Líf í Alheimi

Kennari: Páll Jakobsson

F01 8.1.2013

Material for exam:· Written Handouts· PowerPoint slides· Stu written and said in class.

Exoplanets and intelligent life

α-centauri A/B. ∼ 4 lightyears from the Sun. Planet with mass similar to Earth (M⊕) foundorbiting α-centauri B. The planet is closer to α-centauri B than Mercury is to the Sun. Forthe probability of intelligent life liquid water is nescessary. Therefore life on this planet isunlikely.Proxima centauri orbits the α-centauri Binary system and is the closest star to our Solarsystem.

Exoplanet: planet outside our Solar system.There are 5 methods to be taugth to detect exoplanets(Next lecture).Many exoplanets found are big/massive planets→ Gas giants, most likely no intelligent life.Lighter/denser planets(rocky planets) are much more likely to inhabit intelligent life.

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Tau Ceti: quintet of exoplanets. Their masses are 2 − 6M⊕. One of those planets is in aHabitable zone.Habitable zone: the scientic term for the region around a star within which it is theor-etically possible for a planet with sucient atmospheric pressure to maintain liquid wateron its surface.

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Brief overview of the history of the universe

The age of the Universe is 13.7 · 109 years, which can be approximated as 14 · 109 years.The 3 lightest elements(H, He and Li) formed in the rst 3 minutes after the Big Bang.Decoupling: Photon crashes into electron and kicks the electron away from a proton.

Cosmic microwave background radiation, ∼ 300.000 Yr. Cosmic background radiation iswell explained as radiation left over from an early stage in the development of the universe,and its discovery is considered a landmark test of the Big Bang model of the universe.

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Dark ages: neutral aroms

∼ 500 million years: Galaxies and Quasars begin to form. Heavier elements form in coresand outer layers of stars.

Hubble deep eld

The Hubble telescope looked at one tiny point in space for a long period of time and gatheredlight to make the famous Hubble Deep Field picture. The picture has been updated withmore time gathering light.

Hubble Deep Field 0.5 · 106 sHubble Ultra Deep Field 1 · 106 s

Hubble eXtreme Deep Field 2 · 106 s

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Known exoplanets

· 1992: A few discovered around a neutron star· 1995: The rst around Sun-like star (indirect measurement)· January 2013: 854 (http://exoplanet.eu)

1) Radial velocity + astrometry: 498(386 planetary systems)2) Transits: 291 (236 planetary systems)3) Microlensing: 16 (15 planetary systems)4) Imaging: 32 (28 planetary systems)

1), 2) and 3) are indirect measurements.

Sun's destiny

Life cycle of our Sun: Hydrogen in the core will run out. The Sun's core will begin tocontract, but its outer atmosphere will expand into a red giant star roughly a hundred timeslarger than the Sun. Eventually, the core will collapse and become a white dwarf star, roug-hly the size of Earth. The outer atmosphere will be released into space as a nebula of gasand dust.

Planck time

Planck time is the time required for light to travel, in a vacuum, a distance of 1 Plancklength. Planck length is equal to 1.616199 · 10−35m.Before a time classied as a Planck time, 10−43 seconds, all of the four fundamental forcesare presumed to have been unied into one force. All matter, energy, space and time arepresumed to have exploded outward from the original singularity. Nothing is known of this

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period.In the era around one Planck time, 10−43 seconds, it is projected by present modeling of thefundamental forces that the gravity force begins to dierentiate from the other three forces.This is the rst of the spontaneous symmetry breaks which lead to the four observed typesof interactions in the present universe.

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Líf í Alheimi

Kennari: Páll Jakobsson

F02 10.1.2013

The observable universe

In the observable universe there are

1011 galaxies · 1011 stars/galaxy = 1022 stars = 10 sextillion starsand

1011 galaxies · 1010 Earth like planets = 1021 Earths in the universe

The Fomalhaut b planet

Dark spot in the middle is light from the star blocked by a coronagraph. In the dust ringaround the star a planet, Fomalhaut b, was found. But how was that planet discovered? Bytaking photos and trying to see if the planet/spot moves. By looking at photos from 2004to 2010, with the interval of two years, it is clear that this spot is actually a planet. Thedistance to that system from our Solar system is D ' 25Light years

A coronagraph is a telescopic attachment designed to block out the direct light from astar so that nearby objects, which otherwise would be hidden in the star's bright glare, canbe resolved. Most coronagraphs are intended to view the corona of the Sun, but a new classof conceptually similar instruments (called stellar coronagraphs to distinguish them from

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solar coronagraphs) are being used to nd extrasolar planets around nearby stars.The orbital distance of this planet is d = 50− 300AU , where AU is the average distance

between the Sun and the Earth. This tells us that the orbit is extremely elliptical.If the orbit of Fomalhaut b is on the same plane as the dust ring they will collide and

result in a grand rework show.It is noteworthy that the age of the star in the system, Fomalhaut, is around 400·106Years

while our Sun is 4.57 · 109Years old.

