Transparency of a gas compared to a plasma

[Herbascious J]

TL;DR Summary

Which is more 'transparent' to photons, a gas or a plasma?

I imagine an empty region of space, in between stars, or even galaxies. This void is filled only with a uniform distribution of hydrogen particles. I won't specify how dense this field is, but I would like to know which is more transparent to photons, a gas of hydrogen atoms, hydrogen molecules (H2) or raw protons (ionized hydrogen). In the case of the Ionized hydrogen atoms, I'm assuming the electrons would be in their as well, uniformly distributed, but just not coupled to the proton nuclei, so that the whole field is now an ionized plasma of protons and electrons. I don't have a strong requirement about temperature either. My question is about photons passing through this medium, and if they interact, are dimmed, heat up the medium, appear opaque, etc. What happens to the photons in general. If there are dynamics surrounding temperature or field density, I'm curious to hear about this as well.

I should point out that the drive behind my question is to learn about astronomers trying to measure what is out there, what they can see.

 

[PeterDonis]

Moderator's note: Moved to Astronomy and Astrophysics forum as this question seems more appropriate there.

 

[PeterDonis]

I don't have a strong requirement about temperature either.

You should realize, though, that the differences between your three proposed states for the medium (H2 molecules, H atoms, and proton-electron plasma) already imply different ranges of temperature (roughly speaking, H2 molecules up to about 500-1000 degrees K, H atoms from roughly 500-1000 to 3000-5000 degrees K, and plasma above 3000-5000 degrees K).

My question is about photons passing through this medium, and if they interact, are dimmed, heat up the medium, appear opaque, etc. What happens to the photons in general.

The biggest factor is going to be what the average frequency/wavelength of the light, or, equivalently, average energy of the photons, is. Again, very roughly speaking:

Photons with average energy much less than the ionization energy (energy required to remove an electron completely from an H atom or an H2 molecule) are going to have a variety of possible interactions with H2 molecules or H atoms, corresponding to the differences between the available energy levels in the molecules or atoms. This is the sort of thing we see with ordinary light and ordinary materials: the material will vary in how transparent/opaque it is to light depending on the frequency/wavelength of the light. Such photons won't interact much with free protons or electrons; to the protons or electrons they will simply be seen as an electric and/or magnetic field.

Photons with average energy about equal to or greater than the ionization energy are going to have a simpler set of possible interactions, since they're too energetic to induce energy level transitions; they can ionize atoms or molecules, or scatter off of free protons or electrons. At high enough energies (gamma rays), other processes like pair production become possible, but such photon energies are rare.

 

[Herbascious J]

Photons with average energy much less than the ionization energy ...

... Such photons won't interact much with free protons or electrons; to the protons or electrons they will simply be seen as an electric and/or magnetic field.

This is one of the interests I took in the question. In the strange event that a distance and empty field outside of a galaxy would be full of raw proton and electrons, but not be near a star or high energy system, the star light from the neighboring galaxy would be low energy photons, correct? Does this mean that it would be difficult to detect those free protons out in the abyss? Is there a challenge in directly seeing how much free protons/electrons are out there when they estimate the intergalactic medium? If it's transparent, then do they have to speculate on how much there is based on gravitational behavior of the galaxy and assuming that any free protons/electrons would necessarily be the result of high energy photons, so you'd only expect them to be near a star or something energetic that could strip them apart? I guess what I'm asking is; is it difficult to directly see how many free protons are out there? Its the molicules and atoms that produce the "clouds" that was can see in the telescopes? I hope I'm being clear.

 

[hutchphd]

I honestly don't know the answer here but I will put forth an guess idea (someone will correct me!). Space is pretty much empty (duh!). It may well be that once ionized and ejected from a star or other hot object the probability of recombination is not large and so the lifetime of each ion/electron is large enough to provide nontrivial amounts in steady state.
Maybe.

 

[PeterDonis]

In the strange event that a distance and empty field outside of a galaxy would be full of raw proton and electrons, but not be near a star or high energy system, the star light from the neighboring galaxy would be low energy photons, correct?

