Download Lecture5

II-6 Stellar Mass and Binary Stars
Lec. 5
(Main Ref.: Lecture notes;
FK Sec.4 - 4, 6, 7; 17-9, 10, and 11, Box 4-2, 4-4)
II-6a. Introduction
If the star is isolated, hard to measure mass directly.
But fortunately, ~ half of all stars are in multiple systems – most
in binary systems. Then, can use Kepler’s Law!
Note: However, if the star is on the main sequence, can use the massluminosity (ML) relation, i.e., find L, then get M!
II-6b. Kepler’s Law (Main Ref.: Lecture notes; FK Sec.
4 – 4,6,7; 17-9, Box 4-2, 4-4)
M = M1 + M2 ≤ 4 2 a3 / (G P2 ), Eqn(19) 1
where M1, M2 are mass of Star 1 and Star 2; P = orbital
period;
a = separation between Star 1 and Star 2 = semi-major
axis of one star’s orbit around another.
G = 6.67 x 10–11 Newton m2 /kg2 = Universal constant
of gravitation.
Often elliptical orbit
(see Fig. II-31a, 31b),
but for simplicity, we
will assume a circular
orbit.
Fig II-31a: A Binary Star System
2
Note: the (=) sign only if
the inclination angle  of
the orbital plane to the
plane of the sky is known.
If not, the (<) sign
applies, i.e., only the
maximum mass of the
system M will be found.
Note: in Fig. II-32, M1 <
M2
(See class notes for
explanation.)
Fig II-31b: Elliptical Orbit
M2
2
a2 COM
x
a1
a
Fig II-32: Binary System 3
M1
1
Center of Mass (COM):
In the case of two stars in a binary system, Star 1 and Star 2 move
about COM. See FK for more general definition. See Fig. II-32
and 33.
Fig II-33: Center of Mass
4
Often orbit is elliptical (e.g., see Fig. II-31a, 31b), but for simplicity here we assume
the orbit is circular.
Note: From mechanics and the definition of COM - see Fig. II-32,
a1 M1 = a2 M2
Eqn(20a)
a = a 1 + a2
Eqn(20b)
where a1 and a2 are the distance of Star 1 and Star 2 from COM.
Now, use special unit: then,
M1 + M2
≤ a3 / P2
M = M1 + M2
Eqn(21a)
Eqn(21b)
when M, M1 and M2 in unit of M(solar), P in year, and a in AU.
M = total mass of the system. See Section II-6e for examples.
5
II-6c. Binary Stars
(Main Ref.: Lecture notes; FK Sec. 5-9, 17-9,
17-10, 17-11)
(i) Visual Binaries (FK Sec. 17-9, Lecture notes)
Can visually resolve the two component stars of the binary.
Some coincidence (called `optical double stars’) and the two
stars are not related. But most of them are real double stars – i.e.,
gravitationally bound binary, moving around the COM.
(ii) Spectrum Binaries (FK Sec. 17-10, Lecture notes)
Binary consists of two stars of different spectral types.
Although two stars cannot be visually resolved separately (too far
from us, too close to each other, or both), we can identify the two
stars by the presence of two different classes of spectral lines.
6
(iii) Spectroscopic Binaries (FK Sec. 5-9, 17-10, Lecture notes)
We can identify the binary motion, characteristics of the
component stars, etc., by measuring periodic shifting of
spectral lines due to the Doppler effect as the stars move
around the COM.
In most cases the component two stars cannot be visually
resolved separately, however (too far from us, too close to
each other, or both).
In some cases, the shifting of spectral lines can be resolved for
both stars. In many cases the shifts can be resolved for only
one of the stars – e.g., because the line is too weak, or shift too
small.
See class notes and Fig. II-34 and 35 (next page) for the
detailed explanation.
7
Fig II-34 Radial Velocity Curves
Fig II-35 Shift of Spectral Lines
8
Doppler Effect: See FK Sec. 5-9, Lecture notes
 / 0 = ( – 0) / 0 = v / c
Also,
(v1 / v2 ) = (a1 / a2 )
Eqn(22) (non-rel.)
Eqn(23a)
v = radial velocity. (Derivation not given in this class.)
v = v1 + v2 = 2 π a / P
v1,2 = 2 π a1,2 / P
Eqn(23b)
Eqn(23b)
Method: From the observation and analysis of the spectrum such as Fig. II-35,
measure P, 1, 1, 2 and 2.
1, 1, 2 and 2  Eqn(22) for Star 1 and Star 2 separately(*)
 get v1 and v2  get (v1 / v2 ).
Then, solve Eqns (20a), (20b), (21a), (21b) and (23), and we can find M1 and M2
separately (see EX. 26).
(*) If the lines of only one of the two stars are strong enough for the above analysis,
we do not get M1 and M2 separately.
9
Inclination Angle  - complication
The above analysis applies if we are looking edge on (for
spectroscopic binaries).
Normally it is not. Then, only the maximum mass known.
(See class notes for the explanation.)
If the binary is also an eclipsing or visual binary, we can often
find . Then, can find real mass, not maximum mass (not required for this
course – discussed in Physics 435.)
(iv) Eclipsing Binaries (FK Sec. 17-11, Lecture notes)
When one star of a binary gets behind the other star, eclipse
happens. The intensity (brightness) of the binary vs time curve
is called `light curve’. Then the shape of the light curve can
often determine various useful parameters, e.g., the inclination
angle , nature of the stellar atmosphere, and sometimes even
stellar radius.
See class notes and Fig. II-36 for explanation.
10
11
Fig II-36: Eclipsing Binaries
II-6d. Mass Luminosity Relation (Main Ref.: Lecture notes;
FK Sec.17-9)
For main sequence stars (only!), stellar luminosity L and mass M are related by:
L ~ a M4(*)
Eqn (24a)
a is some constant. Derivation in Physics 435, but not in this course.
