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• If a star is very hot, the electrons will be freed
from the hydrogen atom. (Ionized)
• Once they are free, they act like particles and
emit a continuous spectrum.
• If the star is hot enough that hydrogen is
ionized all the way to the surface, then there
will be no hydrogen Balmer lines.
Here is how line strength depends of
temperature
• Remember, the line strength depends primarily
on two parameters. (1) Surface Temperature and
(2) Number of absorbers.
• If we want learn about the number of absorbers
for a given element (say, calcium, iron, gold, etc)
then we need to know the temperature of the
star. If we know the temperature we can account
for its effect and…
• The line strength will only depend on the Number
of Absorbers.
Spectral typing Summary
• Spectral typing can be used to find the surface temperature
of a star. (Along with color and Wien’s Law)
• Spectral typing can also be used to find out how much of a
given element is in a star.
• HD 161817 has much less of all the elements, other than
Hydrogen and Helium, than the Sun.
• In fact, it has about 0.03 the value of the Sun for all 90
elements. That is 3% the amount in the Sun. The most
deficient star known has about 0.001% the Sun.
• There are also stars with up to 3 times the amount in the
Sun.
A little bit more to do. Stefan-Boltzmann Law.
Imagine we have a cube, that is 1meter X 1meter X 1meter. It
is made of steel and is being heated from inside.
We move a light collector (maybe a huge bundle of fiber optic
cables) that is 1meter X 1meter, right up close to one of the
surfaces so we can capture all the light being emitted from
that side of the cube. The cube begins to glow, and we send
the light collected to a grating which makes a spectrum.
As the cube heats up we continue to collect data.
Detector in front of cube collecting all the light that is
coming from one side of the cube
When the spectrum is recorded, as the cube
heats up, this is what you find.
To find the total energy being emitted from one side of the cube every
second you would need to add up all the intensity at each wavelength.
In other words, find the area underneath the curve.
The result would be:
Lfor one side = σT4
where σ is called the Stefan-Boltzman constant
This means that if you increase the temperature (T) by a factor of 2,
then L will ???
If the temperature increases by a factor of 2
then the Luminosity will ???
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Increase by a factor of 2
Increase by a factor of 4
Increase by a factor of 16
Decrease by a factor of 4
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To find the TOTAL luminosity of the
cube in our experiment we have to..
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Multiply by 2
Multiply by 6
Divide by 16
Do nothing, we
already have the
total luminosity
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• To figure out the total luminosity being
emitted by the box, we would need to
multiply the luminosity of one side by 6. This
is because there are six sides of the cube for
the energy to escape from.
Stars work the same way.
• The total Luminosity being emitted through
the surface of a star is:
L = σT4(4πR2)
Where (4πR2) is the surface area of the star and
R is the radius of the star.
Let’s consider three stars.
• Procyon B, Sirius and Deneb. All three stars
have very similar surface temperatures.
• T ~ 9000 K
• But the Luminosities are not similar. Deneb
has by far the largest, then Sirius, and then
Procyon B.
• Remember that:
• L = σT4(4πR2)
• So we can use the Physics of a radiating object to
determine, not only the surface temperature of a
star (Wien’s Law)
• But also the star’s radius. (Stefan-Boltzman Law)
• Deneb is not more luminous than Sirius because it is
hotter, it is more luminous because it has a bigger
radius.
• Likewise, Procyon B is not less luminous because it
has a cooler temperature, it is less luminous because
it has a smaller radius.
A new distance determination
• Now that we know the way things work on the H-R diagram,
for stars that we can measure distance to using trigonometric
parallax (triangulation) we can make use of these laws to
calculate the distance to stars that are too far away to
measure distance directly.
• Here is what we know.
• B = L/4πd2 and L = σT4(4πR2)
• So
• B = σT4(4πR2)/4π d2 or B = σT4(R2)/d2
• If we measure the apparent brightness, the temperature and
the radius, we can calculate the distance to the star.
• B, the apparent brightness comes from
observing the star at a telescope with a
detector attached to it.
• T can come from either, Wien’s law, the color
of the star, or from spectral typing the
absorption lines.
• But we have to still get a handle on R.
• The strength of an absorption line depends
primarily on two variables.
• The temperature of the star.
• And the Number of Absorbers.
• But there is a secondary effect to the
absorption lines that we can use.
• This effect is called Pressure Broadening.
• Here is how it works.
• The electron is bound to the atom by the
electromagnetic force.
• Different elements have different number of
charged particles.
• This means that the electromagnetic force is
different for each element.
• This causes the electrons to have different
standing waves for different elements.
• So the transition energies between different
excited states are different.
• The absorption lines occur at a different
wavelength for each element.
• BUT, what happens when atoms, with
electrons attached, are packed really close
together?
• The electrons from the neighboring atoms can
have a small effect on the standing wave of
each of the atom’s electrons.
• In a supergiant star (luminosity class I) the star
has a huge volume. That means the atoms are
not close to each other near the surface. They
have virtually no effect on the given energy
levels.
• In a giant star (luminosity class III) the star has
a large radius but not as large as the
supergiant. The atoms near the surface
interact more with each other.
• In a main-sequence star (luminosity class V)
the radius is much smaller and atoms are
packed closely together near the surface.
Luminosity classes in stars.
• The amount of pressure broadening is related to
the radius of the star. When the pressure is high,
(luminsoity class V) the pressure broadening is
large and the radius of the star is small.
• When the pressure is low (luminosity class I) the
broadening is small and the radius of the star is
big.
• We can calculate the radius of the star in this
fashion.
How do we now find the distance?
• We can calculate it from
• B = σT4(R2)/d2
• Here is a visual example of how it works.
• We know the properties of the H-R diagram
for the near by stars.
Quiz #5
• There are two stars, star A and star B. Star A is
approaching the Earth at 100 km/s and Star B
is moving away from the Earth at 200 km/s.
• Compare the Doppler shift for these two stars
by explaining how the spectra will be shifted
and by how much. (I am not looking for a
number here, just a qualitative comparison)