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PH507 Multi-wavelength Professor Michael Smith 1 WEEK 2: LECTURE 4 Standard Candles 1 Spectroscopic Parallax The term spectroscopic parallax is a misnomer as it actually has nothing to do with parallax. Hertzsprung-Russell deduced the main-sequence nearby stars, relating their luminosity to their colour. for PH507 Multi-wavelength Professor Michael Smith 2 Groups of distant stars should also lie along the same mainsequence strip. However they appear very dim, of course due to their distance. On comparison of fluxes, we determine the distance. This works out to about 100,000 pc, beyond which main-sequence stars are too dim. If we take a spectrum of a star we can determine: its spectral class. Using photometry we can measure the apparent magnitude, mV . If we use B and V filters we can also determine the blue apparent magnitude, B and thus determine: PH507 Multi-wavelength Professor Michael Smith 3 the star's colour index, CI = B - V. Knowing either the star's spectral class or colour index allows us to place the star on a vertical line or band along a HertzsprungRussell Diagram. If we also know its luminosity class we can further constrain its position along this line, that is we can distinguish between a red supergiant, giant or main sequence star, for example. Once we know its position on the HR diagram we can infer what its absolute magnitude, M V should be by either reading off across to the vertical scale of the HR diagram or looking it up from a reference table. A main sequence (luminosity class V) star with a colour index of 0.0 (ie A0 V) has an absolute magnitude of +0.9 for example. Now knowing m from measurement and inferring M we can use the distance modulus equation: m - M = 5 log(d/10) to find the distance to the star, d, in parsecs. EXAMPLE Gamma Crucis is an M3.5 III star, a red giant. Measured values: mV = 1.63 and a colour index of +1.60. Using its spectral and luminosity classes we can place it where the red circle is on the HR diagram. Reading across to the vertical PH507 Multi-wavelength Professor Michael Smith 4 axis this corresponds to an absolute magnitude of about -0.8 ….show that the distance is about………? 2 Cepheids as Standard Candles: The Period-Luminosity Relationship Cepheids show an important connection between period and luminosity: the pulsation period of a Cepheid variable is directly related to its median luminosity. This relationship was first discovered from a study of the variables in the Magellanic Clouds, two small nearby companion galaxies to our Galaxy that are visible in the night sky of the southern hemisphere. To a good approximation, you can consider all stars in each Magellanic Cloud to be at the same distance. Henrietta Leavitt, working at Harvard in 1912, found that the brighter the median apparent magnitude (and so the luminosity, since PH507 Multi-wavelength Professor Michael Smith 5 the stars are the same distance), the longer the period of the Cepheid variable. A linear relationship was found. Harlow Shapley recognised the importance of this periodluminosity (P-L) relation-ship and attempted to find the zero point, for then a knowledge of the period of a Cepheid would immediately indicate its luminosity (absolute magnitude). This calibration was difficult to perform because of the relative scarcity of Cepheids and their large distances. None are sufficiently near to allow a trigonometric parallax to be determined, so Shapley had to depend upon the relatively inaccurate method of statistical parallaxes. His zero point was then used to find the distances to many other galaxies. These distances are revised as new and accurate data become available. Right now, some 20 stars whose distances are known reasonably well (because they are in open clusters) serve as the calibrators for the P-L relationship. Further work showed that there