" Basic Astronomy & Astrophysics: with Q's and A's "

Salah Nasri

" Basic Astronomy & Astrophysics: WithQ’s andA’s"

January 2, 2019

In these notes, I discuss a selected number of topics in Astronomy and Astrophysics in the form of questions. For a non-physics major student or layman, very short answers to these questions are provided, whereas for people with special interest in physics and astrophysics more details are given. I also included many appendices with comprehensive discussion relevant to the questions discussed in these notes.

1 1 The Questions

(Q1): How did the ancients knew that Earth was round ?

(Q2): How did the ancients measured the size of the Earth?

(Q3): What is the simplest clear experimental evidence that the Earth spins around its axis?

(Q4): How fast does the Earth spin at the equator?

(Q5): How fast does the Earth move around the Sun?

(Q6): what is the universal law of gravitation ?

(Q7): What is the mass of the Earth ?

(Q8): what is the acceleration of gravity at the surface of an astronomical object of mass M and radius R?

(Q9): What is the escape velocity from the Earth?

(Q10): What is the minimum speed to launch a satellite from Earth ?

(Q11): What is the age of the Earth?

(Q12): What is the critical temperature, Tcritical, of the atmosphere of a planet (of mass M), above which a particular atmospheric component (of mass m) will be lost into space ?

(Q13): What is the ozone and where is it in the atmosphere ?

(Q14): What is the electromagnetic spectrum ?

(Q15): What type of electromagnetic radiation can reach the Earth’s surface ?

(Q16): What is a parallax angle, a parsec, and a light year ?

(Q17): What is the apparent and absolute magnitude of a stellar object ?

(Q18): What is the distance between the Earth and Moon ?

(Q19): What is the mass of the Moon ?

2 (Q20): How was the distance between the Earth and the Sun accurately measured before the 20th century ?

(Q21): How was the mass of the Sun ﬁrst measured ?

(Q22): What is average solar power density reaching the Earth’s atmosphere ?

(Q23): What is the luminosity of the Sun?

(Q24): How to estimate the surface temperature of a star ?

(Q25): What is the size of the Sun?

(Q26): What are the temperature and pressure at the Sun’s center ?

(Q27): What are the layers of the Sun’s atmosphere ?

(Q28): What is the direct method for measuring the mass of a stars ?

(Q29): What is dynamical time scale of a star ?

(Q30): What is Kelvin-Helmholtz time scale of a star?

(Q31): What is nuclear time scale of a star?

(Q32): How does the Sun shine ?

(Q33): What is the chemical composition of the Sun?

(Q34): How does the Luminosity (L∗) of a main-sequence star depends on its mass (M∗) and radius (R∗)?

(Q35): What is the life-time of a star as function of its mass ?

(Q36): What is the lower mass a star can have?

(Q37): What is the maximum mass a star can have ?

(Q38): What is a white dwarf ?

(Q39): How big is a white dwarf ?

(Q40): What is the Chandrasekhar limit?

(Q41): What is a supernovae ?

3 (Q42): What is a neutron star ?

(Q43): What is a black hole ?

(Q44): What is a quasar ?

(Q45): When did the most recent extinction event on Earth occur ?

(Q46): What is the size of the Milky Way galaxy ?

(Q47): How long does it take a spacecraft to reach the planet Mars ?

(Q48): How did astronomers know that the Milky Way is not the only galaxy in the Universe ?

(Q49): What is the age of our universe ?

(Q50): How did we know that our Universe is not static?

(Q51): How did the light elements, hydrogen and helium, form during the evolution of is the universe ?

(Q52): How did the heavy elements form in the Universe? (Q53): which type of telescope is most used by astronomers ?

(Q54): what is Hubble Space telescope ?

(Q55): what is Olber’s paradox ?

(Q56): what are the Greenhouse gases ?

4 The idea of spherical Earth was ﬁrst proposed by Pythagoras, however it was based on aesthetic ground. It a A1): was Aristotle who argued, based on observational facts, ( that the shape of the Earth is spherical. aAncient Greek philosopher and scientist who lived between circa 384 B.C and 322 B.C, and is considered one of the greatest thinkers in history. More Details

The observational facts are: • Stars that are visible in the Northern hemisphere are not for an observer in the Southern hemisphere. Similar argument holds the other way around. For instance, an observer located in the Southern hemisphere can not see the star "Polaris", but once in the Northern hemisphere it becomes visible.

Figure 1: More stars are visible to an observer moving toward the north and looking to the sky toward the North. Similarly,more stars will be visible to her/him when moving toward the south and looking to the sky in the direction toward the south.

• The Earth’s shadow on the moon surface during the different phases of the lunar eclipse always evolves from

5 arched to circular, and this is regardless where you are on the Earth1. Moreover, when the moon was in different positions during an eclipse, the sunlight struck the earth from different angles, and yet the shape of the Earth’s shadow, as it moves across the moon, always looked part of a circle. This means that Earth has a shape that casts a circular shadow in every possible angle. This could only be caused by a spherical shape. • When standing on the shore, and watching a ship moving away, the bottom of ship2 in sea starts to disappear first over the horizon, and then, gradually, the mast.

Around 240 B.C, Eratosthenes of Cyrenea determined the circumference of the Earth by measuring the diﬀer- ence in the angles between the incident light and the (A2): vertical at two diﬀerent locations, and using simple geometry. aModern day Libya. More Details

Eratosthenes [276−194 BC], noticed that at noon during the summer solstice3, the light shone straight down into the bottom of a very deep well4 in the Egyptian city of Syene5, which means that vertical sticks cast no shadow. At the same time in Alexandria, where he lived, it casts a shadow. By measuring the length of the shadow casted by a tower in Alexandria, he determined the angle between the vertical tower and the Sun’s ray6 to be equal to 7.2 degrees. He also knew that the distance between Alexandria and Syene is 5000 Stadia, which was

1 If the Earth had a shape of a disk, then the moon would have to be directly overhead to get a round shadow, which means the sun would be directly below the earth. If that was the case, the entire surface of the Earth would be dark during a lunar eclipse. 2Also called the Hull. 3This is when the Sun reaches its highest position in the sky, which occurs between June 20 and June 22 in the Northern Hemisphere, and between December 20 and December 22 in the Southern Hemisphere. 4 The water was illuminated but not the walls of the well. 5Now known as Aswane. 6Knowing the height of the tower and the length of the shadow one can deduce the angle between the vertical tower and the Sun’s ray.

6 the unit of lengths used at that time7.

Figure 2: Measuring the size of the Earth.

So, Eratosthenes reasoned that if 7◦ decomposes the circumference, C, of the Earth to an arc of length 5000 stadia, then 360◦ C = × 5000 Stadia = 250, 000 Stadia ∈ [40, 000 − 45, 000 km] 7◦ Thus, the radius of the Earth is 40, 000 km R⊕ = ∈ [6370 − 7325 km] 2π 7Unfortunately, there is not an exact conversion of 1 stadia to our unit metric system. However, many historians put it in a range between 0.16 km and 0.18 km.

7 which is close to the true value. This is really remarkable considering the very simple method that he used for this measurement8.

In 1851, the French physicist Leon Foucault, hang up a very long pendulum from a ceiling at the Pantheon (A3): in Paris and let it swing, and showed that its path "appeared " to be slowly shifting, which was a clear evidence that the Earth rotates around its axis. More Details

• To understand how does Foucault’s pendulum demonstrates the rotation of the Earth around its axis, we need to discuss the so called Coriolis force, which was named in honor of the French mathematician Gustave Coriolis9 who proposed the existence of pseudo-force (i.e. an apparent force) that a body experiences when it is moving in a rotating frame. To be more speciﬁc, let us consider a body that is located at a point (r, θ, φ) in spherical coordinates. Its velocity as seen in a reference frame, R, attached at the Earth’s surface (see Fig.3) is ˆ ˆ υ|R = υrrˆ + υθθ + υφφ (1)

Here rˆ, θˆ, and φˆ form an orthonormal basis in the spherical coordinate system. Then, the Coriolis force is given by

Fcoriolis = −2mω × υ|R (2)

where ω = ωzˆ is the angular velocity vector of the Earth. Using the vector identities

zˆ × rˆ = sin θφˆ, zˆ × θˆ = cos θφˆ, zˆ × φˆ = − sin θrˆ − cos θθˆ (3)

8One might wonder what if Eratosthenes assumed that Earth is ﬂat. In this case, for the tower to cast a shadow the Sun has to be be very close to the Earth Surface. With an angle of 7◦, the distance from the Earth to the Sun would be 5000 stadia d = ∈ [6370 − 7325 km] ES tan 7◦ which is of course unacceptable since at that time it was already known (Aristarchus; 270 B.C) that the dES was about 20 times the distance from Earth to the Moon (which we know now that it is smaller by a factor of about 20 than the actual value). 9 Coriolis lived during the 18th and 19th centuries, and published his work on the force caused by the rotation of the Earth in 1835.

8 Figure 3: A schematic of the Foucault pendulum.

Then, Eq (4) reads

h ˆ ˆi Fcoriolis =2 mω υφ sin θ rˆ + υφ cos θ θ − (υr sin θ + υθ cos θ) φ (4)

For example, if a body is moving north, i.e. υr = υφ = 0 and υθ < 0, the Coriolis force acts eastward. • Foucault suspended an iron ball of mass m = 28 kg with a very long wire10 of length l = 67 m from the dome of the Pantheon in Paris. Due to the rotation of the Earth, the plan of oscillations

10Recall that the oscillation period of a pendulum in an inertial frame is ∝ pl/g, and hence with very long wire it takes the pendulum longer time to swing back and forth. Making the pendulum very heavy surpasses the eﬀect of air resistance.

9 will rotate with respect to the surface beneath it. From the view point of an observer in the inertial frame, there are just two forces acting on the bob, the tension T in the string and the force of gravity mg, whereas in the rotating frame of the earth, there are also the centrifugal and Coriolis forces. However, the centrifugal force is a second order in in ω, and so its eﬀect will be neglected. Thus, the equation of motion of the mass m in the earth’s frame (R) is given by

mr¨ = mg + T − 2m ω × r˙ (5) = −mgzˆ + T (zˆ cos θ − xˆ sin θ cos φ − yˆ sin θ sin φ) − 2m ω × r˙

For small oscillations, we can consider θ ' 0, and in this case the component of the tension of the string along the z-axis cancels the weight. Then, in terms of components, Eq (5) reads11

2 2 x¨ − 2ωzy˙ + ω0x = 0, y¨ − 2ωzx˙ + ω0y = 0 (6)

2 where ωz = ω sin ψ, and ω0 = g/L is the natural frequency of simple pendulum. Choosing the initial conditions x(0) = x0, y(0) = 0, x˙(0) = 0, and y˙(0) = 0, the solution to the above system of dif- ferential equations is given by

x0 x(t) = (ωz sin ωzt sin ω0t + ω0 cos ωzt cos ω0t) ' x0 cos ωzt cos ω0t ω0 x0 y(t) = (ωz cos ωzt sin ω0t − ω0 sin ωzt cos ω0t) ' −x0 sin ωzt cos ω0t ω0

The above solution shows that due to the Coriolis force, from the point of view of an observer on the ground it looks as the pendulum’s plane of oscillations precess clockwise at constant angular velocity Ω = −ωz = −ω sin ψ with respect to the Earth. How- ever, from a view point of an observer in a reference frame not attached to the Earth the plane of oscillation stays ﬁxed and it is the Earth which rotates beneath it. • This eﬀect is very small when observed in short time interval, but as the time goes by, the orientation of the oscillation plane changes more and more. Thus, at latitude, ψ, at which the pendulum is placed, the time, T , it takes the oscillation plane of the pendulum to make a full rotation is 11Here x = l sin θ cos φ and y = l sin θ sin φ.

10 2π T⊕ T = = (7) ω| sin ψ| | sin ψ|

where T⊕ is a sidereal period of the Earth, i.e. with respect to ﬁxed stars. For instance, at the Panthéon in Paris, where the latitude is ψ ' 49◦, the period of the oscillation is 0 T |Paris ' 31 hrs 48 (8)

• It is worth mentioning that around 1650, almost two centuries before Foucault, the astronomer Giovanni Battista, suggested that if the Earth is rotating around its axis, then by dropping a ball from a a tower of about 75 meters tall it would land about one centimeter to the side. However, at the time it was not easy to make a measurement with such precision to be able to prove the Earth’s rotation around its axis.

(⊕) (A4): υ = 1675 / ' 0.47 /s Equator km hr km More Details

• The time it takes the Earth to complete one full rotation around its axis is 23 hours and 56 minutes. Note this is the sidereal day, and it is diﬀerent from the 24 hours solar day. The former corresponds to the time that a point on the surface of the Earth return to the same point with respect to to some background stars, whereas the solar day is the time it takes the Sun to return to the same point in the sky. However, since the Earth is orbiting the Sun, it takes 4 minutes for the Sun to return to the same point. Thus, knowing that the Earth’s circumference at the equator is 40075 km (see A 2), the speed with which a point on its surface at this latitude goes in a full circle around rotational axis of the Earth is given by 40075 km υ(⊕) = ' 1675 km/hr (23 + 0.93) hours

• Because the Earth is spinning, we (and every object on it) are subject to the centrifugal force that tries to push us oﬀ into space. However this centrifugal force is just 0.3% the force of gravity, and hence, its eﬀect is negligible.

11 • We do not feel the difference in speed either way because every- thing around us is moving at the same relative speed, even the atmosphere. • If for some reason the Earth suddenly stopped rotating, we and every thing on the Earth will fly off with speed of 1675 km/hr, but not large enough for us be launched into space12. However, the atmosphere would continue moving at the original speed of the Earth’s rotation, producing large wind currents.

(A5): υ(⊕/ ) = 30 km/s More Details

Since it takes approximately 365 days for the Earth to make a full turn around the Sun, and the average Earth-Sun distance is about 1.5 × 108 km (see AXX), then 2π × 1.5 × 108 km υ(⊕/ ) = = 30 km/s (9) 365 × 24 × 3600 s

m m F = G 1 2 g d2 where (A6):

m1,m 2 : masses of the two objects G : Gravitational Constant d: the distance between the two masses More Details

• In 1687, Isaac Newton published the universal law of gravitation and his three laws of motion in "The Mathematical Principles of Natural Philosophy", commonly known as the Principia. 12There are other dramatic eﬀects that make life on Earth unbearable, for example, half the Earth will be permanently facing heat of the Sun, while the other half be exposed to the cold of the space.

