How do Black Bodies Absorb and Emit Radiation?

I have learnt how the gases of elements are able to absorb only certain wavelengths of EM radiation corresponding to the energy transitions between energy levels of orbitals. Furthermore, these excited electrons are able to fall back to lower energy orbitals to release certain wavelengths of radiation.

I have also learnt that perfect black bodies absorb all wavelengths of radiation and emits radiation and how hot, glowing solids and liquids can be modelled as black bodies. However I don't understand how black bodies are able to absorb all wavelengths of EM radiation.

For example, I can't understand how a piece of hot iron (which can be modelled as a black body) can absorb every single wavelength of EM radiation that falls upon it if only certain energy transitions are allowed by electrons. I understand that there are more transitions possible than just electronic transitions, such as vibrational and rotational transitions. However I don't believe that this is able to result in all wavelengths of EM radiation to be absorbed.

Further, for the same reason I am not sure how black bodies are able to produce a continuous spectrum (emit all wavelengths of light) if only certain energy transitions are allowed.

Note: I am aware a similiar question has already been asked: How black body absorbs light?, however it didn't answer my question.

Answer

As you said, vibrational and rotational transitions are also possible, and I believe that the energy differences involved there are enough to have a quasi-continuous absorption spectrum in most real-life scenarios (of course, in real life you will never have perfect absorption at all wavelengths).

In the following picture, you can see two energy wells corresponding to different electronic levels. As you can see, every well contains a great number of vibrational levels and an even greater number of rotational levels (only shown at the bottom of one of the wells for clarity)

This fact greatly increases the number of frequencies that a material can absorb and can give an almost continuous absorption spectrum in many real-life situations. For example, in the following picture you can see the roto-vibrational spectrum of carbon monoxide. As you can see, it is not so different from a continuous spectrum.

Also notice that, as Rob Jeffries pointed out, real-life spectral lines will extend over a range of frequencies (you can see it also in the previous picture: the lines have finite width,i.e. they are not really lines). See enter link description here for more details.

Having said this, we have to remember that a black body is only an idealization. There have been attempts to built nearly-ideal black bodies. Here one of such attempts is described. They have built a "forest" of closely-spaced carbon nanotubes. The single nanotube doesn't have perfect absorbance: it is their structure that is fundamental here. Loosely speaking, incident radiation is partly absorbed and partly reflected within the forest. The reflected radiation propagates in the forest and in every interaction is partly absorbed, until almost all radiation is absorbed. The result is this emissivity (red line in the picture):

Notice that a perfect black body would have unitary emissivity.

Update

To answer your question

There is only a limited number of high energy transitions as there is only a limited number of electronic energy levels. So how would the spectrum for visible, UV, x-rays and gamma rays be continuous?

Actually the number of electronic energy levels is infinite. As you go to higher energy level, they become more and more closely spaced until they form a continuum (see picture below). If you increase the energy of the incident radiation, at a certain point you will ionize the molecule, that is to say you will rip the electron off. So we could say that the limit is given by the ionization energy, the energy needed to ionize the atom.

Helium has the highest ionization energy with 25 eV: since that means ultraviolet radiation, it would seem that radiation above that energy cannot be absorbed.

But that is not true. The point is that there are numerous other (more extreme) processes apart from electronic transitions in which radiation can be absorbed: Compton effect, photoelectric effect, pair production... In the following picture you can see the contribution of the various effects.

Hope this answers your question :-)

thermodynamics - Is air in the refrigerator dry? how much?

I noticed that food gets dry in the refrigerator, so I would like to use this effect to dry some objects otherwise difficult to dry (more specifically: nylon filament for 3D printers, which would requires many hours of heating in an oven).

How can I calculate the humidity inside the refrigerator?

My refrigerator can be set down to -18°C (255 K) so I think the air very close to the cooling pipes reaches, during active operation of the compressor, maybe -20°C (253 K). My assumption is that, at that temperature, air is completely saturated and every excess gets converted into frost/ice.

Since I set the temperature to -12°C (261 K), and looking at the psychrometric chart at http://www.uigi.com/UIGI_SI.PDF, my idea is to move horizontally to the right starting from -20°C, 100% rel. humidity. It looks like air at the end is about 50% rel. humid air -12°C.

Is that reasonable? I would say that at 50% humidity the food should not get that much dry, since it's about the humidity of the external atmosphere (maybe it's 60% outside), so maybe my assumption about -20°C of the cooling element and air 100% saturated around it, or the way I calculated the rel. humidity, is wrong.