are two types of Cepheids, each with its own separate, almost parallel P-L relationship. PH507 Multi-wavelength Professor Michael Smith 6 The classical Cepheids are the more luminous, of Population I, and found in spiral arms. Population II Cepheids, also known as W Virginis stars after their prototype, are found in globular clusters and other Population II systems. Classical Cepheids have periods ranging from one to 50 days (typically five to ten days) and range from F6 to K2 in spectral class. Population II Cepheids vary in period from two to 45 days (typically 12 to 20 days) and range from F2 to G6 in spectral class. Population I and II Cepheids are both regular, or periodic, variables; their change in luminosity with time follows a regular cycle. The empirically derived relationship between PH507 Multi-wavelength Professor Michael Smith 7 a Type I Cepheid's period P (in days), and its absolute magnitude Mv is given by QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. Cepheids are bright and distant. They can be used to determine distances to quite distant galaxies, to about 5 Mpc. HST stretched this to 18 Mpc (Virgo cluster). 3 Tully-Fisher Relation the Tully-Fisher relation, published by astronomers R. Brent Tully and J. Richard Fisher in 1977, is a standard candle that measures the distance to rotating spiral galaxies by the width of the galaxy's spectral lines. The empirically-derived relation states that the luminosity of a galaxy is directly proportional to the fourth power of its rotational velocity, which can be calculated from the width of the spectral line, and uses the distance modulus to find distance from luminosity and apparent magnitude. In a spiral galaxy, the centripetal force of gas and stars balances the gravitational force: mV2/R = GmM/R2. If they have the same surface brightness ( L/R2 is constant) and the same mass-to-light ratio (M/L is constant), then L ~ V4. So, provided we can measure the velocity V, certain galaxies can be used as standard candles. (determine V through the 21 cm line of atomic hydrogen in the galaxy). PH507 Multi-wavelength Professor Michael Smith 8 Axes? Brightest Mb = -19, rotation 100 - 500 km/s 4 Type 1a Supernovae. The peak light output from these supernovae is always about Mb = -19.33 +- 0.25. Therefore we can infer the distance from the inverse square law. Being so bright , they act as standard candles to large distances: to 1000 Mpc. Why are they standard candles? White dwarfs I binary systems. Material from a companion red giant is dumped on the white dwarf surface until the WD reaches a critical mass (Chandrasekhar mass) of 1.4 solar masses. Explosion occurs with fixed rise and fall of luminosity. PH507 Multi-wavelength Professor Michael Smith 9 QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. 5 Other methods include: 1. time delay of light rays due to gravitational lensing, 2. cluster size influences Compton scattering of CMB radiation and bremsstrahlung emission (X-rays). Combining, yields the size estimate (SunyaevZeldovich effect). New Method? Reverse argument: knowing the Hubble constant is 72 km/s/Mpc, (WMAP result), distances can be found directly from the redshift! ( Nearby: Distance = velocity / Hubble Constant ) PH507 Multi-wavelength Professor Michael Smith Questions How do we scale the solar system? How do we find the distance to gas clouds? 