HR 8799

Distance to system,D = 13 lightyears. First, three planets discovered(b, c and d). Later on, afourth planet(e) found near the star(A). The elliptical distance of planet e is d(A, e) = 15AU ,which is similar to Neptune.

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β pictoris b

Distance to system, D = 70 lightyears. Seen edge on. Outside the coronagraph darknessthere is a debris disc, consisting of dust. In that debris disc, planets can be forming. Massof β pictoris b is M = 9MJup. Age of the star, β pictoris, is 10 · 106 yr, which makes itessentially an infant star.

Luminosity

Luminosity is generally understood as a measurement of brightness. In astronomy, luminositymeasures the total amount of energy emitted by a star or other astronomical object in SIunits of joules per second, which are watts.

Comparing the luminosity of the Sun and Jupiter gives

L

LJup

= 109.

This means that the luminosity(optical) of the Sun is a billion times greater than of Jupiter.

Known exoplanets

· 1992: A few discovered around a neutron star· 1995: The rst around Sun-like star (indirect measurement)· January 2013: 854 (http://exoplanet.eu)

Methods to nding exoplanets1) Radial velocity + astrometry: 498(386 planetary systems)2) Transits: 291 (236 planetary systems)

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3) Microlensing: 16 (15 planetary systems)4) Imaging: 32 (28 planetary systems)

Planets discussed above were found by imagingTransit: Measure brightness of star, luminosity drops when planet transits(goes in front)star. Amount of luminosity dip depends on the size of the planet.

The Kepler observer observes transits. If it detects one transit, that object is considereda planet candidate. To be conrmed as a planet, Kepler needs to detect three transist fromthe objects. Furthermore, the duration and dimming must be the same.

A lot of information can be gathered from a transit graph. For example:

Imin

Imax

= 1−(Rexo

Rstar

)2

≈(

99% (Jupiter and Sun)99.99% (Earth and Sun)

)A star's spectrum gives all information about the star, including mass and radius. Thereforeit is possible to calculate the radius of the planet from the equation above. If one is lucky,it is possible to calculate the mass of the planet. That gives the opportunity to nd themass density, ρ, of the planet. The mass density tells whether the planet is a rocky one ora gas/liquid giant.

A transit graph gives information about the period of planet and radius of star. Kepler'sthird law lets one nd the distance between a planet and a star from the planet's period.

Kepler's laws1. The path of the planets about the sun is elliptical in shape, with the center of the sunbeing located at one focus. (The Law of Ellipses)2. An imaginary line drawn from the center of the sun to the center of the planet will sweepout equal areas in equal intervals of time. (The Law of Equal Areas). The planet movesfaster near the Sun so the same area is swept out in a given time as at larger distances,where the planet moves more slowly.

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3. The square of the orbital period of a planet is directly proportional to the cube of thesemi-major axis of its orbit. The third law captures the relationship between the distance ofplanets from the Sun, and their orbital periods. P 2 ∝ a3

Unfortunately, in order to see transits planets must have an edge on orbit. This is the maindrawback of the Kepler observer.

Kepler observer

1% chance to see a transit out of 150.000 stars → 1500 planets expected to be found. InJanuary 2013: 2740 candidates.

Almost all planets found by Kepler have orbits shorter than Mercury. Why is that?Kepler launched in 2009. Planets with long periods need more time to be looked at toobserve 3 transits. With more time planets with longer periods can be conrmed. Hopefullythe Kepler observer will not break down entirely in coming years.

It is easier to nd bigger planets, because they cause more dimming in transit graphs.Also, if a planet's orbit is too elliptical it can move in and out of the habitable zone.

Sizes of planet candidates: Big increase of smaller planets because of better examinationof the noise. Getting signals from the noise.

