Solar EM spectrum, why only upto UV?

(John Park) writes:

You've had a number of good answers. A slightly different way to look at
it is to remember that to a gamma ray, pretty much every electron is a free
electron (since gamma energies are so much higher than light-atom binding
energies). So a gamma making its way though the sun will undergo a lot of
Compton scattering, almost all of it reducing the photon energy.

In addition, at the temperature of the interior of the sun, I think it's
safe to say that every electron is a free electron anyway.

On 27/10/2010 8:32 PM, Sam Wormley wrote:
On 10/27/10 5:14 PM, Yousuf Khan wrote:
However, in my mind, a gamma ray that is absorbed and re-emitted should
be re-emitted in the form of a gamma ray again.

Unless it gives up a bit of kinetic energy to the atom. Where did
you think that core temperature came from?


Where did the core temperature come from? Well, from the gamma rays of
course.

Yousuf Khan

On 10/28/10 4:37 PM, Yousuf Khan wrote:
On 27/10/2010 8:32 PM, Sam Wormley wrote:
On 10/27/10 5:14 PM, Yousuf Khan wrote:
However, in my mind, a gamma ray that is absorbed and re-emitted should
be re-emitted in the form of a gamma ray again.

Unless it gives up a bit of kinetic energy to the atom. Where did
you think that core temperature came from?


Where did the core temperature come from? Well, from the gamma rays of
course.

Yousuf Khan

Initially gravitational collapse provides the temperature. Kirchoff's
law doesn't hold in plasma where electrons are not bound to protons.

On 28/10/2010 3:14 AM, Mike Dworetsky wrote:
Think of it this way: As the energy of the photons slowly makes its way
out, at each radius the area of the sphere inside the sun is larger, but
the total flux is the same. Hence, to conserve flux, the average energy
of the photons emitted and reabsorbed must be lower as you move
outwards. The gamma rays are absorbed thermally (they boost the speed of
the particles, etc), so the re-emitted photons are unrelated to the ones
absorbed (except they have the same energy, minus epsilon). They are not
the same photons. If they were, you would have scattering, not absorption.

Actually, that is the way I was thinking of it, in the macroscopic
sense. However, I was hoping for a little more detailed insight at the
atomic level.

So far, at the atomic scale, the following mechanisms have been
proposed, all of which can be happening to different nucleii at the same
time:

(1) high-energy photon absorbed, boosts ions and makes them move a
little faster, but the ions don't collide due to electromagnetic
repulsion (which involves emitting photons too, which sap energy away):
some of the energy given back in the form of a re-emitted lower-energy
photon.
(2) gamma photons boost ions into each other, creating a fusion process
and the release of still further gamma rays.
(3) In the higher layers of the Sun, super-high energy photons are
rarer, and the high energy photons might arrive in the form of x-rays or
UV, thus giving smaller boosts than gamma rays.

For example, if an electron absorbs a gamma ray, it gets boosted to a
higher orbital. If that electron then drops back down partially, it'll
emit a photon of a lower energy and frequency, and then perhaps later
it'll drop down some more, and emit another lower energy photon. Is
that what's happening?

No, because in the central regions you have full ionization, even for
heavy elements, and certainly for hydrogen, the main source of opacity.
Hence there are no bound quantum states in the hot plasma, only
continuum. Near the surface that is no longer true and it gets interesting.

Understood, electrons are so far apart in the core, it's not even worth
considering them as bound.

Yousuf Khan

Originally Posted by Yousuf Khan[_2_]

On 28/10/2010 3:14 AM, Mike Dworetsky wrote:
Think of it this way: As the energy of the photons slowly makes its way
out, at each radius the area of the sphere inside the sun is larger, but
the total flux is the same. Hence, to conserve flux, the average energy
of the photons emitted and reabsorbed must be lower as you move
outwards. The gamma rays are absorbed thermally (they boost the speed of
the particles, etc), so the re-emitted photons are unrelated to the ones
absorbed (except they have the same energy, minus epsilon). They are not
the same photons. If they were, you would have scattering, not absorption.

Actually, that is the way I was thinking of it, in the macroscopic
sense. However, I was hoping for a little more detailed insight at the
atomic level.