10 PH507 Multi-wavelength Professor Michael Smith MASS MEASUREMENT?? PLANET REVIEW The Solar System 11 PH507 Multi-wavelength Professor Michael Smith 12 PH507 Multi-wavelength Professor Michael Smith 13 In the picture above we see the positions of the asteroid belt (green) and other near-earth objects PH507 Multi-wavelength Professor Michael Smith 14 The material in the plane of the Solar System is known as the Kuiper Belt (50 – 1000AU) or Trans-Neptunian Objects. Surrounding this is a much larger region known as the Oort Cloud (3000 – 100,000 AU), that contains material that occasionally falls in, under the influence of gravity, towards the Sun as comets. The Sun over 1.4 million kilometers (869,919 miles) wide. contains 99.86 percent of the mass of the entire solar PH507 Multi-wavelength Professor Michael Smith 15 system: well over a million Earths could fit inside its bulk. total energy radiated by the Sun averages 383 billion trillion kilowatts, the equivalent of the energy generated by 100 billion tons of TNT exploding each and every second…………. 3.83 1026 W Planetary configurations • Some of the definitions below make the assumption of coplanar circular orbits. True planetary orbits are ellipses with low eccentricity and inclinations are small so the concepts are applicable in real cases. • Copernicus correctly stated that the farther a planet lies from the Sun, the slower it moves around the Sun. When the Earth and another planet pass each other on the same side of the Sun, the apparent retrograde loop occurs from the relative motions of the other planet and the Earth. PH507 Multi-wavelength Professor Michael Smith 16 As we view the planet from the moving Earth, our line of sight reverses its angular motion twice, and the three-dimensional aspect of the loop comes about because the orbits of the two planets are not coplanar. This passing situation is the same for inferior or superior planets. A Retrograde loop occurs when a superior planet moves through opposition, and occurs as the earth's motion about its orbit causes it to overtake the slower moving superior planet. Thus close to opposition, the planet's motion relative to fixed background stars, follows a small loop. PH507 Multi-wavelength Professor Michael Smith 17 Lecture 5: Measuring Mass Mass can be measured in two ways. 1 We could count up the atoms, or count up the molecules and grains of dust and infer the number of atoms. This method can be used if the object is optically thin and we have good tracer: a radiation or scattering mechanism in which the number of photons is related to the number of particles. 2 Otherwise, measuring the mass of an object relies upon its gravitational influence….on nearby bodies or on itself (self-gravity). PH507 Multi-wavelength Professor Michael Smith 18 Newton’s second law states: F = m a, while the first law relates the acceleration to a change in speed or direction (law of inertia). (Third law; action – reaction = 0) Kepler’s empirical laws for orbital motion describe the nature of the acceleration from which masses can be derived. Kepler's Laws First Law: The orbit of each planet is an ellipse with the Sun at one focus p b F C Q r S f q a S = Sun, F = other focus, p = planet. r = HELIOCENTRIC DISTANCE. f = TRUE ANOMALY a = SEMI-MAJOR AXIS = mean heliocentric distance), which defines the size of the orbit. b = SEMI-MINOR AXIS. PH507 Multi-wavelength Professor Michael Smith 19 e = ECCENTRICITY, defines shape of orbit. Ellipse, by definition: SP + PF = 2a (1) e = CS / a (2) Therefore b2 = a2(1-e2) (3) •When CS = 0, When CS = , e = 0, e = 1, b = a, the orbit is a circle. the orbit is a parabola. • q = PERIHELION DISTANCE = a - CS = a – ae q = a(1-e) (4) • Q = APHELION DISTANCE = a + CS = a + ae Q = a(1+e) (5) Second Law: For any planet, the radius vector sweeps out equal areas in equal time intervals • Time interval t for planet to travel from p to p1 is the same as time taken for planet to get from p2 to p3. Shaded areas are equal. • Let the time interval t be very small. Then the arc PH507 Astrophysics 20 from p to p1 can be regarded as a straight line and the area swept out is the area of the triangle S p p1. If f1 is the angle to p1, and f is the angle to p: p1 p2 p3 r1 p r f S Area = 1/2 r r1 Sin (f1-f). Since t is very small, (f1-f) = f r ~ r1 and Sin (f1-f) ~ Area = 1/2 r2 f The rate this area is swept out is constant according