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Last Week Mass-Luminosity relationship Hubbles Law v = H0d Expanding Universe Cosmological principle Olber’s Paradox Meaning of expansion λ / λ0 = (1 + Z) Think in terms of spacetime – we say something 400Mly away has a Lookback time of 400 My Cosmological redshift Cosmological Horizon Big Bang Model vs. Steady State Model Cosmic Microwave background Models of the Universe Must explain - Hubble’s Law - v = H0d - Olber’s paradox – Why is the Sky dark at night? - Why the Universe does not collapse under its own gravity. - diversity of objects seen in our telescopes. It must satisfy -Cosmological Principle and be -Homogeneous and isotropic on largest scale. Cosmological Models • Various models can satisfy these criteria. - Big Bang Model - Steady State Model Is there a way of deciding which is better? Blackbody Spectrum and Cosmic Microwave background What is the spectrum of EM radiation emitted by an object of arbitrary temperature T in thermal equilibrum •Max Planck showed that the spectrum is given by u d = 8hc -5.d [exp(hc/kT) - 1] Wien’s Displacement Law and Stefan-Boltzmann Law • Doubling T increases P by 16 since PA = .T4 where σ = 5.7 x 10-8 Wm-2K-4 • Note that maximum wavelength max shifts with .This can be quantified in Wien’s Displacement Law. • This quantifies the observation max.T = const. that an object changes colour with Temperature e.g.At room = 2.9 x 10-3 mK temp. spectrum peaks in infra-red. • Very important since it allows us to obtain a measure of the SURFACE TEMPERATURE of a star from max.For the Sun max is in blue with bu a lot of radiation in red so it looks yellow.For stars with T = 3000k max is in infrared but significant amount in red.Red Giants are at this T. Cosmic Microwave Background • Essentially only three pieces of evidence underpin the Big Bang Model and the CMB is one of them. • It was discovered by Penzias and Wilson in 1965. They are shown here with the horn antenna they had built for telecommunications via satellites. They found a signal from all directions at = 7.35 cm.The only explanation was that it is real.Work quickly began to measure the background at other wavelengths. Cosmic Microwave Background With the addition of more and more points it became clear that it was a thermal curve at a temp. of 2.7K. Later more precise measurements tell us that it fits a black-body spectrum at 2.726 +/- 0.005K perfectly. Summary-CMB Measurements 1.CMB is seen in all directions as a perfect BB at T = 2.726+/-0.06K 2.At sensitivity levels of 1 in 103 there is a large scale anisotropy in the CMB due to the motion of the Earth through the frame of reference in which the CMB is uniform. Earth is moving at about 350 km s-1 w.r.t this frame.We are moving towards LEO and away from Aquarius.[Galaxy is moving in same direction at 600 km s-1 ] This motion has to be subtracted from the measured spectrum in order to see the real spectrum. Cosmic Microwave Backround For a long time we could see no fluctuations in the Cosmic Microwave bgd. CMB is the spectrum of a body at 2.726 degrees seen whichever direction one looks. It marks the point where the Universe became transparent. Results from COBE Evolution of the Universe in Big Bang Model 1 Assume adiabatic process = No heat into or out of the system[Universe] 2.Assume Universe is an idealised,smooth fluid.A good assumption in the early stages. 3.Constituents are all in thermal eqbm. E = kT 4.First Law of Thermodynamics dU = dQ + dW In an adiabatic system dU = dW = PdV. For a non-viscous fluid P = 1/3 where = energy density. 5.As Universe expands V and must change. We introduce a scale factor R(t) with dimensions of length. So V = 4/3.. R(t)3 6.Now take a very small volume with only one photon in it. R = h/V = hc/V 1/V 1/ R(t)3 1/R(t)4 where depends on redshift and must scale with R(t). 7.Now E = R.V R(t)3 . 1/R(t)4 1/R(t) Thus as Space expands [R(t) increases], T decreases. 8.Hence since E = kT we have T 1/R(t) Big Bang Model Universe has expanded from a singularity -infinitely dense, high T, high pressure “point” The whole Universe expands in line with Hubble’s Law. As it does so T 1/R(t) In other words as it expands it cools. Can we now explain the CMB? Explanation of the CMB When Universe was younger it was not only denser but hotter.If CMB was emitted at a given T then as Universe expanded the wavelength would stretch and effective T goes down.CMB would appear to have lower T. In early Universe it consisted of 75% H:25% He and it was full of radiation.As a result all the H was ionised[a) on left].If an atom formed it soon disintegrated in collision with a photon.Universe was opaque. There was a strong coupling between Matter and Radiation. At some point( 3 x 105 years) T dropped to 4000K.Now the no.of photons with sufficient energy is too small to stop atoms forming. Origin of the CMB Now H atoms form,photons no longer interact strongly with Matter and the Universe becomes transparent. The photons can now spread freely throughout the Universe with a BB spectrum initially at 4000K but with time