12 • In 1749, the French geophysicist Pierre Bouguer tried to measure the density of the Earth using the deflection of a plumb line due to the attraction of a mountain. That would allowed him to determine the strength of the universal gravitational force. Unfor- tunately, he did not succeed as the deviations of pendulum were extremely tiny to be measured. The same method was applied by N. Maskelyne and C. Hutton in 1755 and determined that the average density of the Earth to be between 4.5 − 5 g/cc, which is of the same order of the current measured value. • It was Henri Cavendish who, in 1798, performed an extremely delicate experiment that allowed a precise measurement of the gravitational constant 13. The setup is as follows: A dumbbell consisting of two identical masses m attached to its ends and suspended horizontally from a very thin torsion wire. Two other much heavier balls of identical mass M are placed at the positions shown in the figure(see Fig.4). 14 Theses masses produce attractive forces on the two masses m and cause the dumbbell to twist. The dumbbell will oscillate back and forth before finally settling down at some tiny angle θ away from the initial position due to the restoring torque provided by the wire. With θ being very small, the torque on the dumbbell that arises from the twist takes the form of Hooke’s law

τ = −κθ (10)

Here κ is a constant that depends on the thickness and the material from which the wire is made of. If we denote by d∗ the separation at equilibrium between the centers of the masses in each pair, then gravitational force between each pair of is Mm Fg = G 2 (11) d∗

Hence, the torque on the dumbbell due to the two gravitational 2 forces is (GMm/d∗) L, where L is distance between the small balls of the dumbbell. Applying the equilibrium condition, i.e. the total

13 Actually, Cavendish intended this experiment to be a measurement of the Earth’s density relative to water through the precise knowledge of the gravitational interaction. 14The experiment was originally devised by the geologist John Mitchell around 1780, but he died before he could begin taking measurements. However, Cavendish substantially improved Mitchell’s apparatus.

13 torque equals to zero, we get

2 κθ ∗d G = ∗ (12) MmL

Figure 4: Cavendish’ experimental apparatus taken from [4].

where θ∗ is the angular position of the dumbbell at equilibrium. Note that for θ∗ large enough for it to be measured, the torsion proportionality constant b needs to be as small as possible, and so it is desirable to have l as large as possible15. For that, Cavendish made use of high precision vernier scale and a small telescope to aid the detection of the rotation angle, and measures the displace- 16 ment ∆S∗ of the small ball . So by measuring the

15For a given material, the torsion constant is proportional to the fourth power of its diameter and inversely proportional to its length. 16It is a very challenging task to precisely determine the deflection angle even with the telescope that Cavendish used. In modern experiments, to overcome this difficulty a small mirror is attached to the center of the rod to reflect a light beam on a wall where the angular motion can be more easily observed. If the wall is at a distance D from the mirror, then when the mirror rotates with an angle θ∗, the spot on the wall from the reflected light beam moves a distance x = D tan 2θ∗ ' 2Dθ∗. A nice animation of this type of setup can be viewed at https://www.youtube.com/watch?v=EE9TMwXnx-s.

14 The constant κ can be obtained in terms the moment of inertia of the dumbbell, I, and its oscillation period, T0, as follows. By twisting the dumbbell (in the absence of the large masses) by small angle and let it oscillate around its equilibrium position, the equation of motion reads

Iθ¨ + κθ = 0 (13)

This√ is just the equation of harmonic oscillator with frequency ω = κI, and so by measuring the period of the oscillations one can determine the constant b from the relation 2π 2 κ = I (14) T0 Plugging this expression of b into (12) yields17

2 2 4π Id∗θ∗ G = 2 (15) MmLT0

or, equivalently, in terms of the displacement ∆S∗, we obtain 2 2 8π Id∗∆S∗ G = 2 2 (16) MmL T0

In this experiment, Cavendish used m = 0.73 kg,M = 158 kg, d0 := (d∗ +∆ S) = 22.5 cm,L = 186 cm, and measured the oscillation period T0 = 14.6 min and the displacement of the small ball 18 19 ∆S∗ = 4.1 mm . With these values one ﬁnds

G(Cavendish) =6 .74 × 10−11 N.m2/kg2 (17)

which is very close to present-day measured value.

17If we approximate the moment of inertia by I = m (L/2)2 + m (L/2)2 = ml2/2, then the expression of GN reads

2 2 2π Iθ∗d∗ G = 2 MT0 which is independent on the value of the mass m. 18He determined the displacement with to an accuracy of better than 0.25 mm. 19We should point out that Cavendish did not explicitly derive the expression of G, and in fact it is not even mentioned in his paper. Instead he computed the density of the Earth which is related to the gravitational constant (see A7). The ﬁrst reference to the universal gravitational constant was in 1873 by Alfred Cornu and Baptistin Baile (which was denoted by the letter f), about 75 years after Cavendish’s work, and the modern notation with the letter G was introduced in 1894 Charles . V. Boys (see ref. [5]).

15 24 (A7): M⊕ = 5.97 × 10 kg More Details

• Consider the Earth to be a perfect sphere of radius R⊕, and mass M⊕. , Let a test particle of mass m, located at some height much 20 smaller than R⊕, free fall toward the ground. Then according to the universal law of gravitation (see A6) and Newton’s second law of motion we have

GM⊕m 2 = mg(r) (18) (R⊕ + r) where r is the distance from the ground to the mass m at some instant t, g(r) is its acceleration due to gravity, and G is the gravitational constant. Now after canceling the common factor m from both sides of the equation, and neglecting the contribution of r as compared to R⊕, we obtain 2 R⊕g⊕ M⊕ = (19) G

Here g⊕ = g(r = 0) is the acceleration of gravity at the surface of 21 the Earth . Thus, using R⊕ ' 6370 km (see A2), G = 6.67 × −11 2 −2 2 10 N.m .kg (see A6), and g⊕ ' 9.8 m/s , we obtain 6 2 2 6.37 × 10 m × 9.8 m/s 24 M⊕ = ' 6 × 10 kg 6.67 × 10−11 N.m2.kg−2 • Since we now know the mass of the Earth and its radius, we can can its average density from the formula

M⊕ 3 ρ⊕ = 3 ' 5.5 g/cm ' 5.5 × ρH20 (20) (4π/3) R⊕

Since most of the rocks have density of about 3 g/cm3, we conclude that the Earth’s interior must be more dense than typical crustal rocks in order for the whole thing to average to about 5.5 g/cm3. 20This means that we will neglect the air resistance. 21 The value of g⊕ can be easily measured by either recording the positions, x(t), that a test particle moves at diﬀerent instances, t, during the free fall, or, by measuring the period of oscillations of a simple pendulum of a given length since its period is a function of the acceleration of gravity at the Earth’s surface.

16 • Another method: If we know the distance from the Earth to the Moon, we determine Earth’s mass using Kepler’s third law and the measured value of the gravitational constant. In the same way, one can infer the mass of a stars that form a orbiting binary star-system with another one (see A12 for details)

( ) GM (A8): surface g = R2 More Details

The above expression for the gravitational acceleration can be obtained using Newton’s universal law of gravitation and Newton’s second law of motion, which for a test mass m located at (or near) the surface of the astronomical object22 of radius R, reads GMm − rˆ = ma (21) R2 where rˆ is a unit vector directed from the center of the celestial body to the test mass. Hence, the acceleration of gravity at the Earth’s surface is given by GM a = − rˆ ≡ g(surface) rˆ (22) R2

with g(surface) as given above. In table1, we give the values of the acceleration of gravity for some of the celestial bodies.

(A9): (⊕) υesc = 11.2 km/s More Details

• The escape velocity is the minimum speed υe that an object needs to have to break free from the surface of a large body such 22For simplicity we assume that the object is spehrical.

17 Object g [m/s] g/g⊕ Sun 273.4 28 Mercury 3.7 0.38 Venus 8.9 0.9 Earth 9.8 1 Mars 3.7 0.38 Jupiter 24.8 2.4 Saturn 9 0.9 Uranus 8.7 0.9 Neptune 11 1.1 Moon 1.6 1/6

Table 1: The values of escape acceleration of gravity of some celestial objects.

as a moon, planet or star. Suppose that an object of mass m leaves the surface of stationary object of mass M >> m, with initial speed υ0. If, for simplicity, we consider the object to be a sphere of radius R, and assume it has no atmosphere, then the energy conservation E(r = R) = E(r), ∀r > R, implies that 1 Mm 1 mυ2 − G = mυ2 (23) 2 0 R 2 ∞

where υ∞ is the speed of the particle at r = ∞, i.e. at the point where the mass m has completely escaped the eﬀect of the gravity due to the mass M. Since the right hand side of the above equation can not be negative, the escape velocity correspond to the value of υ0 for which υ∞ = 0, and so we get r GN M υesc = 2 (24) R which is independent of the mass of the escaping object23. Apply- ing the above formula for the escape velocity to the Earth (using 24 M⊕ ' 6 × 10 kg [see A6], and R⊕ ' 6370 km [see A2]), we obtain

(⊕) υesc =6 .9 m/s = 11.2 km/s (25) In table3, we give the values of the escape velocity for diﬀerent planets in the solar system and some other astronomical objects24.

23It is important to note that the above calculation neglects the eﬀect of air resistance. 24Titan is the Saturn’s largest moon, and Europa is one of Jupiter’s moon.

18 Object M/M⊕ R/R⊕ υe (km/s) Sun 333 × 103 109 617.5 Mercury 5 × 10−2 0.4 4.3 Venus 0.8 0.9 10.3 Earth 1 1 11.2 Mars 0.5 0.1 5 Jupiter 318 11 59.6 Saturn 95 9 35.6 Uranus 14.5 4 21.3 Neptune 17 4 23.8 Moon 10−2 0.3 2.4 Titan 2 × 10−3 0.4 2.6 Europa 8 × 10−3 0.3 2

Table 2: The values of escape velocity of some of the astronomical objects.

(A10): (min) υLaunch-Sat = 7.8 km/s

More Details

• Orbital velocity of a satellite In the absence of the atmosphere, a satellite (or spacecraft) in stable circular orbit around the Earth requires a speed υorb given by balance between the centripetal and the gravitational force, i.e.

2 s M⊕m mυorb GM⊕ 7.8 km/s G 2 = −→ υorb = = 1/2 (26) r r R⊕ + h (1 + h/R⊕)

Here m is the mass of the satellite and r is the radius of the orbit. For a satellite at altitude h, above the ground. In table ... we give ...

• Derivation of υLaunch-Sat Assuming no drag force in space, which is in fact an excellent

19 Orbit Orbital Speed (km/hr) Orbital Period Sun 333 × 103 109 Mercury 5 × 10−2 0.4 Venus 0.8 0.9 Earth 1 1 Mars 0.5 0.1 Jupiter 318 ...

Table 3: The values of escape velocity of some of the astronomical objects.

approximation, The conservation of total energy of the Earth- Satellite system reads

1 2 GM⊕m 1 2 GM⊕m mυlaunch − = mυorb − (27) 2 R⊕ 2 r

or, equivalently, 2 GM⊕ 1 GM⊕ υlaunch = 2 1 − ≥ (28) R⊕ 2 (1 + h/R⊕) R⊕

Hence, s (min) GM⊕ υLaunch-Sat = =7 .8 km/s (29) R⊕

(A11): τ⊕ ' 4.5 billion years More Details

• Radiometric dating: The dating of a material, such as a rock or a fossil, is based on the method of radiometric dating, which is the measurement of the radioactive decay25 of an unstable isotope26 (parent) into more stable element. Some of the radioactive elements that are widely used in radiometric dating are listed in table7.

25Radioactivity was discovered in 1896 by the French physicist Henri Becquerel. 26Isotopes are atoms that their nuclei contain the same number of protons but have different number of neutrons. This means they have different mass but their chemistry is unchanged since the outer shell configuration will be the same.

20 • The equation of a radioactive decay process: Consider a system of radioactive nucleus, P, with NP (0) their number at the instant t = 0. We assume that each particle decays into a stable daughter nucleus, D, with equal probability per unit time, λ, called the decay constant for the decay process. Then, during the time interval [t, t + dt], the decrease in the number of the P particles, NP , and the growth in the number of the D particle, ND, are given by

dNP = −λNP dt, dND = λNDdt (30)

Note that d (NP + ND) = 0, which implies that total number of particles (i.e. parents and daughters) is conserved

NP + ND = NP (0) + ND(0) (31) Now the integration of (30), yields

−λt NP (t) = NP (0) e (32) and −λt ND(t) = ND(0) + NP (0) 1 − e −λt = ND(0) + NP (t) e − 1 (33) This exponential distribution of the parent particles can be inter- preted in term of a decay probability, dP(t), for a single particle to disintegrate between the instant t and t + dt, given by dP(t) = λe−λtdt (34) Thus, the probability for particle to decay within a time t is27 Z t P(t) = dP(t) dt = 1 − e−λt (35) 0 Using this probability distribution, the mean lifetime of the radioactive particle P can be calculated as28 Z τ = t dP(t) = λ−1 (36)

27This is also known as Poisson distribution, and it is normalized to unity. 28Equivalently, it can be obtained as Z ∞ Z ∞ Z ∞ Z ∞ 1 1 1 −λt −1 τ = t dNP = t dNP = t λNP dt = λ t e dt = λ NP (0) t=0 NP (0) t=0 NP (0) t=0 t=0

21 29 Another more related quantity is the half-life, τ1/2, deﬁned as

τ1/2 ' 0.7 τ (37)

Now, suppose we know NP (0) or its ratio compared to some stable element in the sample30, P 0, then from Eq (32) the age is given simply by the expression NP (t = 0) (NP /NP 0 )(t = 0) t = τ ln = τ ln (38) NP (t) (NP /NP 0 )(t)

Now let us generalize the above results to the case when the parent nucleus decay to many daughter nuclei, D1,D2, ..., Dn, via diﬀer- ent processes, with decay constants λ1, λ2, ..., λn i.e.

P → D1 + ....[λ1], P → D2 + ....[λ2], . . P → Dn + ....[λn],

Then,

dNP = −λNP dt, (39)

dNDi = λiNP dt, i = 1, 2, ..., n (40) n ! X d NP + NDi = 0 (41) i=1

Pn where λ = i=1 λi is the total decay constant for these processes. By integrating the above diﬀerential equations we obtain

−λt NP (t)= NP (0) e (42) λi −λt ND (t)= ND (0) + NP (0) 1 − e , i = 1, 2, ..., n (43) i i λ n n X X NP + NDi = NP (0) + NDi (0) (44) i=1 i=1 29This relation can be shown as follows:

N0 ln 2 N(τ ) = N e−λτ1/2 := → τ = ' 0.7 λ−1 1/2 0 2 1/2 λ

30For example, one of the stable particle of the decay product.

22 • Carbon-14 dating [≤ 70, 000 years31]: The interaction of high energy cosmic rays (mostly protons, and helium nuclei) with the atmosphere produce showers of many types of subatomic particles, including energetic neutrons32. These neutrons collide with the nitrogen atoms in the atmosphere producing a proton and an atom of carbon-14, i.e.33

n +14 N →14 C + p

14 These C atoms combines with O2 molecules, forming carbon dioxide, which spreads around at low altitudes, where it is used by the plants, and all living things.