to Kepler's second law, so r2 df/dt = h (6) where h, a constant, is twice the rate of description of area by the radius vector. It is the orbital angular momentum (per unit mass. The orbital angular momentum is conserved. The total area of the ellipse is πab which is swept out in the orbital period P, so using eqn (6) 2ab/P = h. PH507 Astrophysics 21 The average angular rate of motion is n = 2/P, so n a2(1-e2)1/2 = h (7) Kepler’s Third Law Kepler's third law took another ten years to develop after the first two. This law relates the period a planet takes to travel around the sun to its average distance from the Sun. This is sometimes called the semi major axis of an elliptical orbit. P 2 = ka3 where P is the period and a is the average distance from the Sun. Or, if P is in years and a is in AU: P 2 = a3 Kepler’s Third Law follows from the central inverse square nature of the law of gravitation. First look at Newton's law of gravitation - stated mathematically this is Gm1 m2 r2 Newton actually found that the focus of the elliptical orbits for two bodies of masses m1 and m2 is at the centre of mass. F PH507 Astrophysics 22 The centripetal forces of a circular orbit are F1 r1 v2 X Centre of M ass m1 m2 v1 F2 r2 2 F1 m1 v1 4 2 m1 r1 r1 P2 and 2 F2 m2 v2 4 2 m2 r2 r2 P2 where v 2r P and since they are orbiting each other (Newton’s 2nd law) r1 m2 r2 m1 Let's call the separation a = r1 + r2. Then PH507 a r1 Astrophysics 23 m1 r1 m1 r1 1 and multiplying both sides by m2 , am2 m1 r1 m2 m2 r1 or, solving for r1 , am2 m1 m2 Now, since we know that the mutual gravitational force is Fgrav F1 F2 Gm1 m2 2 a then substituting for r1, a3 G m1 m2 P2 42 Solving for P: P 2 a3 G M1 M2 Third Law is therefore: The cubes of the semimajor axes of the planetary orbits are proportional to the squares of the planets' periods of revolution. Example Europa, one of the Jovian moons, orbits at a distance of 671,000 km from the centre of Jupiter, and has an orbital period of 3.55 days. Assuming that the mass of Jupiter is very much greater than that of Europa, use Kepler's third law to estimate the mass of Jupiter. Using Kepler's third law: PH507 Astrophysics 24 m jupiter meuropa 4 2 a 3 GP 2 The semi-major axis, a = 6.71 x 105 km = 6.71 x 108 m, and the period, P = 3.55 x 3600 x 24 = 3.07 x 105 seconds Thus: m jupiter meuropa c 4 2 6.71 108 h 3 19 . 1027 kg c6.67 10 hc3.07 10 h 11 5 2 and since mjupiter >> meuropa, then mjupiter ~ 1.9 x 1027 kg. Summary of Kepler’s Laws PH507 Astrophysics 25 Summary: Measuring the mass of a planet • Kepler’s third law gives G(M+m) = a3/P2 (To remember, do a dimensional analysis: GM/R2 and R/P2 are both accelerations) . Since M >> m for all planets, it isn't possible to make precise enough determinations of P and a to determine the masses m of the planets. However, if satellites of planets are observed, then Kepler's law can be used. • Let mp = mass of planet satellite ms = mass of satellite Ps = orbital period of PH507 Astrophysics 26 as = semi-major axis of satellite's orbit about the planet. Then: G(mp+ms) = 42 as3/Ps2 If the mass of the satellite is small compared with the mass of the planet then mp = 42 as3/(G Ps2) • All the major planets have satellites except Mercury and Venus. Their masses were determined from orbital perturbations on other bodies and later, more accurately from changes in the orbits of spacecraft. So: we can determine the masses of massive objects if we can detect and follow the motion of very low mass satellites. That doesn’t lead very far. How can we determine the masses of distant stars and exoplanets? BASIC STELLAR PROPERTIES - BINARY STARS • For solar type stars, single:double:triple:quadruple system ratios are 45:46 : 8 : 1. • Binary nature of stars deduced in a number of PH507 Astrophysics 27 ways: 1. VISUAL BINARIES: - Resolvable, generally nearby stars (parallax likely to be available) - Relative orbital motion detectable over a number of years - Not possible for exoplanets! 