Gases as blackbodies

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  • #1
snorkack
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Moderator's note: thread split off from https://www.physicsforums.com/threads/confusion-about-absorption-spectra-cool-gasses-absorb.964474/

I don't get your point, thermal doesn't mean infrared! It is because the everyday temperatures we encounter here on Earth means that the thermal emission of the everyday objects peaks at infrared part of the spectrum. For hot stars the thermal radiation may peak in visible or UV part of spectrum. Btw. thermal radiation and blackbody radiation are basically synonyms (at least in astrophysics). So that is for the continuous spectrums...
And regarding discrete absorption lines, they are not restricted to visible part of the spectrum only. You can have lines in all parts of the spectrum (UV, IR, MW, radio...).
Gases are not required to be black bodies.
A completely white body can neither absorb nor emit - no matter what its temperature may be.
And objects that possesses spectral lines are neither black nor white - they are coloured.
For example, consider warm He gas.
It possesses spectral lines of absorption... bluewards of 60 nm. Yes, you can cause it to radiate thermally... by making it hot enough to excite appreciable numbers of electrons. In that case, you also excite other transitions of lower energy, including visible, as well as absorb visible light.
But if He is not hot enough to excite electrons, then it has no spectral lines available. It possesses internal excitations of translational movement of atoms relative to each other, but these have no dipole moment and no way of emitting radiation.
So warm He is an example of heat, but no thermal radiation because no absorption.
 
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  • #2
lomidrevo
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Gases are not required to be black bodies.
A completely white body can neither absorb nor emit - no matter what its temperature may be.
And objects that possesses spectral lines are neither black nor white - they are coloured.
The black body doesn't mean that the color of the object is black. I means that all incident radiation is being absorbed, none is reflected. The radiation emitted by the black body is fully described by it's temperature and depending on it, the spectrum is peaking at some particular wavelength, which corresponds to "color" of the object as we observe it (in case it falls into visible part of the spectrum). This description is oversimplified but I hope it provides the main idea.
In nature, you would probably hardly find any ideal black body, but many objects can be indeed approximated as black body very well. For example in astronomy: radiation emitted by atmospheres of stars or (even planets like earth), CMB ... all of that can be approximated as black body radiation. In these real cases, the spectrum of black body radiation provides the "continuous background" of the spectrum on which some finer features like absorption or emission spectral lines can be superimposed.

But if He is not hot enough to excite electrons, then it has no spectral lines available.
Firstly, as I said before, the thermal radiation is characterized by the temperature of the gas. So if its temperature is > 0 K, it will radiate (no matter the number of excited atoms is very low). Secondly, absorption spectral lines are still available. It just depend on the light propagating through the gas at which wavelengths they become significant (recognizable in the spectrum).

So warm He is an example of heat, but no thermal radiation because no absorption.
I don't want to repeat myself, see my above comments.
Btw. "heat radiaton" = "thermal radiation", although the term "thermal radiation" is much more used in texts.
 
  • #3
snorkack
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The black body doesn't mean that the color of the object is black. I means that all incident radiation is being absorbed, none is reflected.
And none transmitted.
The radiation emitted by the black body is fully described by it's temperature and depending on it, the spectrum is peaking at some particular wavelength, which corresponds to "color" of the object as we observe it (in case it falls into visible part of the spectrum). This description is oversimplified but I hope it provides the main idea.
And radiation emitted by a body that is not black is, if thermal, equal or less to that of black body radiation. Including none at all.
A completely white body would be a body which absorbs no incident radiation at all - all is reflected or transmitted. And an absolutely white body could not emit anything either, no matter what its temperature. (Otherwise it would be an easy perpetuum mobile of second kind).
A completely gray body would be a body which absorbs an equal fraction of radiation at any wavelength, and reflects or transmits the rest.
A body which absorbs different fraction of radiation at different wavelength would be coloured. And that includes anybody possessing spectral lines.
In nature, you would probably hardly find any ideal black body, but many objects can be indeed approximated as black body very well. For example in astronomy: radiation emitted by atmospheres of stars or (even planets like earth), CMB ... all of that can be approximated as black body radiation. In these real cases, the spectrum of black body radiation provides the "continuous background" of the spectrum on which some finer features like absorption or emission spectral lines can be superimposed.


Firstly, as I said before, the thermal radiation is characterized by the temperature of the gas. So if its temperature is > 0 K, it will radiate
No, only if the gas were a black body. Which it is not.
 
  • #4
Drakkith
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Snorkack, are you saying a gas above 0K emits no radiation until it gets hot enough to generate spectral lines?
 