14 12 At any particular time, the ratio of C/ C, denoted by R0, is approximately constant in all living organisms and given by

−12 R0 ' 10

However, once an organism dies the carbon-14 starts to decay to nitrogen via the β-process:

14C →14 N + e− +ν ¯

where ν¯ is the anti-neutrino, a neutral spin 1/2 particle with al- 14C(t) most vanishing mass. Thus, by measuring the ratio R(t) = 12C , then compare it to the assumed initial value R0 in a living organism, scientists can determine the time elapsed since it died. Using 14 Eq (38) and the value of τ1/2 of C from table7, we obtain R [t] = 8022 ln 0 years 14C/12C R(t)

• Potassium-Argon dating:[≥ 104 years]: – Potassium is present in Earth’s crust34 with a concentration of about 1.5%, and its naturally occurring radioisotopes, 40K make up 0.12% of natural potassium, and hence it can be used 31Beyond an age of 50, 000 years the amount of 14C left in the sample will be so tiny that it can not be used for a reliable dating. 32This happens at high altitudes around 9 − 15 km in the atmosphere. 33For thermal neutrons, σ(n +14 N →14 C + p) ' 1.8 × 10−24 cm2 ≡ 1.8 barn. 34As well in oceans and all organic materials.

23 Radioactive Element Symbol Half-life Radon-222 222 Rn 3.8 days Carbon-14 14 C 5, 730 years Potassium-40 40 K 1.25 billion years Uranium-235 235 U 0.7 billion years Uranium-238 238 U 4.47 billion years Thorium-232 232 Th 14.1 billion years Rubidium-87 87 Rb 47 billion years

Table 4: Common radioactive elements (found in rocks, water, and air) .

in dating rocks. About 11% of 40K decays to 40Ar and 89% to 40Ca via the following processes:

40 40 − −10 K → Ca + e +ν ¯;[β process; 89%; λβ = 4.96 × 10 years] 40 − 40 −10 K + e → Ar + ν;[electron capture; 11%; λe = 0.58 × 10 years] – However, because 40Ca can be present both as radiogenic and non-radiogenic35, and so the ﬁrst decay process can not be used for dating rocks. Instead, geologists make use of the decay to argon by electron capture and measure the 40Ar/40K ratio in materials that trap argon. Furthermore, since argon is a noble gas it can escape easily from the extremely hot molten magma, and so it is reasonable to assume that there was no argon initially when the rocks formed. Thus, the argon in a rock today is that which was produced by 40K since the rock has cooled down. – Now, applying Eq (43) to the decay processes above, we obtain

λe λt NAr(t) = NK (t) e − 1 (45) λ −10 where λ = λe + λβ = 5.54 × 10 years is the total decay constant of 40K to argon and calcium. Solving for t, we obtain 1 NAr(t) λ [t]Ar/K = ln + 1 λ NK (t) λe N (t) =1 .8 ln 9.55 Ar + 1 billion years NK (t) 35It means can be produced by another radioactive decay process. So it would be impossible to distinguish what proportion of 40Ca that was produced by potassium.

24 – Uranium-lead dating:[≥ 107 years] The naturally occurring uranium consists primarily of two radioisotopes: 235U and 238U, which produce 207 Pb and 206Pb, respectively, with the emission of α particles, via long decay chains. Another isotope of lead is 204Pb which is non- radiogenic, and so using Eq (33) one can get two independent dates from the uranium-lead system: 206 206 238 Pb Pb U λ238t (t) − (0) = (t) e − 1 (46) 204Pb 204Pb 204Pb 207 207 235 Pb Pb U λ235t (t) − (0) = (t) e − 1 (47) 204Pb 204Pb 204Pb

Both these equations is has the form y = mx + y0, where y is the ratio of the amount of lead produced from radioactive decay to 204Pb, and they are known as the isochron equations.Taking the ratio of the above equations, we obtain

h 207Pb i h 207Pb i 204Pb − 204Pb 235 λ235t Now 0 U e − 1 207 207 = Pb Pb 238U eλ238t − 1 204Pb Now − 204Pb 0 1 eλ235t − 1 = 137.8 eλ238t − 1 where we used the present day ratio 235U/238U is 137.8, as it has been measured in terrestrial, lunar, and meteorite sam- ples. The above equation is known as Pb-Pb isochrone equation. For a rock or meteorite, this equation represents a 207 204 straight line with [ Pb/ Pb]Now is, say on the y-axis, and 206 204 [ Pb/ Pb]Now on the x-axis, and the left-hand side of the equation is the its slope. So if the initial ratio of P b are known or inferred from minerals that essentially had no lead initially present in them36, then one can determine their age.

– The oldest known rocks on Earth:

They were dated at 4.28 billion years, making them 250 million years more ancient than any previously discovered rocks. They were discovered in ...... Quebec

36Such as zircon, apatite, or monazite.

25 1 GMm (A12): T = critical 54 kBR More Details

• The most probable speed of a molecule in a gas Whether a planet has an atmosphere depends on the ratio of the molecular speed in the gas and the escape velocity from the planet. So, we will ﬁrst calculate the mean speed of molecules in a gas. For that, we recall that in thermal equilibrium, the velocity distribution of particles of mass m is given by 3/2 m 2 f(υ)d3υ = N e−mυ /2kB T υ2dυ sin θdθdφ (48) 2πkBT

fIntegrating over the angles θ and φ, we obtain the speed distribution function r 3/2 2 m 2 f(υ)dυ = N e−mυ /2kB T dυ (49) π kBT

The most probable speed, υ0, is given by the speed at which the derivative of f(υ) vanishes. We ﬁnd37

r 2 2kBT 1 1 υ υ0 = −→ f(υ) = 3/2 3 exp − 2 (50) m π υ0 υ0

The most probable speed υ0 given above can also be expressed in terms of the molar mass M of the gas molecules as r 2RT υ0 = (51) M where R = 8.314 J/(gram-mol.K) is called the universal gas constant. Thus, at a given temperature, the more massive the molecules of a gas, the lower is their mean speed. This is why it 37Note that the most probable speed is not the average speed. The later is given by

Z ∞ r 8kBT υ = υ f(υ)dυ = = 2υ0 0 m

26 is more easy for hydrogen and helium to escape the Earth rather than does carbon dioxide, when the temperature of the gas is large enough so that υ0 > υesc. With the Earth’s average temperature of 15◦ (i,e, approximately 280 K), the oxygen and nitrogen gases, have average speeds of about 380 m/s and 406 m/s, respectively. These are much smaller than the escape velocity from the Earth (see A(09)), which explains why these gases are still present in the atmosphere.

• Naive estimate A particle can escape the surface of a planet if its kinetic energy exceeds the gravitational energy of the planet. For a gas molecules, this is equivalent to the condition r 2GM υrms ≥ υesc = (52) R which in terms of temperature, yields 2GMm T ≥ (53) 3kBR Hence, for T ≤ 1000 K, a very tiny fraction of the molecules in the gas have such high speed, and one might think this will have no eﬀect on particles loss from the atmosphere.

In reality, even if υ0 < υesc, there will always be particles in the high-speed tail of the distribution in (49) that have speed larger than the escape velocity which leak into space. Although the fraction of such particles is tiny, the remaining particles will redis- tribute them selves so that the high-speed tail get reproduced so that their speeds follow a Maxwell-Boltzmann distribution, and again the particles in the high-speed tail will escape into space. Therefore, a more relevant quantity to estimate is the time scale for the gas molecules to escape the planet’s atmosphere, as will be discussed below. • Jeans thermal escape ﬂux A particles can escape the planet’s atmosphere if the magnitude of the vertical component of its velocity, υz = υ cos θ, exceeds the escape velocity from the planet, and does not collide with other particles in its motion upward. This region where it is unlikely for

27 a gas particle to collide with another one is known as exosphere. 38 If we denote by nexo the density of the gas particle in the exobase , then the escape rate is Z ∞ 2 Φesc = nexo υz f(υ)υ sin θdθ dυ dφ (54) υesc where 2 υesc Ye = (55) υ0 After integrating (54) yields

nexoυ0 −Ye Φesc = √ (1 + Ye) e (56) 2 π The above expression is called Jeans thermal escape ﬂux. Note that the smaller is Ye, the more signiﬁcant is the escape rate.

As an example, consider the gas molecules N2, which is the most common gas in the Earth’s atmosphere. With the exobase at altitude of about 500 km at which T ∼ 1000 K, we have

υ0[N2] = 0.78 km/s −→ Ye[N2] ' 196 (57) Thus, the escape rate of the nitrogen molecules is −56 nexo −2 −1 Φesc[N2] ' 8 × 10 m s (58) 2.5 × 1025 m−3 25 −3 We see that even if we the value nexo = 2.5 × 10 m , which is the number density of the molecular nitrogen gas at the sea level, the escape rate is vanishingly small. To estimate the number of N2 particles escaping from the atmosphere per unit time, we multiply 2 Φesc by the surface area of the exobase, 4π (R⊕ + H) , where R⊕ is the Earth’s radius39. We get dN [N2] nexo Molecule ∼ −10−41 (59) dt 2.5 × 1025 m−3 billion year which shows that the Earth will hold its nitrogen (and oxygen) gas much longer than the age of the universe.

38This is the altitude above which it is unlikely for the particles to collide with each other. 39In reality, at altitudes above 200 km the number density of nitrogen drops signiﬁcantly compared at sea level.

28 • The time scale for particles to escape entirely Let h be thickness of the region in the atmosphere where the probability for particles to collide with each other is extremely small. Then, a rough estimate of the time it takes for such particles to escape entirely from the atmosphere into space is hn τesc ∼ (60) Φesc

where n is the average number density of the gas above the exobase where particles collision is very rare. Substituting the expression of τesc in the above equation gives √ n Ye Ye h 10 km/s τesc ∼ 3 e minute (61) nexo 1 + Ye 500 km υesc

Let us estimate for how long an Earth-like planet can hold an atmosphere of hydrogen gas. We will assume that initially that the atmosphere contained a substantial amount of hydrogen, and the temperature is about 1000 K at the base of the exosphere40. Using Eq (51), we obtain41

H υ0 ' 4 km/s (62)

Now, if we assume that hydrogen is abundant at altitudes between 200 km and 1000 km, i.e. h ' 800 km, we obtain

(H) τesc |⊕ ∼ 12 days (63)

However, from observations, we know that hydrogen is one of the abundant gases in the exosphere, which seems to be in contradic- (H) tion with our estimate above. This is because in obtaining τesc |⊕ we have assumed that there were no source for hydrogen produc- tion in the atmosphere, which is not correct, whereas at latitudes above 500 km, hydrogen is constantly produced from the collision of high-energy photons from the Sun with water vapor and methane molecules.

40Jeans wrongly assumed that the Earth’s upper atmosphere was very cold and con- cluded that no hydrogen leaked into space. 41The molecular mass of hydrogen is M = 1 g/mole.

29 Planet Components of the Atmosphere Comments 42 Mercury 42% O2, 29% Na, 22% H Extremely thin atmosphere ; Surface pressure ' 0 Venus 96% CO2 , 3.5% N2 Runaway greenhouse eﬀect from CO2 Earth 78% N2 , 21% O2 Thin atmosphere −3 Mars 95% CO2 , 2.7% N2, 1.6% Ar Surface pressure ' 7 × 10 atm Jupiter 90% H2 , ∼ 10% He Height from its base > 5000km Saturn 96% H2 , ∼ 3% He Magnetic ﬁeld smaller than Jupiter Uranus 83% H2 , 15% He2, 2.5% CH4 Blue-Green color of atmosphere fromCH 4 Neptune 80% H2 , 19% He2, ∼ 1% CH4 Bright blue atmosphere fromCH 4

Table 5: The gases in the atmospheres of the planets in our solar system.

• A rule of thumb for retention of a planetary atmosphere Since planets formed about the same time as the Sun, we will require that a planet retains it s atmosphere if the escape time scale, τ (esc), is larger than or equal to the life time of the Sun, i.e. √ Ye h 10 km/s 3 eYe min ≥ 1010 years (64) 1 + Ye 500 km υesc which yields (critical) h υesc Ye ≥ Y = 35.8 − ln + ln (65) e 500 km 10 km/s Thus, as a rule of thumb, a gas in an atmosphere is completely bound to the planet if Ye ≥ 36, or, equivalently, if the escape speed of the planet is at least about six times larger than the most probable speed of the molecules in the atmosphere. So, in terms of temperature translates into the condition 1 GMm T ≤ Tcritical = (66) 54 kBR

• Non-thermal ways for atmospheric loss It is important to point out that the Jeans escape gives a lower limit on the atmospheric loss and that there are other ways by which the atmosphere can be lost, such as impact erosion, and the solar wind43. These processes supply the particles in the higher layer of the atmosphere with energy that allows them to escape 43These are particles that are ejected from the Sun at about 400 km/s and sometimes faster during solar storms.

30 from the atmosphere. Here I will not discuss the details of these mechanisms, and instead I refer the interested reader to the references [8] and [6].

a a. It is composed of three atoms of oxygen (O3) .

b (A13): b. 10% in the troposphere, and 90% in the stratosphere . aIt was discovered in 1840 in Munich by the German chemist C. .F. Schonbein. b In the stratosphere, O2 and O3 absorb UV radiation. More Details

• Ozone formation and destruction The formation of ozone molecules in the stratosphere is produced in a two step process: In the ﬁrst step, UV photons from the Sun44 45 strike the O2 molecules in the stratosphere, dissociating some of them into individual oxygen atoms; b) (Almost all) the produced oxygen atoms, O, combine with oxygen molecules to produce O3 molecules. The ozone generation also get destructed by the UV photons as well via its collision with oxygen atoms, which keeps the concentration of the ozone in balance. This cycle of the formation and destruction of the ozone molecules is known as Chapman mechanism46, represented by the following chemical reactions:

Reaction 1 : O2 + γUV[λUV < 240 nm] → 2 O (67)

Reaction 2 : O + O2 + M → O3 + M (68)

Reaction 3 : O3 + γUV[λUV < 320 nm] → O2 + O (69)

Reaction 4 : O + O3 → 2O2 (70) Here M is a non-reactive particle that can take up some of the 47 energy released in the reaction , γUV denotes a UV photon, and λUV is its wavelength. So the net eﬀect of the above process is an equilibrium between atomic oxygen, ozone, and diatomic oxygen, which can represented by the reaction

O + O3 → 2 O2 (71) 44UV radiation constitute about of 5% of the Sun’s radiation. 45which make up about 21% of the atmosphere. 46In honor of Sydney Chapman who proposed it in 1930. 47 M is usually N2 or O2.

31 As a result of this balance between the amount of ozone produced and destroyed, its total concentration remains relatively constant at a given latitude and altitude in the stratosphere. Ozone concen- trations in the stratosphere are greatest at altitude between 15 km and 25 km, forming ozone layer48 and most of it is formed over the equator where the amount of radiation from the Sun is higher.

