Astronomy 2400Astronomy 2400Physics of StarsPhysics of Stars

Examine basic properties of stars: masses, luminosities, temperatures, and chemical compositions, and how

they are established — the observational method.

Examine the Sun as an example of a typical nearby star — the nearest!

7. Binary Stars and Stellar Parameters7. Binary Stars and Stellar Parameters

GoalsGoals:

1. Recognize different binary star types and how each is observed.

2. Learn how stellar masses, luminosities, and radii are measured using specific binary star types.

3. Learn basic observational formulae for the mass-luminosity relation and link individual star types with specific masses.

Types: http://en.wikipedia.org/wiki/Binary_star

Optical double. Two physically unrelated stars sharing a close coincidence in the line of sight.

Messier 20, the best known example of a line of sight coincidence: separation 49".2 (1863), 51".7 (1966). Inferred absolute magnitudes MV for the two stars:

+0.88, +4.0. Also known as Winnecke 4.http://www.seds.org/messier/m/m040.html

Visual Binary. Two resolved, physically-related stars sharing a very close spatial coincidence.

Beta Cygni (Albireo)

Changes in position angle for HD 30869.

Astrometric Binary. A single nearby star revealed as a binary system from its periodic variations in proper motion reflecting orbital motion about an unseen companion.

The wobbling proper motion of Sirius — an indication of orbital motion about a companion.

Eclipsing Binary. A single star revealed as a binary system from the periodic mutual eclipses of the two stars as they orbit each other with an orbital plane lying nearly in the line of sight.

V-band light curve for YY Sgr.

Primary

Minimum

Secondary

Minimum

Spectrum Binary. A single star revealed as a binary system from the presence of two distinct spectral types in its spectrum — also referred to as a composite spectrum binary.

Example spectra for a single hot subdwarf (top), a composite hot subdwarf (middle), and a K1 V

standard (bottom).

Spectroscopic Binary. A single star revealed as a binary system from periodic variations in in the radial velocities measured from its spectral lines: SB1 = one set of spectral lines detected, SB2 = two sets of spectral lines detected, varying in velocity out of phase with each other.

Spectral line shifts for a SB2.

Review: Properties of ellipses.

r = radius vector, foci = F, F', p = semiparameterθ = position angle, P = periapsis, A = apapsise = eccentricity of orbital ellipsePolar equation of orbit, above right.

Properties of ellipses, 2.

a = semi-major axisb = semi-minor axisc = centre distance = ae

Kepler’s eccentric circle, a circle inscribed on an ellipse.

ν = true anomaly (called θ up until now)E = eccentric anomalyM = mean anomaly = 2π × Area in Yellow/πab

Some geometry:

Yellow Area (PFA) = Area (PDA) – Area (PDF)Area (PDA) = b/a × Area (QDA)

= b/a × [Area (QCA) – Area (QCD)]= b/a × [πa2E/2π – ½ × QD × CD]= b/a × [a2E/2 – ½ a sin E × a cos E]= ½ab (E – sin E cos E)

Some geometry:

Area (PDF) = ½ PD × DF = ½b/a × QD × DF= ½b/a × a sin E × (ae – a cos E)= ½ab sin E (e – cos E)

Thus, Area (PFA) = ½ab (E – e sin E)And M = E – e sin E

The equation: M = E – e sin E

is called Kepler’s Equation, which relates the eccentric anomaly E to the mean anomaly M.M is simple to calculate since it represents the amount of orbit swept out over a time interval t relative to the orbital period P,

i.e. M = 2πt/P, where M is in radians.

M is therefore established by the time t. Once the eccentricity e is established for an orbit, one can calculate the eccentric anomaly E for time t using Kepler’s equation. But the equation is not set up for simple calculation. Instead, it is solved iteratively.

Other equations that follow geometrically:

Kepler’s equation: M = E – e sin E

To solve this iteratively…

2tan

1

1

2tan

21

E

e

e

First reformulate: E = M + e sin E

Next rewrite it in an iterative form called the Newton-Raphson Formula:

Note: M and E are in radians.

Adopt E1 = M, solve for E2, substitute that value into the equation for the next iteration, solve for E3, then continue iterations until the series of Ei values converge on a single value.

An example may help to illustrate the technique, which is readily adapted for Excel.

i

iiii Ee

EeEMEE

cos1

sin1

Example: the orbital elements for the Sirius binary system from a 1960 study are:i = 136°.53a = 7.500 arcseconds (*)e = 0.592 (*)ω = 44°.57Ω = 147°.27P = 50.090 years (*)T = 1894.130 (*)πabs = 0.37921 ±0.00158 arcsecond (Hipparcos) (*)What is the separation of the two stars, in A.U., for 2007.1? Solution uses asterisked parameters.Most recent periastron passage = 1894.130 + (2 × 50.090) = 1994.31. So t = 2007.1 – 1994.31 = 12.79 years, or 2π × 12.79/50.090 = 1.604351 radians.