  • #5
snorkack
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Snorkack, are you saying a gas above 0K emits no radiation until it gets hot enough to generate spectral lines?
Not absolutely no radiation. Wien tail of blackbody radiation in principle exists.
But the total radiation of a black body continuum drops off with fourth power of temperature. The radiation of a spectral line with a certain energy to excite drops off exponentially with temperature. Meaning that in total absence of low energy legal lines to excite, there is very little possibility to emit.

Translational collisions between He atoms cannot emit infrared due to lack of dipole moment. Vibrations of H2 are also barred from emitting - but the symmetry is lifted by collisions.
Can He atoms emit infrared in three-body collisions?
 
  • #6
lomidrevo
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A body which absorbs different fraction of radiation at different wavelength would be coloured. And that includes anybody possessing spectral lines.
I am not sure I understand what you want to say. Do you assume that color of bodies is determined only by (emission) spectral lines? I don't agree. The perceived color of an object is a "mixture" of all radiation coming from that object. But the reflected radiation is much more significant then emitted radiation when it comes to the color of everyday objects (except the light sources).

No, only if the gas were a black body. Which it is not.
I already said that in nature you would hardly find any ideal (perfect) black body. But consider optically thick gas: to a good approximation, its radiation corresponds to black body radiation, determined by Planck's law.

Not absolutely no radiation. Wien tail of blackbody radiation in principle exists.
But the total radiation of a black body continuum drops off with fourth power of temperature.
This pretty much describes the black body radiation. (I suppose you are referring to Stefan-Boltzmann equation). Even if the luminosity is much much lower for low temperatures, the law is the same. So now I am confused about what is your conviction. It contradicts what you said in previous posts.

The radiation of a spectral line with a certain energy to excite drops off exponentially with temperature. Meaning that in total absence of low energy legal lines to excite, there is very little possibility to emit.
I think the source of your misconception is that you relates the thermal radiation and emission spectral lines. Thermal radiation is attributed to thermal motion of particles (which leads to acceleration of charges or dipole oscillations producing the continuous EM radiation) whereas the discrete emission spectral lines are caused by bound-bound electron transitions inside an atom. Those are two different mechanisms.
 
  • #7
snorkack
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I am not sure I understand what you want to say. Do you assume that color of bodies is determined only by (emission) spectral lines? I don't agree. The perceived color of an object is a "mixture" of all radiation coming from that object. But the reflected radiation is much more significant then emitted radiation when it comes to the color of everyday objects (except the light sources).
Absolutely black body reflect no light. All radiation coming from it is emitted thermal radiation, and determined by its temperature independent of incident light.
Absolutely white body reflects all light regardless of wavelength. But importantly, it cannot emit any, no matter what its temperature. All light from it is reflected radiation, and it is unaltered incident radiation.
Bodies have colour of themselves if they interact differently with different wavelengths. But then the difference must exist on both reflection and emission.
I already said that in nature you would hardly find any ideal (perfect) black body. But consider optically thick gas: to a good approximation, its radiation corresponds to black body radiation, determined by Planck's law.
Yes. If it is optically thick through spectrum.
I was looking at the other end - gas which is optically thin at spectral lines and especially thin between the lines.
And in that case, Planck black body law provides the upper bound on the optically thick limit - optically thin gas emits much less.
I think the source of your misconception is that you relates the thermal radiation and emission spectral lines. Thermal radiation is attributed to thermal motion of particles (which leads to acceleration of charges or dipole oscillations producing the continuous EM radiation) whereas the discrete emission spectral lines are caused by bound-bound electron transitions inside an atom. Those are two different mechanisms.

No, not fundamentally.
Nuclear vibrations in molecules are quantized just as well. So are molecular rotations. Thermal motion of particles in a thin gas is liable to produce a line spectrum.
 
  • #8
lomidrevo
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I was looking at the other end - gas which is optically thin at spectral lines and especially thin between the lines.
Well, in the discussion about thermal radiation I had mainly dense, optically thick gasses in my mind. (as this case is much common in astrophysics)

Nuclear vibrations in molecules are quantized just as well. So are molecular rotations. Thermal motion of particles in a thin gas is liable to produce a line spectrum.
OK, I see your point now. My understanding is that, the thermal energy constitutes mainly of the kinetic energy of particles inside the gas (as per kinetic theory of gasses). I would say that energy of vibrations is not that significant. But I don't have any quantitative explanation for that, so I am not going to argue, because I might be wrong. Maybe the density of the gas is the crucial parameter in this context and for really dilute gasses at really low temperature, the kinetic energy of particles become less important than vibrations. Maybe? I don't know.
 