No. The intensity of the light decreases as the inverse square of the distance, but the energy of individual photons does not. So, for example, the intensity of light from the Sun is lower on Earth than it is on Mercury, but the energy of individual photons from the Sun is the same; there are just fewer photons per unit area per unit time.

So when trying to figure out how photons emitted by stars in a galaxy will interact with a cloud of gas or plasma far outside of that galaxy, the energy per photon will depend on the stars doing the emitting; roughly speaking, the photons will be mostly visible light photons, since most stars emit light mostly in that range. The decreasing intensity due to distance means there will be fewer interactions of those photons with the gas or plasma, but each individual interaction will be driven by the energy of an individual photon, relative to the temperature of the gas or plasma.

 

[jim mcnamara]

Hmm. This lists both molecules and ions of molecules that have been observed in the interstellar medium.
It implies that ions can exist for at least periods long enough to be observable. Would emission of a photon "de-ionize" the molecule? I cannot google a decent answer ...my bad probably.

https://en.wikipedia.org/wiki/List_of_interstellar_and_circumstellar_molecules

 

[PeterDonis]

is it difficult to directly see how many free protons are out there?

I believe that plasma is significantly harder to see (as in, detect using telescopes that are sensitive to EM radiation from distant regions of the universe) than gas (atoms or molecules), yes. However, "harder" does not mean "impossible". There will still be effects from plasma on light from distant galaxies, they just are likely to be smaller in magnitude (at least, that's my rough guess) and therefore harder to detect.

Large enough amounts of plasma will have detectable gravitational effects and their presence can be inferred from those effects.

 

[PeterDonis]

Would emission of a photon "de-ionize" the molecule?

No. "De-ionize" means the ion has to combine with a free electron; it would be the electron that would have to emit a photon with enough energy to leave the electron bound to the ion, forming a neutral atom or molecule.

(Strictly speaking, it's not impossible in principle for the ion to emit the photon; however, given the much lower rest mass of the electron, it is much, much more likely that the electron will emit a photon that leaves it in a bound state with the ion, than that the ion will emit a photon that leaves it in a bound state with the electron.)

 

[snorkack]

Ionized hydrogen is conspicuous.
Why?
Because the protons and electrons can recombine.
Yes, recombining a proton and electron emits a continuum photon, which is hard to identify. And when hydrogen is recombined to a ground state, the resulting continuum photon is a Lyman continuum one, which can easily be absorbed by next hydrogen atom it meets, so it cannot even be seen.
But proton and electron can be recombined not only to ground state, but excited states. And when hydrogen is recombined to 3rd or higher orbitals then besides the initial continuum (goes to infrared then) the lines emitted by the excited hydrogen will include Balmer series among others.
Balmer series is visible, cold hydrogen is transparent to it, and the narrow lines are easy to recognize.

 

[andrew s 1905]

If there are strong magnetic fields to accelerate free electrons in a plasma then they will emit cyclotron radiation.

Also significant quantities of hot gas (~10% by mass) were missed in galaxies until they were observed in x-rays.

Regards Andrew

 

[snorkack]

No. The intensity of the light decreases as the inverse square of the distance, but the energy of individual photons does not. So, for example, the intensity of light from the Sun is lower on Earth than it is on Mercury, but the energy of individual photons from the Sun is the same; there are just fewer photons per unit area per unit time.

No. The energy of individual photons will decrease, but not as inverse square.
Why?
Near a hot star or high energy system, you have high energy photons that can ionize ground state hydrogen atoms.

When they encounter hydrogen atoms, even a minority of hydrogen atoms in a plasma where protons are a majority, they ionize the atoms.
What next?
The fast electron may meet another proton and recombine to another ground state hydrogen. Then you get a new photon which has the same high energy as the initial photon, plus or minus the relative speed of second proton to initial atom, and a new and random direction.
But sometimes, and often, the fast electron meets a proton and recombines initially to an excited state hydrogen.
What you then get is one photon when the electron initially recombines and one or more extra photons when the excited state decays to ground state. You get several photons from one high energy photon - of course all of them lower energy than the initial photon. Eventually you reach the point where none of the photons individually has enough energy to ionize hydrogen any longer.
High energy photons ionize atoms and lose energy by being converted to more and lower energy photons. Low energy photons, which do not have enough energy to ionize atoms, do not lose energy that way, and propagate without further loss of energy.

