Disctrete vs continuous spectrum

[Getterdog]

I’ve been digging around trying to get at the physical basis for the black body spectrum. Say I have a neutral gas,non ionized. This has its own discrete spectrum. Are the mechanisms of line broadening the reason why we seek continuous spectrum in the back body curve? That’s question 1. No 2 relates to what I’ve read as the cause of the B.B. curve,ie electron acceleration in atoms via collisions. If two atoms inelastically collide,can em radiation be emitted without either atom having a change in the orbitals,say by cloud deformation and relaxation? Thanks

 

[DrClaude]

I’ve been digging around trying to get at the physical basis for the black body spectrum. Say I have a neutral gas,non ionized. This has its own discrete spectrum. Are the mechanisms of line broadening the reason why we seek continuous spectrum in the back body curve? That’s question 1.

Gases usually don't make for very good blackbodies. In the cases where they would, there will indeed be plenty of broadening.

No 2 relates to what I’ve read as the cause of the B.B. curve,ie electron acceleration in atoms via collisions.

Where have you read such a thing?

If two atoms inelastically collide,can em radiation be emitted without either atom having a change in the orbitals,say by cloud deformation and relaxation?

You can't see a solid as made up of atoms that retain their isolated properties. The collection of atoms that is a solid results in continuous energy bands, so it doesn't make sense to talk about orbitals.

 

[Charles Link]

Line broadening of the discrete spectrum of an atom will in principle, never result in generating the continuous Planck blackbody spectrum. Bremsstrahung (radiation from acceleration of charged particles) is coupled sufficiently to the various (many and continuous) photon modes that ionized gases can quite often radiate very nearly like ideal blackbodies. This doesn't occur with two neutral atoms in an elastic collision, but rather with charged particles that get accelerated and/or decelerated when they collide.

 

[Jilang]

If the gas is not a zero kelvin there will be Doppler broadening.
http://www.phy.ohiou.edu/~mboett/astro401_fall12/broadening.pdf
So the answer to your first question is in some part yes.

 

[Charles Link]

If the gas is not a zero kelvin there will be Doppler broadening.
http://www.phy.ohiou.edu/~mboett/astro401_fall12/broadening.pdf
So the answer to your first question is in some part yes.

The atomic transitions are typically so far apart in wavelength that Doppler broadening is not going to make the emission spectrum from an atom anywhere close to continuous. I think it is giving the OP the wrong idea in this case to tell him that he is partly correct.

 

[Getterdog]

Line broadening of the discrete spectrum of an atom will in principle, never result in generating the continuous Planck blackbody spectrum. Bremsstrahung (radiation from acceleration of charged particles) is coupled sufficiently to the various (many and continuous) photon modes that ionized gases can quite often radiate very nearly like ideal blackbodies. This doesn't occur with two neutral atoms in an elastic collision, but rather with charged particles that get accelerated and/or decelerated when they collide.

So can “deformation” of the electron cloud by inelastic collisions that is ,no jumps between quantum levels,just deformation,generate em radiation.?

 

[Getterdog]

Line broadening of the discrete spectrum of an atom will in principle, never result in generating the continuous Planck blackbody spectrum. Bremsstrahung (radiation from acceleration of charged particles) is coupled sufficiently to the various (many and continuous) photon modes that ionized gases can quite often radiate very nearly like ideal blackbodies. This doesn't occur with two neutral atoms in an elastic collision, but rather with charged particles that get accelerated and/or decelerated when they collide.

Additionally, again I am thinking of a dense non ionized gas say,argon. It should have a black body spectrum.which is continuous. So where do all those other frequencies come from that are not included in Aragon’s usual spectrum which is discrete. If you say collisions,then some radiation has to be produced outside of the usual quantum level jumps,by that mechanism,right? I’ve read many different articles on this and have never really got an answer that connected these ideas. Thanks jk

 

[Charles Link]

Additionally, again I am thinking of a dense non ionized gas say,argon. It should have a black body spectrum.which is continuous. So where do all those other frequencies come from that are not included in Aragon’s usual spectrum which is discrete. If you say collisions,then some radiation has to be produced outside of the usual quantum level jumps,by that mechanism,right? I’ve read many different articles on this and have never really got an answer that connected these ideas. Thanks jk

If you are referring to an arc-type discharge lamp, at higher pressures and currents, the argon gas does become somewhat highly ionized, and is a plasma. The spectrum then becomes more like a blackbody, but will still contain some emission lines. At lower pressures and currents, the extent of ionization is less, and atomic emission lines will be much more prominent in the spectrum.