it has been shifted to 2.726K. Since radiation was continuously scattered up to this point it is known as the Last Scattering Surface. We cannot see anything prior to this time because the Universe was opaque. Tiny ripples represent non-uniformity at that time.Universe was 103 times smaller. Uniformity was remarkable but present non-uniformity comes from it. This clinched the Big Bang Model since no other could explain it. The Expanding Universe and the Friedmann equation We have a scale factor R(t) which grows in size with the Universe.In terms of forces we know the Strong and Weak forces are short range and Electromagnetic effects are neutral over large volumes of space.So after about 1 s the expansion has been governed by GRAVITY. Newton wondered why the Universe did not collapse. Hubble solved this by showing that it is expanding and that the kinetic energy of expansion overcomes the gravitational pull. Expansion of the Universe As the Universe expands the volume increases proportionally to the cube of the scale factor. Scale factor - R(t) The Expanding Universe and the Friedmann equation Imagine an arbitrary spherical volume and let us add a thin shell around it. The sphere contains mass m and has a radius d. The matter in the shell is moving away from the centre with a velocity v, given by Hubble’s Law. K.E. = 1/2 msv2 and P.E. = - Gmms d Total energy = K.E. + P.E. ----(A) Now m = V = 4/3d3 and so eqn. (A) becomes H(t)2.d2 _ 4Gd2 = total energy ------(B) ms 2 3 where we have used v = H(t).d [ ] The Expanding Universe and the Friedmann equation 1.This eqn. contains the density () of the Universe which is proportional to the inverse cube of the scale factor R(t). 2.If we replace the total energy by an energy constant k, k = 2ER(t)2 mSd2 3. Eqn. (B) becomes H(t)2 - 8G 3 = k R(t)2 4.This is the Friedmann Eqn. It describes the expansion of the Universe in terms of measurable quantities. 5.It involves H(t) = H0, = 0, and we can set R(t) = 1 since it is an arbitrary measure of the Universe’s size. If we can determine Hubble’s const. and the density of the Universe we can solve this eqn for k. Not so easy! But in principle----- Friedmann Equation From consideration of the interaction of a shell of matter with a core We derived the Friedmann Eqn. H(t)2- 8G 3 k = R(t)2 where k = 2ER(t)2 mSd2 In this form the equation is consistent with Newtonian mechanics. It has to be modified to take account of General Relativity but it ends up in a very similar form. Later we will use it to consider the question of whether the Universe is “closed or open”. In other words will it expand for ever or will it stop and then collapse. Nucleosynthesis in the Early Universe. 1.In the Early Universe the temperature(energy) was so high that all particles are free. In effect we have a “soup” of particles and photons all continuously interacting with each other. 2.As time passes the temperature and density fall and gradually when particles and anti-particles combine they are no longer separated and they disappear as “free” particles. 3.For example quarks and anti-quarks are initially free.In effect we have a “Quark-Gluon Plasma”. Once quarks are no longer free they come together to form nucleons. [Nucleon is the generic name for protons and neutrons] Proton = 2 Up Quarks + 1 Down Quark( 2u,1d) Neutron = 1 Up Quark + 2 Down Quarks( 1u,2d) 4.From this point quarks are no longer free and are forever confined inside nucleons. Big Bang Model So far we have two pieces of evidence for the Standard Big Bang Model -- Hubble’s Law -- Cosmic Microwave Background. The third piece of evidence relates to Nucleosynthesis The Boltzmann Distribution Law In quantum mechanical systems only certain discrete energy levels can be occupied The Boltzmann Distribution gives the probability of finding the system in a particular energy state. [It applies to classical and quantum systems] In a simple system like the nucleon where there are only two states – the neutron and proton – the energy level diagram is E1 mNc2 = 939.573 MeV E2 mPc2 = 938.280 MeV The ratio of neutrons to protons is NN/NP = [mN/mP]3/2 exp( - (mN-mP)c2/kT), where k = Boltzmann constant At T = 109 K this ratio is 1/7 Nucleosynthesis in the Early Universe. 1.Ratio of n/p declines rapidly with t, T and 3/2 2 NN ( m m )c N P mN Exp. = mP kT NP 2.At T = 109K the ratio is about 1/7 and the Universe is cool enough for a sequence of two body reactions in which bound states of protons and neutrons can be made. Now n+p d+ where d represents the deuteron,a nucleus made up of 1 proton and 1 neutron.This is an isotope of Hydrogen. [ ] (- ) 3.The deuteron is fragile.It is bound by only 2.223 MeV. There are still lots of photons around with this energy so that d+ n+p 4.When T drops to a point where there are too few 2.223 MeV gammas the creation process wins. Now we can begin to make He. Nucleosynthesis in the Early Universe. [n p + e- + e] 1.Mass of neutron = 939.573 MeV Mass of proton = 938.280 MeV Difference in rest mass = 1.26 MeV [Remember E = mc2] This means that the numbers of neutrons and protons are not equal. 