phys371_w3Binary_Stars

Binary Star Systems (Chapter 7) Binary Star Systems 1)Classification of binary star systems 1)Optical Doubles 2)Visual binary 3)Astrometric Binary 4)Eclipsing Binary (detached, semi-detached, unattached) 5)Spectrum Binary 6)Spectroscopic Binary 2)Stellar properties measured with binaries 1)Mass 1) From Visual Binaries 2) Complications with Visual Binary method 2)Radii, Surface brightness, Surface temperature 3)Mass functions from Spectroscopic Binaries Classification of Binary Star Systems •Multiple systems are more of a rule than an exception –At least half of all stars are multiple systems •Optical DoubleOptical Double –Lie along the same line of sight and look as though they are companions –They have no physical proximity to one another –Not gravitationally bound –Not a binary star system! •Visual binaryVisual binary –Sufficiently close to Earth and the stars are well enough separated that we can see the two stars individually (resolved) in a telescope and track their motion over a period of time •Astrometric binaryAstrometric binary –Detect the presence of an unseen companion (faint, cool star) by its gravitational influence on the primary star –Deviation in proper motion Classification of Binary Star Systems •Eclipsing binaryEclipsing binary –Orbital planes along the line of sight (or nearly so) –The system is usually not resolved in the telescope (not vis binary) –Photoelectric or CCD photometry over time produces light curve –Q: What can you tell from the shape of the eclipse dips? –Q: How could you tell that the blue star moved left to right? Classification of Binary Star Systems •Eclipsing binary Example - AlgolEclipsing binary Example - Algol –Q: Why is the eclipse of the B8 star deeper than the eclipse of the K2 star? •Smaller blue star emits more light per unit surface area than the larger red star – Q: Why don't the minima have flat bottoms? •During eclipses, neither star gets completely covered up by the other. Classification of Binary Star Systems •Eclipsing binary ExamplesEclipsing binary Examples Classification of Binary Star Systems •Spectrum binarySpectrum binary –System with two superimposed, independent, discernable spectra Classification of Binary Star Systems •Spectroscopic BinarySpectroscopic Binary –Pair cannot be resolved as a visual binary –Measure relative velocities via the Doppler shift of their spectral  lines –Motion is usually seen in the lines of one star Classification of Binary Star Systems •Spectroscopic binarySpectroscopic binary Classification of Binary Star Systems •Spectroscopic BinarySpectroscopic Binary –Double-line spectroscopic binary Δλ λ0=v c Stellar Properties from binaries •Radii •Surface brightnesses •Temperatures •MASSES Mass Determination Using Visual Binaries •Angular separation greater than telescope resolution •Observation of orbits yields ratio of stars’ masses •If distance is known –Individual masses can be determined •angle of inclination, i –Angle between the plane of the orbit and the plane of the sky Mass Determination Using Visual Binaries •Let's assume i = 0o •From the center of mass coordinates –Let's put CM at 0 and find ratio of masses •If the distance is known R=m1r1+m2r2=0 ⇒m1 m2=r2 r1=a2 a1 α1=a1 d,α2=a2 d dα1 α2a1 a2m1 m2com⇒m1 m2=α2 α1 Mass Determination Using Visual Binaries •General Form of Kepler’s third law Where P is the period of orbit and a is the semi-major axis of the reduced mass •Two equations and two unknowns, game overP2 =4π2 G(m1+m2)a3 m1 m2=α2 α1P2 =4π2 G(m1+m2)a3a=a1+a2 Mass Determination… Other complications: •Parallax •Proper motion –Center of mass is at a constant velocity –Can be factored out Non-zero angle of Inclination ̃α1=α1cosi ̃α2=α2cosi m1 m2=̃α2 ̃α1=α2cosi α1cosi=̃α2 ̃α1•The angles subtended by the semimajor axes will be reduced Doesn’t matter for mass ratios Does matter for Kepler’s Law P2 =4π2 G(m1+m2)a3 α=a d⇒a=αd Angle of Inclination •Angle of inclination must be determined •Resolved by noting the Projected focus does not coincide with the observed focus –Center of mass will be off –Inconsistencies will resultP2 =4π2 G(m1+m2)(αd)3 α=̃α cosi m1+m2=4π2 GP2(̃α cosid )3 m1+m2=4π2 GP2(̃αd)3 (1 cosi)3 Spectroscopic Binaries •Usually not a resolvable system •Double line – each star’s spectra is seen •Can determine: –Individual mass, radii, ratio of flux and temperature •Angle of inclination important –Determines the radial velocities v1rmax =v1sini v2rmax =v2sini max zero Spectroscopic Binaries for mass functions •v1r and v2r must both be measurable –Double-line spectroscopic binary vs. single-line spectroscopic binary –Comparable brightness or one star may be overwhelmed m1+m2=P 2πG(v1r+v2r)3 sin3 i=Pv1r3 2πG(1+v2r v1r)3 sin3 i Pv1r3 2πG(1+m1 m2)3 sin3i=Pv1r3 2πG(m2 m2+m1 m2)3 sin3i=Pv1r3 2πG(m1+m2)3 m23sin3im1 m2=v2r v1r m23 (m1+m2)2sin3 i=P 2πGv1r3 Mass function Depends on P and v1r Eclipsing, Spectroscopic Binaries for radii •Estimation of radii •Assume i ≈ 90o •Time duration from first contact ( ta) to minimum light ( tb) yields info about the radius of smaller star, Δt=tb−ta vlvs v=vs+vlRelative velocity To Earth2rs v=2rs Δt⇒rs=v(tb−ta) 2 Eclipsing, Spectroscopic Binaries •Time duration from first contact ( ta) to first exposure ( tc) yields info about radius of the larger star, –This time the distance traveled in-between these events is rl Δt=tc−ta v=2rl Δt⇒rl=v(tc−ta) 2 vlvs v=vs+vl To Earth2rl Eclipsing Binaries •Ratio of effective Temperatures –Obtained from the light curve –Radiative Surface Flux •Both stars visible with no obstructions –Amount of light detected k is a constant that depends on physical constants that will be the same for each star i.e. distance from Earth, detector calibrationFr=Fsurf=σTe4 Bo=Bl+Bs=k(AlFrl+AsFs)=k(πrl2 Frl+πrs2 Fs) Eclipsing Binaries •Primary eclipse minima –Only the large star is visible •Secondary eclipse minima –Part of the large star is covered by the smaller, hotter star •Examining ratios which allow unknown factors to cancelBp=Bl=kπrl2 Frl Bs=Bl−Bls +Bs=k(πrl2 Frl−πrs2 Frl+πrs2 Frs) Bo−Bp Bo−Bs=(Bl+Bs)−(Bl) (Bl+Bs)−(Bl−Bls +Bs)=Bs Bls kπrs2 Frs kπrs2 Frl=Frs Frl= (Ts Tl)4 ⇒Bo−Bp Bo−Bs= (Ts Tl)4