0

Black body is an object, hence it's made of atoms. Depending on what atom it consists of, that's how the emitted spectrum should be in my opinion. If it contains hydrogen, absorbed light on black body would excite electrons and electrons of hydrogen can only produce certain wavelengths. Hence, there must be dark lines in the final, emitted radiation(which is called - black body radiation).

how does perfect black body have continues spectrum (with no dark lines ?) That means when we talk about perfect black body, we shouldn't talk about atoms inside it as there's no atom existing that absorbs or emits light at any wavelength.

Qmechanic
  • 201,751
Matt
  • 367

2 Answers2

1

Black body is an object, hence it's made of atoms.

No, a black body is not made of atoms. It is an imaginary object which has the special (but physically unrealizable) property that it is a perfect absorber - any incident electromagnetic radiation is absorbed, with zero reflection and zero transmission. Under certain conditions, many objects in nature (stars, rocks, people) can be modeled as black bodies (or some modification thereof) to varying degrees of accuracy.

Depending on what atom it consists of, that's how the emitted spectrum should be in my opinion. If it contains hydrogen, absorbed light on black body would excite electrons and electrons of hydrogen can only produce certain wavelengths.

The electromagnetic radiation emitted by an object is not determined exclusively by the electronic transitions which are possible in its constituent atoms. Solid objects have electronic energy levels which form broad, continuous bands, not discrete spectral lines. Stars are comprised of plasma, not neutral atoms; photons are produced via the scattering of charged particles (bremsstrahlung) and their energies are further randomized via Compton scattering.

That means when we talk about perfect black body, we shouldn't talk about atoms inside it as there's no atom existing that absorbs or emits light at any wavelength.

Yes, that's right.

J. Murray
  • 69,036
  • Even if it is a mathematical idealisation, the pre-idealisation prototype necessarily must be made of atoms, and then we take the limit towards idealisation. And no, it is not needed to be instantaneously absorbed. In fact, the original conception is specifically allowing infinite time to slowly absorb. – naturallyInconsistent Apr 24 '23 at 07:58
  • @naturallyInconsistent I disagree. The blackbody radiation spectrum can be obtained by maximizing the entropy of a photon gas. If you want to ask how such a gas could possibly come to equilibrium in a real physical system, then sure you can make reference to atomic oscillators in the cavity walls, but even then taking the requisite idealized limit strips away any traces of atomic structure. I will remove "instantly" from my answer because I did not mean to make reference to any particular time scale, given the assumption of thermal equilibrium. – J. Murray Apr 24 '23 at 14:44
  • Is it not the case that if you assumed that the cavity walls obeyed classical mechanics, you could derive Rayleigh-Jeans law in the idealised limit? This part is one of the atomic structure traces that gets left intact in the limit and has to be taken correctly at the start. – naturallyInconsistent Apr 24 '23 at 14:49
  • @naturallyInconsistent The assumption is required for the derivation of the Planck distribution is that the energy content of a given electromagnetic mode is some integer multiple of $\hbar \omega$, an assumption already built into the photon gas. It's true that classical oscillators do not provide a suitable model for bringing the (classical) radiation in the cavity to equilibrium, which provided historical evidence for the existence of photons in the first place. – J. Murray Apr 24 '23 at 15:03
  • Thanks for the answer. If we say that star is very close to black body, then it means it's able to absorb radiation incident on it very close number to 100%. Question 1: what radiation does the sun absorb that's incident on it and coming from outside the sun ? Question 2: How would the sun for example absorb all wavelenghts(not all, but since it's close, let's go with all) ? when light gets incident on the sun's surface, sun's atoms can only absorb the specific wavelengths. What's the logic here ? – Matt Apr 25 '23 at 00:10
  • @Matt 1) Photons are generated within the interior of the sun and make a very long journey outwards to the photosphere, which is the essentially the layer of the sun from which photons which eventually make it to Earth are released. Roughly, electromagnetic radiation is incident on the photosphere from within and is emitted from the exterior surface. 2) When we say the photosphere absorbs radiation, we don't necessarily mean that it is absorbed by individual atoms, but rather that it becomes trapped in the photosphere, coming to equilibrium with the plasma via e.g. scattering processes. – J. Murray Apr 25 '23 at 02:00
  • I don't understand one thing. we explain black body where light gets incident from it outside the black body and then it absorbs. In sun's case, there's no light incident on it from outside, but from within. So not fully sure how it goes in the same examples as black bodies. – Matt Apr 25 '23 at 08:22
  • @Matt Black body radiation is just electromagnetic radiation which is in thermal equilibrium at some given temperature. It doesn't matter how the light actually comes to this equilibrium - the fact that it does is sufficient to determine exactly what its energy distribution will look like. This is one of the beautiful facts about thermodynamics. In order to actually compute that distribution, you can model the situation in a number of ways - one of which is by imagining a cavity filled with radiation which is in equilibrium with the cavity walls. – J. Murray Apr 25 '23 at 18:11
  • @Matt If you poke a small hole into that cavity and allow some radiation to escape, then that hole (note: not the cavity, but the hole itself) is a model of a black body radiator, which absorbs all incoming radiation and emits radiation at thermal equilibrium. But that's just a theoretical model which we use to figure out what the black body radiation spectrum actually is. When we treat (a region of) the sun as a black body, we aren't literally saying that the sun is a hole in a cavity; we're saying that the radiation being emitted from the sun is at thermal equilibrium with the solar plasma. – J. Murray Apr 25 '23 at 18:15
1

