Solar absorption lines

Hi Thomas,

The directions of the scattered photons are distributed over 4 pi
steradians (i.e. the full sphere), but the incident photons are
distributed only over 2 pi steradians (coming only from the solar
surface), so per se the intensity should be reduced by a factor 1/2.
You can of course ask what happens to the other half (the one that's
being scattered back towards the sun) and if you assume that through
subsequent scatterings this will also eventually go towards the
observer, then you are right that the scatterings should not change the
intensity.

Exactly!

In any case, even if the backscattered photons are lost for
some reason

I can't think of a reason why they would be lost.

the geometrical effect should result at best in a factor
1/2 reduction.

If they are lost, yes. But I don't believe they are.

So the solar absorption lines have essentially nothing
to do with a geometrical effect but are due to other mechanisms:

Precisely! This is my point.

1) the Doppler effect (photons can essentially not penetrate the solar
atmosphere within the line as the opacity here is so high; they only
get through once they have been shifted out of the line in the course
of the scatterings due to the frequency changes by the Doppler effect).

I assume this is a small effect (ie. (2) is the dominant effect) as we
do not see significant emission lines in the solar spectrum?

2) Photoionization (photons within the line are trapped for so long in
the solar atmosphere that they have a high probability of ionizing
excited states of hydrogen; this means they are lost from the line)

What happens to these electrons after they're ionised?

In your previous post in this thread, you said:

"The photoionization process leads then
subsequently again to a photon on recombination of the photoelectron,
but this will have a completely different wavelength"

& Martin said that the electrons will "radiate at a frequency dependant
on their speed and the field strength".

So wouldn't we be seeing emission lines in the solar spectrum for these
2 effects? (radiating & recombination)

I'll read up on photoionization & recombination in the meantime.

Scott.

"Thomas Smid" wrote in message
oups.com...

The primary mechanisms for the formation of absorption lines in the
solar atmosphere are 1) photons are shifted out of the line region due
to the Doppler effect ...

That is certainly true but would Scott's reciprocity
argument not apply here too? It is as likely that a
photon from a frequency f1 offset from the nominal
line would be moved to the line frequency f0 as it is
for one at f0 to be moved to f1. The result should
just be a broadening of the lines.

and 2) photons in the line are lost due to
photoionization of excited states of hydrogen.

Can you tell me what the relative contributions of
these various mechanisms is? I have really been
addressing the reason why Scott's scattering argument
isn't valid rather than the actual causes of the
energy loss which are beyond my knowledge.

George

"Scott" wrote in message
u...
Hi George,

If a patch of gas is illuminated by a source at
some brightness covering 2 pi steradians and then
emits the same energy into 4 pi sterardians it
must appear less bright.

Yes, but if the gas covers 4 pi steradians (surrounds the emitter)
cancellation occurs.

You have it the wrong way round, the emitter
needs to surround the gas. I've marked up your
diagram below.

Why can't there be a balance when the gas surrounds the emitter,
as in the case of the Sun?

The Sun doesn't surround a patch of gas in
the chromosphere, that's the key. (see below)

I'm talking about a situation where the _gas_ surrounds the
_emitter_.

I know, but to get a balance betytween the photons
lost and gained, you need to be talking about the
situation where the _emitter_ surrounds the _gas_.

I have shown that as a second diagram on the sketch,
it should look like a cavity with a small hole called
"Sun" through which we view the chromosphere.

Suppose a photon is scattered away from us:

(Sun) - * Earth
|
v


You suggest another is scattered to replace it:

(Sun) * - Earth
|
^
|

The situation I am referring to is different.

My ASCII-art is not so good, so I drew a diagram & scanned it in.
Please look at:

http://members.optusnet.com.au/scott...mp/scatter.png

Marked up and 'cavity' version added:

http://www.georgedishman.f2s.com/Scott/scatter.png

I now refer to this diagram.

Consider photon A emitted from the photosphere & heading directly
toward the observer. If the photon is of the right wavelength, say
656nm (H-alpha), then it can be absorbed by a H atom in the
chromosphere and re-emitted in a random direction, identified by A'.

OK, so when we look at the point from which A was
emitted, that photon is lost.