Thus, M = 1.604351 radians.Successive iterations give:

E1 = 1.604351E2 = 2.184496E3 = 2.112776E4 = 2.111807 (converged)E5 = 2.111807E6 = 2.111807E7 = 2.111807 (fully converged)

Thus, tan(½ν) = (1.592/0.408)½ tan(½ × 2.111807)= 1.9753381 × 1.7674087= 3.4912297

And, ν = 2 × tan–1(3.4912297) = 148°.03318

r = a(1 – e2)/(1 + e cos ν)a(A.U.) = a(")/π(") = 7.5/0.37921 = 19.777959A.U.

So, r = 19.777959(1 – 0.5922)/(1 + 0.592 cos 148°)= 12.846497/(1 – 0.502226)= 25.807894 A.U.

or

r = a(1 – e cos E)So, r = 19.777959[1 – 0.592 cos (2.111807 rad)]

= 19.777959[1 – 0.592 cos (120°.99763)]= 19.777959 × 1.3048815= 25.807893 A.U.

i.e. r = 25.81 A.U.

Measurement of Visual Binaries:

separation = ρ (in arcseconds)position angle = θ (in degrees)

Application to Orbits of Visual Binaries: Zwier’s method.

Shaded areais observedorbit, also an ellipse.

Ellipse centre C is unchanged by projection, but the orbit focus occupied by star S does not necessarily fall on a focus of the projected orbit.

Line CS is portion of theprojected major axisD1CSA1. Green linesdrawn parallel to D1CSA1intersect apparent orbitat ends of the projectedminor axis B1CE1. Or obtain it by bisecting allchords parallel to D1CSA1.

Project all chords parallel to the projected minor axis B1CE1 by the factor k = 1/(1 – e2)½ to get the auxiliary ellipse. The auxiliary ellipse is the projection of Kepler’s eccentric circle, so it can be used to obtain the inclination angle i.

The auxiliary ellipse has asemi-major axis A2C denotedα and a semi-minor axisB2C denoted β. Since it is theprojection of a circle, itfollows that:

a = α

cos i = β/α

Also:

e = ae cos i/a cos i = CS/CA1

To obtain k = 1/(1 – e2)½ .

So from geometry one canfind 3 orbital parameters:

i = orbital inclinationa = semi-major axise = orbital eccentricity

That leaves 4 additional parameters to establish:

P = orbital periodT = time of periastron passage (most recent)ω = longitude of periastron (angle from node)Ω = position angle of the node (descending usually, ascending denoted by *)

Since the auxiliary ellipse isKepler’s eccentric circle tilted through the angle i, theaxis A2CD2 represents the line of nodes for the orbit.

So Ω = angle measured fromnorth (N) eastwards to the nearest node, A2 as illustrated here (reverse view), otherwise D2. Radial velocity data are needed to establish whether it is the ascending (going into the plane of the sky) or descending (coming out of the plane of the sky) node.

To obtain the longitude of periastron, measure the angle from the node to projected periastron, hereA2CA1 = λBy spherical trig (below, Using 4 parts formula): tan ω = tan λ/cos i

The geometrical analysis therefore yields:

i = orbital inclinationa = semi-major axise = orbital eccentricityω = longitude of periastronΩ = position angle of the node

The 2 remaining parameters,

P = orbital periodT = time of periastron passage (most recent)

are established from a temporal analysis of the basic observations of separation and position angle.

Spectroscopic Binaries: Here it is necessary to establish the component of one star’s orbit position along the line of sight.

The radial velocity is the time derivative of that component, plus the systemic motion of the system, V0:

irdt

dV

dt

dzVVR sinsin00

sin

90sin

sin

sin

sini

x

The various components of the radial velocity are evaluated with reference to the equation for the areal constant, h:

dt

dr

dt

drr

dt

d cossinsin

21

22

2

2

22

1

sin2

1

sin

cos1

sin1

cos1

1

eP

ea

dt

d

ea

er

dt

d

e

eae

e

ea

dt

d

dt

dr

21

22

2 122

constantAreal eP

a

P

ab

dt

dr

dt

dr

So:

and:

2

12

2

21

22

2

1

cos12

1

cos112

eP

ea

eaP

eea

dt

d

r

r

dt

dr

coscos1

sin2

coscos1sinsin1

sin2

21

21

2

2

eeP

ia

eeeP

ia

dt

dz

If the constant terms are denoted as:

then:

is the equation describing the radial velocity variations of one star orbiting about another.

