Solar absorption lines

Hi William,

Your question is really "why are there ever absorption lines?" If
atoms also emit the light they absorb, why do we see absorption lines
at all?

Sort of. I understand how absorption lines occur in situations where
light is being absorbed & then scattered away from an observer.
eg. interstellar gas/dust.

But I don't understand how absorption lines are created in solar
spectra. The absorbed wavelengths can't be scattered away from
the observer by the chromosphere.

Part
of the reason is that direction of the emitted radiation has no
relationship to the direction of the absorbed radiation.

Yep. Like in the case of interstellar gas/dust.

But, the chromosphere can't scatter away selected wavelengths from
an observer. (cos for every photon scattered away, there is another
photon scattered toward the observer.)

Part of the
reason is that the electrons have many possible energy levels and
don't have to emit radiation at the same wavelength they absorbed it -
they could make one jump up, but two jumps down, for example.

Yep. But if this is the cause of the absorption lines, then I would
expect to see significant emission lines in the solar spectra too.
(which I don't) Think "conservation of energy". Where is the energy
from the 656nm H-alpha photons going?

Electrons also can become unbound from atoms and radiate freely.

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

Scott.

Hi George,

Thanks for joining in this discussion.

Atoms in the chromosphere can scatter in all
directions but they are only illuminated from
one side, that facing the Sun.

Yes. But I don't see how that can create absorption lines.

If the photosphere completely surrounded a
patch of gas then you would be correct and
there would be a perfect balance between
absorption and emission,

Why can't there be a balance when the gas surrounds the emitter,
as in the case of the Sun? (Ignore situations where an atom
re-emits at a different wavelength, for the moment - that complicates
things.)

That's why they appear as emission when
viewing the chromosphere since we see
emission against a dark background.

Yes. I can understand why we see this during an eclipse. We're not
viewing the continuum (it's blocked by the moon), all we're seeing
is selected wavelengths of the continuum that are scattered by the
chromosphere.

As has been said, there are also other ways
for an excited atom to lose energy

You're refering to ionisation?

but at
least part of the energy lost in the absorption
lines is being scattered.

I can see how individual photons can be scattered, but I don't see
how that can create an absorption line because, statistically, there
will be some other atom that will scatter an identical photon to
compensate for it.

Does that help?

I sure appreciate you guys helping me out but, alas, I still can't
see how the solar absorption lines are being created.

Scott.

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

Thanks for joining in this discussion.

Atoms in the chromosphere can scatter in all
directions but they are only illuminated from
one side, that facing the Sun.

Yes. But I don't see how that can create absorption lines.

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.

If the photosphere completely surrounded a
patch of gas then you would be correct and
there would be a perfect balance between
absorption and emission,

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)

(Ignore situations where an atom
re-emits at a different wavelength, for the moment - that complicates
things.)

OK.

That's why they appear as emission when
viewing the chromosphere since we see
emission against a dark background.

Yes. I can understand why we see this during an eclipse. We're not
viewing the continuum (it's blocked by the moon), all we're seeing
is selected wavelengths of the continuum that are scattered by the
chromosphere.

As has been said, there are also other ways
for an excited atom to lose energy

You're refering to ionisation?

Thermal transfer in the gas, perhaps rotational
modes. I'm not sure what would be applicable in
detail.

but at
least part of the energy lost in the absorption
lines is being scattered.

I can see how individual photons can be scattered, but I don't see
how that can create an absorption line because, statistically, there
will be some other atom that will scatter an identical photon to
compensate for it.

Suppose a photon is scattered away from us:

(Sun) - * Earth
|
v


You suggest another is scattered to replace it:

(Sun) * - Earth
|
^
|

But the Sun is only on the left hand side, there
is no source at the bottom of the diagram so this
doesn't happen, the photon is not replaced. The
separations are not to scale, the * representing
the scattering atom would be almost on the surface
of the Sun at this scale but from the point of
view of the atom, the Sun fills half the sky, not
all of it, but the scattered photons go in all
directions.

Is that any clearer?

George

"Scott" wrote in message
u...

... Think "conservation of energy". Where is the energy
from the 656nm H-alpha photons going?

Sideways, in directions where it is seen as
an emission line but only when it isn't
overwhelmed by the Sun, i.e. during an
eclipse. You get emission at the limb and
absorption in the disk so the total is
conserved.

George

On Sun, 09 Jul 2006 23:22:05 +1000, Scott wrote:

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

Emit a continuous spectrum.

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

Thanks for joining in this discussion.

Atoms in the chromosphere can scatter in all
directions but they are only illuminated from
one side, that facing the Sun.

Yes. But I don't see how that can create absorption lines.

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.