2.Nucleosynthesis begins when T = 109K and t 90 secs. The Maxwell Boltzmann distribution gives the ratio NN NP = mN mP 3/2 [ ] (- Exp. mN - mP kT ) ---------(A) 3. Point 1 means that free neutrons are unstable and decay with a mean-life of about 14.8 mins. Point 2 means that there are 7 protons for each neutron at the time when nucleosynthesis begins. [ Note:-If the neutron is bound in a nucleus it is no longer unstable.] Nucleosynthesis in the Early Universe So we start with n/p = 1/7 n+p d+ We can now create deuterons from the protons and neutrons but there are so many photons (with energy greater than 2.233 MeV) that the deuterons break up almost immediately. This is known as the Deuteron Bottleneck- it causes a delay before the light nuclei can be formed. Eventually as T falls the number of photons with E = 2.233 MeV becomes smaller than needed to destroy the deuterons that are formed. Now nucleosynthesis can begin. Big Bang Model Universe has expanded from a singularity -infinitely dense, high T, high pressure “point” The whole Universe expands consistent with Hubble’s Law.[1] As it does so T 1/R(t) In other words as it expands it cools. This allows us to explain the Cosmic Microwave Background.[2] It also allows us to predict the ratio of H:He in the Early Universe.[3] The rates of Nuclear reactions in stars Exp[-E/kT] • The probabilities of nuclear reactions in stars are very small because the energies of the particles are very small but there is a very large number of particles. • The number of reactions is just the product of the number of particles of a given energy multiplied by the probability of the reaction in a collision at that energy. RED = Number of particles as a function of energy Green = Probability of a reaction occurring in a collision Blue = Number of reactions as a function of energy GAMOW PEAK Nucleosynthesis in the Early Universe. Now we get a series of fusion reactions d + d 3He++ + n 3He++ + p+d and the triton (3H = t) is produced in the reactions n+d t+ d+d t+p n + 3He t + p Now we might expect d + d 4He + but the following is preferred n + 3He++ 4He++ + d + 3He++ 4He++ + p 4He++ + p+t 4He++ + n d+t The delay before these reactions start is called the deuterium bottleneck In contrast:The p-p chain;the reactions which power the Sun Overall - 4p 4He + 2e- +2 + 26.7 MeV Nucleosynthesis Sequence in the Early Universe - First neutrons and protons form. - Initially n + p d+γ but photodisintegration occurs at same rate as fusion. - When T falls further fusion occurs faster than photodisintegration then deuterons form. - This leads to formation of 4He - Process stops at this stage because there is no stable or long-lived isotope of elements Z = 5 or 8. Heavier elements are made in stars! Creating Helium in the Early Universe. 1. 14p + 2n 4 He + 12p i.e. 75%:25% for H:He in terms of mass [Remember n/p = 1/7 at 109K – the point when nucleosynthesis starts] 2. On the way a lot of deuterium is created but it is very delicate and is easily broken up by interaction with the very high flux of photons. This process is called Photodisintegration. [ n+p d +γ] [ Deuterium bottleneck ] 3. Once we have made Helium we might expect that we could make heavier elements by the interaction of He + p or He + He. However there are no stable nuclei of mass A = 5 or 8. 4.For example 4He + 4He 8Be but it “immediately “ breaks up into two alpha particles[helium nuclei] 5.Nucleosynthesis in the Early Universe comes to an end. The second reason being that the Universe is now not dense enough!! The rates of Nuclear reactions in stars Exp[-E/kT] • The probabilities of nuclear reactions in stars are very small because the energies of the particles are very small but there is a very large number of particles. • The number of reactions is just the product of the number of particles of a given energy multiplied by the probability of the reaction in a collision at that energy. RED = Number of particles as a function of energy Green = Probability of a reaction occurring in a collision Blue = Number of reactions as a function of energy GAMOW PEAK Solar System Abundances Present day abundances! Here we see the abundances of the chemical elements in Solar System. They are plotted on a log scale relative to the abundance of Hydrogen(H) =1 The abundances of everything except He are smaller by at least a factor of 1000. Cosmic Microwave background 7p 1n 75% H & 25%He 98% of known matter Where does the rest come from? The table summarises the similarities and differences between Decoupling and Nucleosynthesis. Note the differences in energy scales appropriate to nuclear and atomic processes. (CMB) Predicted Abundances of the light elements from Big Bang One of the three pieces of evidence for the Big Bang Model is the abundances of the light elements.It depends on the number of types of neutrino which exist and the density of baryons(strongly interacting particles) Here we see the predictions as a function of baryon density[at top] and of the density B. Band shows observations Line shows critical density for H0 = 65 km/sMpc [Note:-H = 1 in diagram] The Big Bang Model What we have seen is that there are three key pieces of evidence supporting the Big Bang Model. 