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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 The Expanding Universe 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 The Expanding Universe 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 The Expanding Universe 28 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.. The Expanding Universe 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. The Expanding Universe 30 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. The Expanding Universe 31 Updated 20140919 NH Question numbers added