Kepler's ndings

The Kepler-20 system. Plantes are large, small, large, small, large. That is consideredunusual compared to our Solar sytem because there small planets are near the Sun while thebigger ones are further away from the Sun.

The range of the radius of the planets if from 0.9R⊕ to 3.1R⊕. The radial border ofrocky/gas/liquid planets is around 2R⊕.

Kepler-47 system. How to dene a habitable zone of a binary system? It rotates with thecenter of mass.

CCD: A charge-coupled device (CCD) is a device for the movement of electrical charge,usually from within the device to an area where the charge can be manipulated, for exampleconversion into a digital value. The CCD is a major technology for digital imaging.

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Kepler 10, 11 and 22. Kepler-10b was the rs rocky exoplanet to be discoverd. Itscomposition is probably more irony than the Earth. Also, its temperature is around thepoint where iron melts, so it is most likely a big, hot ball of liquid iron.

The Kepler-11 system is a record holder for most planets found in a system, totalling ofsix discovered planets.

In the Kepler-22 system a planet in the habitable zone was found, Kepler-22b. It is mostlikely a liquid giant planet but was despite that called Earth 2.0. This was most liekly justa PR stunt by the Kepler team.

Kepler's planet candidates are getting more and more Earth like. That is because withmore time, planets with longer orbitals are being discoverd. Also, with better technology isgets easier to distinguish smaller planets from noise.

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Astrometry(stjörnuvagg)

A system of two(or more) bodies orbit around their mutual center of mass. Planets orbitaround the surface of the Sun. Center of mass of the system is near the surface of the Sun.The change in the stellar position(wobble) gives away the planet mass.

Example: MJ and d = 1AU . Assume the system lies 3 light years from Earth. → Stellarwobble: 1 milliarcsecond. By putting Earth instead and move system to 30 light years →stellar wobble: 0.3 milliarcseconds.

No chance of measuring today. SIM planet quest(cancelled) expected to launch in 2016was supposed to nd exoplanets by astrometry.

In our Solar system Jupiter is at most fault(because of its large mass) for the Sun'swobble because of its mass. From 30 lightyears the Sun's wobble would be measured around1 milliarcsecond.

An arcminute is a unit of angular measurement equal to 1/60 of one degree. An arcsecondis 1/60 of one arcminute. → 1 arcsecond = 1

60·60 of a degree. While looking at the night sky,extending your arm and lifting one's pinky, the pinky is roughly 1 arcsecond.

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Líf í Alheimi

Kennari: Páll Jakobsson

F03 15.1.2013

More on methods of nding exoplanets

Imaging· Unable to measure mass

Transits· Orbital period· Radius· Unable to measure mass· Low probability to see a transit

1% average probability to detect transit0.1% far out from star (∼ Pluto) → 10% (∼ Mercury)

· Kepler has mostly found planets with short orbital periodsNeeds to see three transits to conrm a planet.With more time planets with longer periods will be conrmed.

Astrometry· Very dicult to detect planets (0 exoplanets)· Easier to detect planets further away(and probably in the habitable zone).· Still important method even though no planets discovered yet.

Radial velocity (spectrum)(Astrometry's cousin)· Take spectrum of a star.· Many spectra are taken and spectrum lines are observed.· More massive planets: Star travels faster, easier to detect.· Planet closer to star: Star travels faster (Kepler's laws).· Sensitivity limit today is ∼ 50 cm/s.· Most planets discovered by this method are so called Hot Jupiters,which are gas giants close to their stars.

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Data from astrometry/radial velocityTake shift of spectrum lines → calculate velocityRedshift → away ; Blushift → towardVelocity is plotted as a function of time. If the data shows a sinusodial curve, the planet'sorbit is round. However, if the curve is elongated, the orbit is elliptical and it is possibleto estimate eccentricity(e) of the orbit. If the eccentricity is zero, the orbit is circular.This method gives a planet's lower mass limit.

If solar system is face on: no data can be acquired. If solar system is edge on: Vr = V andcorrect mass of planet can be measured.By combining the methods of transits and radial velocity:⇒ Get everything (mass, period, radius, density...).Maximum shift: maximum velocityMinimum shift: minimum velocity

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Gravitational lensingA gravitational lens refers to a distribution of matter (such as a cluster of galaxies) betweena distant source (a background galaxy) and an observer, that is capable of bending (lens-ing) the light from the source, as it travels towards the observer. This eect is known asgravitational lensing and is one of the predictions of Albert Einstein's general theory ofrelativity. An Einstein ring can appear or 2/4/6 false images. By looking at spectrums fromdierent objects and two identical spectrums are acquired, it is most likely the same object.Gravitational lensing is also a way to detect dark matter.