1. Hubble’s Law and the expansion of the Universe. 2.The isotropy and thermal spectrum of the Cosmic Microwave Backround. 3. The abundances of the light elements. These three pieces of evidence are independent and all find a natural explanation in the standard Hot Big Bang Model of the Universe. Problems with the Standard Big Bang Model Friedmann equation H(t)2 8G 3 = k R(t)2 where k = 2ER(t)2 mSd2 This is the Friedmann equation which describes the expansion of the Universe with k determining the rate of expansion. •When derived from General Relativity the energy constant k has the interpretation of the overall curvature of the spacetime continuum. Later we will use it to consider the question of whether the Universe is “closed or open”. •Overall there are three possibilities which we can see as being similar to the question of “Escape velocity” for an object leaving a planet etc. In other words it is a question of the total mass in the Universe. Curvature or Shape of the Universe. The matter and energy scattered across all of space give the Universe an overall curvature.How curved depends on average mass density.This includes all forms of energy [Remember E = mc2] 1.In a flat Universe two parallel beams of light would stay parallel forever.No curvature. The Universe will expand forever but more and more slowly until it just reaches infinity. Here k = 0. 2.If the two light beams converge Universe has positive curvature then Universe is closed with k +ve. Here there is sufficient matter to slow down the expansion so that eventually the expansion stops and is reversed.Then it will start to shrink and eventually return to the “Big Crunch”. Note:-This is rather like lines of longitude on the Earth’s surface.They are parallel at the equator but meet at the poles.However the Universe would still have no edge or centre. 3.If the two light beams diverge.Here the surface is like a saddle shape.Here k is -ve. Now the expansion goes on forever because there is not enough matter to slow it down. Curvature or Shape of the Universe. Defined in next slide Three possibilities:Geometry Curvature of space Spherical Flat Hyperbolic positive zero negative Type of Universe closed flat open < Ώ0 > Ώ0 > ΏC Ώ0 = ΏC Ώ0 < ΏC Hint from CMB If CMB were truly isotropic then we could not tell which case prevailed. However there are hotspots. Their apparent size depends on curvature. If closed, spots appear larger. If open, they seem smaller. If flat then they appear to be their true size. Calculations suggest they should be about 1 degree in angle, close to what is observed. Critical Density Friedmann equation 8G 2 H(t) 3 k = R(t)2 Flat Universe:- If k = 0 then the R.H.S. is zero. Then C = 3 H02 8G where C is the critical density of the Universe. We can also write it in terms of = .. This is called the density parameter. C [Here we have the ratio of the combined average mass density(sum of for matter,radn & any other form of matter) and critical density 0 ] If CMB were truly isotropic then we would have absolute uniformity from all directions.There are localised hotspots however due to small density variations in the Early Universe.Apparent size will depend on curvature.Calculations suggest 0 is close to 1-Flat Universe.WHY? Critical Density Best estimates from all the matter we can see in galaxies[visible matter] is that = 0.02. From the way galaxies rotate we also have evidence that there is a large amount of matter we cannot see--Dark Matter. This raises the value of to 0.2 or so.This value has a large uncertainty but it is remarkable that it is so close to a Flat Universe. Problem:-There is nothing in the model to suggest any particular value for k but it seems that out of all possible values it is within 1-2 orders-of- magnitude of the critical value. The problem is made worse by the fact that if the value of the density parameter departed from 1 even by a little in the Early Universe we can show that by now it would be different by an enormous factor. The original expansion velocity would have to have been fine-tuned to exactly the right velocity so that the critical density is 1 now. Predicted Abundances of the light elements from Big Bang One of the three pieces of evidence for the Big Bang Model is the abundances of the light elements.It depends on the number of types of neutrino which exist and the density of baryons(strongly interacting particles) Here we see the predictions as a function of baryon density[at top] and of the density B. Band shows observations Line shows critical density for H0 = 65 km/sMpc Horizon Problem Cosmic Particle Horizon:- the Earth is at the centre of an enormous sphere with radius ~ 15 billion light years. The surface of this sphere is called the Cosmic particle Horizon. We can see nothing outside this sphere because there has not been time for the light to reach us. This is part of the reason why the Sky is dark at night. The other reason is that everything is receding