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luminosity / black bodies
/ wien's law and stefan-boltzmann law / stella spectra
Luminosity
The total
power radiated by a star in all directions is known as its luminosity
and, as such, the SI unit for luminosity is watts (W). When you compare
this to the power received by an observer on the Earth, you can see
that the two quantities are quite different. The power received per
unit area on the receiving end is known as the star's brightness
and this is measured in watts per metre squared (W m-2).
If two stars
were at the same distance from Earth, the one that had the greatest
luminosity would also have the greatest brightness. However, because
stars are at different distances from the Earth, their brightness will
depend on the luminosity as well as the distance from Earth. The
brightness of a star will decrease with distance according to the
inverse square law.
The
relationship between luminosity (L), brightness (b) and distance (d) is
given by the following formula:

In essence, as
the light from the star spreads out in all directions it is distributed
over an area equal to the surface area of a sphere of radius d.
It is ,
therefore, possible for two stars with very different luminosities to
have the same brightness as viewed from Earth. This is because the more
luminous star may be located further away.
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Black Body
Radiation
A black body
is an object which absorbs 100% radiation that hits it so it does not
reflect any radiation and therefore appears to be black. It is also a
perfect emitter of radiation, and at a particular temperature, it would
emit the maximum amount of energy for that particular
temperature. The whole spectrum of light emitted by such a body
is known as black body radiation.
Although the
gas making up a star would normally be considered transparent, and thus
a poor absorber of light, the layers of gas are effectively so deep
that they absorb 100% of the light striking the star. None is
transmitted through the star, and none is
reflected. Further contributing to this effect is the fact that
the gas is ionized, which scatters the light randomly and leads to
multiple opportunities for the light to be absorbed before escaping.
Since a star
behaves like a perfect absorber of light, it also behaves like a
perfect emitter.

This is a
black body curve for a temperature of 5800K. This graph shows the black
body radiating energy at every wavelength. As seen from the graph, the
peak wavelength, the point where it emits maximum energy, is around 5 x
10-7 m. Sometimes the y-axis is labelled as 'intensity'. To
be absolutely correct, it should be called the 'intensity function'.
In this way the area under the graph between two wavelengths gives the
intensity of radiation emitted. Said another way, the area under the
graph represents the total power radiated between those two
wavelengths.

Black body
radiation curves at different temperatures.
Notice that as the temperature increases, the peak
wavelength
emitted by the black body decreases, meaning the star is emitting light
of a higher average frequency. The peak shifts into the red, then
orange, then yellow, and so on into the ultraviolet and ultimately the
X-ray end of the spectrum. Since the
graphs do not touch the x-axis, black bodies emit at every wavelength
which means that some radiation is emitted even at low temperatures and
at any temperature above absolute zero, a black body will emit some
visible light. Also, as temperature increases, the total energy emitted
increases as the area under the curve increases.
Since stars act as though they were black bodies, their
radiation spectra can be used to determine their surface temperature.
This is the main application of black body radiation curves in
astronomy. More on this in the next section.
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Wien's Law and the Stefan-Boltzmann
Law
WIen's Law
Wiens’s Law tells us that the temperature of a hot
object determines the peak wavelength in its spectrum. Said
another way, the temperature of a black body (or the surface
temperature of a star) is inversely proportional to the peak wavelength
given by the black body radiation spectrum for that object. Stars at a
very high temperature have a smaller peak wavelength than stars at a
lower temperature. The peak wavelength emitted by a star is
related to its temperature by Wien’s Displacement Law.

Where:
λmax is the wavelength in the
spectrum of the star which corresponds to the peak intensity function
on the radiation spectrum
T is the temperature in Kelvin.
and the constant is 2.90 x 10-3 Km
Hotter objects emit most of their radiation at shorter
wavelengths; hence they will appear to be bluer while cooler objects
emit most of their radiation at longer wavelengths; hence they will
appear to be redder. Wien's law is very useful to astronomers because
by measuring the spectrum of radiation being emitting from a star, the
star's surface temperature can be determined.
Stefan-Boltzmann Law
The Stefan-Boltzmann law connects the luminosity of a
star with its size and temperature. The total power radiated by the
star (luminosity) is directly proportional to its surface area and to
the fourth power of its temperature.