Obviously the observer would never see this photon. You, & several
others, have said as much already.

Yes, I agree your starting point.

Now consider photon B which (were it not for the chromosphere) would
not normally be seen by the observer. There is a small chance that this
photon can be scattered directly toward the observer, identified by B'.

True but as you have drawn it, B' is arriving from
a different part of the Sun so won't arrive at the
same point on its image. To replace the lost photon,
you need another arriving along the path that A would
have taken had it not been scattered. I have shown
that as C', a photon scattered into the observer's
eye. The question then is where C came from. There
are no photons arriving like C form the direction
in which A was scattered.

In order to get a configuration in which the emitter
can replace every lost photon, you would need the
situation shown below your diagram. Obviously the
Sun isn't a hollow sphere but a cavity emitter is
the standard way of producing radiation close to
a black body.

Now integrate this effect over the entire surface of the Sun.
There is basically 3d sphere of H atoms randomly scattering, say,
H-alpha photons over 4 pi steradians. Statistically, the observer
is going to receive a large number of _scattered_ H-alpha photons.

From different angles.

So I believe it's not the scattering effect that is contributing
to the absorption lines, it must be something else. You can't scatter
photons _away_ from an observer when the gas surrounds the emitter.

You can, and that configuration will do, you cannot
produce lines when the _emitter_ surrounds the _gas_.

However, as others have said, there are other ways
energy is lost as well as scattering which may be
more significant in explaining the source of the
lines.

George

In article ,
Scott writes:
What process causes absorption lines in solar spectra (when
measured from above the Earth's atmosphere)?

From your subsequent posts, I suspect you are thinking of low-density
plasmas. (So, I suspect, are several people who have replied.)
Stellar atmospheres are different because they are a lot denser. The
buzzword would be that (to a first approximation) you have "local
thermodynamic equilibrium," not "detailed balance." In other words,
a common physical process is that a photon is absorbed and excites an
atom. The atom then decays _not_ by radiation but by a collision,
imparting the extra energy to an atom or ion nearby. Of course the
inverse process also occurs, but it's less common because there is a
net transfer of energy outward. This gives absorption lines without
creating corresponding emission lines in other directions.

There's also a geometric issue. Most spectra of the Sun are of a
small area of the disk, so you wouldn't see the "balancing" emission
lines in these spectra even if the lines existed. However, this is
not the major effect. There are some "whole-disk" spectra of the
Sun, and of course there are plenty of whole-disk spectra of sunlike
stars that show absorption lines. (Postings about the chromosphere
"flash spectrum" are correct, but the net emission doesn't come close
to balancing the absorption as can be seen in the whole-disk
spectra.)

The best way to start thinking about stellar atmospheres is a
"plane-parallel" model. Imagine a single square centimeter on the
surface of the Sun and a very long column below it. At any depth,
the gas has a fixed temperature and density, the same in all
horizontal directions, but density and temperature both increase as
you go down. (We ignore the temperature inversion; absorption lines
are formed below it, so for this purpose just take the temperature
minimum as the "surface.") Do you see why this model has to produce
absorption lines as viewed from above? If so, your question is
answered. If not, do you understand the concept of "optical depth?"
That's the key to answering your question in more detail than in the
first paragraph.

If you want to understand this at a serious level, you will need to
find a textbook. Dimitri Mihalas and John Jefferies are two authors
who have written good texts, but there may be even better ones around
these days. (No prizes for guessing how long it has been since I
actually studied this stuff!)

--
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)

Steve,

Does my reply to Scott appear to be correct?

-- Jeff, in Minneapolis

Scott replied to Jeff Root:

Electrons also can become unbound from atoms and
radiate freely.

Can you elaborate on what you mean by "radiate freely"?

He means "radiate at any wavelength".

It's still not clear to me what this means. What might the
electron do once it's unbound?

Bump into another electron or atom. When that happens, it
will emit light.

Imagine a thin gas in front of a black background. The
gas is too thin and too cold to have significant blackbody
emission, so all you see is black.

The gas is illuminated by a very strong source of light
off to one side, where you can't see it. This is the same
as looking at the solar chromosphere during an eclipse.
Most of the light of particular wavelengths entering the
gas is absorbed

Yep.

