#### Transcript Lecture 6 - University of Waterloo

```Lecture 5
Binary stars
Binary stars
• 85% of all stars in the Milky Way are part of multiple systems
(binaries, triplets or more)
• Some are close enough that they are able to transfer matter
through tidal forces. These are close or contact binaries.
Examples
Two stars are separated by 3 A.U. One star is three times more
massive than the other. Plot their orbits for e=0.
Example: binary star system
Two stars orbit each other with a measurable period of
2 years. Suppose the semimajor axes are measured
to be a1=0.75 A.U. and a2=1.5 A.U. What are their
masses?
m1 r2 a2
 
m2 r1 a1
4 2 a 3
P 
G m1  m2 
2
Visual binaries: mass determination
A perfect mass estimate of both stars is possible if:
1. Both stars are visible
2. Their angular velocity is sufficiently high to allow a
reasonable fraction of the orbit to be mapped
3. The distance to the system is known (e.g. via parallax)
4. The orbital plane is perpendicular to the line of sight
Example: Sirius
m1 r2 a2
 
m2 r1 a1
4 2 a 3
P 
G m1  m2 
2
Sirius A and B is a visual binary:
 a period of 49.94 yr
 a parallax of p=0.377”
 The angular extent of its semimajor axis is
a=aA+aB=5.52”.
 aA/aB=0.466
a
Assume the plane of the orbit is in the plane of the sky:
R
R is the distance to
the star.
Visual binaries: inclination effects
In general the plane of the orbit is not in the plane of the sky.
Here is the true orbit
Focii
Visual binaries: inclination effects
In general the plane of the orbit is not in the plane of the sky.
Here is the true orbit, which defines the orbital plane
Visual binaries: inclination effects
In general the plane of the orbit is not in the plane of the sky.
Now imagine this plane inclined against the plane of the sky with angle i:
i
Visual binaries: inclination effects
In general the plane of the orbit is not in the plane of the sky.
Now imagine this plane inclined against the plane of the sky with angle i:
i
2acosi
Instead of measuring a semimajor axis length a, you measure acosi
where i is the inclination angle
Visual binaries: inclination effects
This projection distorts the ellipse: the centre of mass is not at the
observed focus and the observed eccentricity is different from the true
one.
This makes it possible to determine i if the orbit is known precisely enough
Visual binaries: inclination effects
In practice we don’t measure a physical distance a, but rather
an angular distance that we’ll call a. If a is the true angular
distance, and a is the measured (projected distance) then:
m1 a 2 a 2 cos i a 2



m2 a1 a1 cos i a1
So the ratio of the masses is independent of the inclination
effect
However, the sum of the masses is
not:
a cosi
R
4 2 a 3
m1  m2 
GP 2
3
4 2 aR 

GP 2
4 2  R  3

 a
2 
GP  cos i 
3
Example
How does our answer for the mass of Sirius A and B
depend on inclination?
a cosi
Sirius A and B is a visual binary:




a period of 49.94 yr
a parallax of p=0.377”
The angular extent of its observed semimajor axis is
a=5.52”.
aA/aB=0.466
R
Ra 
cos i
7.1 10 5

pc
cos i
a
i
2acosi
Example
How does our answer for the mass of Sirius A and B
depend on inclination?
Ra 
cos i
7.1 10 5

pc
cos i
a
mB 
0.40
M Sun
cos 3 i
mB a A

 0.466
m A aB
a
1.5 1012
aB 

m
1  a A / aB
cos i
m A  2.1mB 
4 2
a3
mB 
GP 2 1  aB / a A 
0.84
M Sun
cos 3 i
Thus our answers are a lower limit on the mass of these stars. The
measured inclination is actually i=43.5 degrees. So cos3i=0.38
and mA=2.2 Msun, mB=1.0 Msun
Break
 

z  obs rest 
rest
rest
Spectroscopic binaries
vr
 z if z  1
c
Single-line spectroscopic binary: the
absorption lines are redshifted or
blueshifted as the star moves in its orbit
Double-line spectroscopic binary: two
sets of lines are visible
Java applet: http://instruct1.cit.cornell.edu/courses/astro101/java/binary/binary.htm
Spectroscopic binaries: circular orbits
• If the orbit is in the plane of the sky (i=0) we observe no radial
velocity.
• Otherwise the radial velocities are a sinusoidal function of time.
The minimum and maximum velocities (about the centre of mass
velocity) are given by v max  v sin i
1r
1
v2max
r  v2 sin i
Spectroscopic binaries: circular orbits
• We can therefore solve for both masses, depending only on the
inclination angle i
m1 
m2 
v

max 2
2r
v
2G sin 3 i
max
2r
Pv
v
max
1r

max 2
2r
v
2G sin 3 i
max
1r
Pv
max
1r
• In general it is not possible to uncover the inclination angle.