2. ASTROMETRIC BINARY: only one component detected 3. SPECTRUM/PECTROSCOPIC BINARIES: - Unresolved - Periodic oscillations of spectral lines (due to Doppler shift) - In some cases only one spectrum seen 4. ECLIPSING BINARY: - Unresolved - Stars are orbiting in plane close to line of sight giving eclipses observable as a change in the combined brightness with time (light curves). Some stars may be a combination of these. ************************* DETAILS *************************************************** 1. Visual Binaries • Angular separation ≥ 0.5 arcsec (close to Sun, long orbital periods - years – remember: at 1 parsec, 1 PH507 Astrophysics 28 arcsec corresponds to 1 AU) EXAMPLE: Sirius: Also known as Alpha Canis Majoris, Sirius is the fifth closest system to Sol, at 8.6 light-years. Sirius is composed of a main-sequence star and a white dwarf stellar remnant. They form a close binary, Alpha Canis Majoris A and B, that is separated "on average" by only about 20 times the distance from the Earth to the Sun -- 19.8 astronomical units (AUs) of an orbital semi-major axis -- which is about the same as the distance between Uranus and our Sun ("Sol"). The companion star, is so dim that it cannot be perceived with the naked eye. After analyzing the motions of Sirius from 1833 to 1844, Friedrich Wilhelm Bessel (1784-1846) concluded that it had an unseen companion 1. Visual Binaries • Angular separation ≥ 0.5 arcsec (close to Sun, long orbital periods - years) – example Sirius: HST: PH507 Astrophysics • Observations: Relative positions: = angular separation 29 = position angle Absolute positions: Harder to measure orbits of more massive star A and less massive star B about centre of mass C which has proper motion µ. PH507 Astrophysics 30 Declination N M otion of centre of mass = proper motion µ Secondary E Right Ascension B Primary C A NB parallax and aberration must also be accounted for. • RELATIVE ORBITS: - TRUE orbit: q = peri-astron distance (arcsec or km) Q = apo-astron distance (arcsec or km) a = semi-major axis (arcsec or km) a = (q + Q)/2 - APPARENT orbits are projected on the celestial sphere Inclination i to plane of sky defines relation between true orbit and apparent orbit. If i≠0° then the centre of mass (e.g. primary) is not at the focus of the elliptical orbit. Measurement of the displacement of the primary gives inclination and true semi-major axis in arcseconds a". PH507 Astrophysics 31 i i Incline by 45° Apparent orbit True orbit • If the parallax p in arcseconds is observable then a can be derived from a". Earth B radius of Earth's orbit a a" Sun p r = distance of binary star For i=0°, a = a"/p AU A (In general correction for i≠0 requ Now lets go back to Kepler’s Law … • From Kepler's Law, the Period P is given by 2 3 4 a P = G (mA + mB) 2 For the Earth-Sun system P=1 year, a=1 A.U., mA+mB~msun so 4π2/G = 1 3 a P = (mA + mB) 2 provided P is in years, a in AU, mA, mB in solar masses. PH507 Astrophysics 32 The total system mass is determined: mA + mB = ( a" 3 1 ) 2 p P • ABSOLUTE ORBITS: d c rA * rB B f e B q A Q A Semi-major axes aA = (c+e)/2 aB = (d+f)/2 Maximum separation = Q = c + f Minimum separation = q = d + e So aA + aB = (c+d+e+f)/2 = (q + Q)/2 = a a = aA + aB (1) (and clearly r = rA + rB) From the definition of centre of mass, mA rA = mB rB ( mA aA = mB aB) mA/mB = aB/aA = rB/rA So from Kepler’s Third Law, which gives the sum of the masses, and the equation above, we get the ratio of masses, ==> mA, mB. Therefore, with both, we can solve for the individual masses of the two stars. PH507 Astrophysics 33 3 Spectroscopic Binaries To measure the periodic line shifts requires • Orbital period relatively short (hours - months) and i≠0°. PH507 Astrophysics 34 • Doppler shift of spectral lines by component of orbital velocity in