In observational astronomy an Einstein ring is the deformation of the light from a source(such as a galaxy or star) into a ring through gravitational lensing of the source's light by anobject with an extremely large mass (such as another galaxy, or a black hole). This occurswhen the source, lens and observer are all aligned.

Abell 370 is a galaxy cluster located approximately 6 billion light years away from theEarth (at redshift z = 0.375), in the constellation Cetus. Its core is made up of sever-al hundred galaxies. Abell 370 appears to include several arcs of light, which in fact aremirages caused by gravitational lensing of more distant objects.

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Microlensing

The second bump's amplitude indicates the mass of the planet. By observing the graph, onecan also get the planet's orbital radius.

It is nescessary to watch hundreds of thousand stars because of low probability to nd aplanet by this method. The main drawback of this method is that it is a single event andthe planet will never be seen again.

This was the rst method to detect Earth-like planets and the mass range of planetsfound by this method is

3.3M⊕ ≤M ≤ 3.7MJup.

Where is the limit between a star and a planet? A Brown Dwarf is an intermediate statebetween a planet and a star. Their mass range is

13MJup ≤M ≤ 80MJup.

Rough numbers for Rocky/Gas planet limitsA rocky planet that can sustain life and keep its atmosphere, has the radius range of

0.8R⊕ ≤ R ≤ 2R⊕.

If R ≤ 0.8R⊕ ⇒ M ≤ 0.5M⊕: Not enough mass/gravity to hold onto life sustainingatmosphere.If R ≥ 2R⊕ ⇒ M ≥ 10M⊕: Enough gravity to hold on to Hydrogen, the most commonelement in the Universe. Likely a gas giant.Exception: Titan, one of Saturn's moons has a mass 0.02M⊕ and a thick atmosphere, mainlyof nitrogen. The Cassini probe dropped another probe, Huygens, onto Titan. Huygenslanded on Titan and recorded sound and took images.

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Subatomic particles

The ratio of neutrinos and atoms in the universe; Neutrinos:Atoms ⇔ 109 : 1Neutrino background originates when the Universe was t = 1 s. Neutrinos got decoupled.A proton consists of three quarks: up+ up+ downA neutron also consists of three quarks: down+ down+ upThe strong force holds quarks together in the nucleus. Protons have the same charge andrepel, but the strong force holds them together.

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Líf í Alheimi

Kennari: Páll Jakobsson

F04 17.1.2013

Job of the lecture: create all elements in the periodic table

Age of the Universe What happened10−37 − 10−5 s Quark soup (individual quarks)

10−5 s Hadrons(3 quarks) start to form, e.g. protons and neutrons.Aslo: p+ + e− ↔ n+ νe

100 s Primordial soup mainly e−, p+ and n. T = 109K,universe still expanding and cooling.Still hot enough(high kinetic energy) for nuclear fusion to take place

Now The average temperature of the universe is T ' 2.75K.

Matter : Antimatter = 1.000.000.001 : 1.000.000.000

Asymmetry(→ matterworld). This asymmetry was needed for matter to survive today.Matter and antimatter can clash and annihilate each other. The cosmic microwave background(CMB)got created when this matter and antimatter annihilated each other.

Planck time: tp ≡√

~Gc5

= 5 · 10−44 s

Mass ratio of lightest elements when t = 100 s:

H (∼ 75%)He (∼ 25%)

Miniscule Li, Li/H ∼ 10−10

Some chemical equations. n neutron, p proton, γ photon, D ionized deuterium. Only cores,no electrons bound to core:

n + p −→ D + γD + D −→ 3He++ + np + D −→ 3He++ + γn + 3He++ −→ 4He + γ

D + 3He++ −→ 4He++ + p

CMB: Cosmic microwave background

a: Before recombination, photons in the cosmic background prevented protons and electronsfrom forming hydrogen atoms.b: As soon as hydrogen atoms could survive, the universe became transparent(to photons).This occured roughly 400.000 years after the Big Bang. One can see this as static(∼ 1% ofit) on TVIn physical cosmology, the term decoupling is often used for the moment during recomb-ination when the rate of Compton scattering became slower than the expansion of the uni-verse. At that moment, photons nearly stopped their interactions with charged matter anddecoupled, producing the cosmic microwave background radiation as we know it. The termdecoupling is also used to describe the neutrino decoupling which occurred about one secondafter the Big Bang.