So far, at the atomic scale, the following mechanisms have been
proposed, all of which can be happening to different nucleii at the same
time:

(1) high-energy photon absorbed, boosts ions and makes them move a
little faster, but the ions don't collide due to electromagnetic
repulsion (which involves emitting photons too, which sap energy away):
some of the energy given back in the form of a re-emitted lower-energy
photon.
(2) gamma photons boost ions into each other, creating a fusion process
and the release of still further gamma rays.
(3) In the higher layers of the Sun, super-high energy photons are
rarer, and the high energy photons might arrive in the form of x-rays or
UV, thus giving smaller boosts than gamma rays.

For example, if an electron absorbs a gamma ray, it gets boosted to a
higher orbital. If that electron then drops back down partially, it'll
emit a photon of a lower energy and frequency, and then perhaps later
it'll drop down some more, and emit another lower energy photon. Is
that what's happening?

No, because in the central regions you have full ionization, even for
heavy elements, and certainly for hydrogen, the main source of opacity.
Hence there are no bound quantum states in the hot plasma, only
continuum. Near the surface that is no longer true and it gets interesting.

Understood, electrons are so far apart in the core, it's not even worth
considering them as bound.

Yousuf Khan

The structure of the Sun is important.

Nuclear fusion reaction only occurs in the center of the Sun.

At that time gamma ray is emitted.

This gamma ray is absorbed and transferred into heat in the outside layers.

Finally the heat is emitted on the surface of the sun.

At that time the heat emission only rely on surface temperature.

The equation is Planck's law equation.

Yousuf Khan wrote:
On 28/10/2010 3:14 AM, Mike Dworetsky wrote:
Think of it this way: As the energy of the photons slowly makes its
way out, at each radius the area of the sphere inside the sun is
larger, but the total flux is the same. Hence, to conserve flux, the
average energy of the photons emitted and reabsorbed must be lower as you
move
outwards. The gamma rays are absorbed thermally (they boost the
speed of the particles, etc), so the re-emitted photons are
unrelated to the ones absorbed (except they have the same energy,
minus epsilon). They are not the same photons. If they were, you
would have scattering, not absorption.

Actually, that is the way I was thinking of it, in the macroscopic
sense. However, I was hoping for a little more detailed insight at the
atomic level.

The core regions are optically very opaque and matter and light are in local
thermodynamic equilibrium. That is to say, the distribution of photon
energy is given by Planck's radiation law and the distribution of velocities
of particles is given by the Maxwell distribution. If there are any bound
states their populations are strictly governed by the Boltzman distribution.
All three expressions would use the same thermodynamic temperature T.
However, the density and temperature of the sun's central regions are so
high that there are probably no bound states.

So far, at the atomic scale, the following mechanisms have been
proposed, all of which can be happening to different nucleii at the
same time:

(1) high-energy photon absorbed, boosts ions and makes them move a
little faster, but the ions don't collide due to electromagnetic
repulsion (which involves emitting photons too, which sap energy
away): some of the energy given back in the form of a re-emitted
lower-energy photon.
(2) gamma photons boost ions into each other, creating a fusion
process and the release of still further gamma rays.
(3) In the higher layers of the Sun, super-high energy photons are
rarer, and the high energy photons might arrive in the form of x-rays
or UV, thus giving smaller boosts than gamma rays.

For example, if an electron absorbs a gamma ray, it gets boosted to
a higher orbital. If that electron then drops back down partially,
it'll emit a photon of a lower energy and frequency, and then
perhaps later it'll drop down some more, and emit another lower
energy photon. Is that what's happening?

No, because in the central regions you have full ionization, even for
heavy elements, and certainly for hydrogen, the main source of
opacity. Hence there are no bound quantum states in the hot plasma,
only continuum. Near the surface that is no longer true and it gets
interesting.

Understood, electrons are so far apart in the core, it's not even
worth considering them as bound.

Actually, given the density of the core is higher than that of water, the
electrons are very close together (but moving very fast).


Yousuf Khan

--
Mike Dworetsky

(Remove pants sp*mbl*ck to reply)

Yousuf Khan wrote:

According to this, Sunlight is composed only of the range of the
electromagnetic spectrum from IR to UV, with the majority occurring in
the visible spectrum.

Sunlight - Wikipedia, the free encyclopedia
http://en.wikipedia.org/wiki/Sunlight#Composition

Question, if the Sun generates its photons inside its nuclear core, then
all of that light should be in the form of gamma rays. But by the time
it reaches the surface of the Sun, it doesn't go much over UV in energy.
So what mechanism is in place on the Sun that steps its photon energy
down from the gamma ray range to the lower frequencies?