line of sight (nominal position is radial velocity of system): wavelength Time wavelength Time 2 Stars observable 1 Star observable See: http://instruct1.cit.cornell.edu/courses/astro101/j ava/binary/binary.htm • Data plotted as RADIAL VELOCITY CURVE: PH507 Astrophysics Professor Glenn White 35 the line of sight (i<90°), the shape is unchanged but velocities are reduced by a factor sin i. • Take a circular orbit with i=90° a = rA + rB v = v A + vB Orbital velocities: vA = 2π rA / P vB = 2π rB / P v = 2πr/P Since mA rA = mB rB Ratio of masses: mA/mB = rB/rA vB/vA (2) = rB vA •Shape of radial velocity curves depends on orbital eccentricity and orientation r v = rA v B • If the orbit is tilted to • In general, measured velocities are vB sin i and vA sin i, so sin i terms cancel: so we measure the mass ratio accurately. • From Kepler's law mA + mB = a3/P2 (in solar units). PH507 Astrophysics Dr. S.F. Green Observed quantities: 36 vA sin i => rA sin i vB sin i => rB sin i } a sin i So can only deduce (mA + mB) sin3 i = (a sin i)3/P2 For a spectroscopic binary, only lower limits to each mass can be derived, unless the inclination i is known independently. DETAILED DERIVATION when only one radial velocity is known (eliminate VB using mass ratio) **************************************************** 1. Assume a planet and star, both of considerable mass, are in circular orbits around their centre of mass. Given the period P, the star’s orbital speed v* and mass M*, the mass of the planet, Mp is given by Mp3/ (M* + Mp)2 = v*3 P / (2 G) Note that there are 9 unknowns: P, a*, ap, M*, Mp, v*, vp, a, M - 9 variables However, a = a* + ap M = M*+ Mp Centre of mass; M* a* = mp ap PH507 Astrophysics Dr. S.F. Green 37 Kepler’s law relates: P, a, M P = 2 a*/v* P = 2 ap/vp ….so that is 6 equations. Note: we usually only know vr* = v* sin I and we assume the planet mass is small. 4 Eclipsing Binaries • Since stars eclipse i ~90° PH507 Astrophysics Professor Glenn White 38 • For a circular orbit: 1, 1' FIRST CONTACT 2, 2' SECOND CONTACT 3, 3' THIRD CONTACT 4, 4' FOURTH CONTACT v 4' 3' 2' 1' 1 2 3 4 Observer in plane • Variation in brightness with time is LIGHT CURVE. • Timing of events gives information on sizes of stars and orbital elements. • Shape of events gives information on properties of stars and relative temperatures. If smaller star is hotter, then: PH507 Astrophysics 39 Case 1 Smaller star is hotter Case 2 Larger star is hotter F or magnitude Secondary minimum Primary minimum time Case 1 t'1 Case 2 t 1 t'2 t2 t'3 t' 4 t3 t4 t1 t'1 t2 t'2 t3 t'3 t4 t'4 PH507 Astrophysics 40 • If orbits are circular: minima are symmetrical ie t2-t1 = t4-t3 = t2'-t1' = t4'-t3'; minima are half a period apart; eclipses are of same duration. Asymetrical and/or unevenly spaced minima indicate eccentricity and orientation of orbit. • For a circular orbit: DistanceR=S/R velocity x time = L t3 v(t2 t 4 – t1) 2RS =t 1 t 2 and 2RS + 2RL = v (t42R-L t1) => 2RL = v(t4 - t2) 2RS • Light curves are also affected by: Non-total eclises No flat minimum Limb darkening (non-uniform brightness) "rounds off" eclipses Ellipsoidal stars (due to proximity) "rounds off" maxima Reflection effect (if one star is very bright) PH507 Astrophysics Professor Glenn White 41 5 Eclipsing-Spectroscopic Binaries • For eclipsing binaries i ≥ 70° (sin3i > 0.9) • If stars are spectroscopic binaries then radial velocities are known. So: masses are derived, radii are derived, ratio of temperatures is derived Examine spectra and light curve to determine which radius corresponds with which mass and temperature: • Since Luminosity L = 4 R2 T4, (Stefan-Boltzmann equation) the ratio of luminosities is derived from LA LB = RA RB 2 TA TB 4 PH507 Astrophysics Professor Glenn White How else to determine mass? 1. Mass-Luminosity relation binaries. 2. Lensing/micro-lensing. 3. Virial expression 42 derived from