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Formation of heavier elemetns

H, He, Li are not enough to create life. Need stars to create life(fusion in center of stars). Inthe centre of the Sun it is hot enough(high pressure) for particles to overcome the Coulombforce.

hydrogen?→ helium

There are two main nuclear reactions(energy source of stars)[TEST ]1) p− p chain (proton-proton)2) CNO cycle (Carbon-Nitrogen-Oxygen)There are three variations of the p − p chain. In two of them certain elements(Be and Li)act as catalysts. Let's look at one variation:PP − I:

4 11H −→ 4

2He + 2 e+ + 2 ν e + 2 γ

11H + 1

1H −→ 21H + e+ + νe (x2)

21H + 1

1H −→ 32He + γ (x2)

32He + 3

2He −→ 42He + 2 1

1H

All those elements are ionized(11H is the same as p on p.20). e+ is a positron and νe is anelectron neutrino. The multiplication by two in the eqs. above are to match up the boxedeq.

Bottom line: energy gets liberated since 1 helium atom is lighter than 4 protons (E = mc2).Change mass into energy(in the Sun). The Sun loses ∼ 4million tons/sec.

Energy released in a single reaction(boxed eq.)

Etot = Ereaction ·NHeatoms = 4 · 10−12 J · 1056 = 4 · 1044 J

From this energy, with the approximation that the whole sun can burn H to He, one canestimate how long the Sun will live.

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E/L = 4 · 1044 J/4 · 1026 Js≈ 1018 s ≈ 30 · 109 yr

However, the Sun will only spend 12 · 109 yr on the main sequence. This overestimation isbecause only in the center of the Sun is it hot enough to burn up helium.

The life of a star is a battle between gravity and gas pressure.Smaller stars than the Sun: less pressure → slower nuclear reaction → long lived.Massive stars(short lived): M = 50M → t ∼ 1million years. Can be called rock stars -live fast, die young. Planets and life do not stand a chance of forming.

η Carinae: 50− 100M. Is on the brink of exploding(as a supernova)

HR− diagram: Hertzprung-Russell diagramx-axis: surface temperature(hot to cool) in Kelvin.y-axis: brightness (solar units).90% of all stars are on the main sequence. In main sequence stars Hydrogen burns to Helium.Stars spend 90% of their lives on the main sequence.Giants are stars that have drifted from the main sequence.First stars that formed are probably still alive(small and cold).

When H gets depleted → fusion stops → gravity takes over → pressure and temperature ↑(Helium ash) → Helium burns to Carbon → star turns to Red Giant

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Destiny of the SunWhen all the hydrogen gets consumed into helium, there will be no more energy that willpush outward and prevent the Sun from collapsing. As a result, it will collapse. The collapsewill put tremendous pressure on the helium core and cause the helium nuclei there to fuseinto heavier elements just like before.Once again the fusion will push the outer regions outward. This time, the outward push willallow the Sun to expand much larger than it was earlier in its lifetime. At this point in thelife cycle of th Sun, it will then become a Red Giant. While the outer regions continue toexpand, the helium nuclei in the core will continue to fuse into carbon.Now, carbon will no longer fuse further. As such the Sun's core will stabilize while the restof its outer regions will proceed with their expansion and will eventually be shed o. Thecore will be known as a white dwarf, while the outer regions will be known as a planetarynebula. In the end of the life cycle of the Sun, the white dwarf will release all of its remainingheat and will evolve into a cold dark mass known as a black dwarf.

Formation of heavier elements revisited

The CNO cycle is not the primary source of energy for our Sun. It uses carbon, nitrogenand oxygen as catalysts. Importance increases as the Sun ages.If the CNO cycle is operative in the Sun, where did the C, N and O come from?The sun does not belong to the rst generation of stars (4.5Gyr vs. ∼ 14Gyr). It formed ina molecular cloud that previous generations of stars created(e.g. via supernova explosions).

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Stellar populations

Population I Stars of the 3rd generation, like our Sun.Have high metallicity.Contain the youngest stars in the Universe.