The result is a low-intensity blackbody glow from the gas,
with a bright-line spectrum superimposed on it.

Where is the energy coming from to heat the gas to have
a blackbody glow?

If it absorbs a H-alpha photon & then re-emits it, there's
no energy left behind.

When the gas absorbs light, it gains kinetic energy, which
heats it. But it is still cooler than the lower parts of
the photosphere, where the highest-energy light originates,
so the wavelength of the emitted light is usually longer
than that of the absorbed light, and the excess energy it
has is usually lost in multiple photon emissions.

I intend to address the remainder of your post later.
It is involved.

-- Jeff, in Minneapolis

Scott wrote:

1) the Doppler effect (photons can essentially not penetrate the solar
atmosphere within the line as the opacity here is so high; they only
get through once they have been shifted out of the line in the course
of the scatterings due to the frequency changes by the Doppler effect).

I assume this is a small effect (ie. (2) is the dominant effect) as we
do not see significant emission lines in the solar spectrum?

Well, the effect does not exactly result in emission lines, but merely
in a slight enhancement of the continuum near the absorption lines,
something like this:



1.4 *** ***
1 ****** * * ******* continuum level
* *
* *
* *
* *
* *
* *
0.2 ***


This effect must in principle necessarily occur. If you assume for
instance an infinite opacity within the line, no photon remaining
within the line will ever be able to penetrate the scattering layer. It
can only reach the observer once its frequency has been shifted into
the region where the opacity is small (i.e. outside the line).
So this mechanism would actually produce absorption lines without that
any photons are lost. However, it is likely that the photons within the
line are already destroyed before this happens, e.g. due to
photoionization of excited levels, so in this sense it might not be
significant.



2) Photoionization (photons within the line are trapped for so long in
the solar atmosphere that they have a high probability of ionizing
excited states of hydrogen; this means they are lost from the line)

What happens to these electrons after they're ionised?

In your previous post in this thread, you said:

"The photoionization process leads then
subsequently again to a photon on recombination of the photoelectron,
but this will have a completely different wavelength"

& Martin said that the electrons will "radiate at a frequency dependant
on their speed and the field strength".

So wouldn't we be seeing emission lines in the solar spectrum for these
2 effects? (radiating & recombination)

Yes, when the electrons recombine, they produce emission lines, but not
necessarily at the wavelength of the absorption lines. If you assume
for instance that an H-alpha photon (656 nm) photoionizes an excited
atom which is in the n=3 state, and then assume that the photoelectron
produced this way recombines into the n=2 state, this will lead to a
photon at 328 nm (and then a further photon at 122 nm when the atom
decays into the ground state n=1). So the original photon within the
656 nm line will be lost for good in this case. If the recombination
goes into the n=3 state and then subsequently into the n=2 state, the
656 nm photon is of course recovered, but this is only the case for a
fraction of the recombination events. The situation is even worse for
absorption lines other than hydrogen, because as hydrogen is by far the
dominant element, electrons will in all likelihood recombine with
hydrogen ions (protons) rather than with ions of the minor elements. So
for the latter, the losses within the absorption lines due to
photoionization will be close to 100% as subsequent recombination
produces radiation at the hydrogen frequencies rather than at the
frequencies of the element considered.

Thomas

George Dishman wrote:

"Thomas Smid" wrote in message
oups.com...

The primary mechanisms for the formation of absorption lines in the
solar atmosphere are 1) photons are shifted out of the line region due
to the Doppler effect ...

That is certainly true but would Scott's reciprocity
argument not apply here too? It is as likely that a
photon from a frequency f1 offset from the nominal
line would be moved to the line frequency f0 as it is
for one at f0 to be moved to f1. The result should
just be a broadening of the lines.

No, the point is that photons within the line will not be able to
penetrate the scattering layer if the opacity is high enough. On the
other hand, once their frequency has been shifted outside the line,
they will leave the scattering layer straight away. So effectively, all
photons originally within the line are eventually observed outside the
line (if they are not destroyed before that).
(see also my reply to Scott above).


and 2) photons in the line are lost due to
photoionization of excited states of hydrogen.