• Impact of ozone depletion Although the UV-C radiation, which is the most dangerous among the UV radiations, are completely absorbed by the ozone layer in the stratosphere, a depletion in the concentration of O3 in the ozone layer will have serious eﬀect on life on Earth. For instance, more of the UV-B radiation49 (280 - 315 nm ) will reach the Earth’s surface, which, for us human, can cause skin cancer, cataracts50, and suppress the immune system. One of the major causes of ozone depletion in the stratosphere is chloroﬂuorocarbons, CCl2F2, usually abbreviated as CFCs, which is used in many applications, including refrigeration and air conditioning. The reason for the CFC being a source of ozone- depletion is the presence of chlorine51 atoms that catalyzes the breakdown of ozone to diatomic oxygen via the reaction

CCl2F2 + γUV → C ClF2 + +Cl

CCl3F + γUV → C Cl2F + +Cl

O3 + Cl → ClO + O2

ClO + O → Cl + O2

A CFC molecule does not interact with anything in the lower atmosphere and eventually reaches the upper atmosphere where the Sun’s radiation is intense enough to break it apart producing chlorine. Moreover, since the CFC has a lifetime of about 20 to

48The ozone layer was discovered in 1913 by the French Physicists C. Fabry andH. Buisson. 49The UV-A radiation have relatively long wavelength (320 -400 nm) and accounts for approximately 95% of the UV radiation reaching the Earth’s surface (the other 5% is UV-B radiation). Excessive exposure to UV-A will hurt the central vision of the eye, cause premature aging, and can lead to skin cancer. 50It is the clouding the thickening of the eye natural lens. This is the leading cause of vision loss. 51Actually, the radicals nitrogen oxide (NO), hydrogen oxide (HO), and bromine oxide (BrO), are also eﬀective ozone-killers.

32 Figure 5: The three types of UV radiations and ozone layer.

100 years, from the chain reaction above, at very low temperature, as it is the case over the Antarctic, a single chlorine atom can destroy hundred thousands of ozone molecules before it form one of other chlorine compounds and exit the stratosphere. In fact, in 1985, British scientists discovered that the ozone layer over the Antarctic was thinning. This ozone decline over the south pole was called ozone hole52. Fortunately, in 1987 many industrial nations came together and agreed to phase out the CFCs and signed the Montreal protocol, and over the next decades most of the world’s nations have signed up.

52The ﬁrst ozone hole over the Arctic was observed in 2011 after an unusually cold Arctic winter.

33 • Gamma rays −−−−−−−−−→ [< 10 pm] • X-ray −−−−−−−−−−−−−−→ [10 pm − 10 nm] • Ultraviolet radiation −→ [10 − 400 nm] (A14): • Visible light−−−−−−−−−→ [400 − 700 nm] • Infrared radiation −−−→ [700 nm − 1 mm] • Microwave radiation −→ [1 mm − 10 cm] • Microwave radiation −→ [10 cm − km] More Details

See ﬁgure.5 and table 5.

Figure 6: The electromagnetic spectrum.

34 Type of radiation Temperature [K] Typical sources Gamma rays > 108 matter falling into black body X-rays 106 − 108 gas in cluster of galaxies Ultraviolet 104 − 106 very hot stars Visible 103 − 104 stars; hot planets Infrared 10 − 103 very cools; planets; cool cloud of dust Radio waves < 10 cosmic microwave background ; electron in moving in magnetic ﬁeld

Table 6: Examples of astronomical sources emitting in each range of the EM spectrum [14].

A15): All visible light, most of the Radio waves, and ( some infrared light. More Details

• For a summary see Figure.6.

• Optical astronomy – Refractor telescope (uses lens) [Galileo (1609)] – Reflector telescope (uses mirror) [Newton (1668)] – Reflector telescope is more advantageous than refractor because mirror is free from chromatic aberration, which affects a lens. Also, making a large lens of high quality s much more difficult than making a large mirror.

35 Figure 7: Atmospheric absorptions at diﬀerent wavelengths[19].

• A parallax is half the angular displacement of nearby star as seen from the Earth at one time of year, and six months later. • A parsec, denoted bypc, is the distance to an object (A16): that has a parallax angle of 1 arsecond. • A light year is the distance that light travels in one year, and it is given by 1pc = 3.26 ly More Details

36 Figure 8: Parallax.

• Parsec (pc), astronomical unit (AU), and light years (ly):. Let D be distance to an astronomical object, and p be the the parallax angle, then average distance between the Earth and the Sun D = p 1 AU arsec 1 = = 1 AU p[in radian] p arsec 60 × 60 × 180 arcsec = × 1 AU π p

Thus, arcsec D = 206, 265 AU (72) p

Now by deﬁnition 1 pc is the distance that corresponds to a parallax angle p = 1 arcsec, we obtain

1pc = 206, 265 AU (73)

37 To express distance in terms of light years, we use the fact that 1 ly = speed of light × one year time interval = 2.99 × 108m/s × 3.15 × 107 s ' 9.42 × 1015 m and so, 206, 265 AU arsec 3.07 × 1016 m arsec D = ly = ly 9.42 × 1015 m p 9.42 × 1015 m p arsec =3 .26 ly p Setting p = 1arcsec in the above expression results in a distance of 1 pc, which implies that 1pc =3 .265 ly (74) • In 1838, the German astronomer Friedrich Bessel, was the ﬁrst to successfully measure the parallax of a star, and that was for the star 61 Cygni, and obtained p(61Cygni) = 0.286 arcsec (75) corresponding to a distance of 11.4 light years from the Earth. • Measuring star distances ⇒ Earth Based Telescopes53 that use adaptive optics to re- move the distorting eﬀect due to the atmosphere, one can measure a parallax angle as small as 0.01 arcsec, which corresponds to a distance of 100 pc.

⇒ Space Based Telescopes can measure distance to objects up to about few × 100 pc away. The Hipparcos satellite54 was able to measure the parallax of more than 100, 000 stars with a precision of 0.001 arcsecond, corresponding to a distance of a kpc, which is about 1/30 the size of our galaxy (see Q9).

In table ... we show the parallax angles and the corresponding distances to the ﬁve nearest stars to our solar system.

53With ordinary telescope, one can observe stars at distances about 30 pc ' 100 ly away from us. 54It was launched by the European Space Agency (ESA) in 1989 and operated until 1993.

38 Star Name Parallax [arcsec] Distance [pc] <—–> [ly] Comments Sun xxx 1.3 <—–> 4.22 Alpha-Centauri xxx 1.3 <—–> 4.22 Uranium-235 235 U 0.7 billion years Uranium-238 238 U 4.47 billion years Thorium-232 232 Th 14.1 billion years Rubidium-87 87 Rb 47 billion years

Table 7: The eleven nearest stars to us.

• Apparent magnitude m = −2.5 log F + constant. (A17): • Absolute magnitude is the apparent magnitude at distance d = 10 pc More Details

• The Greek astronomer Hipparchus (2nd century BC) divided stars into 6 classes or magnitudes, where the 1st magnitude is the brightest, and the 6th is the faintest, based on the fact that human eye sensitivity is not linear55. However this classification was vague until around 1850 when N. Pogson made it more quan- titative and defined the faintest stars visible to the naked eye are 100 times fainter than the brightest stars. Thus, two successive classes should differ in apparent brightness by a factor56

1/5 (100) = 102/5 = 100.4 (76)

We will use the letter m to denote the magnitude of a stellar object. Suppose two stars have apparent brightness F1 and F2, and their magnitude classes are m1 and m2, then

F2 2 (m −m ) = 10 5 1 2 (77) F1

Note it is by convention (Hipparchus) fainter objects have higher value of magnitude than brighter ones. Applying the logarithm 55Similar to the human ear, the eyes sensitivity is logarithmic. 56The largest ground-based telescopes, such as the Keck telescope, can detect stars and galaxies at (apparent) magnitudes between 24 and 26, using adaptive optic techniques.

39 on both sides of the equation yields F2 m2 − m1 = −2.5 log (78) F2 Thus, the magnitude and the ﬂux of a stellar body are related by

m = −2.5 log F + C (79)

where C is a constant ..... • If the apparent brightness is based on the radiation in all wave lengths, then the magnitude deﬁned from it is called the bolometric magnitude, and we write Z ∞ mbol = −2.5 log Fλ dλ + Cbol (80) 0

where Fλ is the apparent brightness at wave length λ. • Unfortunately it is not easy at all to determine the bolometric magnitude since the measuring devices of the intensity of light are have diﬀerent sensitivity to diﬀerent wavelengths. A more convenient system is called UBV, which stands for ultraviolet- blue-visual wavebands, which was introduced by H. Johnson and W. Morgan in the mid-20th century.

(A18): The average distance is approximately 385, 000 km More Details

• Simple way to get a rough estimate ...... • The most accurate measurement of the Earth-Moon distance uses the Lunar Laser Ranging experiment, where lasers on Earth are aimed at retroreﬂectors planted when the astronauts visited the moon more than 40 years ago. Because of the precise measured value of the speed of light, the distance to the moon is measured down to a milimeter accuracy. As a result astronomers found that the moon is drifting from us at a rate of 3.8 cm per year. • the moon travels around the Earth in an elliptical orbit. Its closest point, known as the perigee, to the Earth is at 363, 000 km away, and the most distance point, known as the apogee is at at 363, 000 km away.

40 22 (A19): 7.3 × 10 kg ' 1% M⊕ More Details We can get a rough estimate of the mass of the moon as follows. Con- sidering the Moon as a sphere of radius RM and uniform mass density ρM , its mass reads

3 4π 3 RM ρM MMoon = ρMRM = M⊕ (81) 3 R⊕ ρ⊕

Using the fact that the size of the Moon is approximately 27% that 57 of the Earth , i.e. RM ' 1730 km and assuming that the density of the moon-rocks is about the density of a typical Earth’s crust, i.e. 3 ρM ∼ 3 g/cm , we ﬁnd

22 MMoon ∼ 1 % M⊕ ' 7 × 10 kg (82)

• [1672; Giovanni Cassinia, and Jean Richter]: By measuring the distance from Earth to Mars when in opposition with the use of the parallax method, and then applying Kepler’s third law. A20): ( • [1771; Jerome Lalande]: By measuring the transit of Venusb a An Italian-French astronomer. bIt was the famous English astronomer Edmund Halley (with a comet bearing his name) who in 1716 pointed out that transits of Venus across the Sun (would take about 7 hours) would occur on June 1761 and 1769, and if these are missed, the next one will be more than a century after. More Details

57The estimate of the size of the Moon was known since the ancient Greek (more than 2000 years ago) from the timing the lunar eclipse.

41 • The ﬁrst attempt to measure the distance from Earth to the Sun was carried out by Aristarchus in 3rd century B.C, which h e determined to be approximately 20 times the distance between the Earth and the Moon. We know that this estimate is about a factor of 20 smaller than actual value. • Cassini and with his colleague Jean Richer, made a simultaneous measurement of the position of Mars, from Paris and Cayenne, respectively, and determined that the parallax angle of Mars with respect to Paris-Cayenne baseline to be 24 × π p[Mars] = 24arcsec = = 1.16 × 10−4 radian 180 × 3600 Given that the distance between Paris and Cayenne is 6437 km, he deduced that the distance between Earth and Mars is distance between Paris and Cyenne DEarth→Mars = p[Mars] 180 × 3600 = 6437 × ' 55 million km (83) 24 × 2π

• What is the transit of Venus? The idea of using the transit of Venus to measure the distance between Earth and VenusJ was proposed in 1663 by James Gregory......

In 1771, Lalande used the combined the data of the transit that occurred in 1761 and 1769 to derive a distance of 153 million kilometers, which was just 2% higher than the actual value. • The most precise measurement of the Earth-Sun distance is by reflecting radar pulses off the surface of Venus. This is done as follows: we send a radar pulse at Venus when it appeared at first or third-quarter phase, so that the line of sight is perpendicular to the Venus-Sun line (see Fig...). If ∆t is the time it takes between sending and receiving the pulse, then the Earth-Venus distance is c∆t d = (84) 2 By measuring the angle θ between the Earth-Sun line and the line of sight, one can calculate the Earth-Sun distance, given by d 1 AU = ' 149.6 × 106 km (85) cos θ

42 [1798]: By measuring Newton’s universal constant of gravitation and using the, then known, value of the dis- (A21):tance between the Earth and the Suna. aNewton estimated the ratio of the mass of the Earth to that of the Sun and obtained 1/169, 282, which is about a factor of 2 larger than the actual value. More Details

• The value of this constant has been measured for the ﬁrst time in 1798 by Henry Cavendish58 using a method based on torsion balance59, and he obtained

G = 6.67 × 10−11 N.m2/kg2

is within about 1% of the actual value. • One of the implications of Newton’s universal law of gravitation is Kepler’s third law, given by 4π2 P 2 = a3 G (M + mp)

where M is the mass of the Sun, mp the mass of the planet orbiting it, P the planet’s orbital period around the Sun, and a is its mean distance from the Sun. Since the Sun is much heavier than all the planets in the solar system, we can can approximate (M + mp) ' M and write

2 3 2 3 4π a 4π a⊕ M = 2 = 2 (86) GP GP⊕

7 where P⊕ = 3.1 × 10 seconds is the orbital period of the Earth 8 around the Sun, and a⊕ ' 1.5 × 10 meters is the mean distance Earth-Sun60. So the only quantity that needs to be determined in order to calculate the mass of the Sun is Newton’s gravitational constant61. Inserting the value of G above in Eq (86), yields

30 M ' 2 × 10 kg 58Actually Cavendish’s goal was to estimate the average density of the Earth. 59 For more details about this experiment, see my lecture notes on classical mechanics in my webpage "faculty.uaeu.ac.ae/snasri". 60From the measurement of the transit of Venus. 61In fact, the constant G was not measured until about 70 years after Newton’s death.

43 (A22): 1368 W/m2

More Details

The number above is known as the solar constant, and it is usually denoted by GS. If there was no atmosphere, the total power of solar radiation reaching the Earth’s surface would be the solar constant times the cross sectional area of the Earth62, which yields 1.7 × 1017 W. Thus, in one year, the top of the Earth’s atmosphere receives on average approximately 5.46 Yotta Joule =1 .52×1018 kWh63. Hence, the mean irradiance at the surface of the Earth is

(0) Amount of sunlight incident on the Earth G |Earth = S Area of the Earth facing the Sun 1.7 × 1017 W = ' 680 W/m2 [2 × 3.14 × (6, 400 × 103 m)2]

Because Earth has an atmosphere, around 30% of the solar energy gets reﬂected by the clouds into space and 20% is absorbed by water vapor, dust and O3 molecules in the air. In this case the solar energy incident on the Earth64 per day is65

(with Atm) HS |Earth = 50%×

Thus, even if just 0.1% of the solar energy that reaches the ground is used it can power the entire world.

26 (A23): L = 3.9 × 10 Watt More Details

• Luminosity is the energy ......

62 2 That is πRE, with RE ∼ 6, 400 km is the radius of Earth 63 Units conversions: Yotta = 1024; kWh = 3.6 × 106 Joule. 64Also called the insolation. 65Assuming that the day is an average 12 hours at any location on the Earth.