(*) Note: it deviates from 4 for both low and high mass, but ~ o.k.
See Fig. II-37.
By dividing Eqn (24a) for the star by that for the sun, we get:
L / L☉ ~ ( M / M☉ )4
Eqn (24b)
**********************************************
EX 23
1 M☉ Star  ~ 1 L☉ ;
10 M☉ Star  ~ 104 L☉ ;
0.1 M☉ Star  ~ 10-4 L☉
12
Note: ms mass changes
from ~ 0.2  60 M☉.
FK
FK
Fig II-37 Mass Luminosity Relation
Fig II-38 Mass on the H-R Diagram
13
II-6e. Examples
EX 24
Two stars in a binary system have the same mass of 2M☉
and they are separated by 4 AU. What is the orbital period?
Ans: 4 years. (see class notes.)
*************************************************************************************
EX 25
Two stars in a binary system are separated by 4 AU and their
orbital period is 1 year. What is the total mass of the system?
Ans: 64 M☉. (see class notes.)
*************************************************************************************
EX 26
In problem EX 25, the ratio of the distance of Star 1 from COM to that of Star 2
is 3. What are the mass and distance from COM of Star 1 and Star 2?
Ans: Star 1 is 16 M☉, and Star 2 is 48 M☉. Distance of Star 1 and 2 from COM are
3AU and 1AU, respectively.
(See class notes.)
14
• How to find Mass? (Summary)
• If stars are in a binary system, use Kepler’s Law ((Eqn(19) or
Eqns(21a)(21b))  If we can measure a and P, we get
the total mass of the system M.
P can be measured directly for all binary kinds if it is not too long.
a can be measured if the system is looked edge on (both for
spectroscopic and eclipsing binaries) or if inclination angle is
found (o.k. for some visual and eclipsing binaries).
Otherwise, only the maximum mass can be found.
• If both a1 and a2 (or their ratio a1/a2) are found (can happen for
some of all binary kinds), then Eqns (20a)(20b) give mass of both
stars separately.
• If it is a spectroscopic binary, the relative wavelength shifts of the
lines of the two stars (if both can be resolved) will give ratio of the
radial velocity of the two stars, which gives the ratio a1/a2 from
Eqn(23).
• If a star is a main sequence star, the mass luminosity relation
15
Eqn(24b) and measurement of luminosity give mass.
II-7 Stellar Structure and Energy Source
(Main Ref.:
Lecture notes; FK Sec. 16-1, 16-2)
II-7a. Energy Source for Main Sequence Stars
(Main Ref.: Lecture notes; FK Sec.16-1)
Gravitational Source: too fast! tG = 2.5 x 106 years!
But: geological and biological evidence, radioactive decay time, etc.

Age of the Earth t ~ 4.6 x 109 years ~ Age of the sun t☉.
So,  Sun must be as old!
Chemical Source: even faster! tC ~ 104 years!
Nuclear Source: o.k.! tN ~ 1010 years!
Nuclear Fusion: Main Sequence stars 
4 H  He + energy
Eqn(25)
16
Low mass stars pp chain, high mass stars CNO cycle (details not required for this class).
Idea: Mass m can be converted to energy E, by Einstein’s:
E = m c2
Eqn(26)
By the `Hydrogen (H)-Burning’ (i.e., H  He conversion = Eqn(25)),
0.7 % of total stellar mass will be converted to energy (Derivation optional)
But H-burning stops in the central region of the star when ~ 10  30% of the mass
changes from H to He (see Sections II-7b and 8b).
Then, we get: duration of H-burning = tN ~ 1010 years!
Derivation: use Eqn(26) and relation
EN ~ L☉ tN ~ m c2.
Eqn(27)
EN = nuclear energy released by H-burning. m is the fraction of stellar mass converted to energy
by H-burning = 0.007 x (0.1 – 0.3) M☉.
See HW V-4 and class notes for the details.
Note: 1 year ~ 3 x 107 sec., c = 3 x 108 m/sec, L☉ = 4 x 1026 J/sec., M☉= 1.9891 x 1030 kg
17
II-7b. Stellar Structure-Models (Main Ref.: Lecture
notes; FK Sec.16-2)
Stellar interior – hot, e.g., central core temperature
Tc ≥ ~ 107 K for stars like the sun (- compare with the surface
temperature Ts = 5800 K). H and He (main composition) all
ionized at such high T.
Note: Thermonuclear reactions (e.g., H-burning) take place only in the
central high T core.
Stellar structure found by solving basic stellar structure equations:
(i) Hydrostatic Equilibrium: mecahnical balance,
(ii) Energy Balance:
Total energy production/sec = Total energy loss/sec,
18
(iii) Energy Transport: the manner in which energy produced in the
central core is transported to the surface.
(iiia) Conduction: energy carried by free electrons (e.g., metals).
(iiib) Radiation: energy carried by photons (`random walk’ –
takes long time from interior to surface).
(iiic) Convection: energy carried by particles, by gas motion (e.g.
boiling watter).
•
•
By solving these basic equations, can find the stellar structure 
Results: see Fig. II-40, 41.
Note: Temperature T, pressure P and density  decrease, while
luminosity L and mass m increase, from center to surface.
Note: T > ~ 107 K only for r < 0.20 R☉, where H-burning can take
place.
(See class notes for detailed explanation.)
19
Fig II-39: Pressure Balance
Fig II-40: Sun’s Interior Structure
Table II-4: Model of Solar Structure
20
Fig II-41: Interior
Structure of a Sun-like
Star – radial distribution
of L, m, T, and .
21