Today in Astronomy 241: My First Stellar Atmosphere Model

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Today in Astronomy 241 My First Stellar
Atmosphere Model

• Todays reading Carroll and Ostlie Chap.
  9.3-9.4
• Plane-parallel atmospheres
• Gray atmospheres
• The Eddington approximation
• Solution for the temperature structure of a gray,
  plane-parallel atmosphere in thermal equilibrium

Illustration of the Eddington approxi-mation
(Carroll and Ostlie, fig. 9.15)
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Plane-parallel and gray atmospheres
z

• Plane-parallel atmosphere
• where is called the vertical optical
  depth between point at which light is emitted,
  and surface. The visible stellar photosphere is
  plane parallel, to good approximation.
• If opacity is independent of wavelength (e.g. if
  electron scattering or H- photoionization
  dominates), this becomes

0
?
dz
ds
Radiative transfer for gray atmospheres
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Some useful relations for gray atmospheres

• From the first two moments of the plane-parallel,
  gray transfer equation integrate over solid
  angle and one getsMultiply through by cos?
  and integrate over solid angle, and one gets
• Furthermore, assume thermal equilibrium
  everywhere, and one obtains

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The Eddington approximation

• The intensity of the radiation field can be
  broken into outward and inward parts, Iout and
  Iin, travelling in the z and -z directions.
• But Iin 0 at the top of the atmosphere, where
  the vertical optical depth is zero. Thus

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Important results for a plane-parallel gray
atmosphere in the Eddington approximation

• That is, the effective temperature is equal to
  the physical temperature at an optical depth of
  2/3. But the effective temperature determines the
  flux we see thus our view is determined by the
  value of the source function at an optical depth
  of 2/3.

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In-class problem

• A. Suppose that the source function the ratio
  of the emission coefficient and the opacity is
  equal to the Planck function, for a path through
  a constant-temperature cloud. Show that the
  solution of the radiative transfer equationfor
  light passing through the cloud is where
  is the total optical depth ofthe cloud. How
  does this simplify in the limits of large or
  small optical depth?

Cloud
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Todays in-class problems

• Problem 9.17 i.e. derive the equations on slide
  4.
• Answers and/or secrets of the problems we did
  last class
• Problem 9.4 substitutethe bounds run from ? to
  0, which makes up for the minus sign. Note
  thatdefineto arrive at the final result.

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Todays in-class problems (continued)

• Problem A. The Planck function is of course
  independent of optical depth thus the source
  function is a constant (at given wavelength), and
  the differential equation can be solved by direct
  integration

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Todays in-class problems (continued)

• If If, on the other hand,The latter
  represents optically-thin emission from the
  cloud, a common case for bright nebulae.