A blackbody must be capable of absorbing (and emitting) light at all wavelengths. Anything in nature that approximates to an ideal blackbody (a tiny hole into a large cavity) has this property.

You can find out what processes lead to this absorption in the linked question.

What are the various physical mechanisms for energy transfer to the photon during blackbody emission?

In your example of a hydrogen gas (and I guess to are talking about stars), the "continuum absorption" in the visible part of the spectrum is provided by photoionisation of the the H$^{-}$ ion, which has an ionisation energy of just 0.7 eV and absorbs photons across the visible part of the spectrum. This means that stars like the Sun have spectra approximating to blackbodies in the visible part of the spectrum.

The Sun does of course have "dark lines" (Fraunhofer lines) in its spectrum. This is because both: (a) absorption is enhanced at the wavelengths of atomic transitions as you suggest; and (b) there is a temperature gradient in the Sun which then means we see to depths occupied by cooler plasma at those wavelengths. Thus the lines are not totally dark, they are just less bright than the surrounding continuum. The non-isothermality of the Sun's atmosphere is why it doesn't emit an ideal blackbody spectrum.

ProfRob
  • 130,455
  • Re (a) both absorption and emission are enhanced, re (b) the gas temperature gradient is not the reason for the Fraunhofer absorption lines. They are present because the gas causing them via absorption/scattering is optically too thin to produce enough radiation to fill in the absorbed/scattered intensity. This would be the case even if it had the same temperature everywhere, but its density was too small. To produce radiation with intensity close to that of a black body radiation, the emitting matter has to be dense enough. – Ján Lalinský Apr 24 '23 at 17:20
  • The solution of the radiative transfer equation for a very thick slab (with any density profile) is $I_\lambda = \int^{\infty}{0} B\lambda \exp(-\tau')\ d\tau' $ @JánLalinský If the slab is isothermal then this just equals $B_\lambda$ at that temperature. – ProfRob Apr 24 '23 at 18:23
  • In that result, in addition to uniform temperature, you're assuming Kirchhoff's law is valid for arbitrarily thin gas and also you're assuming that observation of radiation happens at a distance big enough so that the atmosphere column becomes opaque. This seems to be a specific case; for thin enough gas, scattering is relevant and Kirchhoff's law is not necessarily accurate. Length of the gas column is finite and it is not clear it can be approximated by infinity. – Ján Lalinský Apr 24 '23 at 23:19
  • Are we sure Fraunhofer lines are entirely formed in the deep layers of the atmosphere where the assumption of LTE and Kirchhoff's law are valid? I would expect some lost intensity in those lines is due to diluted gas, possibly out of LTE. – Ján Lalinský Apr 24 '23 at 23:22
  • Thanks for the answer. so, by photoionisation of the H− ion, you mean that previously it was H-(with 1 proton, 2 electrons), but photon came, hit it, hydrogen atom absorbed and absorbtion caused 1 electron to be removed. I understand that, but if we know that atoms can only absorb specific wavelengths(photon energies), how in your example would it be able to absorb continous spectrum - that means all wavelengths. Not all wavelengths would have enough energy to cause electron removal and not discrete energy to cause electron shell change, hence not every photon would be absorbed. Thoughts ? – Matt Apr 25 '23 at 00:21