44 • The Sun’s luminosity can be estimated as follows. We know the Earth’s atmosphere receives approximately 1370 Watt per square meters, also called the solar constant, and the distance between Earth and the Sun is about 150 million km. Thus, the amount of power radiated66 by the Sun the solar constant times the area of a sphere of radius qual to the distance from the Sun to the Earth, i.e. 2 9 2 L = 1370 Watt/m × 4π × 150 × 10 m ' 3.9 × 1026 Watt

A24): (surface) ( T ' 5800 K More Details

• The observed electromagnetic radiation emitted by the Sun has approximately a blackbody spectrum, and hence we can apply Wein’s displacement law67

λmaxT = 0.29 cm. K (87)

where λmax is the wavelength in the spectrum with the maximum intensity. For instance, the sun emits maximum at the visible or yellow wavelength, λ ∼ 500 nm, then we find 0.29 × 10−2 T ' K ∼ 5800 K (88) 500 × 10−9 • This method can be also applied to stars since the the spectrum of the radiation emitted from them is a Planck function. • The light spectrum from stars has absorption lines, and so it may not be possible to observe a peak in the spectrum. Moreover, the interstellar dust may distort the spectrum light, an effect known as the "interstellar reddening", which can produce erroneous tem- perature69. Instead, astronomers use absorption lines to determine spectral classification. 66Power is the energy per unit time, which has unit of Joule per second, which we call the Watt. 67One can measure the light spectrum emitted by the star68, fit it with Planck’s distribution and deduce the temperature. 69Applying Wien?s law to a reddened spectrum will yield a temperature that is too cool.

45 • We can measure only the star’s surface temperature, not the interior temperature. A more accurate method to measure the temperature of a star is from the analysis of the absorption lines, where their strength (how ‘dark? they are) depend on the stars temperature.

A 15 (A25): Using the Earth- Sun distance (see ) and very elementary geometry. More Details

• A rough estimate using elementary geometry: 8 Knowing the Earth-Sun distance (D⊕−odot ∼ 1.5 × 10 km) and ◦ the Sun’s angular diameter θ ' 0.5 ∼ 0.0088 radian as seen from the Earth is half degree, we have tan θ R = D × ⊕− 2 θ ∼ D⊕− × ∼ 660, 000 km 2

• A better estimate using the Stefan-Boltzman formula: The electromagnetic radiation emitted by the Sun can be approximated by a black body distribution. in this case, the luminosity, L , is given by the Stefan Boltznmann formula s 2 4 L L = 4πR σTS =⇒ R = 4 4πσTS

Here σ = 5.67 × 10−8 W.m−2.K−4 is the Boltzmann constant, and 70 TS is the Sun’s surface temperature . Using the measured values of the Sun’s luminosity (see A 19) and the estimate of the Sun’s surface temperature, TS = 5800 K, (see A 23), we obtain s 3.9 × 1026 R = ' 690, 000 km 4 × 3.14 × 5.67 × 10−8 × (5800)4 70To be more precise, this is the eﬀective temperature of the Sun, i.e. the temperature that the Sun would have if it radiated as a BB.

46 11 7 (A26): Pc ' 2.5 × 10 atm, Tc ' 1.5 × 10 K More Details

Obviously we have no way to measure directly the temperature or the pressure at the center of the Sun, but we can infer their values from the equations that govern the evolution of stars. Here, I will give a rough estimate for these quantities using two properties that most of the stars obey during their life time; namely that they are in hydrostatic equilibrium71 and the particles they are made of behave as an ideal gas.

• Estimate of the central pressure of the Sun: Consider an a star of mass M∗ and radius R∗ which is hydrostatic equilibrium. With this assumption the the pressure is given by the diﬀerential equation (see the Appendix for details) dP GM(r) = − ρ(r) (89) dr r2

Integrating both sides of the equations from r = 0 to the radius of the star r = R∗, gives

Z R∗ GM(r) Pc = − 2 ρ(r) 0 r

where we used P (R ) = 0 is taken as the deﬁnition of the Sun’s surface. To get an order of magnitude estimate we will take ρ(r) to be constant and equal to the star’s average mass density, i.e. M ρ ∼ ∗ 4π 3 (90) 3 R∗

Now we can perform the integration easily and we get72

2 2 4 3 GM∗ 9 M∗ R Pc ∼ 4 ∼ 10 atm (91) 8π R∗ M R∗

71We should note that there are types of stars that expand and contract during their life time, and hence are not in hydrostatic equilibrium. This includes, 721 atm = 105 N/m2 ≡ 1 Pa.

47 which is smaller than the true value by almost two order of magnitudes, due to the underestimated value that we used for the mass density. Nevertheless, this order of magnitude estimate provides a valuable information about the condition at the interior of the Sun, which otherwise had no direct mean to measure it. • Estimate of the central temperature of the Sun: To estimate the temperature we need an equation of state that relates pressure and the mass density of the gas. To a good approximation gas an ideal gas we have

P = nkBT (92)

−23 −1 where kB = 1.38 × 1 J.K , and n is the number density of the particles in the gas. We can express n in terms of the mass density of the gas, ρ, as ρ n = m with m is the mean mass per particle73. If we assume that the gas 74 inside the star is mostly hydrogen, then n = ρ/mH , so that −1 mH Pc gram.mol Pc Tc = = (93) kBρc R ρc

73The mean mass per particle is deﬁned as P j njmj + neme m = P ≡ µmH j nj + ne

Here nj is the number density the atoms (or ions) of of type j and of mass mj that are present in the gas, ne is the number density of electrons with me the corresponding mass, and µ is a dimensionless quantity called the average molecular weight which is just the mean mass per particle in unit of hydrogen mass also . However, the electron is many order of magnitudes lighter than proton, we can neglect their contribution in the expression of the mean mass particle. We also know that mj ' AjmH , with Aj is the atomic mass number and mH is the mass of the hydrogen atom, and so we can write P j njAj µ = P j nj + ne Example:Let us consider an ionized hydrogen gas. In this case, due to the conservation of electric charge we have ne = np, with p denotes the ionized hydrogen, i.e. the protons. Thus

np × 1 1 ρ µ = = =⇒ nH = 2 np + np 2 mH . 74 −1 −1 Here we used the fact that NAmH = 1 g, and R = kBNA ' 8.314 J.mol .K , where 23 NA ' 6 × 10 is the Avogadro number.

48 Note that ρc is the mass density at the center of the Sun, which is obviously much larger than the average density. On the other hand, the nominator contains the central pressure which we have underestimated as compared to its true value, and hence for the sake of order of magnitude we can again take ρ = ρ and use the 75 value of Pc to estimated the central temperature, and we ﬁnd ,

−1 1 gram.mol GM∗ 7 M∗ R Tc ∼ ∼ 10 K (94) 2 R R∗ M R∗

(a) Photosphere (b) Chromosphere (A27): (c) Transition zone (d) Coronoa More Details

• Photosphere ∗ This is the layer we observe, and is about 500 km thick. ∗ Average temperature of 5800 K. ∗ Mass density at of about 2 × 10−4 kg/m3. ∗ Highly opaque. ∗ Absorbs and re-emit radiation produced in the Sun. ∗ Energy can be transported via radiation. ∗ Contains dark areas called the sunspots76. ∗ Contains granulations produced by convection currents of hot gas rising from below.

75The corresponding thermal energy in the center of a star will be of order (thermal) M∗ R Ec ∼ 0.87 keV M R∗

Note that this energy falls in the range [0.1 − 100] keV, and hence the corresponding radiation emitted will be X-ray. However, due to the high dense medium around the core

49 Figure 9: Components of the Sun and its atmosphere (Credit: NASA).

• Chromosphere It is the layer above the photosphere. It was first observed at the edge of the solar during the lunar eclipse, and is about 1500 km thick and has mass density 5 × 10−6 kg/m3. The temperature increases from about 4000 K in the lower chromosphere to about 10, 000 K in the upper chromosphere. • Transition zone Above the chromosphere, 8500 km thick, temperature 8000 K with rapid increase, mass density 2 × 10−10 kg/m3. • Corona Very hot and thin upper atmosphere, temperature > 1, 000, 000 K, mass density 2 × 10−12 kg/m3. The ratio of the magnetic pressure to the gas pressure in the corona is much larger than unity, and hence Its dynamic is dominated by the solar magnetic field. Visible in the far-UV and -ray wavelengths. of the star, these radiation loose most of its energy before reaching the surface. 76They ormed by concentration of large magnetic fields (B > 1000‘G). They look dark because they are cooler (with T ∼ 4300 K) than the other areas in the photosphere.

50 By measuring how strong is the pull of the star on another stellar object. Except for the Sun, one can not measure directly the mass for a single star. For A28): that, astronomers study the motion of binary stars. ( By measuring the size of the orbit, the stars’ orbital speeds, and their orbital periods, we can determine accurately the masses of the individual stars in the binary system. More Details

• Binary stars and their classiﬁcations About 1/3 of all stars form binary systems (i.e. system of two stars orbiting each other), and ∼ 2/3 of all the stars are multi- ple star systems. The ﬁrst binary star to be discovered (in 1650) was the star Mizar and its fainter companion Alcor in the Big Dipper’s handle77. Usually astronomers denote the brighter of the binary system by the letter A, and the other one by B78.

There are four type of binary stars, as will be discussed below, depending on the way we are able to observe them. However, they are not necessarily mutually exclusive, i.e. a binary could be of two types at the same time. (a) Visual Binaries: They tend to be relatively close to us with the two stars well enough separated that they can be resolved by optical means79. Typically, the apparent motion of these systems takes over decades of observation to image their orbits. An example of binary system is the two stars Alpha Centauri A and Alpha Centauri B, at 1.34 pc away from the Sun, separated by a distance of about 23 AU, and orbit each other with a period of 80 years. (b) Spectroscopic Binaries:

77they are 12 arcminutes apart, nearly half the angular diameter of the moon, and so it is possible to see it as a double star even with the naked eye. 78Other letters, such as C, D,ect may be used if the system comprises more than two stars. 79About 1000 visual binaries have been detected.

51 They appear as a single star when observed visually through telescope, i.e. they can not be resolved, but which the lines in the spectrum are double with the spacing of the components of the lines vary periodically, and at the lines even become single line. In most spectroscopic binaries, the separation distance between their members is less than 1 AU and have short orbital period in the range of few days to a couple of months. Thus, they will have large orbital velocities compared to the visual binaries, which enhance the chance of observing their velocity curves80 from the Doppler eﬀect on the their spectra (see more below). (c) Astrometric binaries: For some star systems only one star is visible, but one can detect that it wobbles (i.e. wavy motion) about an invisible center of mass of the two stars. For example, the star Sirius is a binary system, but the component Sirius A is so brighter than Sirius B that astronomers could only see Sirius A. Sirius B was detected by observing the motion of the Sirius A81. (d) Eclipsing Binaries: In this case the orbital plane of the two stars lies along the line of sight, so that the stars eclipse each other. Such systems can not be resolved in the telescope, and so can not be also visual binaries. An example of eclipsing binary star is Algol (discovered in the 1600s) in the constellation Perseus, which has a period of 2.8 days. • Mass determination using binaries Let m1 and m2 be the masses of the stars in the binary system orbiting each other around their center of mass, they trajectory of each mass is an ellipse with semi-major axes a1 and a2, respectively, and with (equal) orbital period P given b (for the derivation see the Appendix) 4π2 P 2 = a2 (95) G(m1 + m2)

Here a = (a1 + a2) is the total separation between the two stars. Moreover, from the deﬁnition of center of mass, m1r1 = m2r2, 80This is the graph of the radial velocity of a star versus time. 81It was F. Bessel who in 1844 discovered that the bright component of Sirius (i.e. Sirius A) displayed a wavy motion with a period of 50 years. In 1862, Alvan G. Clark found the fain companion of Sirius A which turns out to be a white dwarf (will be discussed in ....).

52 with r1 and r2 denotes the distance of each star from the center of mass. Since the orbit of the two stars have the same eccentricity 82 (see Appendix ...), r2/r1 = a2/a1, and hence we have

m1 r2 a2 = = (96) m2 r1 a1 The method for determining the mass of the stars in the binary depends on the type binary system: – The stars’s masses in visual binary systems For this type of systems, astronomers monitor the motion of the individual stars over the years and even decades and map their orbits. In addition, they can measure the period, P , the angles θ1 and θ2 subtended by the semi major axes a1 and a2, and the distance, to the binary d, if the stars are close enough so that it would be possible to use the method of parallax83 With this data. In this case, we can express the ratio of the masses and the total mass of the system in terms of observables as

m1 dθ2 θ2 = = (97) m2 dθ1 θ1 4π2 d3 (m + m )= (θ + θ )3 1 2 G P 2 1 2

Combining these equations yields84. We obtain

2 3 4π d 2 m1 = θ2 (θ1 + θ2) (98) G P 2 4π2 d3 m = θ (θ + θ )2 2 G P 2 1 1 2

In the discussion above we assumed that our line of sight lies in the orbital plane of the stars. However, if the perpendicular to the orbital plane is inclined by an angle i to the line of sight, then in this case what we observe is the projection of the motion on the plane along the line of sight, and so the

82 It is also equals to the ratio υ2/υ1, where υ1 and υ2 are the speed of the m1 and m2. respectively, at a given instant. 83If the stars are within a distance less than 100 pc or so. 84It is important to note that astronomers are able to deduce the masses of the stars only after taking into account the parallax of the binary star and the motion of the center of mass of the system.

53 measured angles subtended by the stars are related to the true angles by

(obs) (obs) θ1 = θ1 sin i,θ 2 = θ2 sin i (99)

Note that the unknown inclination does not aﬀect the ratio of the masses since in this case

(obs) θ θ2/sin i θ2 m1 2 = = = (100) (obs) θ /sin i θ m θ1 1 2 2

However, the expression of the total mass in Kepler’s third law becomes

3 (obs) (obs) 4π2 d3 θ1 + θ2 (m1 + m2)= (101) G P 2 sin3 i

The angle of inclination can be determined by noting that the projected focus does not coincide with the observed focus, and as a result the center of mass85 will be oﬀ which results in consistencies. Hence for visual binary systems it is possible to determine the masses of the stars86.

– The stars’s masses in spectroscopic binary systems Although we may not be able to resolve the two stars in the binary, one can analyse their motion by studying the Doppler shift in each stars’ spectrum as the stars move in their orbits: the the spectral lines shift to shorter wavelengths when the star is moving toward the observer, and to longer wavelengths when the star is moving away. There are two kind of spectroscopic binaries: ∗ Double-Lined spectroscopic binary (SB 2): In this type binaries, the spectra of both stars are visible, and

85The center of mass can be found by deﬁning a straight line in space for which the ratio of the angular distances of each star is the same at all time. 86There are situations where astronomers can not measure distance to the binaries. However, even in this case it may still be possible to deduce m1 and m2 if it one can measure a1, a2, and the stars’ radial velocities (i.e the projections of the velocities along the line of sight). The later can be done by analyzing the doppler shift of the spectral lines (see discussion on the spectroscopic binaries).

