PHAS 1102 Physics of the Universe 3 – Magnitudes and distances Brightness of Stars • Luminosity – amount of energy emitted per second – not the same as how much we observe! • We observe a star’s apparent brightness – Depends on: • luminosity • distance – Brightness decreases as 1/r2 (as distance r increases) • other dimming effects – dust between us & star Defining magnitudes (1) Thus Pogson formalised the magnitude scale for brightness. This is the brightness that a star appears to have on the sky, thus it is referred to as apparent magnitude. Also – this is the brightness as it appears in our eyes. Our eyes have their own response to light, i.e. they act as a kind of filter, sensitive over a certain wavelength range. This filter is called the visual band and is centred on ~5500 Angstroms. Thus these are apparent visual magnitudes, mv Related to flux, i.e. energy received per unit area per unit time Defining magnitudes (2) For example, if star A has mv=1 and star B has mv=6, then 5 mV(B)-mV(A)=5 and their flux ratio fA/fB = 100 = 2.512 100 = 2.512mv(B)-mv(A) where !mV=1 corresponds to a flux ratio of 1001/5 = 2.512 1 flux(arbitrary units) 1 6 apparent visual magnitude, mv From flux to magnitude So if you know the magnitudes of two stars, you can calculate mv(B)-mv(A) the ratio of their fluxes using fA/fB = 2.512 Conversely, if you know their flux ratio, you can calculate the difference in magnitudes since: 2.512 = 1001/5 log (f /f ) = [m (B)-m (A)] log 2.512 10 A B V V 10 = 102/5 = 101/2.5 mV(B)-mV(A) = !mV = 2.5 log10(fA/fB) To calculate a star’s apparent visual magnitude itself, you need to know the flux for an object at mV=0, then: mS - 0 = mS = 2.5 log10(f0) - 2.5 log10(fS) => mS = - 2.5 log10(fS) + C where C is a constant (‘zero-point’), i.e. C = 2.5 log10(f0) UBV Johnson system (1) So, different types of stars emit strongly at different wavelengths, thus will have different brightness depending on the filter (i.e. the wavelength band) used to observe them. Harold Johnson (1921-1980) pioneered the standard UBV system of filters for measuring magnitudes in various colours. U B V 3600A 4400A 5500A transmission (arbitrary units) Wavelength (Angstroms) 3000 4000 5000 6000 Absolute magnitudes (1) Knowing how bright a star is on the sky is very useful – but the stars all lie at very different distances from the Earth ! we really want to know a star’s intrinsic brightness – i.e. its luminosity (total energy emitted per unit time, in Watt). Astronomers have two ways of quantifying this: Absolute magnitudes and Luminosities Absolute magnitude: magnitude a star would have if it were placed 10 parsec = 3 x 1017 m away from the observer 10pc Absolute magnitudes (2) The flux from any source falls off as the inverse square of the distance, i.e. Recall: Example: a star lies at distance d (in parsecs) with apparent magnitude m and flux fm. If this star was 10 parsecs away, so that its flux was fM, then (because of the inverse square law): But from the definition of magnitude: Distance modulus So, since then where is known as the distance modulus The absolute (V band) magnitude of the Sun is +4.82 Extinction and reddening reddening extinction Any gas and dust which lie between an observer and a star will scatter and absorb the star’s light, making it dimmer. This is extinction. Dust scatters blue light more than red, which makes the star look redder (although, strictly speaking, ‘less blue’) and this is called reddening. This effect makes sunsets and sunrises red. Reddening and colour excess The effect of scattering is made easier to estimate because it is wavelength-dependent, so it will manifest itself as a colour change: Observed colour = (B – V) Intrinsic colour = (B – V)o Reddening is measured by the colour excess which is defined as: E(B – V) = (B – V) – (B – V)o It is measured in magnitudes Correcting for reddening and extinction Spectral features (e.g. Spectral type absorption and emission lines, red continuum) Estimate (B – V)o + In general, extinction is given by: Observe (B – V) A = m – mo Deduce E(B-V) where mo would be the apparent magnitude if there was no extinction. In the V band, studies show that AV = V – Vo ~ 3.1E(B – V) Including extinction in the distance modulus Remember the distance modulus equation: m – M = 5log10d – 5 BUT, m in this equation has been assumed to be unaffected by dust and gas, so it should read: mo – M = 5log10d – 5 From A = m – mo => mo = m – A m – M = 5log10d – 5 + A Astronomical Distance Ladder Direct method – Trigonometric parallax (1) distant stars ! is the parallax angle nearby star !" d 1AU Larger the distance, smaller