 

[Getterdog]

If you are referring to an arc-type discharge lamp, at higher pressures and currents, the argon gas does become somewhat highly ionized, and is a plasma. The spectrum then becomes more like a blackbody, but will still contain some emission lines. At lower pressures and currents, the extent of ionization is less, and atomic emission lines will be much more prominent in the spectrum.

As I understand it, all “body’s “ give off black body radiation,regardless of composition. So simply put,in a dense gas with no free electrons,what accounts for the continuous spectrum? Thanks

 

[Charles Link]

As I understand it, all “body’s “ give off black body radiation,regardless of composition. So simply put,in a dense gas with no free electrons,what accounts for the continuous spectrum? Thanks

Gases give off "blackbody radiation, but in the case of a "dense" gas, the emissivity $\epsilon$ can be very small=as it typically will be for a gas. Emissivity is normally a function of wavelength, but assuming it is independent of wavelength, you will have $M=\epsilon \sigma T^4$, with emissivity $\epsilon <<1$, as the radiated power per unit area, instead of $M_{bb}=\sigma T^4$, with emissivity $\epsilon=1$. For a gas that is virtually transparent, by Kirchhoff's law, the result will be that the emissivity $\epsilon$ is nearly zero, and $M$ will be much smaller than that of a blackbody. $\\$ Meanwhile, the optical absorption properties of a plasma are much different. Plasmas are not optically transparent, so, by Kirchhoff's law, a plasma can have emissivity $\epsilon$ very nearly 1.0 in some cases.

 

[Getterdog]

Gases give off "blackbody radiation, but in the case of a "dense" gas, the emissivity $\epsilon$ can be very small=as it typically will be for a gas. Emissivity is normally a function of wavelength, but assuming it is independent of wavelength, you will have $M=\epsilon \sigma T^4$ as the radiated energy per unit area, instead of $M_{bb}=\sigma T^4$. For a gas that is virtually transparent, by Kirchhoff's law, the result will be that the emissivity $\epsilon$ is nearly zero, and $M$ will be much smaller than that of a blackbody. $\\$ Meanwhile, the optical absorption properties of a plasma are much different. Plasmas are not optically transparent, so, by Kirchhoff's law, a plasma can have emissivity $\epsilon$ very nearly 1.0 in some cases.

Ok,I looked up emmisivity and note that you mentioned it is a function of wavelength which is still continuous,not discrete. So can anyone really explain a continuous spectrum without also assuming free electrons running around? Thanks jk

 

[Charles Link]

Ok,I looked up emmisivity and note that you mentioned it is a function of wavelength which is still continuous,not discrete. So can anyone really explain a continuous spectrum without also assuming free electrons running around? Thanks jk

Without free electrons, there isn't much of a mechanism to generate a non-zero emissivity. There are other electromagnetic mechanisms that can result in very small values of emissivity, but nothing that will make it close to 1.0. $\\$ In the case of molecules, you do get some of these additional sources=vibrations, and rotations, etc. A gas of neutral atoms is largely transparent, and thereby has low emissivity.

 

[Getterdog]

Having reviewed everything on this both here and in stack exchange,here is what I’ve come up with. There is no one mechanism to explain the full frequency span of black body radiation,and different frequency regimes will have different mechanisms.However,if the body is to be in thermodynamic equilibrium with the ambient em surroundings,it must have this distribution. My mistake was to assume collisional processes could account for it all. This sound about right?

 

[Daz]

I think there’s a misapprehension here that gases emit black body radiation. They do not. I suspect that the OP’s misapprehension that they do stems from lax terminology; some people write “black body radiation” when what they should really have written is “thermal radiation.”

In fact, nothing emits a true black body spectrum. Put another way, the only thing that emits true black body radiation is a true black body. (The clue is in the name.) And they don’t exist.