  • #9
Tom.G
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Absolutely white body reflects all light regardless of wavelength. But importantly, it cannot emit any, no matter what its temperature.
(bold added)
Then it would follow that if a White body were heated via contact with a hot body, then the hot body could be removed and the White body would maintain its temperature until at least the end of the Universe. Since we haven't yet found an "Absolutely white body", I guess we'll never know.
Sigh. :frown:
 
  • #10
snorkack
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Well, in the discussion about thermal radiation I had mainly dense, optically thick gasses in my mind. (as this case is much common in astrophysics)
But so is the case of tenuous, optically thin gases.
OK, I see your point now. My understanding is that, the thermal energy constitutes mainly of the kinetic energy of particles inside the gas (as per kinetic theory of gasses). I would say that energy of vibrations is not that significant. But I don't have any quantitative explanation for that, so I am not going to argue, because I might be wrong.
Simple quantitative explanation is equipartition of energy between degrees of freedom.
In steam, a water molecule has 3 atoms.
Therefore, just 1/3 of the thermal energy of steam is the 3 translational degrees of freedom of water molecules.
Second third is 3 rotational degrees of freedom, and third third is 3 degrees of freedom for internal vibrations.
Maybe the density of the gas is the crucial parameter in this context and for really dilute gasses at really low temperature, the kinetic energy of particles become less important than vibrations. Maybe? I don't know.
Not so much for the stockpile of energy, but for radiation, and absorption.
O and H have different polarity, so water is allowed to radiate during internal vibrations and rotations - vibrations and rotations cause acceleration of charges.
Yes, water molecules might also radiate during collisions - translational energy is a third of all energy of steam.
But in case of translational energy, charges are only accelerated and able to radiate during collisions. Between collisions, the molecules are not accelerated and cannot radiate.
Whereas a molecule which is left vibrating or rotating is accelerating charges all the time between collisions and therefore allowed to radiate.
The lower the density of the gas, the less important radiation during collisions would be relative to radiation between collisions.
(bold added)
Then it would follow that if a White body were heated via contact with a hot body, then the hot body could be removed and the White body would maintain its temperature until at least the end of the Universe. Since we haven't yet found an "Absolutely white body", I guess we'll never know.
Sigh. :frown:

In astronomy, we see a lot of objects giving off feeble, but high energy radiation - starting with crown of Sun. These cannot possibly be black. What they are is tenuous, optically thin gases that are able to cool only slowly, which is why they are still hot.
And dark matter emits nothing. Seems that it qualifies as "absolutely white body" and therefore we have found such.
 
  • #11
DrClaude
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Absolutely black body reflect no light. All radiation coming from it is emitted thermal radiation, and determined by its temperature independent of incident light.
Absolutely white body reflects all light regardless of wavelength. But importantly, it cannot emit any, no matter what its temperature. All light from it is reflected radiation, and it is unaltered incident radiation.
I would just like to point out that there is an entire "spectrum" (pun intended) of grey bodies in-between the two extremes, and that there are not actual black bodies (emissivity = 1) in nature (we always find emissivity < 1).

Bodies have colour of themselves if they interact differently with different wavelengths. But then the difference must exist on both reflection and emission.
Yes, Kirchhoff's law of thermal radiation.

Simple quantitative explanation is equipartition of energy between degrees of freedom.
In steam, a water molecule has 3 atoms.
Therefore, just 1/3 of the thermal energy of steam is the 3 translational degrees of freedom of water molecules.
Second third is 3 rotational degrees of freedom, and third third is 3 degrees of freedom for internal vibrations.
And this depends on also on temperature. For instance, at room temperature, vibrational modes are "frozen": there is not enough energy to get a significant population in the first excited vibrational state, such that only translational and rotational degrees of freedom play a role.
 
  • #12
lomidrevo
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But in case of translational energy, charges are only accelerated and able to radiate during collisions. Between collisions, the molecules are not accelerated and cannot radiate.
Whereas a molecule which is left vibrating or rotating is accelerating charges all the time between collisions and therefore allowed to radiate.
But as per Lorentz law, there will always be a force acting on the charge in a nonzero EM field. Thus the moving charges would have nonzero acceleration even between collision, wouldn't they? We know that EM field in the gas cannot be exactly zero.

A completely white body would be a body which absorbs no incident radiation at all - all is reflected or transmitted.
And dark matter emits nothing. Seems that it qualifies as "absolutely white body" and therefore we have found such.
That would mean that dark matter works as a perfect mirror, which apparently contradicts observations. Dark matter doesn't interact with light at all - it cannot be a white body as per definition you provided in previous post.
 