21

21

sin2K

eP

ia

coscosK0 eVVR

The radial velocity VR reaches a maximum value when cos (ν + ω) = 1, i.e. ν + ω = 0° when the star is entering the plane of the sky, and it reaches a minimum value when cos (ν + ω) = –1, i.e. ν + ω = 180° when the star is exiting the plane of the sky.

Measured with respect to V0 maximum VR is given by:

α = VR – V0 = K (1 + e cos ω),

while minimum VR is given by:β = V0 – VR = K (1 – e cos ω).

So:K = (α + β)/2 and e cos ω = (α – β)/(α + β)

Orbital Solutions: Lehmann-Filhés Method

Measured relative to V0:

V0 is therefore found either from planimetry or mathematical integration software.

Points where VR – V0 = K[cos(ν + ω) + e cos ω] = 0 have cos(ν + ω) = –e cos ω, so:

BAreaAAreasoBut

BArea

AArea

13

23

12

3

2

2

1

zz

zzdtdt

dz

zzdtdt

dz

z

z

z

z

βα

βαcoscos

e

Now integrate the velocity curve relative to the points of inflection, which correspond to nodal passage of the star: ascending node for maximum VR (1) and descending node for minimum VR (3).

At 2, cos (ν + ω) = –e cos ω. Designate the true anomaly at that point as ν1.

121

232

121

AAreaAAreaSo

0AArea

0AArea

3

2

2

1

ZZzzdtdt

dz

ZZzzdtdt

dz

z

z

z

z

A positive sign applies since point 2 lies between the ascending and descending nodes where z reaches a maximum, i.e.

z2 = r1 sin (ν1 + ω) sin i

A minimum value for z is reached at point 4 where

z4 = r2 sin (ν2 + ω) sin i

2

sinAnd 2

2

1

22

11

2

1

12

222

111

sinsin

sinsin

sinsinBut

sinsinArea

sinsinArea

1

4

2

1

r

r

ir

ir

irdtdt

dz

irdtdt

dz

z

z

z

z

which, combined with the equation for e cos ω yields a solution for both e and ω.

12

12

11

22

1

2

1

2

2

1

2

1

2

2

21

2

1

2sinwhichFrom

onsubstitutibysin2

sin2

sinsincoscos1

sinsincoscos1

cos1

cos1

cos1

cos1

cos1

1and

cos1

1Since

e

e

e

ee

ee

e

e

e

e

r

r

e

ear

e

ear

Periastron passage T occurs for ν = 0, in other words for:

VR = V0 + K(1 + e) cos ωwhich can be determined since K, e, ω and V0 are known.

The possible ambiguity in where T falls in the velocity curve is resolved by noting that (ν + ω) = 0° at point 1 and (ν + ω) = 180° at point 3.

The orbital period P is established by curve fitting to the radial velocity data, typically by Fourier analysis or other similar techniques.

The half amplitude of the velocity curve is

when P is expressed in days (= 86400 seconds) and α and β are in km/s.

21

21

21

21

2

2

2

2

1600,21

2

1

2

1sin

21

sin2K

eP

eKP

ePia

eP

ia

Some examples to illustrate how the velocity curve depends upon the orbital parameters:

Parameters Obtained:Double-lined systems.

Single-lined systems:

32128

321

23

321

1

2

1

2

1

2

2

1

KK11038.10

KK12

1sin

K

K

sin

sin

23

23

eP

eiMM

ia

ia

a

a

M

M

2

31

20

221

23323

1

sin10993.3

functionmassthe,sin

sin

P

ia

MMM

PiMia

Eclipsing Binaries:

Eclipse terminology:Denote the larger star (usually the cooler star) as star 1 and the smaller star (usually the hotter star) as star 2.A transit is an eclipse of star 1 by star 2, and can be either annular or partial.An occultation is an eclipse of star 2 by star 1 (the deeper, primary eclipse) and can be either total or partial.The exact situation depends upon the inclination of the orbit, i, and the radii of the two stars.

Complicating factors: Stellar limb darkening.

Eccentric orbits.

Irradiation and ellipticity.

At present the light curves of eclipsing binaries are analyzed using sophisticated computer models that incorporate limb darkening,

ellipticity, and irradiation effects in a single package, i.e. the Wilson-Devinney code.

A total eclipse (occultation): W Delphini.The hotter star (more radiant) is completely covered at mid-eclipse, leading to a flat-bottomed light curve.

An annular eclipse (transit): YZ Cassiopeiae.The hotter star (more radiant) is larger than the cooler star (secondary), so is not completely covered at mid-eclipse, leading to a round-bottomed light curve.The point α = 1 corresponds to what we would term third contact.