Black, remember, means the object absorbs all light at all wavelengths. In turn, this implies that within that object there must be electronic transitions, vibrational and rotational transitions which, with the help of some line broadening perhaps, are able to absorb photons of any energy. By the same token, if an object can absorb light at any energy, it can also emit light at any wavelength.

If an object — a sample of argon gas, for example — possesses no mechanism by which it can absorb light at a particular wavelength then it will emit no thermal radiation at that wavelength either. (Alternatively, the emissivity is zero at that wavelength.)

So the misapprehension is believing that “all bodies emits black body radiation regardless of composition” that is categorically not the case.

 

[Getterdog]

Well,we’re told that the sun emits light as a black body at a certain temperature,and has a peak emission just a a black body would. But I thought even a cooler non ionized gas has some emmisivity that follows the black body curve. So how should they approach this in beginning astrophysics,so that it’s more accurate? Thanks jk

 

[Daz]

I suppose that I should have pointed out that it is possible to get very, very close to the ideal black body curve and in fact it gets easier as the object gets hotter. That is for several reasons: (1) the “things” with which the light interacts (electrons, vibrational modes, etc) will follow either Fermi-Dirac or Bose-Einstein distributions (maybe even Maxwell-Boltzmann distribution as well for some things but I can’t think of any right now.) The distribution of energy states becomes more like a continuum at higher energies so the higher the temperature the more your F-D or B-E distribution encroaches into the continuum-like region. (2) The line-broadening mechanisms are temperature dependent and increase with T.

So please don’t think that the black body spectrum is just a theoretical curiosity — I think I may have given that impression. In reality many things (including the sun) are very good approximations to it.

As an aside it is precisely the fact the black body spectrum is seen so precisely in nature that spurred the development of quantum mechanics. It was known for centuries that blacksmiths could tell the temperature of their forges just by looking at the colour of the light, and the coals and the iron all looked the same. That sparked interest in the black body curve and ultimately led Plank to hypothesise quanta.

Despite that, you shouldn’t go looking for an explanation for transitions where none exist. Consider the following thought experiment. Take a sample of argon gas - we know it is completely transparent in the green region of the spectrum. Now take that sample and place it inside a perfectly reflecting container (a perfect hohlraum) with a window filled with a filter which let's green light pass but reflects all other wavelengths. We know that any green light entering through the window will not be absorbed so we don’t care about that. But if, as you believe, your sample of argon emits black body radiation then some of that emission will be in the green region and that green light will bounce around inside the hohlraum until it eventually escapes through the window. In other words, there would be a net outflow of energy and the sample would spontaneously cool down.

Doesn’t happen!

 

[Getterdog]

I suppose that I should have pointed out that it is possible to get very, very close to the ideal black body curve and in fact it gets easier as the object gets hotter. That is for several reasons: (1) the “things” with which the light interacts (electrons, vibrational modes, etc) will follow either Fermi-Dirac or Bose-Einstein distributions (maybe even Maxwell-Boltzmann distribution as well for some things but I can’t think of any right now.) The distribution of energy states becomes more like a continuum at higher energies so the higher the temperature the more your F-D or B-E distribution encroaches into the continuum-like region. (2) The line-broadening mechanisms are temperature dependent and increase with T.

So please don’t think that the black body spectrum is just a theoretical curiosity — I think I may have given that impression. In reality many things (including the sun) are very good approximations to it.

As an aside it is precisely the fact the black body spectrum is seen so precisely in nature that spurred the development of quantum mechanics. It was known for centuries that blacksmiths could tell the temperature of their forges just by looking at the colour of the light, and the coals and the iron all looked the same. That sparked interest in the black body curve and ultimately led Plank to hypothesise quanta.

Despite that, you shouldn’t go looking for an explanation for transitions where none exist. Consider the following thought experiment. Take a sample of argon gas - we know it is completely transparent in the green region of the spectrum. Now take that sample and place it inside a perfectly reflecting container (a perfect hohlraum) with a window filled with a filter which let's green light pass but reflects all other wavelengths. We know that any green light entering through the window will not be absorbed so we don’t care about that. But if, as you believe, your sample of argon emits black body radiation then some of that emission will be in the green region and that green light will bounce around inside the hohlraum until it eventually escapes through the window. In other words, there would be a net outflow of energy and the sample would spontaneously cool down.