Where σ is the Stefan-Boltzmann constant : 5.67 x 10-8
W m-2 K-4
A is the surface area of the star in metres squared (m-2)
T is the temperature in Kelvin (K)
For the full derivation of this formula please see the
following website: http://en.wikipedia.org/wiki/Stefan-Boltzmann_law
Make sure that if you are using these formulae that your
temperature is in Kelvin and the wavelengths are in metres (and not
nanometres).
In terms of stars, A is the surface area of the star,
The radius of the star, r, is linked to its surface area
using the equation A = 4πr2 . This means that if we know the
luminosity of a star and its temperature ((from Wien's Law), we can
work out its radius or size.
Practice Questions
-
Two stars have equal radii and the temperature of
the first star is twice that of the second. What is the value of
the ratio of the luminosity of the first to that of the second?
1:16
-
The brightest star in Orion is Betelgeuse.
This star has a surface temperature of 3250K and a distance of 500 pc
from earth. Second brightest is Rigel. It has a surface
temperature of 11,600K and is a distance of 400 pc from earth.
Assuming they are the same size, find the ratio of apparent
brightness. (The unit "pc" stands for "parsec" which we will
define later. 1 pc = 3.09 x 10^16 m)
253.6:1
-
The sun has a surface temperature of 5800 K, what is
the peak wavelength in the solar spectrum?
5.00 x 10-7
m
-
The star Betelgeuse has a surface temperature of
3250 K, what is the peak wavelength and What colour would it be?
8.90 x 10-7 m, Red giant
-
Sirius B is a white dwarf, which is smaller than
earth, has a very high density of a temperature of 30,000 K, what is
its peak wavelength?
9.67 x 10-8 m
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<>Stella
Spectra
When hot ionized gas is viewed against a dark background, one observes
line spectra corresponding to the emission lines of the elements on the
gas. You may have observed these lines in your chemistry class
using a diffraction grating.
However, since
a good emitter is a good absorber, this means that a cold gas will
absorb at exactly the same frequencies it would emit if it was hot.
The discovery
of spectral lines was made by Fraunhofer who, in the early 1800's,
magnified the Sun's spectrum and discovered dark lines which could be
identified with particular elements (based on spectra in the
laboratories).

Spectrum
from the Sun
Laboratory spectra generally look like the inverse of
the Sun's spectrum: instead of dark lines, it has bright lines
superimposed on a faint continuum. But the bright lines show up at the
same wavelengths as the dark lines, and can be used to identify
elements in stars.

Absorption spectrum of helium

Another way of conceiving
absorption spectrum
There are three types of spectra:
-
Continuous spectrum - a solid or
liquid body radiates an uninterrupted, smooth spectrum (called a Planck
curve).
-
Emission spectrum - a radiating
gas produces a spectrum of discrete, bright spectral lines (like helium
above).
-
Absorption spectrum - a continuous
spectrum that passes through a cool gas and has specific spectral lines
removed (inverse of an emission spectrum). This is what most stellar
spectra look like (like the sun above).
The missing wavelengths in a star's absorption spectra correspond to
the absorption specturm of a number of elements in the star. The
absorption is taking place in the outer layers of the star, which means
we have a way of telling what elements exist in the star - at least in
its outer layers. The spectra of stars are also used to classify them
into spectral classes.

Ways of obtaining different
types of spectra
Spectral Classes
Stars are divided into a series of spectral types based
on the appearance of their absorption spectra. Some stars have a strong
signature of hydrogen, others have weak hydrogen lines, but strong
lines of calcium and magnesium.
There are 7 basic spectral classes, which by tradition
are called O, B, A, F, G, K and
M. (A common mnemonic for remembering the classes is "Oh be
a fine girl(guy) and kiss me.") Note that the spectra classes
are also divisions of temperature such
that O stars are hot, M stars are cool. Between the classes there were
10 subdivisions numbered 0 to 9. For example, our Sun is a G2 star.
Sirius, a hot blue star, is type B3. The syllabus, however, alludes to
the notion that you only have knowledge of the principle spectral
classes (OBAFGKM).
Why do some stars have strong lines of hydrogen, others
strong lines of calcium? The answer is not composition (all stars are
95% hydrogen) but rather surface temperature. As temperature increases,
electrons are kicked up to higher levels by collisions with other
atoms. Large atoms have more kinetic energy, and their electrons are
excited first, followed by lower mass atoms.
Notice in the graph below, that the coldest stars (M class) can
actually contain
molecules!!! As the temperature increases, molecules break apart
into
constituent atoms. If the collision is strong enough (high
temperatures)
then the electron is knocked off the atom and we say the atom is
ionized. So as we go from low temperatures in stars (a few thousand K)
we see heavy atoms, like calcium and magnesium, in the stellar spectra.
For stars with higher temperatures, we see lines from lighter atoms,
such as hydrogen. The heavier atoms are all ionized by this point and
have no electrons to produce absorption lines.

Relative strengths of
absorption lines for different spectral classes

Absoprtion spectra for
different types of stars classified according to spectral class
Follow the link below to complete a
tutorial on determining which elements are present in a star from its
spectra.
http://www.learner.org/teacherslab/science/light/color/spectra/spectra_1.html
The characteristics of the different
spectral classes are summarised in the table below.
| Class |
Effective Surface Temperature/K |
Colour |
|
O
|
28 000 - 50 000
|
blue
|
|
B
|
9900 - 28 000
|
blue-white
|
|
A
|
7400 - 9900
|
white
|
|
F
|
6400 - 7400
|
yellow-white
|
|
G
|
4900 - 6000
|
yellow
|
|
K
|
3500-4900
|
orange
|
|
M
|
2000-3500
|
orange-red
|
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