54 hence we can measure both the velocity curves87 which will be mirror of each other but diﬀerent in amplitude if their masses are not the same. Because of the inclination angle i, the observed maximum radial velocities88 read

(max) (max) (max) (max) υ1r = υ1 sin i,υ 2r = υ2 sin i (102)

(max) (max) where υ1 and υ2 are the actual maximum speeds of the stars m1 and m2, respectively, given by (see Appendix ..) (max) m2 2πa υ1 = 1/2 (103) m1 + m2 P (1 − e2) (max) m1 2πa υ2 = 1/2 (104) m1 + m2 P (1 − e2)

Here e is the eccentricity of the ellipse89 which can be determined from the measurement of both velocity curves. We deduce that

(max) m1 υ = 2r (105) m (max) 2 υ1r This means that from the observed maximum amplitudes of the radial velocities of the stars we can determine the ratio of their masses90 Moreover, by adding the two equa- (max) tions in (102) and using the expressions of υ1 and (max) υ2 given above, we can solve for the semi-major axis a and obtain ! P (1 − e2)1/2 υ(max) + υ(max) a = 1r 2r (106) 2π sin i

87This can be done by measuring the Doppler shift in their spectral lines using the relation υ λ − λ r = obs e c λe where υr is the radial velocity (i.e. the velocity along the line along the line of sight), λe and λobs are the emitted and the observed wavelengths, respectively. 88It Corresponds to the maximum redshift or blue shift in the spectral lines. 89It is the same for the orbits of both stars. 90As well the ratio of their semi-major axes which follows from Eq (96).

55 which after substituting in Eq (95) yields the total mass as 3 2 3/2 (max) (max) ! P (1 − e ) υ1r + υ2r (m1 + m2)= (107) 2πG sin i

Solving for the masses we obtain, we get

3 m1 sin i = .... (108) 3 m2 sin i = .... (109)

Thus, with the inclination angle unknown, we can not determine the absolute masses of the stars, and instead we can measure a lower limit on them. But usually astronomers consider < sin3 i >, which is the average value of sin3 i over he inclination angle between 0 and π/2. As- suming that the orbits for such binary systems to be ran- domly oriented, i.e the angles i are distributed according to the (normalized) probability distribution P (i) = sin i di, we find .91 Z π/2 3π < sin3 i >:= sin3 i P (i) di = ' 0.589 (110) 0 16 However, since no Doppler effect will be observed for an inclination angle i ∼ 0, astronomers find that it is more plausible to consider larger < sin3 i > than the value given above, and, usually, they take to be 2/3, instead. ∗ Singled-lined spectroscopic binary: This correspond to the case where one star, say star #2, is faint and so that it may not be possible to detect it spectrum lines. How- ever, still we can see that the spectral lines of the brighter 91 The integral can be evaluated as follows:

Z π/2 Z π/2 1 − cos 2θ 2 sin3 θ P (θ) dθ = dθ 0 0 2 " # 1 Z π/2 Z π/2 Z π/2 = dθ + cos2 2θ dθ − 2 cos2 2θ dθ 4 0 0 0 1 π 1 Z 1 3π = + (1 + cos 4θ) dθ = 4 2 2 0 16

56 star exhibit oscillations due to the Doppler eﬀect, indi- cating that it is part of a binary system. Thus, similar to the case of the doubled-line spectroscopic binary, astronomers can measure the velocity amplitude υ1r of the star #1, the period P , as well as the eccentricity e of its orbit. Since υ2r is not observable, we can use Eq (105) (max) to solve for υ2r in terms of the ratio of the masses, the maximal radial velocity of star #1, and the inclination angle. After substituting into (107), we obtain

3 3 m2 3 P (max) 2 sin i = υ1r (111) (m1 + m2) 2πG

The quantity on the left-hand side of this equation is called the mass function. We see that even though we can not determine the masses of the stars in the single- lined binary, the mass function sets a lower limit on the mass of its fainter component (i.e. star #2 in our case):

3 P (max) m2 > υ (112) 2πG 1r This limit is specially useful as it could be a signature for a black hole. if m1 m2, then the mass function simpliﬁes and we obtain the following relation between the masses

3 3 3 P (max) 3 m2 sin i = υ m1 (113) 2πG 1r

This can correspond to a star-planet system, where m2 is the mass of the invisible planet. – The stars’s masses in eclipsing binary systems Most of the eclipsing binaries are relatively close to each other which tend to circularize their orbits. Such eclipsing binaries are often spectroscopic binaries and hence their radial velocity curves can be measured. In addition, astronomers also measure the light curve, i.e. the ﬂux of light observed as function of time, from which a number of useful informations can be extracted. We consider that the binary consists of a smaller and larger star, and denote their associated parameters by the suﬃx s, and l, respectively. When the smaller star passes behind the larger gets completely hidden we say that the eclipse

57 is total, otherwise it is a partial eclipse. Whenever, one star passes behind its companion, the light from system decreases and reaches a minimum during the eclipse of each star. Al- though during the eclipse, the same area of each star will be covered by its companion, in general, the two minima in the light curve may not equal. The relative amount of light drop depends on the relative surface brightness of the two stars, or, equivalently, their eﬀective temperatures. The deepest minimum, also called primary minimum, occurs when the hotter star is eclipsed and the secondary minimum occurs when the cooler star is eclipsed. Here we will assume that the smaller star is hotter (see Fig. 10 below). Taking the normal to the orbital plane to be approximately perpendicular to our line of sight92, we can calculate the radii of the stars as υ Rs = (tb − ta) (114) 2 υ υ Rl = (tc − ta) = Rs + (tc − tb) (115) 2 2

where ta is the time when the small star is the point of ﬁrst contact (a.k.a the time of ingress), tb is the beginning of the 93 phase of its total eclipse , tc is the instant at which it starts to to emerge, and υ = (υs + υl) is the relative velocity of the two stars. One can also extract information about the temperatures of stars of the eclipsing binary. Let F0 be the total measured ﬂux of light from the system when the stars are not eclipsed. Then we have 2 4 2 4 F0 = k πRs σTs + πRl σTl (116)

where Ts and Tl are the eﬀective temperatures of the small and large star, respectively, and k is a constant that depends on the distance to the binary and the eﬃciency of the detector. Now when the hotter star passes behind the cooler one, i.e at the primary, the amount of light detected is

(primary) 2 4 F = kπRl σTl (117) 92Note that even if the correct angle was say i = 800 instead of i = 900, we will be making an error of less than 5% in our estimates. 93 The extent of the duration (tb − ta) depends on the relative size of the two stars. In particular, if the stars have similar size, the shape at the minima of the light curve will be have a sharp edge.

58 Figure 10: Eclipsing binary star[16].

whereas the secondary minimum (i.e. when the hotter star in front of the cooler), the detected brightness is

(secondary) 2 2 4 2 4 F = k πRl − πRs σTl + kπRs σTs (118)

Since k is not known with precision, we use the relative depths of the primary and secondary brightness minima to determine the ratio of the temperatures of the stars, and obtain 94

(primary) 4 F0 −F Ts (secondary) = (120) F0 −F Tl

Therefore, by measuring the primary and secondary minima of the light curve, we can determine the ratio of the eﬀective temperatures of the two stars.

94If the larger star is also the hotter, then we get

F − F (primary)F − F (secondary) T 0 0 l (119) = Ts

59 3/2 −1/2 (A29): τ = 30 R∗ M∗ dyn mins R M More Details

The dynamical time scale (or the free time scale) of a star is the time it takes for it to collapse under its own gravity, i.e. in the absence of pressure.This time scale t can be estimated using the equation of energy conservation95/r2. 2 GM∗ 1 dr GM∗ = + (121) R∗ 2 dt r

Here M∗ is the mass of the star of initial radius r = R∗. Then, we can solve for dt and then integrate. We get

−1 Z τdyn Z 0 1 1 τdyn : = dt = − 2GM∗ − 0 R∗ r R∗ R3 1/2 Z 1 x 1/2 = ∗ 2GM∗ 0 1 − x π2R3 1/2 R3 1/2 = ∗ ' ∗ 8GM∗ GM∗

Equivalently, we can express τdyn in terms of the average mass density, 3 96 ρ, by using the relation R∗ = 3M∗/4πρ, and obtain

1/2 3π 1.4 g/cc τdyn = ' 30 min 32Gρ ρ

So in the case of our Sun97, the dynamical time scale is about 30 min. For star of smaller size and about solar mass98, the dynamical time scale can be of order a second.

Now let us suppose that the pressure force is not totally absent, and 95It can be easily shown that we obtain the energy conservation equation in Eq eq:Econs:1 equation if we start from Newton’s equation of motions r¨ = −GM∗ 96Actually the higher densities in the interior of stars yield smaller value for the dynamical time scale. 97 30 8 M = 2 × 10 kg, (seeA20), R ' 7 × 10 m → ρ = 1.4 g/cc. 98 For example, a white dwarf has a mass of order the solar mass and radius RWD ∼ −2 6 (WD) 10 R → ρWD ∼ 10 ρ , we get τdyn ' 3 sec

60 instead it is equal to some fraction δ < 1 of the gravitation force. Then in this case the equation of motion reads

GM∗ r¨ = −(1 − δ) (122) r2 Multiplying both sides of the equation by r˙ we obtain 2 1 dr GM∗ GM∗ = − + (1 − δ) (123) 2 dt r R∗

which has the form as Eq (123), and so the time scale for the collapse with such a level of imbalance between gravity and pressure is

2 3 1/2 −1/2 π R∗ τdyn(δ) = (1 − δ) 8GM∗

So in the case of our Sun99, even a 1% imbalance between gravity and pressure, the Sun would have collapsed in about 5 hours. Since we know that the Sun was around for about 4.5 billion years, we conclude that the pressure and gravitational must cancel each other to an incredible degree of accuracy. Similar argument holds for all the stars in the main sequence and some other type of stars that continue their life oﬀ the main sequence100. Hence from the stability of stars we can conclude that they are in a state of hydrostatic equilibrium or near perfect hydrostatic equilibrium.

−1 2 −1 (A30): τ = 1.5 R∗ M∗ L∗ KH million years R M L More Details

• The Kelvin-Helmholtz timescale, denoted by τKH , it is the time it takes for a star to radiate all its stored up thermal energy if the nuclear reactions in it were suddenly cut oﬀ. So if L∗ is the

99 30 8 M = 2 × 10 kg, (seeA20), R ' 7 × 10 m → ρ = 1.4 g/cc. 100For the white dwarfs, as we will discuss later, what holds the them from collapsing under their own gravity is the so called degeneracy pressure which depends on the density of the electrons and not the temperature. This is due to the exclusion principle for the electrons in quantum mechanics.

61 luminosity of the star, i.e. the rate of the energy radiated away, assumed to be constant, then,

Ethermal τKH = (124) L∗

To get an estimate of τKH, we make use of Virial theorem, which states that the average thermal energy stored in the star is equal to half the magnitude of its gravitational potential energy (for the derivation see the Appendix), which means that the star’s reservoir of thermal energy is about the same order of magnitude. Hence, we obtain

2 GM∗ τKH = 2R∗L∗ −1 2 −1 R∗ M∗ L∗ ' 1.5 × 107 Yrs (125) R M L

• Since we know101 that the Sun was shining for more than 4 billion years, which is many order of magnitudes larger than the The Kelvin- Helmholtz time scale, implies that there must be other mechanism than converting the gravitational potential energy into radiation. The new source of energy can not be due to some chemical com- bustion process, since the typical energy per reaction is of order of few eV which is insuﬃcient to power the Sun for more than some thousands years102. The new mechanism turns out to be nuclear fusion which was discovered after Einstein derived his famous formula E = mc2 in 1905, and showed that a tiny amount of mass of matter can be converted into a huge amount of energy in the form of radiation. Hence, τKH is a time scale that determines how quickly a star contracts before nuclear fusion in the star begins, and so it gives an order of magnitude of the time scale of the pre-main sequence.

−1 (A31): τ ' 10 M∗ L∗ nuc billion years M L

101Based on radiometric dating of rocks in our solar system. 102An estimate of the time scale associated with chemical reactions can be estimated as follows.....

62 More Details

• The nuclear time scale, denoted by τnuc, is the time it take a star exhaust its supply of nuclear fuel. The heat released from the fusion of hydrogen into helium is approximately 0.7% the total mass of hydrogen103 in the core of the star which is about 10% of the the total mass of the star. If assume that the burning nuclear fuel in the star’s core occur at the current rate, L∗, then

10% × 0.7% × M∗ τnuc = L∗ −1 M∗ L∗ ' 1010yrs (126) M L

• We note that τdyn τKH τnuc, which implies that over the time scale of nuclear fusion the stars are in hydrostatic and thermal equilibrium throughout most of their lives.

• Denoting by dMH /dt the rate of converting hydrogen into helium inside the Sun, we have

dMH 2 L = 0.7% × c (127) dt

Using the measured value of L (see A22), it follows

dMH ' 600 million tons/second (128) dt In other words, the Sun converts 600 million tons of hydrogen into 596 million tons of helium every second, and hence about 4 million tons of hydrogen is transformed into energy. • Since we know that the age of the solar system is about 4.5

103The energy released by one nuclear fusion process of 4 H → He is

2 2 h −24 8 2i h −24 8 2i ∆E = 4mH c − mHec = 4 × 1.673 × 10 g × 3 × 10 m/s − 6.645 × 10 g × 3 × 10 m/s =4 .23 × 10−12 J = 26.4 MeV

Thus, the hydrogen burning in a star results in 6.7 MeV per nucleon, or, equivalently, about 0.7% of the mass of the hydrogen.

63 The energy radiated by the Sun is generated via the A32): process of thermonuclear fusion of hydrogen into he- ( lium nuclei, which occurs in several steps (for details, see below). More Details

• pp and CNO cycles in stars: In 1939, Hans Bethe showed that there are two mechanism for nuclear fusion inside stars. (a) proton-proton cycle: This is the main fusion process for stars with mass about that of our Sun, in which the temperature is around 107 K. It proceeds in the following stages: Stage 1

1 1 2 + 1H +1 H → 1H + e + ν Qe/ν = 0.42 MeV

Here the fusion of two protons forms the deuteron104 and pro- duces a positron (e+) and a neutral elementary particle, ν, called the neutrino105. The emitted e+ annihilates almost in- stantly with an electron in the plasma, producing two photons (stage 1) 106 with Qγ = 1.02 MeV of energy . We should point out 2 that two protons can not form a stable bound state 2H, and instead a stable deuteron is formed107. is unstable and so This 104Some textbooks use the word "deuterium" when describing the above process, where as it is more correct to use "deuteron". The deuterium is the atom and deuteron is its nucleus. 105Neutrinos are elementary particles that interact very weakly with matters and travels with almost the speed of light. They are about million times lighter than the electron. 106So the total energy emitted in this ﬁrst stage of the pp cycle is 1.44 MeV 107The instability of the diproton is due to Pauli exclusion principle, which forces the two spins of the protons to be anti-aligned (called the singlet state). In this case the attractive nuclear force between the two nucleons in the singlet state is weak and not enough to form a stable bound state, even in the absence of the repulsive Coulomb force (so this argument also applies for the dineutron system, which misses by just 60 keV energy). This is not the case for the deuteron where n and p can have their spin aligned (called triplet state) , and hence the attractive nuclear force between them is stronger than that of the singlet state. Theoretical calculations show that if the strong nuclear force was about 2% stronger, a stable diprotons (as well dineutron) would form.