The geometrical effect does thus reduce here the absorption lines only
to half the intensity of the continuum (as for subsequent scattering
events the radiation field is practically isotropic).

The explanation for the Fraunhofer lines must thus lie elsewhere. The
primary mechanism should be:

1) In the course of the scatterings, photons are shifted from the line
center into the line wings due to the Doppler effect.

2) Due to multiple scattering, photons within the line stay for such a
long time in the solar atmosphere that they have a high probability of
ionizing excited states of hydrogen (the density of the latter is too
small to affect any radiation that is going straight through (i.e. the
continuum), but for photons within the line the effective pathlength is
several orders of magnitude larger and photoionization of excited
states will be important). The photoionization process leads then
subsequently again to a photon on recombination of the photoelectron,
but this will have a completely different wavelength, i.e. like for
mechanism (1) photons will be lost from the line.

Thomas

Scott wrote:
Hi William,

Your question is really "why are there ever absorption lines?" If
atoms also emit the light they absorb, why do we see absorption lines
at all?

Sort of. I understand how absorption lines occur in situations where
light is being absorbed & then scattered away from an observer.
eg. interstellar gas/dust.

OK so you can understand that a gas cloud in the way allows photons to
be scattered out of your line of sight to a distant star.

But I don't understand how absorption lines are created in solar
spectra. The absorbed wavelengths can't be scattered away from
the observer by the chromosphere

It happens exactly the same way in the chromosphere. The geometry is a
little more complicated since the light source behind the cool gas is
extended.

There is a massive flux of continuum photons streaming away from the
surface of the sun or photosphere. In the chromosphere the only ones
that really matter are those that will get through in the direction of
the observer. (ie at the distance of the earth a narrow acceptance
angle of about 1/2 degree or 0.01 radians.

You still effectively have a bright continuum source sat behind a cool
envelope of gas that can absorb specific wavelengths and then re-emit
them isotropically.

When a photon gets absorbed it gets re-emitted in a random direction.
The photons that you would have seen at the absorbtion frequency are
therefore mostly lost and a dark band appears. You can see the same
sort of self-absorption of a terrestrial HPS light source where the
relatively cool sodium vapour near the glass envelope cuts into the
pressure broadened semi-coninuum emission of the main light.

Any shovelware CD will allow you to see the Fraunhofer absorption lines
in a solar spectrum if it is held at glancing incidnce.

Part
of the reason is that direction of the emitted radiation has no
relationship to the direction of the absorbed radiation.

Yep. Like in the case of interstellar gas/dust.

But, the chromosphere can't scatter away selected wavelengths from
an observer. (cos for every photon scattered away, there is another
photon scattered toward the observer.)

There isn't and that is where your misunderstanding lies. You can
clearly see the bright chromospheric emission at the solar limb if you
have a suitable narrowband filter or at a solar eclipse (but it is much
dimmer than the total continuum emission)

Part of the
reason is that the electrons have many possible energy levels and
don't have to emit radiation at the same wavelength they absorbed it -
they could make one jump up, but two jumps down, for example.

Yep. But if this is the cause of the absorption lines, then I would
expect to see significant emission lines in the solar spectra too.
(which I don't) Think "conservation of energy". Where is the energy
from the 656nm H-alpha photons going?

The ones that you would have seen (in a 1/2 degree acceptance angle)
have been re-radiated isotropically. That gives a roughly 10000
attenuation in brightness on absoprtion lines.

Electrons also can become unbound from atoms and radiate freely.

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

Free electrons in a magnetic field are forced to spiral and will
radiate at a frequency dependant on their speed and the field strength.

Regards,
Martin Brown

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.

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_.

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

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

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'.

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.

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.

Scott.

Hi Jeff,

Thanks for joining in too.

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?

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.

The bright-line
spectrum isn't especially bright. Only a small fraction
of the light of a particular wavelength which is absorbed
is then re-emitted at one of the particular wavelengths
toward your eye. The vast majority of it is emitted either
in other directions or at other wavelengths, or both.

Yep.

Now move the source of light behind the gas. You now see
a very strong, high-temperature blackbody radiation, with
the cooler gas again absorbing particular wavelengths of
that light, which causes a dark-line spectrum.

Well that works if the gas doesn't surround the emitter.
(Such as the case of interstellar gas/dust.) Please see my
reply to George in this thread. It has a diagram & an
explanation for why I think this doesn't apply in the
case of the chromosphere.

Scott.

Scott wrote:

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.

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.

Scott.

Scott,

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. In any case, even if the backscattered photons are lost for
some reason, the geometrical effect should result at best in a factor
1/2 reduction. So the solar absorption lines have essentially nothing
to do with a geometrical effect but are due to other mechanisms:
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).
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)

Thomas