Population II Much older stars with lower metallicityPopulation III Stars containing pure hydrogen and helium and nothing else.

The rst stars to form in the Universe.Have never been observed.

Population I - last generation (counter intuitive denitions).Metallicity: everything heavier than He.Population I/II have been observed. Population III should still exist (possibly formed whenthe Universe was 200− 400 million years old)

Crab Nebula (1054) rst documented Supernova. Cassiopeia A (1947) is a supernova remn-ant in the constellation Cassiopeia and the brightest extrasolar radio source in the sky atfrequencies above 1GHz. A supernova can be as bright as their galaxy(which consists of100 billion stars). In the center of supernova remnants → black hole or neutron star.

(a) Crab Nebula (b) Casssiopeia A

Supernova remnants

Formation of heavier elements re-revisited

Stars start to form heavier elements when they move o the main sequence(HR diagram),i.e. when they stop fusing H into He. More massive stars can create silicone(Si), phosphor-us(P), sulfur(S), magnesium(Mg), aluminum(Al), etc. However, the nuclear reactions stopat iron(Fe, atomic mass 56). Energy is needed to create heavier elements in this way.

When stars stop fusing H → HeStart fusing: He −→ C

C + He −→ O∗ ↓ Higher temperatureO + He −→ N

∗ Sun stops and becomes a white dwarf with a Carbon-Oxygen core.

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Interiors of an old massive star

What about more massive elements than iron?s-process(slow): Takes place at a relatively low neutron density, e.g. inside certain typesof stars(mainly asymptotic giant branch). Time scale(adding 1 neutron to element) isaround 10− 105 years. Adds neutrons to isotopes. Walk up the periodic table. Occurs atintermediate temperature conditions in stars.

r-process(rapid): High neutron density, e.g. during supernova explosions. Time scale isonly a few seconds. (νe is an anti-neutrino.)

β decay: In nuclear physics, beta decay is a type of radioactive decay in which a betaparticle (an electron or a positron) is emitted from an atomic nucleus.

β− decay ⇒ AZN −→ A

Z+1N′ + e− + νe

β+ decay ⇒ AZN −→ A

Z−1N′ + e+ + νe

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Líf í Alheimi

Kennari: Páll Jakobsson

F05 22.1.2013

Exoplanet detection problems

Problem 1 - Radial velocity

[This method is biased to Hot Jupiters]How sensitive do instruments of some aliens have to be(i.e. how small velocities should theybe able to measure) in order to detect:

a) Jupiter located 1AU from the Sun?b) Earth located 5AU from the Sun?

NB : How does velocity (~v) depend on distance (r)?

Fgravitational = Fcentripetal ⇒ F =mMG

r2=mv2

r

⇒ v =

√MG

r;[v ∝ 1√

r

]M is the star's mass, m the planet's mass, G is the gravitational constant, r is the distancebetween planet and star and v is the planet's velocity.

It can easily be shown:

m

M=Vrvr

⇒ Vr =m

Mvr

If Earth is at 1AU :

Vr =M⊕M

√MG

r=

M⊕√M

√G

r= 9 cm/s

The equation above shows that this method is ideal to nd big planets near a star(HotJupiters). Unable to measure today (can measure ∼ 50 cm/s)a)

Vr =MJup√M

√G

1AU= 28m/s

Easily observable by our methods (detection limit ca. 1995).b)

Vr =MJup√M

√G

5AU= 4 cm/s

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Problem 2 - astrometry

Same aliens, Determine instrument sensitivity(in arcseconds) to detect the Sun's wobble ifEarth is at r = 5AU . Assume the aliens are 5 pc away(1 pc ' 3.26 lightyears). It caneasily be shown:

m

M=R

r

⇒ R =m

Mr =

M⊕M· 5AU = 2.25 · 106m

R is the radius of star's period around a common center of mass. How big of an angle doesR cover?

tan θ ≈ θ =R

5 pc=

2.25 · 106m

5 · 3.09 · 1016m≈ 1.46 · 10−11 rad = 3 · 106 ′′

2π rad = 360; 2π rad = 360 · 60 · 60 ′′

Conclusion: 6 microarcseconds (Diameter of star's orbit)

ESA: Gaia in October 2013 [1 billion stars] aims at sensitivity of 20micro ′′ (predicted tond tens of thousands exoplanets).

