1. Dec 14, 2017

### Arup Biswas

When I study any book of Quantum Mechanics like Resnick or Beiser etc all start with blackbody radiation! But how this radiation is produced? Google says due to increased collision of particles causing the acceleration and em wave but what particles? How they are accelerated from what? Like if we heat a iron rod! At 1st it will become red. Is this due to the fact that electron jump to the next state in the orbits and then come back radiating energy? The more heat i give to the rod more the frequency shifts towards blue(white)! Can this be reasonable explanation(i.e. google quora says all trush) or I got wrong deep inside?

2. Dec 14, 2017

### Drakkith

Staff Emeritus
The particles of the material. When heated, the atoms and molecules tend to oscillate or vibrate back and forth, which means that they accelerated over and over, generating EM radiation. In conductors, many of the electrons are freed from their atoms by metallic bonding and can move about the material. These electrons are subject to collisions which accelerate them and produce more EM radiation. Electronic transitions (electrons moving up and down between energy levels) still take place, but the majority of the radiation produced is from all of this microscopic motion.

3. Dec 14, 2017

### Drakkith

Staff Emeritus
Whoops, forgot to answer this in my previous post:

Heating the rod increases the amplitude of the vibrations and the magnitude of the thermal motion of the atoms, molecules, and electrons. As a result, the interactions between the particles tend to produce larger accelerations and thus a higher average frequency of the emitted radiation.

4. Dec 14, 2017

### Arup Biswas

Let us put a lightweight iron ball in a place at heat it! Will it oscillate producing em wave? Then why do atoms or molecules oscillate? Thus to in such a beautiful fashion to radiate a particular colour?

5. Dec 14, 2017

### Drakkith

Staff Emeritus
Heating something adds energy to it. This energy takes the form of several things, one of which is the motion of its constituent atoms and molecules. If you're trying to find the fundamental reason for why this happens, it's because the way this energy is transferred to the material is either through collisions with other particles or EM radiation. Both of these accelerate the particles in the material, adding energy to it and heating it.

Iron balls don't oscillate when heated because they are made from huge numbers of atoms whose individual motions average out to near-zero. Hence the ball doesn't jump about like popcorn when heated.

6. Dec 14, 2017

### .Scott

Atoms and electrons.
Yes.
At any given temperature, the wavelength components of Black Body radiation will follow a curve: intensity vs wavelength. In 1900, the formula for these curves was proposed by Planck and confirmed experimentally.
At that point, the question became what must be going on at the atomic level to create such a curve? The electron energy levels actually interfere with the smooth Black Body radiation curve - and offer more clues about what is going on at the atomic level.

7. Dec 14, 2017

### Delta²

Good question!. The reason an iron ball no matter how light will not macroscopically oscillate when heated is that:

When we offer heat we increase the average kinetic energy of the molecules. But each molecule oscillate in "random" direction, loosely speaking heat offers kinetic energy in a different random direction for each molecule, if there was a Demon (Maxwell's Demon for example) that could make the heat we offer to go as energy of oscillation in the same direction for all molecules then we would see the ball oscillate macroscopically as well.

The colours we receive from a heated body are from electrons of atoms changing energy states, the random oscillation motion of the molecules is not responsible for this. I see now your point, if all random oscillations are cancel out on average, how is there em radiation due these random oscillations(shouldn't be cancelled out as well??)

Last edited: Dec 14, 2017
8. Dec 14, 2017

### Arup Biswas

Firstly I would like to go in more details how all the oscillations cancels out? I think oscillation occurs but the radiation emitted by one of them must be cancelled out by another particle's radiation, if we take light to be wave! Now in answer to Delta, should I assume every radiation to be radiated uniformly in all direction? Then in every possible direction there comes a radiation in opposite phase to cancel it out! Here I would like to bring the liquid's Surface Tension analogy! The oscillating particles on the surface of the material! Radiation from any one of them cancels out in every possible direction except outside the material as there is no wave incoming! Thus we get the radiation occurs from the surface of the material! I don't know I made what from what Funny

9. Dec 14, 2017

### Delta²

I don't think it matters whether the radiation is uniformly radiated or not. Given 1cubic mm of a macroscopic object , there is a huge number of molecule in it so for every molecule oscillating in one direction there will probably be another molecule within a cube of 1mm that oscillates in the exactly opposite direction so the radiation will tend to cancel out though we will not have perfect cancellation (unless both molecules were oscillating on the same exact place which of course cant be the case).

And no I don't think your reasoning for the radiation coming only from the thin surface layer of a body is correct. (ok to be honest I am not completely sure).

@Drakkith in his post claims that the majority of black body radiation is from the microscopic movement (I suppose he means the oscillations) and not from electron state transitions...Drakkith what have you got to say about what I said in the first paragraph in this post?