Scattering.


I assume that there is also a multiplication effect, to conserve energy
balance, where they trade fewer gamma photons for a greater number of
lower energy photons?

Or, far more likely, the gamma rays are scattered down into oblivion
[Compton scattering] while the thermalized result glows in a nice little
blackbody that peaks in the green.


Yousuf Khan

On Oct 29, 7:30*pm, "Y.Porat" wrote:
it seems to me thatr no one of the above
solved the inhtereting question!!

how about some QUANTITATIVE considerations??

How about some? For starters, you can look some up. Try Iglesias, C.
A. & Rogers, F. J., "Opacities for the solar radiative interior",
Astrophysical Journal 371, 408-417 (1991) for a start. Note the
opacity. From the opacity and the depth, you can get an idea of the
optical thickness. Note that you can't see through the sun, according
to the theoretical prediction. This agrees with observation.

the light emisin process is done mainly
ie *Alpha particle creation
now
the *enrgy involved * *in it is
about say * 20 Mev
that is far more than light s * energy !!!
it is gamma energy

and please dont forget
*it is done at the outer surface of sun !! *not inside the core

now the *question remains
why is that energy
not emitted until our globe !!!???
or at * least some of it ??

Because, at depth, the Sun is opaque. Basically, we see the outmost
opaque layer. Since at the depths where the solar atomosphere is
opaque, the atmosphere and the radiation are in thermal equilibrium,
we see radiation that is close to a blackbody spectrum of the
temperature of the outermost opaque layer (which is the layer that we
see).

What else would you expect? Put a light bulb inside an opaque ball,
and you see thermal radiation from the outer layer of the ball, warmed
up by the internal light bulb.

On Oct 29, 1:58*pm, Timo Nieminen wrote:
On Oct 29, 7:30*pm, "Y.Porat" wrote:

it seems to me thatr no one of the above
solved the inhtereting question!!

how about some QUANTITATIVE considerations??

How about some? For starters, you can look some up. Try Iglesias, C.
A. & Rogers, F. J., "Opacities for the solar radiative interior",
Astrophysical Journal 371, 408-417 (1991) for a start. Note the
opacity. From the opacity and the depth, you can get an idea of the
optical thickness. Note that you can't see through the sun, according
to the theoretical prediction. This agrees with observation.

the light emisin process is done mainly
ie *Alpha particle creation
now
the *enrgy involved * *in it is
about say * 20 Mev
that is far more than light s * energy !!!
it is gamma energy

and please dont forget
*it is done at the outer surface of sun !! *not inside the core

now the *question remains
why is that energy
not emitted until our globe !!!???
or at * least some of it ??

Because, at depth, the Sun is opaque. Basically, we see the outmost
opaque layer. Since at the depths where the solar atomosphere is
opaque, the atmosphere and the radiation are in thermal equilibrium,
we see radiation that is close to a blackbody spectrum of the
temperature of the outermost opaque layer (which is the layer that we
see).

What else would you expect? Put a light bulb inside an opaque ball,
and you see thermal radiation from the outer layer of the ball, warmed
up by the internal light bulb.

------------------
OK
yet let me give you some hints:
1
what is the gravitational force at the surface of sun
iow
at a distance equal to the suns radius ??!!

2
photons have mass !!!
including the gamma photons ..... bigger than light ...
3
think about the black hole ...
--------------

TIA
Y.Porat
---------------------

On 27/10/2010 6:55 PM, dlzc wrote:
Now to your original question, "sunlight" is filtered by the
atmosphere, and UV-C and more energetic radiation is completely
blocked by nitrogen, oxygen, and ozone pitches in as well (but most
important for absorbing in UV-B).

Lest ye think the Sun does not emit X-rays and such...
http://www.asr.ucar.edu/2004/HAO/tiso.html
... it does. Just not much.

David A. Smith

Well, the graph in the following article led me to believe that, it
finishes around the high UV:

Sunlight - Wikipedia, the free encyclopedia
http://en.wikipedia.org/wiki/Sunlight#Composition

One graph showed the sunlight composition at the "top of the
atmosphere", which I assume means completely unfiltered solar light,
while the other one showed what remains at sea level.

Yousuf Khan