from us. Horizon Problem Perhaps the biggest problem for the Big Bang Model is the Horizon Problem. It involves communication of information between different parts of the Universe.Since the Universe has a finite age in this model even light can only have Cosmic Particle horizon travelled a finite distance in this time. This distance defines the OBSERVABLE UNIVERSE.This is what we can see and is finite whatever the size of the Universe. As we see in the picture radiation from the Big Bang has just reached A and B. Radiation from A and B cannot have reached each other.In particular this means that the two sides of the sky cannot have interacted to create thermal eqbm and a common T. Very Strange!! Cosmic Particle horizon Horizon Problem CMP Problem is even worse.The radn. has been travelling uninterrupted since decoupling so thermalisation would have had to occur before then when the Universe would have been much smaller. So places that look quite close in the CMP sky would not have been able to interact to establish eqbm. So how does it happen that the CMB is essentially isotropic without this communication between the different parts of the Universe? Worse still how can we create the very small fluctuations, the seeds from which galaxies etc grow? An Inflationary Universe The most widely adopted soluton to the Horizon and flatness problems is to introduce a period of very rapid expansion in the Early Universe. This idea was introduced by Alan Guth in 1981. In essence the inflation is defined as a very short period in which the scale factor grows exponentially.It is accelerating with d2R > 0 dt2 From Friedmann’s eqn we can show that 1. d2R = -4G + 3p R dt2 3 c2 + 3p 2 < 0 i.e p < -c c2 Cosmological Constant 3 Eventually analysis leads to R(t) = exp [(/3)1/2.t] Thus we have a very dramatic expansion dominated by this constant [ ] An Inflationary Universe In simple terms the figure shows Guth’s idea of an inflationary Universe. At the top we see normal expansion. Below we see a very rapid expansion over a very short time near t = 10-35s. Now a small part of the Universe, small enough to achieve thermalisation before inflation can expand to be much bigger than our presently observable Universe. An Inflationary Universe 1.Flatness Inflation neatly solves the flatness problem.Before inflation the Universe was very curved. Afterwards it would have expanded so much that our little bit,the observable Universe would appear flat Note:-It is very like looking at a part of the surface of a balloon before and after you inflate it. An Inflationary Universe 2.Horizon Problem. In the Inflationary Model of the Big Bang the observable Universe is said to have evolved from a volume some 1050 times smaller than in the Standard Big Bang Model. Accordingly, before expansion,the Universe was much smaller than its horizon distance so all of it could reach the same temperature.Then inflation made it larger and preserved the uniform temperature that we see today in the CMB. 3.Why did inflation occur? One possibility is that it represents a phase transition and in such transitions there is often a large release of energy. A new model encompassing this is needed-Supergravity? 10-43secs.--gravity froze out from other forces. 10-35secs--strong Force froze out from Electroweak force-Inflation 10-32secs--Standard Big Bang. Expansion of the Universe As the Universe expands the volume increases proportionally to the cube of the scale factor. Galaxies 1.We already saw that Hubble settled the debate on the nature of nebulae with his study of Cepheid variables in Andromeda.M31 is 2.2 Mly away. 2.He went on to study many other galaxies and from the measurements of redshift and distance he produced Hubble’s Law. 3.He also classified the galaxies he saw by their shapes ellipticals spirals barred spirals irregulars then he further subdivided them M74 Gemini M31-Andromeda Galaxy-2.2Mly from Earth, Part of our Local cluster of galaxies. Photo-mosaic picture of the Sombrero Galaxy taken with the Hubble Space Telescope over several orbits.Glowing central bulge of stars surrounded by pancake shaped disc. Classification of Galaxies Here we see the main types of galaxy both in plan view and in side view. All three types can vary greatly in detailed shape,size etc. Hubble arranged them in his so-called Tuning fork diagram and they are then sub-classified.For example Ellipticals are subdivided from E0 to E7 according to how elliptical they appear.The E0 are closest to circular. Hubble’s Tuning Fork Diagram Formation of galaxies 1.Formed from contraction of huge clouds of gas under gravity.The rate of star formation then decides elliptical or spiral.If it is low then gas has plenty of time to settle into a flattened disc.Star formation continues in the disc because it has lots of H.It becomes a spiral. If it is high then all the gas is used up in star formation before disc forms and we get an elliptical. 