Last edited: Dec 14, 2017
10. Dec 14, 2017

### Khashishi

Not to nitpick, but metals like iron aren't made up of molecules, but of atoms. The outer electrons can move fairly freely between atoms, so it's sort of like a sea of electrons surrounding a lattice of positive ions. For high frequencies of random motion, like optical frequencies, the electrons act like a plasma. Anyways, you have collisions between electrons and ions which cause emission and absorption via bremsstrahlung or some other processes.

Typically, the Planck spectrum is derived for a photon gas in a box, so you might wonder why a hunk of metal would have the same spectrum. It doesn't, exactly. There is an emissivity factor in there, which depends on frequency. But the emission at any frequency has to scale with temperature in the same way as the Planck spectrum due to the principle of detailed balance.

11. Dec 14, 2017

### Drakkith

Staff Emeritus
Not much really. I don't know enough to say whether it's accurate or not.

12. Dec 14, 2017

The mechanisms responsible for a high emissivity in a material seem to be quite subtle. Some materials can be transparent and have low emissivity, and others can be reflective, and also thereby having low emissivity. Other materials such as some paints can be made to be reflective at some wavelengths and absorbant at others, versus black paint that has a high emissivity throughout the visible region of the spectrum. This can be caused by the molecular and electronic properties of the materials such as the dye in the paint. Meanwhile, some materials are found to also have changes in their emissivity with temperature. If a material has a high emissivity, sometimes, all that is necessary to achieve that high emissivity is a very thin film of that material. Meanwhile, a roughened surface of a reflective material will be less reflective than a highly polished surface, and thereby will have higher emissivity. $\\$ Additional comment: Metals, which in some ways can be modeled as plasmas with an atomic lattice, often are found to be highly reflective (reflectivity close to 1.0) throughout much of the spectrum and thereby exhibit low emissivity. (For metals, a large complex term (an absorbant part) in the index of refraction, makes them good reflectors. This follows from the formula for reflectivity (at normal incidence) $R=\frac{|\tilde{n}-1|^2}{|\tilde{n}+1|^2}$). Meanwhile, crystals, with well ordered atomic lattices, are often transparent, and thereby also have low emissivity. $\\$ One way of artificially achieving high emissivity is to put a small aperture in an enclosure. Light that enters the aperture, regardless of the material on the inside walls,(assumed to be opaque and partially reflective), will undergo multiple reflections and very little will reflect back out. By Kirchhoff's law, the emissivity of the aperture is necessarily very close to 1.0. Thereby, this aperture can be considered to be a nearly ideal blackbody, and if the material (i.e. the inside walls) is heated, the aperture behaves as an ideal blackbody.

Last edited: Dec 14, 2017
13. Dec 15, 2017

### Delta²

You are dead sure that the blackbody radiation comes from the thermal motion of the molecules/atoms and not from electrons switching state within molecules/atoms?

14. Dec 15, 2017

I think it is clearly a combination of both. The electron contributions to some processes can be at times separated from the ion contributions, but here I think both species make contributions to the result.

15. Dec 15, 2017

### Drakkith

Staff Emeritus
Yes, I'm quite certain that the majority of the radiation comes from thermal motions and not from electronic transitions.
You can read plenty of this on the wiki page.

The colors from something like a sodium flame are sometimes from electronic transitions, yes, but in general the color of a hot object like a bar of iron or a star is not heavily influenced by the electrons moving between energy states.

16. Dec 15, 2017

### Cthugha

You are confusing the field and the intensity. As the intensity is proportional to the squared modulus of the field, the average field will be zero as it can also take negative values, but the average intensity will be larger than zero because at any instant it can only have positive values or a value of 0. In fact you can easily simulate this yourself. Take 100 harmonic oscillators with the same amplitude and random phase, add their vectorial amplitudes and get the instantaneous intensity. Then simulate a time series, where each oscillator may receive a small phase kick with a certain probability within each time step and monitor the intensity over time. If the phases are not fluctuating too strongly, the photon number distribution you will get is the Bose-Einstein distribution for blackbody radiation.

17. Dec 15, 2017

### Arup Biswas

Eisberg Resnick also says that most of the radiation occurs due the accelerated electrons! He has also a proof of it!

18. Dec 15, 2017

The question still remains, what exactly is the arrangement of the molecules, such as the dye in a paint, that allows thermal effects to set these electrons in motion=(i.e. accelerate) and radiate? The arrangement of the ions, and perhaps thermal motion of the ions, must play an important role or all substances would show similar emissivity.

19. Dec 15, 2017

### Arup Biswas

I did not checked the detailed proof what they have done! Probably iy would help what is their arrangement and how acceleration occurs! #charles!

20. Dec 19, 2017

### Khashishi

I think the physics is quite different for metals (which tend to be reflective over a large range) and organic dyes (which have strong absorption peaks). Crystals tend to be transparent. Electrons can be accelerated in metals, but in dielectric crystals, they are more like oscillators.