Doesn’t happen!

Thanks ,it’s becoming clearer. I should really focus on thermal radiation. My thought experiment is of a cold substance,so that thermal motion is insufficient to cause “orbital jumps’. Nevertheless there should be thermal radiation. Is this solely because of acceleration/deceleration of the electron clouds during collisions? This would then give a continuous distribution of radiation?

 

[Daz]

No, it is not solely because of one thing (acceleration / deceleration of electrons, etc.) and you will not get a “continuous distribution of radiation” from a cold gas. As you correctly observed earlier there are a variety of mechanisms contributing to the observed spectra, whether that be black, blue or purple with yellow spots.

The term “thermal radiation” describes how an object emits and absorbs radiation at various wavelengths (all wavelengths if it’s truly black) in order to reach thermal equilibrium with its surroundings.

At extremely high temperatures x-rays are the dominant component. But you only get x-rays from nuclear transitions, right? At lower temperatures far-IR is the dominant component. But you only get far-IR photons from vibrational or rotational transitions, right? Between these two (arbitrary) extremes (near UV, visible, near IR) electronic transitions dominate, right? And further out, radio waves would dominate at the low-energy end, while gamma rays would dominant at the high-energy end.

The emission and absorption of these disparate objects (radio, microwave, IR, visible, UV, X-ray, gamma) photons proceeds via very different mechanisms, which begs the question how do these various mechanisms conspire to generate the required spectrum and reach thermal equilibrium? One might be tempted to imagine that there’s some sort of sub-atomic conspiracy going on:

Electrons: Oi Nucleus! We need another of those high-energy photons PDQ or the humans are going to figure what were up to!
Nucleus: Gimme a break I’m dancing with another nucleus to generate a low-energy photon as per your request.

Actually, there’s no conspiracy. The required spectra are brought to you courtesy of statistical physics. Everything that can happen will happen with a probability determined by Maxwell-Boltzmann, Fermi-Dirac or Bose-Einstein* and while in any given interval the spectrum won’t look exactly like what you expected, on average — ON AVERAGE — it will.

* Yea yea, density of states an' all that, as well.

 

[Getterdog]

No, it is not solely because of one thing (acceleration / deceleration of electrons, etc.) and you will not get a “continuous distribution of radiation” from a cold gas. As you correctly observed earlier there are a variety of mechanisms contributing to the observed spectra, whether that be black, blue or purple with yellow spots.

The term “thermal radiation” describes how an object emits and absorbs radiation at various wavelengths (all wavelengths if it’s truly black) in order to reach thermal equilibrium with its surroundings.

At extremely high temperatures x-rays are the dominant component. But you only get x-rays from nuclear transitions, right? At lower temperatures far-IR is the dominant component. But you only get far-IR photons from vibrational or rotational transitions, right? Between these two (arbitrary) extremes (near UV, visible, near IR) electronic transitions dominate, right? And further out, radio waves would dominate at the low-energy end, while gamma rays would dominant at the high-energy end.

The emission and absorption of these disparate objects (radio, microwave, IR, visible, UV, X-ray, gamma) photons proceeds via very different mechanisms, which begs the question how do these various mechanisms conspire to generate the required spectrum and reach thermal equilibrium? One might be tempted to imagine that there’s some sort of sub-atomic conspiracy going on:

Electrons: Oi Nucleus! We need another of those high-energy photons PDQ or the humans are going to figure what were up to!
Nucleus: Gimme a break I’m dancing with another nucleus to generate a low-energy photon as per your request.

Actually, there’s no conspiracy. The required spectra are brought to you courtesy of statistical physics. Everything that can happen will happen with a probability determined by Maxwell-Boltzmann, Fermi-Dirac or Bose-Einstein* and while in any given interval the spectrum won’t look exactly like what you expected, on average — ON AVERAGE — it will.

* Yea yea, density of states an' all that, as well.

LOL, I hope you’re a teacher! Thanks ,I’ve got to mull on this awhile. Jk