Can you tell me what the relative contributions of
these various mechanisms is? I have really been
addressing the reason why Scott's scattering argument
isn't valid rather than the actual causes of the
energy loss which are beyond my knowledge.

I don't know what the relative contributions are. These are just two
mechanism that I think should be of relevance here. It may depend
anyway on the circumstances.

Thomas

Scott wrote:

1) the Doppler effect (photons can essentially not penetrate the solar
atmosphere within the line as the opacity here is so high; they only
get through once they have been shifted out of the line in the course
of the scatterings due to the frequency changes by the Doppler effect).

I assume this is a small effect (ie. (2) is the dominant effect) as we
do not see significant emission lines in the solar spectrum?

Well, the effect does not exactly result in emission lines, but merely
in a slight enhancement of the continuum near the absorption lines,
something like this:



1.4 *** ***
1 ****** * * ******* continuum level
* *
* *
* *
* *
* *
* *
0.2 ***


This effect must in principle necessarily occur. If you assume for
instance an infinite opacity within the line, no photon remaining
within the line will ever be able to penetrate the scattering layer. It
can only reach the observer once its frequency has been shifted into
the region where the opacity is small (i.e. outside the line).
So this mechanism would actually produce absorption lines without that
any photons are lost. However, it is likely that the photons within the
line are already destroyed before this happens, e.g. due to
photoionization of excited levels, so in this sense it might not be
significant.



2) Photoionization (photons within the line are trapped for so long in
the solar atmosphere that they have a high probability of ionizing
excited states of hydrogen; this means they are lost from the line)

What happens to these electrons after they're ionised?

In your previous post in this thread, you said:

"The photoionization process leads then
subsequently again to a photon on recombination of the photoelectron,
but this will have a completely different wavelength"

& Martin said that the electrons will "radiate at a frequency dependant
on their speed and the field strength".

So wouldn't we be seeing emission lines in the solar spectrum for these
2 effects? (radiating & recombination)

Yes, when the electrons recombine, they produce emission lines, but not
necessarily at the wavelength of the absorption lines. If you assume
for instance that an H-alpha photon (656 nm) photoionizes an excited
atom which is in the n=3 state, and then assume that the photoelectron
produced this way recombines into the n=2 state, this will lead to a
photon at 328 nm (and then a further photon at 122 nm when the atom
decays into the ground state n=1). So the original photon within the
656 nm line will be lost for good in this case. If the recombination
goes into the n=3 state and then subsequently into the n=2 state, the
656 nm photon is of course recovered, but this is only the case for a
fraction of the recombination events. The situation is even worse for
absorption lines other than hydrogen, because as hydrogen is by far the
dominant element, electrons will in all likelihood recombine with
hydrogen ions (protons) rather than with ions of the minor elements. So
for the latter, the losses within the absorption lines due to
photoionization will be close to 100% as subsequent recombination
produces radiation at the hydrogen frequencies rather than at the
frequencies of the element considered.

Thomas

George Dishman wrote:

"Thomas Smid" wrote in message
oups.com...

The primary mechanisms for the formation of absorption lines in the
solar atmosphere are 1) photons are shifted out of the line region due
to the Doppler effect ...

That is certainly true but would Scott's reciprocity
argument not apply here too? It is as likely that a
photon from a frequency f1 offset from the nominal
line would be moved to the line frequency f0 as it is
for one at f0 to be moved to f1. The result should
just be a broadening of the lines.

No, the point is that photons within the line will not be able to
penetrate the scattering layer if the opacity is high enough. On the
other hand, once their frequency has been shifted outside the line,
they will leave the scattering layer straight away. So effectively, all
photons originally within the line are eventually observed outside the
line (if they are not destroyed before that).
(see also my reply to Scott above).


and 2) photons in the line are lost due to
photoionization of excited states of hydrogen.

Can you tell me what the relative contributions of
these various mechanisms is? I have really been
addressing the reason why Scott's scattering argument
isn't valid rather than the actual causes of the
energy loss which are beyond my knowledge.

I don't know what the relative contributions are. These are just two
mechanism that I think should be of relevance here. It may depend
anyway on the circumstances.

Thomas