2.Alternatively galaxies form by merger of several gas clouds rather than from a single gigantic cloud of gas. 3.Many small gas clouds form and then merge. Which idea is correct is not clear!!! What is clear is that they form under the action of gravity If the dust cloud starts with some small rotational velocity then as the dust cloud shrinks this velocity will increase in order to conserve angular momentum. There are two forces on a mass at the surface of the dust cloud – the gravitational force towards the centre and the centrifugal force maintaining its rotation around the axis. Resolving the forces we see that one component tends to push the mass towards the “equator”. We will end up with a disc. The exact form is going to depend on the timescale. Coma Cluster - 20Mly across containing thousands of galaxies(300 seen here). It is about 270 Mly away.Two supergiant galaxies seen in centre. Hubble Deep Field-A very narrow sample of the sky looking as far back as 10 10 years in some cases.Data taken over 10 consecutive days. Galaxies All Spirals Ellipticals Irregular Mass(solar masses) 109 - 4x1011 105 - 1013 108 - 3x1010 Luminosity(solar) 108 -2x1010 3x105 - 1011 107 - 109 Diameter(kpc) Types of star % of observed galaxies 5-250 Disc-Young Centre-Pop2 and old pop 1 1-200 Pop 2 and Old Pop 1 77% Note:-Pop 1=significant heavy elements Pop 2 =few heavy elements-Old 20% 1-10 mostly Pop 1 3% Galaxies Ellipticals have little dust and interstellar gas-hence few new stars. As a result they are made mostly of old, red, population 2 stars. Note:-Pop 1=significant heavy elements Pop 2 =few heavy elements-Old Within ellipticals star motion is at random. Irregulars- they do not fit into any scheme. Irr 1-look like underdeveloped spirals Irr 2-odd shapes probably due to collisions or violent activity in centre. Examples:- Large and Small Magellanic Clouds. Both have substantial amounts of interstellar gas. Hence active star formation. Clusters of Galaxies 1.Galaxies are not found at random but are observed in clusters with the members of the cluster in constant motion under gravity. 2.They are classified as Poor or Rich dependng on number of galaxies they contain. 3.Milky Way is part of a poor cluster-some 30 galaxies,some of which were only observed recently. They are mostly dwarf ellipticals. It is called the Local Group. 4.They are also classified as regular or irregular depending on overall shape. The Local group is irregular.The Coma Cluster is regular. 5.Shape is related to dominant type.Rich,regular clusters contain mostly ellipticals and lenticular galaxies.Irregulars have a more even mixture. 6.Clusters are grouped together in superclusters.A typical supercluster contains dozens of individual clusters spread over a region up to 30Mpc( 100 Mly ) across. 7.Overall there are huge voids in space-they may contain gas or very dim galaxies. Voids on the largest Scale The Evolution of Stars Ideally we would follow a star’s “life” from its genesis in a cloud of gas and dust to its “death”. However the timescale is way beyond our life span. The alternative approach is to realise that the numbers of stars we can see is very large. We assume that a) the lifetimes of stars are shorter than the age of the Universe and b) we can observe stars at all stages of development. We have seen now how to measure stellar distances, masses, surface temperatures, luminosities etc. We can now ask whether these quantities are correlated in any way. One way to look at this is the Hertzsprung-Russel diagram. Hertzsprung-Russell Diagram Key tool in our understanding of star formation and of star “birth” and “death” Proposed independently by Hertzsprung and Russell It links stellar luminosity and surface temperature Note:- You will find it plotted in slightly different ways and for selected groups of stars. Hertzsprung-Russell Diagram 1.The H-R diagram plots Luminosity against Surface Temperature. Note:-Log Luminosity is used because of the large range and it is plotted against decreasing temperature. 2.Each star is represented by a point on the diagram. 3.The results depend to some extent on the sample of stars.They could be from stars within a limited volume round the Solar system, members of a cluster,stars of apparent brightness above a certain limit,etc. 4.Any H-R diagram shows that only a limited combination of values of T and L are allowed. 5.Most stars lie on a thin strip running diagonally across the diagram This is the Main Sequence. 6.Top right is also populated with brighter stars with lower T.Red Giants 7.Lower left is also rich in stars.They are bluish-white and small.The size is comparable to that of the Earth but with approx. the same mass as the Sun. They are White Dwarfs. Hertzsprung-Russell Diagram 1. Temperature By restricting the range of stars plotted one can test ideas of stellar evolution. Here we see stars from a particular globular cluster. These groups of stars are very old and different from open clusters. The ages are such that only stars of 1 solar mass or less are left.They are close to the age of the Universe. Most other stars not on the main Sequence here are White dwarfs or brown/black dwarfs.In general they are Population 2 stars with less than 1% of heavy elements compared with Population 1 stars where it is 2% or so. [Contrast with stars in the disc.] Stellar Birth 1.Seeing the early stages is difficult. It starts with a collapsing cloud of gas and dust and it is not hot enough to shine so we don’t see it. As it collapses half of the potential energy is turned into kinetic energy [Heat]. [Virial Theorem] Triggering of such collapses is not fully understood. 