“Interpreting astronomical spectra” -- Continuum origins

I’ve been reading “Interpreting astronomical spectra” and am still confused about the micro processes contributing to the continuum. In particular,the sun’s photosphere. I’m not interested in the black body thermodynamics approach because some processes should cause this.
Bound-bound, Photoionization,recombination and free-free transitions do not seem to spread photon energy’s. So is the continuous spectrum solely caused by line broadening from various factors ? How about a star which is almost all hydrogen, how can broadening create a continuum?

Getterdog said:

So is the continuous spectrum solely caused by line broadening from various factors ?

Good question. First and foremost, do the spectral lines actually form a continuum? Here's a 10 nm slice of the solar spectrum:


As you can see there are a LOT of spectra lines. Somewhere on the order of 25,000 between 250 and 1000 nm. Depending on the width of each line it is conceivable that they could overlap enough to cover the entire visible and near-visible spectrum. However, from the picture above it looks like they either don't overlap enough to do so, or most of them aren't absorbing enough light to show up in this image.

Getterdog said:

How about a star which is almost all hydrogen, how can broadening create a continuum?

What proportion of hydrogen are we talking about here? The Sun is already something like 92% hydrogen (by particle number, not mass), with almost all of the remaining 8% being helium. Obviously removing everything but hydrogen and helium would give you a spectrum with barely any absorption lines at all, making it impossible to have a continuum from them.

Why do you say "free-free transitions do not seem to spread photon energy"? Consider a gas of photons and electrons only. It will have a continuous spectrum because the photons scattered off the electrons can have any energy.

Thanks,
I was under the impression that Thomson scattering doesn’t change the photon frequency,correct me if I’m wrong. So you’re saying that the continuum is solely due to free free transitions? It always seems that when discussing stellar atmospheres that the photo sphere provides A smooth black body curve ,more or less, and the atmosphere above is where we get all the absorption lines. The book doesn’t get into details except to say that a free electron cannot absorb a photon and increase its energy and it cannot emit a photon and loose some of its kinetic energy because both momentum and energy cannot be conserved. But the process can occur if there is an atom or ion interacting with the electron which can take up the recoil. ( pg 52). So you’re saying this changes the photon frequency?

Getterdog said:

Thanks,
I was under the impression that Thomson scattering doesn’t change the photon frequency,correct me if I’m wrong.

It does not except for Doppler shift. And the limit of Compton scattering.

Getterdog said:

So you’re saying that the continuum is solely due to free free transitions? It always seems that when discussing stellar atmospheres that the photo sphere provides A smooth black body curve ,more or less, and the atmosphere above is where we get all the absorption lines. The book doesn’t get into details except to say that a free electron cannot absorb a photon and increase its energy and it cannot emit a photon and loose some of its kinetic energy because both momentum and energy cannot be conserved. But the process can occur if there is an atom or ion interacting with the electron which can take up the recoil. ( pg 52). So you’re saying this changes the photon frequency?

Thomson scattering does not change photon frequency, but it also does not change photon count. Free-free transitions change photon count. Like the braking radiation. A fast electron collides with a nucleus and emits a new photon, which was not incoming into the interaction. Since the interaction is free-free, the electron bounces back with some energy left, the photon is emitted with continuum energy - anything between zero and the kinetic energy of incoming electron.
Another source of continuum emission is free-bound transitions. Like formation of hydride ions. A free electron collides with hydrogen atom and forms a hydride ion. The energy of hydride ion is defined - it actually even has no excited states - but the incoming electron can have different energies. So the photon energy is a continuum depending on the energy of incoming electron.

I’d like to know a lot more about these two mechanisms. Do you have any suggested books,articles on this?
Thankyou ! John

I would add this regarding your comment on hydride. The author states that in free bound transitions, the excess electron energy is converted to kinetic energy. I feel I have a long way to go before I fully understand this , given that even cooler stars with smaller free electron density we still see a a continuum. And free free transitions are only a factor in hotter stars.

Getterdog said:

I would add this regarding your comment on hydride. The author states that in free bound transitions, the excess electron energy is converted to kinetic energy. I feel I have a long way to go before I fully understand this

No. There are two different processes:

1. Free-free transition. Free electron (with a continuous energy) in the beginning, free electron and a photon in the end. In that case, the initial kinetic energy of electron (itself continuum to begin with) gets split in two parts - residual electron energy and emitted photon energy, both with continuum distribution.
2. Free-bound transition. Free electron (with a continuous energy) in the beginning, bound state and a photon in the end. In that case, there is no outgoing electron with kinetic energy, so the whole kinetic energy of incoming electron plus the attachment energy (a fixed value) get emitted as a photon.