In the vicinity of a hot star, you have high energy photons because the few hydrogen atoms nearby have not been able to absorb them all, and conversely the photons have been able to ionize most hydrogen atoms there.

 

[PeterDonis]

The energy of individual photons will decrease, but not as inverse square.

If you include the effects of interactions of photons with matter, yes. I was talking about the idealized case where there is no matter for photons to interact with between being emitted and reaching the particular cloud of matter (gas or plasma) that the OP was interested in.

 

[andrew s 1905]

The spectra of the light emitted at the photosphere of the sun , as an example, matches that received at the top of our atmosphere very accurately with small additions from light from the chromosphere and corona.
Regards Andrew

 

[snorkack]

You should realize, though, that the differences between your three proposed states for the medium (H2 molecules, H atoms, and proton-electron plasma) already imply different ranges of temperature (roughly speaking, H2 molecules up to about 500-1000 degrees K, H atoms from roughly 500-1000 to 3000-5000 degrees K, and plasma above 3000-5000 degrees K).
The biggest factor is going to be what the average frequency/wavelength of the light, or, equivalently, average energy of the photons, is. Again, very roughly speaking:

Photons with average energy much less than the ionization energy (energy required to remove an electron completely from an H atom or an H2 molecule) are going to have a variety of possible interactions with H2 molecules or H atoms, corresponding to the differences between the available energy levels in the molecules or atoms. This is the sort of thing we see with ordinary light and ordinary materials: the material will vary in how transparent/opaque it is to light depending on the frequency/wavelength of the light. Such photons won't interact much with free protons or electrons; to the protons or electrons they will simply be seen as an electric and/or magnetic field.

Low energy photons can undergo Thomson scattering off charges - electrons or protons, but protons have much smaller cross-section because it is inversely proportional to mass. They also undergo Rayleigh scattering off polarizabilities... and Raman scattering, too.
H atom has NO excited states between 121 nm (Lyman alpha) and 21 cm (spin alignment of proton and electron). H₂ molecule has lowest vibrational state around 2200 nm and lowest rotational state near 85 000 nm. But here the thing is that hydrogen atoms are both neutral... they are accelerated but not charges. And therefore cannot radiate, nor absorb radiation.

 

[PeterDonis]

H atom has NO excited states between 121 nm (Lyman alpha) and 21 cm (spin alignment of proton and electron).

Sure it does:

https://en.wikipedia.org/wiki/Hydrogen_spectral_series

hydrogen atoms are both neutral... they are accelerated but not charges. And therefore cannot radiate, nor absorb radiation.

Sure they can. They won't emit or absorb radiation due to their acceleration, but that doesn't mean they won't emit or absorb radiation at all.

 

[snorkack]

Sure it does:

https://en.wikipedia.org/wiki/Hydrogen_spectral_series

But all these series are between different excited states, except Lyman series. And Lyman series ends with Lyman alpha at 121 nm. A photon longer than 121 nm cannot excite a H atom out of ground state.

 

[PeterDonis]

A photon longer than 121 nm cannot excite a H atom out of ground state.

Ah, got it.

 

[Fredbonyea]

All radiant energy obeys Chandrasekar's equation for radiation transfer. Gases in space are often ionized because of interactions with high energy photons, and there is a net radiation background that we are all aware of.

"I guess what I'm asking is; is it difficult to directly see how many free protons are out there?"

Very difficult; and an entirely open field for careful and unbiased research. For example, you have to be aware of photon multiplier effects that can occur in space as easily as in a vacuum tube.

We were not aware of the 'bending' moment of the sun until we flew a Ranger craft beyond the solar horizon and tried to use the measured curvature to collaborate the relative gravitational lensing determinations used by Eddington. The radio bandwidth used was totally uncooperative; and it is only when corrections for a magnetic field are included that a relativistic signal can be isolated. Even then, it requires some iffy assumptions about the magnetic gradient.

We need several space telescopes to really sort this, and we need some of them to be a long ways from our planet.