2.If the temperature of the gas cloud reaches high enough temperature the particles[protons]will have enough energy to interact and nuclear reactions will begin at about 8 million Kelvin .As we will see this releases energy which heats the gas and raises its pressure. 3.If heated enough, the gas pressure will countermand the gravitational contraction and the star will stabilise under these two opposing forces. 4.At this stage the star will be moving to the left on the H-R diagram and will end up on the Main Sequence. GAS Gravitation Heat The rates of Nuclear reactions in stars • The probabilities of nuclear reactions in stars is very small because the energies of the particles are very small but there is a very large number of particles. • The number of reactions is just the product of the number of particles of a given energy multiplied by the probability of the reaction in a collision at that energy. RED = Number of particles as a function of energy Green = Probability of a reaction occurring in a collision Blue = Number of reactions as a function of energy GAMOW PEAK Hydrogen burning Gravitación 4p + + + + 2+ 2e-+26MeV Hydrogen Helium or Our friend the Sun (6000ºK, yellow) The p-p chain;the reactions which power the Sun Overall - 4p 4He + 2e- +2 + 26.7 MeV The CNO-Cycle: In stars where we already have C,N and O we can get hydrogen burning 4p + 2e- + 2 +26.4 MeV The C,N and O nuclei act as catalysts for the burning process Hans Bethe-1938 Reactions and Beta decays compete. • The probabilities of nuclear reactions in stars is very small because the energies of the particles are very small but there is a very large number of particles. At each step in the CNO-cycle there is a competition between beta decay and another proton capture. Which “wins” depends on the average time between captures or decays. Since the no. of reactions depends critically on the temperature so does the competition. [T(Sun) = 1.55 x 107 K] ● Cross-over (p-p to CNO) is for stars slightly more massive than Sun The CNO-Cycle: In stars where we already have C,N and O we can get hydrogen burning 4p + 2e- + 2 +26.4 MeV The C,N and O nuclei act as catalysts for the burning process Hans Bethe-1938 Life Cycle of Stars and Nucleosynthesis 1. Formation from large clouds of gas and dust. 2. Centre of cloud is heated as it collapses under gravity 3. When it reaches high enough temperature then nuclear reactions can start. 4p 4He + 2e + 2ν + 26.7 MeV 4. This raises temperature further and star eventually reaches equilibrium under heating internally and gravitational collapse. 5. The process of making heavier nuclei occurs in the next stage. After the Main Sequence 1.Once a star’s hydrogen is used up its future life is dictated by its mass. 2.During the H-Burning phase the star has been creating He in the core by turning 4 protons into a He nucleus plus electrons and neutrinos. Once the H burning stops in the centre the star contracts and some of the potential energy is turned into heat. If the core temperature rises far enough then He-burning can begin. Coulomb(electrostatic) barrier is 4 times higher for two He nuclei compared with protons. 3.Now we face again the problem of there being no stable A = 5 or 8 nuclei. 4.It turns out that we can bypass these bottlenecks but it depends critically on the properties of the properties of individual levels in Be and C nuclei. The Creation of 12C and 16O • H and 4He were made in the Big Bang.Heavier nuclei were not produced because there are no stable A = 5 or 8 nuclei. There are no chains of light nuclei to hurdle the gaps. • How then can we make 12C and 16O? • Firstly 8Be from the fusion of two alphas lives for 2.6 x 10-16 s cf. scattering time 3 x 10-21 s. They stick together for a significant time. • At equilibrium we get a concentration of 1 in 109 for 8Be atoms in 4He. • Salpeter pointed out that this meant that C must be produced in a two step process. • Hoyle showed that the second step must be resonant.He predicted that since Be and C both have 0+ s-wave fusion must lead to a 0+ state in 12C close to the Gamow peak at 3 x 108K. • Experiment shows such a state at 7654 keV with = 5 x 10-17s The 7654 keV state has / 1000 A rare set of circumstances indeed! 