After looking up everything I could on free-bound transitions,I think you’re right. Either Dr. Emerson is wrong or his information was incomplete. Interesting. His comment on the impossibility of the electron-photon without a third ion is in question as an oscillating electron emits radiation or am I confusing things?

Getterdog said:

Either Dr. Emerson is wrong or his information was incomplete. Interesting. His comment on the impossibility of the electron-photon without a third ion is in question as an oscillating electron emits radiation or am I confusing things?

A "free" electron is by the Newton law of inertia moving in straight uniform motion, or sitting in place. Either case it is not oscillating and therefore cannot emit radiation. You need a third particle to accelerate the electron and allow it to emit radiation.

I should have said accelerated or deaccelerated. Again, my questions pertain to stars less than 10,000K, below the ionization of hydrogen And in the visible range only. Bremstalung is soft Xray, vibrational and rotational modes are infrared. Free bound transitions should be minimal with low free electron density. And yet there is a continuum. One article suggested it was due to poor resolution of the instruments. I just don’t feel I have a good explanation for this.. We are talking about low metalicity stars.

Here's what I think happens. Deep inside the star, the temperatures are quite high and we have a fully ionized plasma. The electrons and photons scatter through Compton scattering, and the result is an electron gas and photon gas in thermal equilibrium, each with a continuous thermal spectrum. As the photons propagate outward to regions of lower temperature, they continue to collide with the cooler free electrons, losing energy so the photon temperature drops. This is the source of the continuous spectrum. Then in the cooler atmosphere, emission and absorption lines modify the spectrum to give the final spectrum we see.

Getterdog said:

And yet there is a continuum.

What continuum are you referring to? The thermal spectrum, or the spectral lines? My interpretation from your original post was that you were not interested in the thermal spectrum, which I thought was odd since the absorption lines aren't a continuous spectrum. Perhaps I was mistaken in what you were asking?

Drakkith said:

What continuum are you referring to? The thermal spectrum, or the spectral lines? My interpretation from your original post was that you were not interested in the thermal spectrum, which I thought was odd since the absorption lines aren't a continuous spectrum. Perhaps I was mistaken in what you were asking?

My interpretation was that the question is, what is the mechanism that makes a gas black? Why should a gas (of nonuniform absorptivity at different wavelength) have a blackbody thermal spectrum?

My question is this. The photosphere is considered close to a continuous black body spectrum. What we observe is the stellar atmosphere doing is job of absorption and in some case emission. I am asking is what are the micro-mechanisms causing the thermal curve in a cool star, less than 10000 K and low metalicity.,which means low free electron density. I am aware of the Planck derivation but this seems inadequate because it already assumes a continuous radiation field. Phyzguy has come up with some more plausible. Maybe I need to head over to the university. This has been bugging me for a long time.

I just ran across this, that thermal motion causes dipole oscillation and causes charge acceleration. With low free electron density the dipole assumes greater importance.so the dipole oscillation takes place completely within the atom? How does that work? Do collisions change the dipole moment of the atoms?

snorkack said:

Why should a gas (of nonuniform absorptivity at different wavelength) have a blackbody thermal spectrum?

This requires a thermal recevoir with very particular definitions, and was a nontrivial part of the exploration of this subject by Boltzmann (and Gibbs and a host of other very good brains....). The nature of the gas of matter is far from uniform but it provides the primary means by which the electromagnetic energies can interact and equilibrate.
Various layers and segments in the star are at different temperatures...the segregation comes largely from gravity and magnetic effects. The photosphere is what we typically think of as the visual surface....i.e. the star.

My photonics text gives an interesting explanation of collisional broadening in that collisions change the phase of normal lifetime emissions and in doing so creates Fourier transform effect with spreading of the spectrum. How this plays into this I’m not sure. The atomic dipole oscillation sound promising.

According to some grad lectures I’ve been watching, line transition strength is due to the dipole moment between orbitals. So the frequency spread may be due to interruption of this due to collisions?

Getterdog said:

TL;DR Summary: Micro processes responsible for continuum

Photoionization,recombination and free-free transitions do not seem to spread photon energy’s. So is the continuous spectrum solely caused by line broadening from various factors ? How about a star which is almost all hydrogen, how can broadening create a continuum?