Planet transits of the star HD 209458.Note that the eclipses are round-bottomed, a result of limb darkening on the star.

Summary:Visual binaries give the sum of the masses of the stars in a system. If a binary is resolved and close enough for astrometry to detect the motion of the system barycentre, the individual masses for the stars can also be established. Luminosities can be derived for systems of established distance.Spectroscopic binaries place constraints on the masses of stars in the system. SB1s give only a mass function, while SB2s give mass ratios. An eclipsing SB2 yields the masses of both stars, since i is established.Eclipsing binaries give the luminosities of both stars in the system (from R and Teff), but only yield masses if they are also SB2s.

Result:A relationship between mass and luminosity of stars, the ML relation, is established from the best studied systems (Popper, 1980, ARAA, 18, 115).

When calibrated (Smith, 1983, Observatory, 103, 29) one obtains:log L/Lsun = (3.99 ±0.03) log M/Msun M/Msun > 0.43i.e. L ~ M4

log L/Lsun = (2.26 ±0.20) log M/Msun – (0.64 ±0.20) M/Msun < 0.43, i.e. L ~ M2¼

A better calibration from Griffiths, Hicks & Milone (1988, JRASC, 82, 1):log L/Lsun = 4.20 sin (log M/Msun – 0.281) + 1.174for angle argument in radians.

Note turnover

Results for typical stars:Main Sequence:B0 V ~14 Msun

B5 V ~4 Msun

A0 V ~2.1 Msun

F0 V ~1.5 Msun

G2 V ~1.0 Msun

K0 V ~0.8 Msun

M0 V ~0.4 Msun

M supergiants ~15-25 Msun O5 V ~ 32 Msun

K giants ~ 1-2 Msun

The most massive stars? Perhaps ~60 Msun

Note that the ML relation exists only for stars lying near the main sequence.

Postscript:Dynamical parallax, a technique used for binary systems with dwarf components to estimate the distance to the system.

Iterate beginning with (M1 + M2) = 1 or 2, estimate πdyn from the above equation, evaluate the luminosities of the stars in the system, estimate new masses from the ML relation, and continue until convergence.

3

13

2

21

dyn

2

3

21

so

1

MMP

"a"

P"

"aMM

Sample Problem:Calculate the dynamical parallax for a visual binary system consisting of two stars, both classified as spectral type G5 V with magnitudes V = 6.26 and V = 6.36, having an orbital period of P = 25.0 years and a semi-major axis a = 0".67. Assume bolometric corrections of BC = –0.05 for G5 V stars and that the system is close enough to be unreddened by interstellar dust.Solution:From the equation for dynamical parallax,

So MV(1) = 6.26 + 5 log 0.0621972 + 5 = 5.23, Mbol(1) = 5.18MV(2) = 6.36 + 5 log 0.0621972 + 5 = 5.33, Mbol(2) = 5.28

log L1 = [Mbol(Sun) – Mbol)]/2.5 = (4.79 – 5.18)/2.5 = –0.156log L2 = [Mbol(Sun) – Mbol)]/2.5 = (4.79 – 5.28)/2.5 = –0.196log M1 = log L1/3.99 = –0.156/3.99 = –0.0391, M1 = 0.914log M2 = log L2/3.99 = –0.196/3.99 = –0.0491, M2 = 0.893

0621972.000.20.25

67.03

13

23

13

2

21

1dyn

MMP

"a"

Our new estimate for M1 + M2 = 0.914 + 0.893 = 1.807 Msun and

So MV(1) = 6.26 + 5 log 0.0643371 + 5 = 5.30, Mbol(1) = 5.25MV(2) = 6.36 + 5 log 0.0643371 + 5 = 5.40, Mbol(2) = 5.35

log L1 = [Mbol(Sun) – Mbol)]/2.5 = (4.79 – 5.25)/2.5 = –0.184log L2 = [Mbol(Sun) – Mbol)]/2.5 = (4.79 – 5.35)/2.5 = –0.224log M1 = log L1/3.99 = –0.184/3.99 = –0.0461, M1 = 0.899log M2 = log L2/3.99 = –0.224/3.99 = –0.0561, M2 = 0.879Our new estimate for M1 + M2 = 0.899 + 0.879 = 1.778 Msun and

Further iterations give:π1 = 0.0621972, π2 = 0.0643371, π3 = 0.064685, π4 = 0.0647336,π5 = 0.0647824, π6 = 0.0647824 (converged)

So the dynamical parallax of the system is πdyn = 0".065.

0643371.0807.10.25

67.03

13

23

13

2

21

2dyn

MMP

"a"

064685.0778.10.25

67.03

13

23

13

2

21

3dyn

MMP

"a"