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Dimensions of Stars

Having determined that stars are at
large distances, high temperatures and
very luminous, the next question to ask
is how large they are.

An Example:
Dimensions of Stars
– The star Betelgeuse (a Orionis) has the
following parameters:
– Distance 150 pc
– Apparent magnitude 0
– Absolute magnitude -6
– Relative luminosity 104 L
– Temperature 3000 K (I.e., a class M star).
Dimensions of Stars
– The luminosity depends on the surface
area of the star (R2) and T 4 (from
Stefan’s law).
– Hence, the ratio of the radii of Betelgeuse
and the Sun is:
– R / R = (T / T )2 (L / L)1/2 = 370
– This corresponds to almost 2 AU – I.e.,
larger than the orbit of Mars (1.5 AU)!
106
Dimensions of Stars
Supergiants
104
102
Luminosity (L)
– Applying the
same
technique to
other stars
allows lines of
constant radii
to be plotted
on the HR
diagram.
Giants
1
Sun
10-2
10-4
40,000 20,000 10,000
5,000
Temperature (K)
2,500
Dimensions of Stars

Note:
– Most stars on the main sequence have radii
similar to the Sun, but that, as we have
shown, the luminous red stars have large
radii.
– White dwarfs have very small radii –
comparable to that of the Earth!
Luminosity and Spectral Class

Can the luminosity be deduced directly
from the spectral class?
– No - e.g. Betelgeuse and Barnard’s Star are
both class M, but the latter is 18 stellar
magnitudes fainter

Details of the spectrum provides a
handle:
Luminosity and Spectral Class

Absorption line shapes:
– Narrow lines  Low density, hence large
dimensions
– Broad lines  High density, hence a
compact object
Rigel, a B8 supergiant
Hg
The B8 main sequence component of Algol
Hd
Luminosity and Spectral Class
106
Such studies lead to
the idea of
104
II
III
Luminosity Classes
– Labelled I to V in
order of decreasing
luminosity
– Provides another
handle on distance

“Spectroscopic
Parallax”
102
IV
Luminosity (L)

I
V
1
10-2
10-4
40,000
20,000
10,000
5,000
Temperature (K)
2,500
Spectroscopic Parallax

Example:
– The star a Leonis (Regulus) has an
apparent magnitude of 1.4 and its
spectrum shows it to be a B7V star.
– This gives an absolute magnitude of -0.6.
– The distance modulus is therefore 2, giving
a distance of 25 pc.
Spectroscopic Parallax

Note:
– The luminosity classes are broad and
poorly defined. Hence, a distance
determined by this method should be
regarded as an estimate
– This method has nothing to do with
Parallax!
Masses of Stars

We have now established methods of
measuring :
– Distance
– Brightness
– Composition
– Temperature
– Size
Masses of Stars


we require the mass to gain a complete
picture of the properties of stars.
For example:
– are giant and dwarf stars proportionatly
more massive than the Sun, or are they
about the same?
– How does mass vary along the main
sequence?
Masses of Stars

It is impractical to measure the mass of
an isolated star.

Many stars occur in gravitationally
bound multiple systems – in particular,
binary stars.

By observing the orbits of such stars
(and knowing the distance), the total
mass can be found
Masses of Stars

Application of Kepler’s Third Law
a3
M1  M2  2
P
– Mi in Solar Masses, a in AU and P the
orbital period in years
M1
a
M2
Masses of Stars

Example:
– The orbit of the binary star 70 Ophiuchi
(parallax 0.2 arc sec) has been plotted over
many years.
– The period is 87.7 years
– The semi-major axis is 4.5 arc sec.
– What is the sum of the masses of the two
stars?
Masses of Stars
– 70 Ophiuchi

Distance = 5pc
– ~ 106 AU.

Hence, the semi-major axis is ~ 22 AU.

The period is 87.7 years

The total mass is ~ 1.4 M
Masses of Stars

The ratio of the masses:
– The two stars orbit a common centre of
mass
– The major axes of the two ellipses are in
the inverse ratio of the masses.
M1
a1
a2
M2 M1/M2 = a2/a1
Masses of Stars

Together with the sum of the masses,
the individual masses can now be
obtained.
– Orbital period  Total mass
– Ratio of major axes  Ratio of masses
Masses of Stars

Binary systems containing red giants:
– show masses of red giants fall into the
same range as main sequence stars


confirming our conjecture they have low
densities)
White dwarfs are found to have masses
about 1 M
– indicating these are very dense objects!