64 ﬁrst stage of the pp is governed by the weak interaction108, and so the process is extremely slow, which set the pace for the conversion of hydrogen to hydrogen nuclei. For temperature inside the Sun, the time scale for this process is about 8 × 109 yrs. Stage 2

2 3 (stage 2) 1H + p → 2He + γ Qγ = 5.49 MeV

It is worth noting that even though in principle the reaction 2 2 4 1H +1 H → 2H is possible, the probability for it to occur is negligible due to the fact that the density of deuteron in the plasma is extremely small as compared to to protons109. The time scale for this process is about 1second. Stage 3-a (PP I) [happens ∼ 83% of the time]

3 3 4 (stage 3-a) 2He +2 He → 2He + 2 p + Q = 12.86 MeV

The time scale of this ﬁnal stage is 2.4 × 105 years110.

Therefore, the proton-proton cycle generates energy that can be described by the net reaction

4 (net) 4p → 2He + 2 p + 2ν + Q

where

(net) (stage 1) (stage 2) (stage 3-a) Q = 2Qe/ν + 2Qγ + 2Qγ + Q = 26.7 Mev (129)

which is the net energy released in the PP I chain reaction.

108There are four fundamental interactions in nature: Gravitational, Electromagnetic, Weak, and Strong. 109Approximately there is 1 deuteron for every 1018 protons. 110 3 1 4 Note that although 2He could in principle reacts with 1H, the resulting 3Li is unstable, 3 1 and It decays back to 2He+1H.

65 Stage 3-b [happens∼ 17% of the time]111

3 4 7 2He +2 He → 4Be + γ (130)

∗ Stage 3-b1 (PP II) [happens 99.88% of the time]

7 − 7 4Be + e → 4Li + ν (131) 7 4 4Li + p → 2 2He (132)

∗ Stage 3-b2 (PP III) [happens 0.02% of the time]112

7 8 4Be + p → 5B + γ (133) 8 8 ∗ + 5B → 5Be + e + ν[Qe/ν = 15 MeV] (134) 8 ∗ 4 5Be → 2 2He (135)

Although the PP III is energetically negligible, it is very important in the detection of the neutrinos as their energy can be as large as 15 MeV, which makes them easier to detect than the ones emitted in the PP I chain.

(b) Carbon-Nitrogen-Oxygen cycle This is another way to reach the same end as the p-p cycle, i.e. producing helium-4, which is the dominant reaction for stars with mass M∗ > 1.5 M , as their interior is hotter and the pressure is higher than in the Sun. Here the carbon, nitrogen, and oxygen act a catalyst, and there will be no net produc- tion of these elements in this cycle. In such environment, the dominant source of energy generation is via the CNO cycle, in which hydrogen nuclei could fuse using ordinary carbon (6 C) as a catalyst in a set of reactions, and the net of the cycle is the fusion of 4 protons into a single helium. This cycle proceeds in 6 stages • Two nuclei can fuse to form a larger nucleus if they approach one another within the distance scale of the strong nuclear force, i.e. −13 dstrong = 10 cm. Because the nuclei have the same (positive)

111It dominates the pp cycle for stars with core’s temperatures between 14 million K and 23 million K. Because the temperature in the core of our Sun is close to 15 K, this branch occurs up only 17% of the time. 112It dominates the pp cycle for stars with core’s temperatures greater than 23 million K.

66 electric charge, there is an electrostatic potential barrier113 of order 1 MeV preventing them from getting close enough for the fusion to take place, unless the temperature inside the star is very high so that their kinetic energy can overcome the Coulomb barrier. So, if the temperature of the proton gas inside (core of ) a star is T , the average kinetic energy of a proton is 3 T < Kp > = kBT ' keV (136) 2 107 K

This means that for a typical star, such our Sun, where the temperature is of order 107 K, the protons can not fuse 114. In fact, even for T ∼ 109 K, the particles (protons) will not have energy to penetrate of the Coulomb barrier, and hence the nuclear fusion can not occur; at least classically. • The solution to the above problem is quantum mechanics, where microscopic particle can have wave-like properties; known as wave- particle duality. Then, there is a non-zero transmission probability through the electrostatic barrier, even if the thermal kinetic energy is order of magnitudes smaller than what is required to overcome it. This phenomenon is known as the quantum tunneling. Let see how it works. The expression of the probability of tunneling through a Coulomb barrier is given by115

2 2 2 1 4π Z1Z2e − µ Ptunneling ∝ √ exp (−2G) exp − e 2kT (138) E hυ

113 The Coulomb barrier between two nuclei of (positive) charges Z1e and Z2e, respectively, and with an interaction range rc ∼ dstrong is

2 Z1Z2e Ecb = ∼ Z1Z2 MeV rc

114Although the gas of particles in the Sun’s core follow the classical Boltzmann distribution, and hence there are nuclei with larger thermal kinetic energy (in the tale of the distribution), the probability for having a particle with Kp ∼ 1000 < Kp > is practically zero. 115It is proportional to the nuclear cross section, which for non-resonant reaction is given by

S(E) σ(E) = √ e(−2G) (137) E with G is the Gamow factor (given in (139)), and S(E) is called the astrophysical S factor, which is usually a weakly varying function of energy.

67 with G is the so-called Gamow-Factor116,

r 2 m 2πZ1Z2e G = √ (139) 2 h E

Here Z1e and Ze2, are the (positive) charges of the two nuclei, respectively, is their relative velocity., and µ is the reduced mass of the two nuclei. Since the particles are in thermal equilibrium, the above expression of Ptunneling has to be folded with the Maxwell Boltzmann distribution, and so the probability for fusion is

2 2 2 4π Z1Z2e − µ Pfusion ∝ exp − e 2kT (140) hυ

We see that the ﬁrst exponential increases for higher thermal velocities, where the second exponential (coming from the Boltz- mann distribution of velocities) decreases. Thus, the fusion reaction rate is important only in some energy range, known as Gamow peak. We also note that for heavier nuclei, for the fusion to occur requires much higher temperature than for light nuclei. For the proton-proton fusion in the core of the Sun, one ﬁnds

( ) −20 Ptunneling ∼ 10 (141)

For more details about the cross sections of the diﬀerent processes in the pp chain and their reaction rates see Appendix XX.

116It was derived in 1929 by G. Gamow. See [21] for details.

68 2 Appendices

2.1 Keplerian motion of binaries

Setting the origin of coordinates in the center of mass, the positions of the two masses read m2 m1 r1 = r, r2 = − r, (142) M M Taking the time derivative, we obtain the velocities with respect to an observer in the CM m2 m2 υ1 = r˙ = − υ2 (143) M m1 m1 m1 υ2 = − r˙ = υ1 (144) M m2

2.2 Speciﬁc Intensity and Radiation Pressure

Consider a surface element dA, and an incoming or outgoing radiation ﬁeld along some direction eˆ. Then, the energy that ﬂows through this element of area in time dt from a solid angle dΩ with frequencies lying in a range ν to ν + dν is proportional to dt, dν, dΩ, and the projection of dA normal to eˆ, i.e. dA⊥ = eˆ. nˆ dA = cos θ dA. Thus, we can write

dEν = Iν(r, t, eˆ) cos θdAdtdΩdν (145) The quantity I(r, t, eˆ) ≡ I(t, eˆ) has unit of J.s−1.m−2.ster−1.Hz−1, and 117 is called speciﬁc intensity . Here we will assume that Iν(t, eˆ) = Iν(eˆ), i.e. independent of time. The integration of Iν(r, t, eˆ) over all the frequencies is called the surface brightness, i.e. Z dE −1 −2 −1 S(Ω) = Iν(r, t, eˆ) dν = ←− [erg.s .cm .ster ] (146) dtdA⊥dΩ The energy ﬂux of the radiation associated with frequency ν is the energy radiation coming from all direction per unit time, per unit area , and per unit frequency so that118 Z Z dEν Φν = dΩ = Iν(eˆ) cos θ dΩ := 2πFν (147) dAdtdν

117In the cgs system it has the unit erg.cm−2.s−1.ster−1.Hz−1. 118We can deﬁne the ﬂux energy vector per unit time per unit frequency as

dΦν = Iν dΩ ˆe

69 and the bolometric ﬂux can be obtained by integrating Fν over all frequencies. During the inﬁnitesimal time interval dt the radiation travels a distance dl = cdt, and so we can rewrite the expression in (145) as

Iν(eˆ) Iν(eˆ) dEν = dldA⊥dΩdν = dV dΩdν (148) c c

where dV = dldA⊥ the element of volume through which the radiation passes. Thus, the energy density of radiation at speciﬁc frequency ν, denoted by Uν, can be obtained by integrating dEν/dνdV over radiation coming from all over directions, i.e. 1 Z 4π Uν = Iν(eˆ) dΩ ≡ Jν (149) c c

Where Jν represents the average intensity over all directions. In the particular case of an isotropic radiation ﬁeld, such as black body radiation, the above relation becomes

(isotropic) 4π U = Iν (150) ν c In the frequency range ν to ν + dν. , this radiation field carries a mo- mentum dpν = (dEν/c) ê (recall this is a special case of the Einstein equation E2 = p2c2 + m2c4 when m = 0), which during the infinitesi- mal time interval dt its normal component ˆn. dpν = cos θdEν/c exerts a force on the surface element given by

cos θdEν/c Iν 2 dfν = dν = cos θdAdΩdν (151) dt c Thus, dividing the above expression by dA and integrating over all solid angles, the radiation pressure at a speciﬁc frequency ν reads119 Z Iν 2 Pν = cos θdΩ (152) c and the ﬂux received at some surface is given by Z Z Φν = dΦν . ˆn = Iν (eˆ) cos θ dΩ

119In general one can deﬁne the radiation pressure tensor as Z I P ij = ν einjdΩ ν c where ei and nj are the components of the unit vectors eˆ and nˆ in the cartesian coordinates.

70 which for isotropic radiation ﬁeld ( Eq. (150)), we get 1 Pν = Uν (153) 3 If we integrate over the frequencies, we get the radiation pressure of black body120 Z ∞ 1 1 4 1 Prad = Uν dν = aT ≡ urad (154) 3 0 3 3 where a = 4σ/c, with σ = 5.67 × 10−8 J.m−3.s−1.K−4 is the Stefan- Boltzmann constant.

Now one important question one should ask is "does the speciﬁc intensity change along the ray path ?" To answer this question let us (s) consider a surface element dA (e.g. a source), located at a point rs, from which a radiation ﬁeld is emitted in some direction eˆ that goes (d) through an the element of surface dA (e.g. a detector), at a point rd. Then, the amount of energy that propagated from dAs to dAd during the time interval dt in the frequency range ν to ν + dν must be the same, i.e.

(s) (d) Iν(rs, eˆ) dA⊥ dtdΩddν = Iν(rd, eˆ) dA⊥ dtdΩsdν (155)

(d) (s) where dΩd and dΩs are the solid angles subtending dA and dA from the points rs and rd, respectively, given by

(d) (s) dA⊥ dA⊥ dΩd = 2 , dΩs = 2 (156) |rs − rd| |rs − rd|

(s) (d) (s) (d) where dA⊥ and dA⊥ are the projections of dA and dA normal to eˆ and "−eˆ", respectively. Substituting the above expressions of the solid angles into Eq (155) yields

Iν(rs, eˆ)= Iν(rd, eˆ) (157)

Thus, in empty space, the speciﬁc intensity along path ray is independent of distance from the source. 120We used the fact that for black body radiation the speciﬁc intensity is given by

2hν3 1 Iν = 2 hν c e /kBT − 1 where h ' 6.62 × 10−34 J.s is the Planck constant.

71 Before we close this section, we deﬁne nth the moments of the radiation ﬁeld, which we will use later when we discuss the solution to the radiation transport equation: Z π (n) 1 n Mν := Iν (cos θ) cos θ dθ (158) 2 0

(0) (1) For instance, Mν = Jν (average intensity), Mν =Φ ν/4π := Hν (called the Eddington ﬂux), and for the 2nd moment, usually denoted Kν (called the K-integral), it can be shown that it is related to the radiation pressure.

2.3 Radiation Transport

In the presence of matter, the speciﬁc intensity changes along the the ray path due to three possible eﬀects:

• Emission121 → Addition of photons to radiation beams.

(emission) (dIν)emission = Eν d s −→ dEν = EνdV dtdνdΩ (159)

with s is the length parameter of the ray path and Eν is the emission coeﬃcient of the matter which has unit erg.cm−3.s−1.Hz−1.ster−1. • Absorption by matter → Removal of photons from the ray path, and the decrease of the speciﬁc intensity is proportional to the incident intensity:

(absorption) (dIν)absorption = −αν Iν ds −→ dEν = αν Iν dV dtdΩ (160)

and αν is the absorption coeﬃcient of the medium, and has unit cm−1. This coeﬃcient can be expressed in terms of the number density of matter particles in the medium, n, and the cross section of the interaction cross section σν of the radiation and matter, as

αν = nσ ν (161)

• Diﬀusion (i.e scattering) → Change of the direction of photons, i.e. a photon initial in a beam along a direction ˆe is removed from 121for example as a result of charged particles acceleration, or from excited atomic and molecular states

72 it from it and added to another one along ˆe0. The decrease in Iν(ˆe) caused by the scattering can be written as

(decrease) (dIν)scattering = −σν Iν ds (162)

−1 with σν is the diffusion coefficient with the unit of cm . The increase in Iν(ê) from photons that scattered off other beams to one along ê is given by122 Z 0 (increase) 0 dΩ (dIν) = σν Iν(ê ) ds = σν Jν ds (163) scattering 4π

Thus, including all the diﬀerent contributions above, the transport equation of radiation, also known as the transfer equation, reads

dIν = Eν − ανIν − σν (Iν − Jν) (164) ds A quantity that is relevant when studying the propagation of the radiation in a medium is the combination of the absorption and diﬀusion coeﬃcients, given by

αν + σν 2 −1 κν = ←− [m .kg ]. (165) ρ where ρ is the mass density of the matter in the medium. This coefficient is known as opacity of the medium. A related parameter is the mean free path of photons, defined as 1 1 λν = = (166) αν + σν ρκν which is the average distance that a photon can travel before being absorbed (with a probability αν/αν +σν) and diffused (with a probability σν/αν + σν).

The simplest case where Eν = σν = 0, i.e. purely absorptive medium, the equation of radiative transfer has the solution

−τν Iν(s)= Iν(s0) e (167) 122Here I am assuming that the diﬀusion is isotropic. If this is not the case we write

(increase) Z 0 dIν 0 0 dΩ = σν (ê, ê )Iν (ê ) ds scattering 4π

73 where Z s 0 0 τν = αν(s )ds (168) s0

is called the optical depth. A medium that has τν 1 is called optically thick (or opaque) at the frequency ν, whereas if τν 1 it is said to be optically thin or transparent.