  • #13
lomidrevo
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And this depends on also on temperature. For instance, at room temperature, vibrational modes are "frozen": there is not enough energy to get a significant population in the first excited vibrational state, such that only translational and rotational degrees of freedom play a role.
Correct me if I am wrong, but doesn't it suggest that in gasses at low temperatures the translational movement of particles makes the significant contribution to the thermal energy? Isn't this leading to a continuous spectrum as in case of a black body radiation?
 
  • #14
DrClaude
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Correct me if I am wrong, but doesn't it suggest that in gasses at low temperatures the translational movement of particles makes the significant contribution to the thermal energy?
Rotation is also present from a relatively low temperature.

Isn't this leading to a continuous spectrum as in case of a black body radiation?
For dilute gases, there is very little thermal emission related to translation. I wouldn't consider a blackbody model to apply to a dilute gas.
 
  • #15
lomidrevo
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For dilute gases, there is very little thermal emission related to translation. I wouldn't consider a blackbody model to apply to a dilute gas.
So can we say that the density of the gas is the only key parameter here? According to the kinetic theory, the mean free path doesn't depend on temperature.
 
  • #16
snorkack
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But as per Lorentz law, there will always be a force acting on the charge in a nonzero EM field. Thus the moving charges would have nonzero acceleration even between collision, wouldn't they? We know that EM field in the gas cannot be exactly zero.
But gas particles are overwhelmingly neutral at low temperatures.
A completely white body would be a body which absorbs no incident radiation at all - all is reflected or transmitted.

That would mean that dark matter works as a perfect mirror, which apparently contradicts observations. Dark matter doesn't interact with light at all - it cannot be a white body as per definition you provided in previous post.
I was careful to specify that all is reflected or transmitted. A body that reflects no light but transmits all fits within the phrase "all is reflected or transmitted" and therefore within the definition of a "white body".
 
  • #17
Tom.G
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I was careful to specify that all is reflected or transmitted.
But that definition of a White body is in conflict with all the definitions I've been able to find. Here are a couple examples.

Emissivity
Emissivity values can range from 0 to 1. A blackbody has an emissivity of 1, while a perfect reflector or whitebody has an emissivity of 0.

http://gsp.humboldt.edu/OLM/Courses/GSP_216_Online/lesson8-1/emissivity.html

A material with an emissivity value of 0 would be considered a perfect thermal mirror.
https://www.thermoworks.com/emissivity_table
 
  • #18
lomidrevo
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I was careful to specify that all is reflected or transmitted. A body that reflects no light but transmits all fits within the phrase "all is reflected or transmitted" and therefore within the definition of a "white body".
Ah sorry, I was acting too quickly... I wanted to quote another of your statement: (I admit I should rather double-read my response before posting)
Absolutely white body reflects all light regardless of wavelength.
I've searched on google for definition of white body, and usually it is described as body that "reflects all incident radiation ", so the transmittance is implicitly excluded. Maybe just here, they used "zero absorptivity" which would include a body transmitting all incident radiation as well.
But none of the definitions looks authoritative enough to me. Do you have a reference to be recommended? Is the term white body actually helpful in science? Honestly I've heard of it for the first time in this thread.
 
  • #19
snorkack
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Maybe just here, they used "zero absorptivity" which would include a body transmitting all incident radiation as well.
But none of the definitions looks authoritative enough to me. Do you have a reference to be recommended? Is the term white body actually helpful in science? Honestly I've heard of it for the first time in this thread.

The important point here is to pay attention to implication of Kirchoff´s law. Emissivity is necessarily proportional to absorptivity. Bodies with low absorptivity necessarily have low emissivity and low emission even at high temperature.
And it is then irrelevant whether the rest of incident radiation is reflected, refracted, scattered or transmitted. Optically thin gases are more common than objects which cause strong reflection/refraction/scattering unaccompanied by absorption.
 
  • #20
lomidrevo
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Emissivity is necessarily proportional to absorptivity. Bodies with low absorptivity necessarily have low emissivity and low emission even at high temperature.
That is clear, no discrepancy on this one.

But with respect to the description of dark matter, it is not irrelevant whether the incident radiation is reflected or transmitted. As pointed out, the most common definition of a white body seems to be "a body which reflects all incident radiation regardless of wavelength". For example, consider the definition on wikipedia (sorry, I am not able to find the definition in some more serious sources):
A white body is one with a "rough surface [that] reflects all incident rays completely and uniformly in all directions.", with reference to Planck, M. (1914). The Theory of Heat Radiation. Masius, M. (transl.) (2nd ed.).​
With this definition, you cannot regard dark matter as an example of a white body.

You yourself used two different definitions of the white body, see posts #3 and #7. Maybe there is not only a single definition of the white body being used. In such case the term might not be very useful, as different people mean by it bodies with a ("slightly") different properties. Just my opinion...
 

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