the parallax angle. Trigonometric parallax (2) In one year, a nearby star will trace out an ellipse on the sky due to parallax: Definition 1 parsec is the distance of an object for which ! = 1 arcsec d (parsec) = 1/! (”) 1 parsec = 206,265 AU = 3.086 x 1016 m = 3.26 light years Parallax was first measured by Bessel in 1838 who found ! = 0.294” for 61 Cygni. Our closest star is Proxima Centauri (# Cen): ! = 0.764”, d = 1.31pc Trigonometric parallax (3) Earth 1AU star !" Sun d Ground-based telescopes – measure ! to ~ 0.01” (d=100 pc) Hipparcos (HIgh Precision PARallax COllecting Satellite) – measured ! to ~ 0.002” (d=500 pc) for some 120,000 stars Gaia – will measure ! to 2x10-5 arcsec (d=50,000 pc) Proper motion µ is measured in arcseconds per year. Largest proper motion known is for Barnard’s Star, where µ = 10.34 arcsec/year. d µ" vt space velocity vt = tangential speed (in km/s) d = distance (in km) (SI units; µ in radian/sec) if vt in km/s, µ in arcsec/year and d in parsec e.g. Nova expansion So, if a star has proper motion µ (in arcsec/year), the tangential velocity of the star is vt = 4.74µd (where d is in parsec and vt in km/s). nova shell d µ" vr vt Instead of a star, consider a nova – an explosion from a star. We assume that the shell thrown off is spherically-symmetric. If we can observe the expansion and measure its velocity, we can determine its distance. Nova expansion... nova shell d µ" vr vt Using spectroscopy and measuring the Doppler effect (c = speed of light): If µ is the proper motion of the shell due only to its expansion: and since vt = |vr|, then (in parsec, if vr in km/s and µ in arcsec/year) Indirect method: Spectroscopic distances – HR diagrams (1) -5 0 Main sequence MV +5 +10 +15 30000 20000 10000 6000 4000 2000 temperature, K If an HR diagram is well-calibrated, the luminosity, and thus the absolute magnitude MV, of any star of known colour or spectral type can be derived. PHAS 1102 Physics of the Universe 4 - Stellar energy generation PARTIAL notes Stellar energy generation Stars are generally very stable. Assuming that only the inward gravitational force and the outward (thermal) pressure are at work in the gas, a star is in hydrostatic equilibrium: Gravitational acceleration [N m-3] dP dr M(r) r !(r) where P: outward pressure (force per unit area) r: radius from the centre of the star G: gravitational constant M(r): mass within radius r !(r): density at radius r The four fundamental forces of Nature Strength Range (m) Carrier Particles acted on quarks (make up nuclear p and n) Strength Range (m) Carrier Particles acted on any particle with charge Strength Range (m) Carrier Particles acted on quarks and leptons nuclear (e, µ, !, ") Strength Range (m) Carrier Particles acted on any particle with mass Fusion: light bind together to form heavy Fission: break up heavy to form light Shown is the energy released per fusion/fission event; this is called the binding energy. Note that a relatively large amount is released from hydrogen to helium. Still, it takes 1038 reactions per second to support the sun! Mass deficit and binding energy The simplest fusion reaction, becoming significant at T ~ 107 K, involves 2 protons (H nuclei) and 2 neutrons combined in the next stable nucleus, 4He (! particle). Atomic mass of proton: 1.0078 amu Atomic mass of 4 protons: 4.0312 amu Atomic mass of 4He nucleus: 4.0026 amu ! mass deficit of 0.0286 amu amu: atomic mass unit (1/12 of the mass of a 12C atom, = 1.66 x 10-27 kg) Mass deficit = nuclear binding energy that holds nucleus together Mass deficit converted into energy according to Einstein’s equation: Total nuclear energy in the Sun Assume 10% of H in the Sun (temperature and density are high enough only in the core) converts into 4He. Fraction of mass liberated into energy: (i. e. efficiency ~ 0.7%) So Then the Sun can continue to radiate for Solar system age: 5 x 109 years ! Sun is ~ halfway through its H-burning phase Fusion processes for H ! He Simultaneous collision of 4 protons very unlikely! Two thermonuclear processes lead to conversion H ! He: 1) Proton-proton chain (PP) Dominates at T < 2 x 107 K 2) Carbon cycle (CNO) Dominates at higher temperatures (contributes 2% of solar energy) Particles involved: Protons, neutrons, electrons, positrons, e+ (= electrons, but charge = +e) neutrinos, ! (very small mass, no charge, only energy and spin) Proton-proton chain (PP1) e+ 1) Weak nuclear force (5 x 109 yr) deuteron ! 0.42 + 1.02 = 1.44 MeV 2) Electromagnetic force (1 s) ! 5.49 MeV 3He 3) Strong nuclear force (3 x 105 yr) ! 12.86 MeV Dominates in stars the size 4 He of the Sun, or less.