding.
Moving towards observer – blueshift λo<λs
Line spectrum at rest
Moving away from observer – redshift λo>λs
If light is travelling towards an observer the wavelength would appear to
contract and move towards the ‘blue’ end of the spectrum – hence Blue
Shift.
Redshift is more often observed as a spectral tracing as shown in this
diagram
From:
http://www.faculty.umb.edu/gary_zabel/
Courses/Parallel%20Universes/Texts/Re
mote%20Sensing%20Tutorial%20Page%2
0A-9.htm (12th June 2014)
Wavelength (nm)
Redshift spectral tracing
Re
dsh
ift
Changes in the frequency is also observed as a change in the wavelength
(remember that v = fλ) This can be observed in both stars and galaxies.
Redshift (z) can be calculated using the equation
z = λobserved – λsource
λsource
Where
= Δλ = vg
λ
c
remember z has no units
z = red shift (no units)
Δλ = λobserved – λsource
λobserved = wavelength measured by the observer (m)
λsource
= wavelength measured at source (m)
vg = velocity of galaxy (ms-1)
c = speed of light (ms-1)
The Expanding Universe
21
ODU Problem booklet 2
P14 Q 19 - 20
Red Shift
Example 13
An astronomer observes the spectrum of light from a star. The
spectrum contains the emission lines for hydrogen.
The astronomer compares this spectrum with the spectrum from a
hydrogen lamp. The line, which has a wavelength of 656 nm, from the
lamp is found to be shifted to 663 nm in the spectrum from the star.
The redshift of the light from the star is
z = λobserved – λsource = 663 – 656 = 0.011
λsource
656
Measuring Distances in Space
1. Parallax
Distant objects appear to be located at a different angle relative to near
objects depending on where you position when you make the observation. This is
called parallax. This is used to calculate astronomical distances.
Earth
Sun
Nearby
star
Distant
stars
Parallax angle
One Parsec (pc)
The distance to a star that subtends an angle of 1 second at an arc of length 1
astronomical unit (the distance from Earth to the Sun)
2. Comparison of absolute to apparent luminosity
Cephid variable stars have a predictable variation of luminosity (the amount of
electromagnetic energy radiated per second). The absolute luminosity is
converted to apparent luminosity, which can then be used to calculate the
distance using the inverse square law
The Expanding Universe
22
ODU Problem booklet 2
P14 Q 19 - 20
Hubble’s Law
The graph shows
the relationship
between the
velocity of a galaxy,
as it recedes from
us, and its distance
(Hubble’s Law).
1
Recession velocity km s-
Edwin Hubble noticed that the light from some distant galaxies was red
shifted. By examining the redshift of galaxies at different distances
from Earth he realised that the further away a galaxy was the faster it
was travelling.
0
v = H0 d
Distance (Mpc)
v = recession velocity
H0 = Hubble constant
d = distance to galaxy
Current estimate for
H0 = 2.34 x 10-18 s-1
Example 14
An observatory collects information from a distant star which
indicates that a redshift has taken place. The wavelength of the light
from the source is 656 nm, while the wavelength of the observed light
is 676 nm.
(a)
Calculate the recessional velocity of the star.
(b)
If the recessional velocity of a distant galaxy is
1.2 x 107 ms-1, show that the approximate distance
to this galaxy is 5.2 x 1024 m.
(SQA SQP Q6 adapted)
(a)
z = λobserved – λsource = 676 – 656 = 0.03
λsource
656
z = v v = zc = 0.03 x 3 x 108 = 9 x 106 ms-1
c
(b)
v = H0 d  d = 1.2 x 107
2.3 x 10-18
The Expanding Universe
= 5.2 x 1024 m
23
ODU Problem booklet 2
P15 Q 1 - 13
Estimating the Age of the Universe
Hubble’s Law can be rearranged to give an estimate for the age of the
universe.
Since
H0 = v
d
and
time = d
v
so time = 1
H0
Example 15
Calculate the age of the Universe given the estimate for the Hubble
constant is 2.3 x 10 -18 s-1.
time = 1
H0
=
1
2.3 x 10 -18
= 4.2 x 1017 s roughly 13.8 billion years
Hubble’s Law suggests that galaxies further away from us are moving
away faster than galaxies closer to us, which suggests that the universe
is expanding.
Model of the Expanding Universe – the balloon analogy
Small balloon with
stars drawn on
Inflated balloon. Surface and
stars have expanded.
The model is a good analogy
because
The model is not a good
analogy because
Distance between stars
expands
Expansion is in more than one
dimension
The bigger the gap the faster
the expansion.
Stars do not expand at this
rate
Centre of the balloon is not
the centre of the universe
This is a 2D representation of
a 3D model
The Expanding Universe
24
Importance of Expansion
Expansion or Explosion?
Explosion
Expansion
Different parts fly off at different
speeds
Expansion explains the large scale
symmetry we see in the distribution
of galaxies
Fast parts overtake slow parts
Expanding space explains the
redshifts and the Hubble Law
Difficult to imagine a suitable
mechanism to produce the range of
velocities from 100kms-1 to almost
the speed of light
Expansion also explains redshifts
and the Hubble law even if we are
not at the centre of the universe
It seems likely that velocity would be
related to some physical property.
E.g. if given the same energy, less
massive galaxies would be moving
faster
If this was the case a definite
correlation between mass and
velocity would be expected – this is
not observed.
Hubble’s law works well even if we
only plot data for galaxies of similar
mass
Balloon analogy – every galaxy
moves away from every other as
the space expands
No galaxy is located at the centre
Not only are we not at the centre
of the universe, it doesn’t even
need to have a centre.
Faster galaxies would leave slower
ones behind, resulting in those near
the centre (start) being more closely
packed than those on the periphery
(finish), but this is not observed.