Answer

I think that your reasoning is correct. You're right in using the temperature of the cooling pipes at -20˚C rather than the freezer temperature of -18˚C to determine how much moisture was squeezed out of the air at the lower temperature setting. Also, as far as I can tell your interpretation of the humidity chart is correct at the air humidity should be around 50% at -12˚C.

As for food in the freezer getting dry or getting freezer burn, although 50% humidity can be considered high for many purposes such as the relative humidity of your living room, food will still dry out at 50% humidity. You need to reach 100% humidity to stop all drying action.

quantum mechanics - How many different formulations of QM currently exist?

I read some while ago that, currently, eleven different formulations of quantum mechanics exist. Is this correct / accurate? If yes, can someone provide a pointer(s) (i.e. link(s)) to the various formulations?

Answer

The mathematical formulation of QM, as defined by Dirac, is a closed thing--- there is always a Hilbert space of states, and operators which act on the Hilbert space to produce physical changes. This mathematical scheme, however, is very general, and when you write down the description of specific quantum systems, you have to make some assumptions about the Hilbert space structure and the form of the linear operators. There are general classes of systems which are best defined in different formal schemes, I suppose that these formalisms are the quantum mechanical formulations you are talking about.

The historical formulations of quantum mechanics were matrix mechanics and wave mechanics, which are two pictures of the same mathematical structure. The two are unified and contained in Dirac's transformation theory formalism, which is what people learn today as "Quantum mechanics." This formulation includes all the others in a certain sense, because it defines quantum mechanics.

The path-integral formulation came later, and it is also mostly equivalent to the ordinary formulation, but as it is generally presented, it makes the extra assumption that the Hamiltonian is quadratic in the momentum, so that the Lagrangian can be expressed simply in terms of the trajectory. This assumption is not absolutely necessary, but it is mathematically convenient, and it is correct for nearly all applications, and when it is not, like the string world-sheet action in Nambu-Goto form, it can often be made correct using auxiliary variables. The path integral makes unitarity nontrivial, it is trivial in Dirac's formulation.

The path integral links quantum systems to non-quantum systems by analytic continuation, and this is surprising when looking at the pure Dirac formalism, so the path integral should count as a second formulation, truly different from Dirac's. The path integral formulation produces a natural description of gauge theories, which is very inconvenient in Dirac form.

The third formulation is more recent, and this is the PT-symmetric quantum mechanics. In abstract terms, PT symmetric QM is again just ordinary Dirac QM, just as path-integral QM is also Dirac QM. But it should count as a new formulation, because the metric on Hilbert space is defined dynamically, from the Hamiltonian, and you would never find it starting with a naive metric. The naive metric on Hilbert space makes the Hamiltonian seem to be non-Hermitian, and to verify that there is a Hermitian Hamiltonian requires work.

This formulation is only about 10 years old, and is very actively studied, but I think it is probably the most fundamental formulation, considering.

• Dirac QM


• Path integral


• PT-symmetric QM

You also could formulate PT symmetric QM as a path integral, maybe you would count this as a fourth. But this gives three options. In addition, you can choose to formulate each of these as either acting on the Hilbert space of states, or on the space of all density matrices

• States



• Density matrices

The density matrix formulation is arguably more fundamental. It subsumes the formulation in terms of Wigner functions. So this is a choice of two options. Multiplying gives 6.

Different theories

There are different theories which are closely related to QM, which are generally not viable. These are most of the objective collapse theories, or nonlinear state evolution theories. But one deformation of quantum mechanics at least, is not ruled out by anything except theoretical principles, and this is the superoperator formulation

• Superoperator formulation

The superoperator formulation generalizes the notion of Hamiltonian to the most general operator on the density matrix, rather than the state-vector. This formulation allows for a description of instantaneous decoherence, and it is developed in the 1970s in an obscure book which I read a long time ago, and whose author escapes me (I am sorry, it doesn't google for me).

Hawking also advocated a direct density matrix formulation as part of his program of information loss in black holes, but I don't think he cited the earlier book I am talking about (probably because he wasn't aware of it--- I stumbled across it in a library years ago). The name "superoperator" is not standard--- it is the name given in quantum information theory to the operator which multiplies the density matrix to give its time evolution. This is the quantum analog of the Fokker-Planck equation.

Here is a refernce with "nine formulations", with somewhat different ideas of what constitutes a formulation: http://www-physique.u-strasbg.fr/cours/l3/divers/meca_q_hervieux/Articles/Nine_form.pdf

special relativity - A relative time dilation paradox.