1010 years Red Giant (3000ºK Red) H burning The Earth will be engulfed!! ++ 12C + 16O Path of Solar Mass Star on Hertzsprung Russell Diagram White Dwarf H, N, O ¡¡only!! (Hubble) Fluorescence Helix Planetary Nebula in the constellation of Aquarius Binding energies in Nuclei • E = mc2 • The curve below shows the Binding Energy per nucleon as a function of the mass of the nucleus. A = 56 B.E. per nucleon in MeV Mass of the nucleus (A) The End of Fusion Reactions in Stars A = 56 Binding Energy per nucleon as a function of Nuclear Mass(A) •When two nuclei fuse together energy is released up to mass A = 56 Beyond A = 56 energy is required to make two nuclei fuse. •As a result we get the burning of successively more massive nuclei in stars.First H, then He, then C,N,O etc. •In massive stars we eventually end up with different materials burning in layers with the heaviest nuclei burning in the centre where the temperature is highest. •When the heaviest(A = 56) fuel runs out the star explodes-Supernova If theEtoile star is eight times more massive than the Sun massive supergéante H He C O Ne Na Mg Al Si P S SUPERNOVA Gravitación Fe C. THIBAULT (CSNSM) Death of a Red Giant: SUPERNOVA October 1987 1056 Joules of energy This happened 170000 years ago in the nearest galaxy The Destiny of the Stars… White Dwarf Main Sequence AÑOS Density/ 109 109 Red Giant Brown Dwarf Massive Stars Supernova Algún 109 segundo 100 kg C. THIBAULT (CSNSM) The probability of Reactions in Stars The particles in the stellar gas have an energy distribution given by the Maxwell-Boltzmann distribution(seen on left). The probability of penetration of the Coulomb Barrier is given by the expression on the right. The product of the two gives the probability of the two nuclei fusing. The resulting peak is called the Gamow window. Spectrum of Cassiopeia We see here the remnants of a supernova in Cassiopeia.This radio telescope picture is taken with theVery Large Array in New Mexico. From the measured rate of expansion it is thought to have occurred about 320 years ago. It is 10,000 ly away. With optical telescopes almost nothing is seen. The inset at the bottom shows a small part of the gamma ray spectrum with a clear peak at 1157 keV,the energy of a gamma ray in the decay of 44Ti. Principe de nucléosynthèse Principle of laNucleosynthesis protons 63 65 28 Ni 58 59 60 61 62 64 27 Co 26 Fe 59 29 Cu 54 55 56 57 58 35 30 40 neutrons Il y a compétition Competition between twoentre processes ••Capture Captureof d’un neutron a neutron •• Radioactivité Radioactivity– n p + e-+ C. THIBAULT (CSNSM) Part of the Slow Neutron Capture Pathway In Red Giant Stars neutrons are produced in the 13C( 4He,n) 16O reactions. The flux is relatively low.As a result there is time for beta decay before a second neutron is captured. The boxes here indicate a stable nuclear species with a particular Z & N. Successive neutron captures increases N. This stops when the nucleus created is unstable and beta decays before capture. The pathways for the s- and r-processes S-process:Neutron flux is low so beta decay occurs before a second neutron is captured.We slowly zigzag up in mass. R-process:Neutron flux is enormous and many neutrons are captured before we get beta decays back to stability. The Abundances of the Elements for A = 70 - 210 Note the double peaks at N = 46/50, 76/82, 116/126 They are due to production by the two separate processes Abundance Predictions The Reaction Pathways in Stellar Nucleosynthesis Remember Elements above Iron(Fe) are made in s- and r-processes. • The reaction pathways in the r- and rp-processes of nucleosynthesis lie a long way away from the line of stability. Red Giant:Betelgeuse Orion Nebula e.g., Diehl et al., Astron. Astrophys 97, 181 (1993); Publications of the Astr. Society of the Pacific 110:637 (1999) Full-sky Comptel map of 1.8 MeV gammas in 26Mg following 26Al GS -decay. (a) Spin traps, eg. 26Al, (N=Z=13) 0+ state -decaying spin-trap. 0+, T=1 5+, T=0 (decays direct to 26Mg GS 228.3 keV, T1/2=6.3 secs via superallowed Fermi +…forking in rp-process + 26 0 keV, T1/2=7.4x105 yrs (decays to 2 states in Mg via forbidden, Dl=3 decays). Cosmic Microwave Background Cosmic Microwave Background Explorer( COBE), a satellite launched by NASA in 1989 was the first to show that there are very small fluctuations in the CMB at the level of one part in 105. Summary of CMB Measurements Summary of CMB Measurements Cosmic Microwave Background Penzias and Wilson tried hard to make their signals go away. Unknown to them a group at Princeton had been working on the question of what sort of radiation would have been left over from the Big Bang. Once the two groups got together it was clear that they had observed just this radiation. Summary – Last Week. Big Bang Model:- Three pieces of evidence - Hubble’s Law - Cosmic Microwave Background - Abundance of the light elements in the Early Universe Problems with the model - Flatness Problem - Horizon Problem Standard Big Bang cannot explain either. ”Solution” – Inflationary Big Bang - This “solves” the two problems but no satisfactory explanation. Hubble’s classification of galaxies. Superforce reigns GUTs Electroweak era QGP/Hadron phase transition Universe becomes transparent Last Week Big Bang Model – Three major pieces of evidence - Problems – Horizon and Flatness problems hence Inflationary Big Bang – introduced by Guth (1981) - solves these problems History of Universe in terms of the Big Bang. Finally – Formation of stars and galaxies. Nuclear Radii Stable Nuclei R = R0.A1/3