If the universe is expanding now then
 Going backwards in time it must have been a lot smaller.
 At one time everything in the universe must have been
concentrated at one spot (singularity) which had zero
and contained all the energy
in the universe
 This leads us to believe that
the Universe started from a
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25
size
Big Bang Theory
The Big Bang Theory
suggests that the
Universe was initially
in a very hot and dense
state, and then rapidly
expanded.
What evidence would
back this up?




http://en.wikipedia.org/wiki/Chronology_of_the_universe#mediaviewer/File:History_of_the_Universe.svg

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26
Redshift of
galaxies
Expansion
(suggested the
idea in the first
place)
Temperature of
the universe
Composition of
elements making
up the universe
Darkness of the
sky
Temperature of Stellar Objects
The thermal emission
peak graph on the left
shows the relationship
between the intensity of
radiation emitted from
stars of different
temperatures is related
to the wavelength of
the light emitted from
the star.
http://en.wikipedia.org/wiki/File:Wiens_law.svg
Stars emit radiation from the entire electromagnetic spectrum, but we
can measure the temperature of a star by looking for a peak in the
wavelengths emitted.
Stars which have a higher temperature emit more radiation, with shorter
wavelength, than stars which are ‘colder’.
This can be used to help find the average temperature of the universe,
which in turn could help provide evidence for the Big Bang.
This is also called black body radiation.
If the Big Bang took place it should be possible to detect its effect by
measuring the thermal emission of radiation in the Universe. This
radiation should have a longer wavelength, with a peak corresponding to
the temperature the Universe has cooled to.
The Expanding Universe
27
ODU Problem booklet 2
P17 Q 1
Cosmic Microwave Background Radiation
The diagram shows the
spectrum obtained by the
COsmic Background Explorer
(COBE) satellite over three
years. The results were exactly
in line with predicted values.
Cosmic Microwave Background (CMB) radiation:
 is the dominant source of radiation in the Universe
 is isotropic (spread uniformly throughout space)
 shows the characteristics of black body radiation (as illustrated by
the results above)
 has a temperature of approximately 3K due to cooling on expansion
 corresponds to a redshift of 1000, so the early temperature of
this radiation was approximately 3000K
CMB radiation has been described as the ‘afterglow’ of the Big Bang,
cooled to a faint whisper in the microwave region of the electromagnetic
spectrum.
The Wilkinson Microwave Anisotropy Probe (WMAP), launched in 2001,
carried out a more sensitive measurement of the radiation. The picture
showing the temperature of the universe provides information on the
history of the universe and helps shape predictions for the future of the
universe.
http://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe#mediaviewer/
File:Ilc_9yr_moll4096.png
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Additional Evidence for the Big Bang
Hydrogen and Helium (Nucleosynthesis)
If the Big Bang took place the lightest elements should have been formed
soon after the event.
Theory suggests that, fractions of a second after the Big Bang, the
Universe was very hot and filled with a mixture of particles. As it cooled
down Hydrogen, Helium and trace amounts of Lithium should be produced.
The proportion of Helium detected in the Universe (approximately 24%)
provides evidence for this aspect of the Big Bang Theory and helps
predict the rate of expansion of the Universe.
Heavier elements were not produced at this time because temperature
had fallen too far and most of the neutrons had been used up. These
elements were created in stars.
Olbers’ Paradox
The sky is full of stars. Stars emit light. Why are there dark bits in the
sky?
1.
The light from any stars further than the age of the universe away
will not have had time to get to us.
2. There is a ‘dark horizon’ beyond which light from the stars can never
reach us because the Universe is expanding at the rate the light
travels towards us. The distance the light must travel is always
increasing..
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29
How will it end?
Big chill
http://sci.esa.int/education/3
5775cosmology/?fbodylongid=1706
Big crunch
The shape of the Universe depends on how much mass it contains because
this is related to the pull of gravity. There are three possible outcomes.
1. Closed Universe
If the mean density exceeds a critical value then the universe will
expand to a finite size and then begin to collapse back in on itself,
slowly at first, and then at an ever-increasing rate until all the
galaxies collide and the universe ends in a Big Crunch.
2. Open Universe
If the mean density is less than the critical value, gravity will slow
the rate of expansion towards a steady value and the expansion will
continue forever.
3. Flat Universe
If the mean density equals the critical value, the universe will be
able to expand forever with the recession velocities tending
towards zero.
The critical density is about 5 hydrogen atoms per m3.
Ω = actual density of the Universe
critical density for a flat Universe
Ω < 1 Open Universe – Big Chill
Ω = 1 Flat Universe – Big Chill
Ω > 1 Closed Universe – Big Crunch.
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Updated 20140919 NH
Question numbers added
How will it end?
1.Dark Matter
Problem – Calculation of the amount of matter in the galaxy suggests
there is more matter than astronomers can detect.
Evidence - Orbital speed of the Sun mass of galaxy inside its orbit
Speed of rotation of an object is determined by the size of
the force maintaining its rotation. For the Sun the central
force is due to gravity, which is determined by the amount of
matter inside the Sun’s orbital path.
Conclusion – There must be a significant amount of mass that we cannot
see.
This invisible matter has been termed dark matter.
2.Dark Energy
Problem – The universe is expanding at an increasing rate.
Evidence – Hubble’s Law and the evidence from red shift shows that
expansion is continuing to increase. A force must be acting
against the force of gravity, pushing matter apart.
Conclusion – Dark Energy is a source of energy, which produces the force
that acts against gravity.
Roughly 68% of the Universe is Dark Energy, Dark Matter makes up
about 27% and less than 5% is normal matter.
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Updated 20140919 NH
Question numbers added