In the case where σν = 0, it is customary to caste Eq (164) in the form123

dIν d τν τν = Sν − Iν → (Iν e ) = Sν e (169) dτν dτν

Here dτν = ανds is an inﬁnitesimal optical depth, and Sν = Eν/αν called the source function. Note that if Iν > Sν, the speciﬁc intensity tends to decrease along the ray; whereas if Iν < Sν, it tends to increase along the ray. This means that Iν tries to approach the source function, and hence in this respect, the transfer equation describes a relaxation process. Integrating Eq (169) over the optical length from 0 to τν yields

τν Z 0 −τν −(τν −τν ) 0 0 Iν(τν)= Iν(0)e + e Sν(τν) dτν (170) 0 If the emission and absorption coeﬃcients are constant, we can carry out the the integration above to get

−τν −τν Iν(τν)= Iν(0) e + Sν 1 − e (171) A type of radiation that is of great of importance in Physics and Astro- physics is the one radiated by a system that is in thermal equilibrium at temperature T , or locally in in thermal equilibrium. For example, the atmosphere is not in thermal equilibrium because the temperature is not the same through out all its volume, but locally it can be considered in thermal equilibrium. For such systems the speciﬁc intensity of radiation is given by the Planck function (for derivation see next section) 2hν3 1 Bν(T )= 3 hν (172) c e kT − 1

123 In the presence of scattering, i.e. σν 6= 0, we obtain an equation of the same form as (169), where Sν is replaced with a new source function Seν , given by

Eν + σν Jν Seν = αν + σν

74 For instance consider a cavity at temperature T and on its side we make a tiny hole to measure the intensity of the radiation inside it without disturbing its thermal equilibrium state. This enclosure is the best approximation of a black body. Then, we place an object at same temperature T inside the cavity. The addition of the object will not change the nature of the system "cavity + Object", and so it is also a black body enclosure at the same temperature T . This implies that

Sν = Bν(T ) and Iν(0) = Bν(T ) (173)

Substituting the above expressions in Eq (171), we obtain

Iν(τν)= Bν(T )= Sν −→E ν = αν Bν(T ) (174) which is just the statement that the object emits the same amount of thermal radiation as it absorbs at the same frequency. This is known as Kirchhoﬀ’s law.

Now suppose we have an object in thermal equilibrium at some temperature T , i.e. Sν = Bν(T ), in some radiation background, I0, which is not necessarily a blackbody radiation. Then, the speciﬁc intensity reads

−τν −τν Iν(τν)= Iν(0) e + Bν(T ) 1 − e (175)

Thus, in general the speciﬁc intensity is diﬀerent from the black body radiation. But for optically thick object, the exponentials in the above equation can be approximated by unity and we get

(thick) Iν = Bν = Sν (176) and so the speciﬁc intensity from objects with τν 1 has Planck function. This is the case of the radiation emitted by hot piece of iron or in the center of a star. On the other hand, for a optically thin object, i.e. τν < 1, the solution in Eq (175) can be approximated by

(thin) Iν = Iν(0) + Bν(T ) τν (177)

In particular, if there is no radiation background (Iν(0) = 0), the spe- ciﬁc intensity reads

(thin) Iν = Bν(T ) τν = αν Eν (178) Hence for optically thin medium the intensities are large at frequencies where the absorption coeﬃcient is large. This occur at the frequencies

75 of the spectral lines, and hence we expect an emission line spectrum at these frequencies. An example of this system is nebula.

If the there is a thermal radiation at temperature Tin, i.e. Iν(0) = Bν(Tin), traveling towards the object which is in thermal equilibrium at temperature T2, then the speciﬁc intensity in (171) reads

(thin) Iν ' Bν(Tin)+ τν [Bν(Tout) − Bν(Tin)] (179) There are two cases: (a) if Tout > Tin ⇒ The outgoing speciﬁc radiation is enhanced at frequencies where τν is the largest. Thus, the spectrum will contain emission lines on top of the continuum. For instance, the observed emission of UV and X -ray from the solar corona implies that the temperature there is much higher than that of the photosphere.

(b) if Tout < Tin ⇒ The outgoing speciﬁc radiation is reduced at frequencies where τν is the largest. Thus, the spectrum will contain absorption lines superimposed of the continuum. An example 7 the radiation traveling from the centre of the Sun (Tin ∼ 10 K) 3 outward, exists the photosphere (Tout ∼ 5 × 10 K) consists of a continuum with a lot of absorption lines.

2.4 Plane-Parallel Symmetry Often the speciﬁc intensity is function of only the polar angle124, and the vertical coordinate, z. Writing the element of distance ds as dz ds = (180) µ where µ cos θ, Choosing the z axis in the outward direction with the origin at the center of star, the optical depth will be deﬁned as

dτν = −ανdz (181) where the minus sign above is introduced in accordance with our choice of the direction of the z-axis, i.e. as one go deeper inside the star, we expect the the optical depth to increase . Then, Eq (169) becomes d −τν /µ −τν /µ µ Iν e = −Sν e (182) dτν 124The angle between the normal to the plane, which we chose to be the z-axis.

76 which upon integration yields

Z τν −tν /µ τν −tν /µ dtν Iν e ]τν,0 = − Sν e (183) τν,0 µ

We are interested in two cases: i) τν,0 = ∞: when the ray path is moving in the stellar material toward the outward direction, i.e. 0 ' µ ' 1, and for this we choose, and ii) τν,0 = 0: when the ray path is moving in the inward direction of the stellar material, and with the boundary condition that Iν(0, µ) = 0, i.e. speciﬁc radiation vanishes at the surface of the star (where we take τν|surface = 0) for a downward ray. We get

( ∞ − R S e−(tν −τν )/µ dtν , 0 ≤ µ ≤ 1 τν ν µ for Iν(τν, µ)= (184) R τν −(τν −tν )/µ dtν − 0 Sν e µ for − 1 ≤ µ ≤ 0

Assuming that τν 1, the local thermal equilibrium inside the star, and so

Sν ' Bν(T (τν) := Bν(τν ) (see Eq (173)). At a depth tν, around τν, the source function can be expressed as a Taylor series of Bν(τν), i.e.

∞ n n X (tν − τν) d Bν(tν) Sν(tν) = |τ (185) n! dnτ ν n=0 ν Substituting this expression into Eq (184) for µ ≥ 0, and making the change 125 of variable x = (tν − τν) /µ, we obtain

∞ n X n d Bν(tν) Iν(τν, µ)= µ |τ (186) dnτ ν n=0 ν It is straightforward to show that for µ ≤ 0, one finds exactly the same expression for the specific intensity. Now, the average intensity, the Eddington flux, and the K-integral read 2 1 d Bν Jν = Bν + 2 + ... (187) 3 dτν 3 1 dBν 1 d Bν Hν = + 3 + .. (188) 3 dτν 5 dτν 2 1 Bν 1 d Bν Kν = 2 + .. (189) 3 + 5 dτν 125In obtaining this result we used the fact that Z ∞ xn e−x dx = n! 0

77 Thus, at large optical depth, we can approximate Iν, Jν, Hν, and Kν, by the expressions

dBν Iν = Bν + µ (190) dτν Jν = Bν (191) 1 dBν Hν = (192) 3 dτν 1 Kν = Bν (193) 3

Expressing the optical as dτν = −κν dz (see Eq (166)), the Eddington ﬂux can be written in terms of the gradient of the temperature as

1 dBν dT Hν = − (194) 3κνρ dT dz

2.5 Einstein Coeﬃcients

2.6 Average Molecular Weight

2.7 Virial Theorem

2.8 The Stellar Equations

Consider a star of total mass M∗ and an inﬁnitesimal cylindrical element of mass between the radius r and r + dr.

1. The equation of conservation of mass Let m(r) is the mass contained be the mass within the star at a radial distance r from its center. If ρ(r) is the mass density of the material at r, then the change of mass dm at a point r is

dm = ρ dV (195)

where dV is an inﬁnitesimal volume element that contains dm. Taking dV to be the volume of a small cylindrical slab of area dA, we can write the above equation as dm(r) = ρ(r) dA (196) dr

78 This diﬀerential equation as is the This equation is often known as the continuity equation. Assuming spherical symmetry, the above equation reads dm = 4πr2 ρ (197) dr

2. The equation of hydrostatic equilibrium The element of mass dm is subjected to type of forces. The first one is a downward gravitational force, given by Gm dm G m ρ dr dA dFg = − êr = − êr (198) r2 r2

where ˆer is a unit vector along the radial distance and directed outward. The second force arises from the diﬀerence in pressure at the top of the mass element and on a small cylindrical slab, i.e.

dFP = P (r) dA êr − P (r + dr) dA êr ' dP dA êr (199)

Now applying Newton’s second law, yields Gm(r) ρ dP d2r − − ˆer = ρ (200) r2 dr dt2

If we demand that the star is in hydrostatic equilibrium, then the left hand side vanishes, and we obtain the condition dP Gm(r) = − ρ (201) dr r2 Note that the gradient of the pressure is negative, and that implies that the pressure increases towards the center of the star. A rough estimate of the central ppressure can be obtained using the above equation (see A 25).

We can also use the mass m as the dependent variable instead of the radial distance by applying the chain rule dP/dr = (dP/dm)×(dm/dr) and using Eq (195) and Eq (201), to obtain dP Gm = (202) dm 4πr4 Thus, by solving the system of equations (197) and (201) we can determine the mechanical structure of the star in hydrostatic equilibrium.

79 However, there are three variable functions {m(r), ρ, P (r)}, and thus an extra relation is needed. Indeed, for most of the stars126, their stellar material is locally in thermal equilibrium and behaves as an ideal gas, where there is an equation of state for the pressure in terms of the density and the temperature (see below). But then we will have four variables {m(r), ρ, P (r),T }, and so an additional equation for the temperature must be supplied in order to be able to solve this system. This is provided by the transport equation which we will discuss below. It is worth pointing out that there special stellar objects (e.g. white dwarfs) for which the the equation of state is independent of the thermal properties of the star, i.e. P = P (ρ), and in such cases the system can be solved.

4π 3 Now let us multiply both sides of Eq (201) by 3 r and then carry- ing out the integrating for each side. We get

4π R∗ Z R∗ 1 Z R∗ Gm(r) r3P − P (r) 4πr2 dr = − dr (203) 3 0 0 3 0 r

The ﬁrst term vanishes as P (R∗) = 0 deﬁnes the surface of the star. The term on the right hand side of the equation is the total gravitational potential energy, Eg, of the star. Dividing both sides by the volume of the star, V∗, we obtain 1 < P >V = − Eg (204) 3

R Vs where < P >V = 0 P dV /Vs is the volume-averaged pressure. This is one form of the Virial theorem. For an ideal gas, the equation of state reads ρ P = nkBT = kBT (205) µmH Here n = ρ/m is the number density of the particles in the gas, and µ is the mean molecular weight of the gas (see Appendix 2.4 for details). Using this expression for the pressure in Eq (204), we have M∗ Eg = −3kBT = −3NkBT (206) µmH with N the number of particles in the gas. We also recall that from the kinetic theory of gases that the internal thermal energy per particle 126The stars that are still on the main sequence.

80 per degree of freedom is

(d.o.f) 1 = k T (207) th 2 So for N particles, each with β degrees of freedom, the total internal energy of the gas reads127 β Eth = NkBT (210) 2 Thus, Eq (206) becomes 6 Eth + Eg = 0 (211) β

This equation is one form of Virial theorem. In many situations where it is possible to estimate only Eg or Ethermal, but both, and so we can use this relation to determine the other. Now, writing the total energy, ETotal, as the sum of the total internal energy and the gravitational energy, we have β − 6 (β − 6 ETotal = Eth = Eg (212) β 6

In particular, for an ideal monatomic gas (or fully ionized gas), β = 3, and we obtain

(mono) (mono) Eg E = −E = (213) Total th 2 We see that the total energy is negative (i.e. the star is bounded) and it is equal to half its gravitational energy or the negative of its internal energy. So the more tightly bound the star is the hotter must be to 127Often the total internal energy is given in terms of the ratio of the speciﬁc heat at constant pressure to the speciﬁc heat at constant volume, which is related to the number of degrees of freedom per particle as β by

β + 2 γ = (208) β and we write 1 E = Nk T (209) th γ − 1 B

81 maintain hydrostatic equilibrium. However, if the star contacts128, the change in its gravitational energy, δEg, and we can write

δEg δEg δEg (mono) δEg = + = + −δE (214) 2 2 2 th where we made use of the Virial theorem in the second equality. So half the change in the gravitational energy (due to the contraction of the star) gets radiated away, and the other half is added to the internal of the star, making it hotter. So, there are three consequences of gravitational contraction of star: 1) Its internal energy increases, and hence gets hotter, 2) its total energy decreases, and hence becomes more tightly bound, and 3) radiate away energy. 3. The equation of conservation of energy Let (r) be the rate of energy generation129 per unit mass at the points on the sphere of radius r, L(r) the total energy per unit time that ﬂows across the spherical of radius r, and L(r + dr) at (r + dr). Then, the change in luminosity dL := (L(r + dr) − L(r)) should be equal to the total amount of energy generated in the spherical shell of radius r and thickness dr, ie. dL = 4πr2 drρ (r) (215) or, equivalently, dL = 4πr2 drρ (216)

4. The energy transport equation The energy created within the interior of a star must be transported outwards to its surface. There are three types of heat transport: a) conduction, b) convection, and c) radiation. It turns out that in most stars, transfer of energy by conduction is not relevant, and hence we will concentrate on the other two types.

First we consider heat transfer by radiation.

128One may ask why does the star contracts in the ﬁrst place and not stay in static equilibrium. The reason for that is because the surface of the star is at some temperature it will radiate some amount of energy, and that (even if it is very small) implies more negative gravitational energy. 129By gravitational contraction and nuclear energy. Note that in general is a function of the mass density, ρ, and the temperature, T , but since both of them are function of r we use = (r).

82 2.9 Degeneracy Pressure

2.10 Doppler Eﬀect

2.11 Virial Theorem

2.12 Cosmological Redshift

• The ﬁrst peak: emitted at te and detected at t0

• The second peak: emitted at (te + δte) and detected at (t0 + δt0)

The line element of light is ds2 = 0, and so cdt dr = −√ (217) a(t) 1 − kr2

Since the right hand side depends only the comoving coordinate r, the integration of the left hand side between te and t0 or the between (te + δte) and (t0 + δt0) is the same thing since both are equal to an integral from 0 to r, i.e.

Z t0 cdt Z (t0+δt0) cdt Z te+δte cdt Z (t0+δt0) cdt = ⇐⇒ = (218) te a(t) (te+δte) a(t) t0 a(t) t0 a(t)

Since δti/ti 1, the above integrals can be approximated by

cδte cδt0 a(te) = −→ δte = δt0 < δ0 (219) a(te) a(t0) a(t0)

Using the fact that cδ0 = λ0 and cδe = λe, we ﬁnd the relation

λ0 a(t0) = (220) λe a(te)

This means that in an expanding universe the wavelength of the radiation received will longer than the emitted by a factor equal to the ratio of the scale factors of the universe at the two times. 2.13 Gamow Peak

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