Is there a
relationship
between
luminosity
and mass?
Luminosity, L (Solar Luminosities)
Mass-Luminosity Relationship
L = M 7/2
Mass, M (Solar Masses)
Mass-Luminosity Relationship

L = M 7/2
– Approximate relationship
– Valid for low mass stars up to about 1 solar
mass
– L = M 3 gives better fit for high mass stars
– Note for Main Sequence stars only
Stellar Interiors

Basic Ideas
– Hydrostatic Equilibrium
– Classical and Quantum Gases

Electron and photon gases
– Ionisation

The Saha Equation
– Heat Transfer Mechanisms
– Fusion
Hydrostatic Equilibrium

Since main sequence stars are stable,
and are neither expanding or
contracting, they are clearly in
equilibrium
– Internal pressure supports against gravity
Hydrostatic Equilibrium

We wish to examine:
– The relationship between mean pressure
and potential and kinetic energies

The Virial Theorem
– The consequences of equilibrium of nonrelativistic and ultra-relativistic gases
Hydrostatic Equilibrium

Consider a volume element in a spherical
system of mass M and radius R
R
r
P.DA
g(r)DM
(P+DP)DA
Mass DM
Hydrostatic Equilibrium

Net force due to the pressure gradient
dP
DP  DA 
Dr  DA
dr
– Noting that the mass, DM, of the element
is: DM = r(r)Dr DA, and including the
gravitational force, g(r) we have the
acceleration of the volume element:
d 2r
1 dP
 2  g r  
dt
rr  dr
Hydrostatic Equilibrium

In equilibrium, net acceleration is zero.
– Hence:
 g r  

1 dP
rr  dr
We can write g(r) as
Gm (r )
g r  
2
r
Where m(r) is the
mass contained
within a radius r
Hydrostatic Equilibrium

Therefore, if the star is to be in
hydrostatic equilibrium, the pressure
gradient at any distance r from the
centre is:
dP
Gm r rr 

2
dr
r
Hydrostatic Equilibrium

We can use this expression to relate the
average pressure to the total
gravitational potential
– Multiply both sides by 4pr3
– Integrate from 0 to R

R
0
2
R Gm r 4 pr rr dr
dP
3
4 pr
dr   0
dr
r
= dm
Hydrostatic Equilibrium

The right hand side is simply the
gravitational potential:
M
E g   0

Gm r dm
r
The left hand side can be evaluated
using integration by parts:
df x 
dg x 
 g x  dx dx  f x g x    f x  dx dx
Hydrostatic Equilibrium

The left-hand side then becomes:
P r   4pr
=0

3 R
0
R
2


 3 P r  4 pr dr
0
= -3<P >V
= dV
Hydrostatic Equilibrium

Hence, the average pressure in a
system in hydrostatic equilibrium and
with gravitational potential Eg is:
1 Eg
P 
3V
 The Virial Theorem
Hydrostatic Equilibrium

Non-relativistic (p << mc) vs. ultrarelativistic (p >> mc) gases.
– Fundamental relationships:
P 
n
3
p v
Ep  p c m c
2
2
2
2
4
Hydrostatic Equilibrium

Non-relativistic gases:
– p.v. = mv2
– Hence:
P 

n
3
mv
2
2n 1

mv 2
3 2
Pressure = 2/3 kinetic energy density
Hydrostatic Equilibrium

Consequences:
2 E KE 
1 Eg 
P 
 

3 V 
3V 

Hence:
2E KE  E g  0
Hydrostatic Equilibrium

Since:
E total  E KE  E g
E total

 1 
 E KE   E g 
 2 
i.e., the system is bound with binding
energy -Etotal = internal kinetic energy
Hydrostatic Equilibrium

Implications:
– Tightly bound systems are hot
– Energy loss from a gas cloud causes contraction
and heating


1/2 loss of gravitational energy goes into radiation, 1/2
into raising the temperature
Increase in temperature provides pressure to counter
gravity
– Excess energy production in a star causes
expansion and cooling
Hydrostatic Equilibrium

Ultra-relativistic gases:
– p.v. = pc
– P >>mc, so EKE = pc
– Hence:
P 

n
3
pc
Pressure = 1/3 kinetic energy density
Hydrostatic Equilibrium

Consequences:
– EKE = -Eg
– System only in equilibrium if Etotal = 0
– Hydrostatic equilibrium of ultra-relativistic
particles (e.g., photon gas, high energy
electron gas) easily disrupted.
Next Lecture:
Classical and Quantum Gases
 The Saha Equation

– Derivation
– Consequences for ionisation and
absorption