  • @JánLalinský LTE atmospheres are usually used to analyse stars. They certainly produce Fraunhofer lines and include scattering. NLTE effects are second order. I am also assuming you are observing the star from a distance, which is the normal case? The atmosphere column integration always goes from zero to infinity as the path goes into the star. There are no transparent stars. – ProfRob Apr 25 '23 at 05:49
  • @Matt ionisation can result in absorption of a photon of any energy above the ionisation energy. The ionisation energy of the H$^{-}$ ion is 0.7 eV, so photons with wavelength less than 1.77 $\mu$m can be absorbed. – ProfRob Apr 25 '23 at 05:54
  • The RT equation you seem to assume is valid for Kirchhoffian source, it does not take into account scattering. Scattering atmosphere has different source function and it does not lead to such simple expression in terms of $B_\lambda$. – Ján Lalinský Apr 25 '23 at 07:07
  • See section 1.2 Kirchhoff's law in https://ipag.osug.fr/~bacmanna/radiativetransfer/rt4-equilibrium.pdf, "It is then possible to have a line even if the temperature is constant, as long as $S_{\nu,fg} \neq I_{\nu,bg}$". – Ján Lalinský Apr 25 '23 at 07:09
  • Optically thick atmosphere is present near the photosphere, but it is relatively small part of the light path; most of the path is in the dilute atmosphere where assuming LTE is dubious. So it is not clear that effect of the latter is negligible. – Ján Lalinský Apr 25 '23 at 07:32
  • @JánLalinský I use stellar atmospheres in my day-to-day research. Isotropic scattering, where $S_\nu \simeq I_\nu$, has no influence on that equation other than to make optical depths larger. NLTE effects are negligible and ignored for the vast majority of stellar spectroscopic analysis. They are important corrections in some scenarios. They have nothing (little) to do with the formation of Fraunhofer lines. The same notes you refer to extensively describe how absorption lines are formed due to the temperature gradient.- slide 19 explicitly says so. – ProfRob Apr 25 '23 at 10:17
  • I agree that your assumptions and the corresponding RT model can reproduce some absorption lines. However, slide 19 also says absorption lines can be formed without presence of temperature gradient, which confirms my doubts about general applicability of your claim "non-isothermality of the Sun's atmosphere is why it doesn't emit an ideal blackbody spectrum". Maybe for Sun you're right, but for other stars not. Do you know of any article/book where it is shown that all observed absorption lines from Sun can be described by your model and non-LTE parts of solar atmosphere can be ignored? – Ján Lalinský Apr 26 '23 at 12:08
  • @JánLalinský They produce all absorption lines. You can look at the MARCS models, or the Kurucz models, both of which are LTE and both of which are used to determine the abundances of chemical elements in stellar atmospheres as a matter of routine. You can synthesize a spectrum at the effective temperature of the Sun and see all the absorption lines. NLTE effects are an important correction for some lines. NLTE effects are not the fundamental cause of absorption lines in stellar atmospheres because stars are not isothermal. – ProfRob Apr 26 '23 at 14:49
  • Mr. Kurucz wrote in 2001 in A Few Things We Do Not Know About Stars and Model Atmospheres: "In computing the detailed spectrum, it is possible to adjust the line parameters to match many features, although not the centers of the strongest lines. These are affected by the chromosphere and by NLTE. Since very few lines have atomic data known accurately enough to constrain the model, a match does not necessarily mean that the model is correct." In one paper on MARCS from 2008 I found, authors say that consistent NLTE models are complicated due to lack of data. Thank you for the discussion. – Ján Lalinský Apr 27 '23 at 01:26