Getterdog said:

The photosphere is considered close to a continuous black body spectrum.

The usual blackbody is a continuum limiting case. Cyclic boundary conditions on a finite box produce a normalized solution and the volume is expanded providing the continuum limit. Its close enough to reality to provide answers. A continuum is in the eye ("resolution" has a less poetic ring to it )of the beholder. So if I am not sure what the hangup is in your contemplation. Everything, it seems, is quantized (and discreet), often it doesn't matter and the continuum math is very much easier. The nexuses (nexes?) where there are strict selection rules and conservation rules that directly affect the physics can highlight such sloppiness . Such opportunities have explained the mendeleev table and some other trifles.

Getterdog said:

According to some grad lectures I’ve been watching, line transition strength is due to the dipole moment between orbitals. So the frequency spread may be due to interruption of this due to collisions?

FYI: The transition strength goes by "Fermi's Golden Rule" but this was first written down by P.A.M. Dirac I believe. It is a very important idea, but the explanetion is a bit more nuanced and will be part of any good Quantum treatment.

I guess simply put,how does thermal motion create a more or less continuous radiation field. The answer at least in wiki is that it causes dipole oscillation in the atoms I presume.,assuming the are no free electrons. That is or was my ”hang up”. Now what that means in the quantum treatment I have yet to discover. Maybe you could direct me to the relevant texts. I mean oscillation of a dipole moment.

From Fermi’s rule there must be a continuum of eigenstates into which the transition can occur. This requires an oscillating Hamiltonian. How do we get that? Is this referring to the radiation field? A full treatment would involve the tensor product of the radiation field and the atom. So the eigenstates of th field need not be occupied just present. Am I on the right track?

The golden Rule is the fundamental result for time dependent perturbation theory in any Quantum text. I like Gasiorowicz, Baym, Griffiths Gottfried Sakurai. The time dependent potential is the dipole component of the electromagnetic f field from an excitation thereof. which is the leading strength interaction with a neutral atomic. Energy will be conserved by the process in the end.

Must the perturbing potential be a sinusoidal ‘‘em field or could it be collisions?

It is the interaction of the quantized EM field with the charges on the atom, solved typically via perturbation theory. Please look in a book. Pictures in your head are not sufficient (particularly somebody else's pictures......) The EM field is writtten typically via the vector and scaler potentials.....this is in every book, and I will not spend my morning writing a book chapter! The books do it better! Read and learn them

Sorry to be a sort of dog with a bone .I do astro-spectroscopy (non-professional) and do a lot of outreach for our astronomy organization. I realized I need a much deeper understanding of thermal radiation beyond the usual cavity black body thermal equilibrium approach. I do have several books on this ,most just assume a ever present EM field. I have seen explanations from classical,semiclassical,quantum and even QFT. At this point the core idea seems to be that after adding energy,heat to a system, the density of states in the atoms increases and this provides more frequencies to fill the modes of the field.eventually coming to equilibrium. If this is roughly correct I’ll stop here. Thankyou for you patience.

Dog with a bone is not a bad thing. I do not really like the explanation but am unfamiliar with the particular scenario. The number of total states for the atoms will remain the same regardless....if you squeeze them together the interaction potential will shift the energy from the "isolated atom" lines. Effectively the perurbation expansion is a method for systematically including these small interactions. Each peak which for N isolated atom was N fold degenerate will become N states within a peaked spectrum the width of which is proportional to the interaction. It is very pretty physics.
The semiclassical approximation should be fine but for the Blackbody radiation, the ultraviolet catastrophe did require partitiong phase space in an Ad Hoc fashion using a new constant .

Thanks. I’ll drop the bone for nowI found a nice set of lectures on MIT open courseware on the golden rule. So that should help. Thanks again.

It should also be mentioned that the main source of the Sun's thermal continuum is a particular free-bound transition, that which creates the H minus ion. You start with a free electron over a wide range of energies (that's why it is a continuum) and it gets captured by a previously neutral H atom (which are in copious supply near the surface of a cool star like the Sun). It turns out there is a single bound state in the resulting H minus atom, which exists because the newly bound electron kind of shepherds the previously bound electron over to the opposite side of the proton, allowing the new electron to be bound (weakly) to the proton. Since the final state has definite (low) energy, and the initial state had a range of (higher) energies for the electron, the photon that is emitted comes over a range of energies. This photon is what sunlight is, most of the time.