Re: sun's spectrum - riposte

[Leigh Palmer <palmer@sfu.ca>]

> At 07:29 1/30/98 -0800, Brian wrote:
> > > As Kirchhoff pointed out in connection with thermal radiation, good
> > > absorbers are in general good emitters. (In fact Prevost [1792] said
> > > as much.)
> > > So these 'absorption' lines do not in fact represent a loss of flux -
> > > rather a contrast effect with the continuum radiation surrounding.
> >
> > I'm not sure what point Brian is making here. The chromosphere acts
> > as a filter, absorbing light from the photosphere so that it does
> > not get through to us. It does result in a reduction of flux at those
> > wavelengths. There is still some flux left at those wavelengths, from
> > the chromosphere itself, but it is much attenuated with respect to
> > that at nearby wavelengths. In fact this morning I will introduce my
> > students to the Saha equation so they can understand those lines in
> > some detail.
> >
> > Leigh
>
> > In the case of the chromosphere the radiation
> > is incident over a hemisphere, crudely speaking, and almost as
> > much is scattered into the line of sight as is scattered out.
> > Thus it is the effectof difference in temperature that dominates.
>
> > Is that good for insight?
>
> > Leigh
>
> In that Leigh is locating the absorption mechanism in the
> chromosphere his insight appears to falter.
>
> With the arrival of Roger Freedman's rather handsome solar post
> this morning, I believe Leigh is now better placed to examine
> objections to his explanation.
>
> It is not the hot chromosphere doing the absorbing, but rather
> the outer, cooler layer of the photosphere.

Quite so. I misidentified the coolest layer as the chromosphere.
The absorbing is being done in that layer, if that is the way
one looks at it. I now understand Brian's point and will get back
to it later.

Originally the chromosphere was defined as the red layer of the
solar atmosphere, directly above the photosphere, visible during
a total solar eclipse.

The old definition of the photosphere was that it was the extent
of the visible limb of the Sun, the optically thick part, as in
Frank Shu's textbook which I used to use. (I'm now using Carroll
and Ostlie.) I would like a citation of the new definition so I
can be on the same page as my colleagues, of course. As some of
you know, even mass has been redefined since I first learned
about it. Astronomy used to be more conservative in its ways;
I am a bit disappointed to learn that has changed, but as

I did not know that the chromosphere is defined by the positive
temperature gradient. That is nomenclature, the rest of the
explanation stands; the absorption occurs in the cooler part of
the solar atmosphere. The physics is the same, even though it
appears that the chromosphere is now defined differently.

The physics of the situation requires some more clarification.
I'll take another bash at lending some insight. The problem here
is that the answers are not simple; they really require
mathematical modelling and do not lend themselves well to mere
verbal assault, but a teacher must try.

In the first place, describing the process which produces an
absorption spectrum in a star by the absorption process is a
little misleading. The spectrum is called an absorption spectrum
because its lines are relatively less intense than the continuum
at adjacent wavelengths. This is also true of the spectrum of a
continuous source observed trough an intervening nebula which
scatters or absorbs those wavelengths, a straightforward process.
It is certainly possible to describe the process as an absorption
process, for the radiation which is produced at those wavelengths
at greater depths and higher temperatures is absorbed and does
not penetrate the cooler layer which represents the last optical
thickness or so for that wavelength on the way out. The picture
is complicated somewhat, however, by the radiation from inside
being uncollimated. This last optical thickness for H-alpha is in
the coolest part of the photosphere, it turns out

In the case of the Sun or any other star, however, as Brian
pointed out, there is another way to look at the phenomenon. In
observing the Sun one looks *down to* a depth where the optical
thickness becomes appreciable. Whatever the temperature at that
depth is, the intensity of the radiation emerging will have an
intensity representative of that temperature. In the case of an
"absorption line" like H-alpha, the temperature at the depth to
which one looks is cooler than that to which one looks at adjacent
wavelengths in the continuum, where the photosphere is transparent
to a greater depth, and consequently a higher temperature.

By the old definition of the photosphere the top of that zone
would be wavelength dependent, so perhaps a new definition is
welcome. The minimum temperature boundary would do nicely, but I
can't find that definition in any of the books I have here at
home. The only quantitative statement I can find (Zeilik, Gregory
& Smith, 3rd ed.) states that the temperature begins to rise at
500 km into the chromosphere. A profile is plotted in Fig. 10-6
which shows that very clearly. The coolest part of the solar
atmosphere lies clearly within the chromosphere in this source,
which is copyright 1992. The citation on the figure is J. Verazza,
E. Avrett, and R. Loesser. Unfortunately I am unable to trace
that reference farther (the text has no bibliography). In the
glossary in this text it clearly states that the chromosphere is
hotter than the photosphere, so one might well be confused!

Roger's point about the H-minus contribution to the continuum
opacity is well taken. I had previously attributed that to free
electron opacity, but that is incorrect. Ionization of H-minus
to the free electron continuum is the predominate process.

Leigh