1. Stellar Masses

2. Physics Kepler’s Third Law m m m v G = ( r + r ) r m m m v G = ( r + r2 G 1 2 = 1 1 m1 m2 ( r + r ) 2 r 1 2 1 r1 r2
Center of Mass m m m v 2 G 1 2 = 2 2 ( r + r ) 2 r
Pv1 = 2πr1 ==> v1 = 2πr1/P
Pv2 = 2πr2 ==> v2 = 2πr2/P
m1r1 = m2r2
The Period is the same for both so r1/v1 = r2/v2.
So r1/r2 = m2 / m1 = v1/v2
This is for circular orbits but unequal masses. 1 2 2 m 4 p 2 r G 2 = 1 ( r + r ) 2 P 2 1 2 m 4 p 2 r G 1 = 2 ( r + r ) 2 P 2 1 2 ( m + m ) 4 p 2 ( r + r ) G 1 2 = 2 1 ( r + r ) 2 P 2 1 2 4 p 2 ( r + r ) 3 P 2 = 2 1 G ( m + m ) 1 2

3. Types of Binaries Visual Binaries Astrometric Binaries
A Shopping List
Visual Binaries
Line-of-Sight
Physical
Astrometric Binaries
Common Proper Motion Binaries
Spectroscopic Binaries
Single Line
Double Line
Eclipsing Binaries

4. A Word of Warning The various types are not exclusive!
One could have a visual physical spectroscopic astrometric binary!
Eclipsing and spectroscopic are inclusive
Eclipsing must be spectroscopic
Spectroscopic need not be eclipsing

5. They are close together in the sky
Visual Binaries
They are close together in the sky
HD 38
Possibility 1: Two Stars at different distances lying along the “same” line of sight.
They have nothing to do with each other.
Possibility 2: A Physical Binary system close enough to us that both stars can be resolved as individual objects.
They are related to each other.
So how do you tell the difference?

6. Physical Visual Binaries
What is the Chance that ...
Consider 2 F5 dwarfs of the same apparent magnitude (7.36) 5 arcseconds apart in the sky.
They have the same absolute magnitude and are thus at the same distance.
Their parallax is so their distance is 59 parsecs. (Absolute magnitude = +3.5).
What is their physical separation?
D=r θ where θ= 5" = 0.42(10-5) radians. r = 59 pc so their separation is pc = 295 AU.
Their period: P2 = (295)3 / 3 ==> P ~ 2900 years.
The probability of two stars this close being unassociated is essentially 0.

7. Astrometric Binaries

8. Astrometric Systems
These are physical systems which display their motion on the plane of the sky.
If you wait long enough you will see the entire projected motion.

9. Common Proper Motion Binaries
Two stars which are close.
Neither show apparent orbital motion.
They have common proper motions.
The radial velocities must be consistent but need not be identical.
If the implied absolute magnitudes and apparent magnitudes are consistent with the same distance.
It might be a physical pair.

10. Spectroscopic Binaries
Suppose you have a physical pair which are at a distance of 100 parsecs (from us) and whose separation is 20 AU. What is their apparent separation on the sky?
d = rθ where d = 20 AU and r = 100 parsecs = 100 * * θ  θ = 0.2 arcseconds.
The resolution of a telescope is R = 0.1(105)λ/D (λ and D in cm, R in arcseconds.
At 5000A (5(10-5)) cm this would take a 52.5 cm telescope to resolve from a theoretical point of view
The Earth’s atmosphere at best allows a resolution of ~0.5 arcseconds without active optics.
An observer will measure the combined light of the two stars

11. Spectroscopic Binary: The Movie

12. Spectroscopic Binaries
This is a Double Line System

13. So What Do Real Velocity Curves Look Like?

14. Another Type of Spectroscopic Binary
Actually More Common than Double-Line Systems
The star shows a systematic radial velocity variation but there are only 1 set of lines. Why?
Suppose the brighter star in the system is 5 magnitudes brighter than the other. Then for every 100 photons from the brighter you get 1 from the fainter. The contrast is such that the fainter cannot be directly seen but it’s influence on the motion of the brighter is still seen.

15. The Problem is Geometry
Orbital Parameters
Observer
The Problem is Geometry
The orbital plane is oriented is an arbitrary direction with respect to us.
Consider an object in a circular orbit of period P and radius R. Angular speed = v/R = ω
What does an observer see: v = vocos(ωt)
What if the plane is tilted with respect to us? v = vo sin(i)cos(ωt)
i is the angle of inclination: π/2 means the system is edge on. R

16. Let us Go Back to the 3rd Law
Let us Rewrite R
This can be simplified if we introduce the total distance R between the two stars.
R = r1 + r2
R = r1(1 + r2 / r1)
But r1/r2 = m2 / m1 = v1 / v2
R = r1 (1 + m1 / m2 )
R = (r1/ m2)(m1 + m2 )
4πR3 / G = (m1 + m2 ) P2

17. Solve for the Mass We can determine P - the period
Easier Said Than Done
We can determine P - the period
Somehow determine R - the separation
You now have (m1 + m2 )
The ratio r1/r2 = m2 / m1 = v1/v2 so you have the individual masses given the velocities.
But what about the inclination? R is the true separation and the v’s are the true v not r1sin(i) and v1sin(i).

18. Eclipsing Binaries

19. What Can You Get If It is Not An Eclipsing Binary?
If the orbit is inclined at an angle i, then the Doppler shifts only measure the components vr =vsin(i). In terms of the radial velocities v1r and v2r , and we get (P/2πG)(v1r+ v2r)3 / sin3i = m1+ m2.
If i is unknown all we can do is solve the above equation with i = 90° This will give us a value of m1 + m2 that is a lower limit to the true value.

20. The Mass Luminosity Relation

21. Mass and Luminisity
The previous slide shows a linear relation between two logarithmic quantities (Mbol and Log (Mass)).
This means there is a power law relation between the linear quantities luminosity and mass:
For the Main Sequence L  Mα
α = 1.8 for M/M < 0.3
α = 4.0 for 0.3 < M/M < 3
α = 2.8 for M/M > 3
This has a large impact on stellar evolution
Consider a 10 M star - it is expending energy at times the rate of the Sun but has only 10 times the mass ==> its lifetime is (0.016) that of the Sun!