Let us assume that there are two astronauts A and B who are floating in space. A sees B passing by and vice versa. A sends signals to B every minute. According to A since B is moving his clock will be slower. So B will receive the signals prior to the appointed minute. The same argument can be applied for B who will conclude A's clock is running slow. Who is right?

Answer

Both are right. Any moving clock is slower than a clock at rest, from the perspective of the frame at rest.

Maybe this simplified freehand graphic (apologies for its lack of precision) helps to see that both A and B feel the same about each other's time dilation:

Let's say that the red axis represents A and its proper time measured in minutes (first eight minutes are showed). Green axis and its numbers represents B observer.

Light or radio signals from A to B, represented in red oblique lines, are fired on a minute basis. Six of them are showed, that took six minutes of A proper time. However, these six signals from A to B take some eight minutes in B proper time. B concludes that A clock is slower. The same holds if we invert the situation (green lines from B to A). Well, almost the same (the last green line is intended to go from green 6 to red 8, blame my trembling fingers).

cosmology - Can we observe changes in the fine-structure constant?

The fine structure constant is a number of constants rolled into one equation. Brian Cox mentioned in the April edition of Focus magazine that it is possible that the speed of light was once faster, say, in the earlier universe - hence I would conjecture this constant must then have been different.

If any or each of the constants that make up the fine structure constant are said to be changing, then is there anywhere in the universe where we can reliably observe this; and if so, what are the consequences in such a case? What of our known laws of physics in such cases, those that are used so broadly to garner great results in the field of astronomy to this day, do they become questionable or blatantly break down?

For reference, this is an excerpt from the magazine concerning what was said regarding observations of the speed of light:

What the astronomers are seeing in their study of the distant gas clouds is the last cosmic moments of that decline.

He further mentions gravity with hints of possibilities of change:

Others look for variations in big G as this could be used to develop a new understanding of gravity.

These are fitting statements for the source of them, but seemingly fanciful without further detail.

Answer

There are some scientists working on this. Here is a page with some references. From a quick reading, it seems that there is some evidence that the fundamental constants might be time-dependent. From one of the papers:

In modern higher-dimensional extensions of the standard model of particle physics, low-energy fundamental constants like the fine structure constant α, the proton-electron mass ratio μ = m_p / m_e, etc, are expected to be dynamical quantities that show spatio-temporal evolution.

The paper then goes on to review the (then) current state of knowledge, including observations and laboratory experiments used for finding this out. Since the paper was published in September 2010, it should be fairly current.

standard model - Is a quark‘s constituent mass affected by the chiral limit?

The up- and down quark’s constituent mass is usually taken to be around $300\,\text{MeV}\approx \tfrac{1}{3} m_\text{proton}$. Is this quantity affected by the chiral limit, where we let the quarks’ running mass go to zero?

If so, how?

Answer

Well, yes and no, but essentially no.

The constituent masses for the d and u, respectively, are 336 and 340MeV.

Their respective current masses are 4.3-5.2MeV and 1.8-2.8MeV.

So the current masses are negligible w.r.t. the constituent ones, of the order of 1%. Moreover, as you see, the current u is lighter than the d, but after chiral symmetry breaking the u is ever so slightly heavier than the d. There is a lot of slop in these determinations, and also electromagnetic effects, etc.

The takeaway expectation, then, is that setting the current masses equal to 0 would not take you far away from 330MeV.

computational physics - Minimal Extension of Wave Equation to Include Dispersion

Let's say you are modeling some process with the wave equation $\frac{1}{c^{2}}\frac{\partial^{2}\psi}{\partial t^{2}} = \nabla^{2}\psi$. You wish to improve your model by including dispersive effects, but you want your model to be as simple as possible for computational tractability.

What is the minimal appropriate model for phase velocity, say $c \approx c_{0} + c_{2}\omega^{2}$? and how should the wave equation be altered?

Answer

What you want to do is change the wave equation into a Klein-Gordon equation:

$$\frac {1}{c^2} \frac{\partial^2 \psi}{\partial t^2} - \nabla^2 \psi + \alpha^2 \psi = 0,$$

where $\alpha$ is a constant of appropriate dimension and usually (in quantum theory) given by

$$\alpha=\frac {m c}{\hbar}.$$

Inserting an ansatz of the form

$$\psi=e^{i(kx-\omega t)}$$

yields the dispersion relation

$$\omega^2=c^2(k^2+\alpha^2),$$

from which one can deduce an expression for the phase velocity, given by

$$v_{phase}=c\sqrt{1+\frac{\alpha^2}{k^2}}.$$

You might consider reading these lecture notes for more insights.