Problem 3 - transists

A gas giant with a radius R = 2RJup. To detect it, determine the ux(brightness) sensitivity.Disc area: A = R2π. ∗ denotes a Sun-like star.

AJup = R2Jupπ ⇒

AJup

A∗≈ 1%

A = (2RJup)2π ⇒ A

A∗=

4AJup

A∗≈ 4%

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Cosmic abundance of the elements

Between peaks of elements are 4 atomic mass units. That is because in the core of a star a42He is added to elements.

Star formation

Mainlytakes place in so called molecular clouds

· Contain H2.· The largest: 104− 106M.· Size: Hundreds of light years.· Contain hot cores: powerful infrared sources.· T-Tauri stars: About to enter the main sequence.

Surrounded by a circumstellar accretion disc.

Powerful infrared source: A lot of cosmic dust (microns → nanometers). Absorbs opticallight and emits infrared light.T-Tauri: stars being formed. Almost always surrounded by a protoplanetary disc, whereplanets can form.

Left: optical, right: infrared. Hiding behind the dust are a lot of stars

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Carina Nebula

In the infrared part is it possible to see Herbig-Haro objects, which are the white jets nearthe center of the picture.HerbigHaro objects (HH) are small patches of nebulosity associated with newly born stars,and are formed when narrow jets of gas ejected by young stars collide with clouds of gas anddust nearby at speeds of several hundred kilometres per second. HerbigHaro objects areubiquitous in star-forming regions, and several are often seen around a single star, alignedalong its rotational axis. They last no more than a few thousand years and are only foundin molecular clouds.

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Eagle nebula - Pillars of creation

Molecular cloud. Massive stars being formed.

Orion nebula

Protoplanetary disks around stars. Solar systems being formed.

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Formation of solar systems

Need spin(angular momentum) to form solar systems, otherwise only a star will form.

Galaxy is spinning. Dierent parts in molecular cloud will spin with dierent velocitiesbecause of radius from CM in galaxy → Molecular cloud acquires internal spin.

Angularmomentum = moment of inertia · angular velocity~L = I~ω = constant

Note that the angular momentum is only equal to a constant if there is no torque(no extraforce acting on system).

Assuming ~L is a constant, by decreasing I, ~ω must increase. Therefore when molecularclouds contract(I ↓) they spin faster (~ω ↑).

A rough schematic how a solar system is formed

Formation of the Solar system

i) Particles attract each other (electric charge). Starting point of formation.ii) They grow(by collisions) and separate from the gas in the disc.iii) Move at diereng velocity → collide and stick together (∼ 100 km).iv) Computer simulations show that objects with masses in the range between MMoon andM⊕ attracts signicant amount of gas (start to suck in gas)→ Jupiter, Saturn, Uranus, Neptune

v) In the inner Solar system, either giant planets did not form or the proto-Sun swallowedthem.vi) Gravitational inuence of the gas giants is great: planitesimals are thrown out of the Solarsystem, thousands of AU away. They now form the Oort cloud(origin of long-period comets).

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The Oort cloud: indirect measurements from comets. Extends ∼ 1 light year from the Sun.

The content of the Universe

The content of the universe

Impossible to see dark matter, ∼ 80% of materialin universe(ignoring dark energy).

Ω0 =ρ0ρc,0

Ω0 is the total density parameter, ρ0 density ofthe Universe and ρc,0 the critical density.

Ω0 = 1 for a critical density(at) Universe, theexpansion approaches a constant(an asymptote)at innite time.

Ω0 = 2 results in a closed Universe, where thereis enough stu to pull the universe back togetherand resulting in a Big crunch.

Ω0 = 0.5 is an open universe, where it expandsforever because of too little material.

15 years ago: Dark energy→ old model obsolete.Dark energy changes everything. At some po-int dark energy takes charge and will expand theuniverse exponentially. Galaxies in the sky willdisappear and each galaxy will be alone in theuniverse. Worries are not needed though becauseat that point in time our galaxy will most likelybe consumed by black holes.

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(a) Old model (b) Current model

Fate of the Universe

Problem 4

Look at the critical density, needed to close the universe:

ρc,0 ≡3H2

0

8πG= 10−26 kg

m3

Where H0 = 70km/sMpc

is the Hubble constant(expansion of the universe).

How many H−atomsm3 is equivalent to ρc,0?

10−26 kg/m3

1.67 · 10−27 kg≈ 6 H−atoms

m3

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