However, for large samples of a given type of star it may be
appropriate to take the average inclination to determine the
average mass.
Spectroscopic binaries: non-circular
If the orbits are non-circular, the shape of the velocity curves
becomes skewed in a way that depends on the orientation
e.g. e=0.4, i=30°,
axis rotation=45°
•
•
•
A sinusoidal light curve means orbits are close to circular
From analysis of light curve it is possible to determine the eccentricity and orbit
orientation, but not the inclination.
In practice most orbits are circular because tidal interactions between the stars
tend to circularize the orbits
Java applet: http://instruct1.cit.cornell.edu/courses/astro101/java/binary/binary.htm
Single-lined spectroscopic binaries
In general, one star is much brighter than the other
(remember faint stars are much more common than
bright stars). This means only one set of absorption
lines is visible in the spectrum.
 The Doppler motion of this single set of lines still indicates
the presence of a binary system.
 We can still solve for a function of the two masses:
 
m23
P v1max
3
r
sin
i

2G
m1  m2 2
3
This is the mass function
Eclipsing binaries
A good estimate of the inclination i can be obtained in
the case of eclipsing binaries, separated by distance d:
R1+R2
To observer
i
cos i 
d
R1  R2
d
If d » R1+R2 (which is usually
the case) then i~90
degrees
Eclipsing binaries
A good estimate of the inclination i can be obtained in
the case of eclipsing binaries, separated by distance d:
R1+R2
To observer
cos i 
i
R1  R2
d
If d » R1+R2 (which is usually
the case) then i~90
degrees
d
Assume i=90 degrees when
in reality i=75 degrees.
What is the error in sin3i?
sin3(75)=0.9
So the error on the masses
is only 10% if d > 3.9(R1+R2)
Eclipsing binaries
In the system just described, the eclipse just barely
happens:
To observer
Face on
So the amount of light blocked is not constant, and the light curve (total
brightness as a function of time) looks something like this:
Eclipsing binaries
However, in the case of total eclipse the smaller star is completely
obscured. In this case it is even more likely that the inclination is
close to 90 degrees
Face on
To observer
And the light curve shows constant minima:
Eclipsing binaries
In the case of a total eclipse we can also measure the radii of the
stars, and the ratio of their effective temperatures
If we assume i~90 degrees and circular orbits that
are large relative to the stellar radius, then the
radius of the smaller star is:
And for the larger
star:
v
rl  tc  t a 
2
v
rs  tb  t a 
2
Where v is the relative
velocity between the
two stars
Eclipsing binaries
Ratio of effective temperatures
(0)
(1)
(2)
L  4R 2Te4
Ltotal  Rs2Ts4  Rl2Tl 4
L1  Rl2Tl 4


L2   R  R Tl  R T
2
l
2
s
Ltotal  L1 Rs2Ts4
 2 4
Ltotal  L2 Rs Tl
 Ts 
  
 Tl 
4
4
2 4
s s
Note that (1) will be the
deepest minimum if Ts>Tl.
(often the case since the
brightest, largest stars are the
cool supergiants)
Alternatively (2) will be the
deepest minimum if Tl>Ts
Stellar masses
• For select star systems,
we can therefore
measure the mass
directly.
• Luminosity is closely
correlated with stellar
mass
 Energy production rate is
related to stellar mass
 If the available energy is
proportional to mass, how
depend on their main
sequence location?
L  M 2.5
L M5
The main sequence revisited
• The main sequence is a
mass sequence
 More massive stars
are closer to the
top-left (hot and
bright)
M=30MSun
M=MSun
M=0.2MSun
The main sequence revisited
• The main sequence is a
mass sequence
 More massive stars
are closer to the
top-left (hot and
bright)
• Stars on the main
sequence have radii 13 times that of the
Sun
 Supergiants have
R>100 RSun
 White dwarfs have
R~0.01 RSun
M=30MSun
M=MSun
M=0.2MSun
Densities
Since we know the stellar masses and radii, we can
compute their average densities
Sun:
M

4 / 3R 3

1.99 1030 kg
4 / 36.96 10
 1409 kg/m
3
8
m

3
Recall water has a density of
1000 kg/m3
Dry air at sea level: 1.3
kg/m3
Densities
Since we know the stellar masses and radii, we can
compute their average densities
Supergiants (Betelgeuse):
M  10M Sun
R  1000 RSun
   Sun
M  R 


M Sun  RSun 
3
 1409101000  1.4110 5 kg/m 3
3
or 1/100,000 times less
dense than air.
Densities
Since we know the stellar masses and radii, we can
compute their average densities
White Dwarfs (Sirius B):
M  0.6M Sun
R  0.01RSun
  6 105  Sun
or 850,000 times denser than water.
```