# Red-hot materials

1. Nov 10, 2009

### Infrasound

Here is a problem that has been bothering me for some time...

What is the source of electromagnetic radiation (usuall red visible light) coming from materials (such as metals) that are hot enough.

An example that has always stuck with me is a demonstration that I witnessed at a car show of an exhaust manifold on an engine becoming red hot.

Is it the oscillation of whole molecules that initiates an electromagnetic wave, or is it something slightly more complicated?

Perhaps it's the collision of two molecules and the subsequent shaking of outer electrons or maybe even the nucleus from the force. Im assuming that we are not involving any quantum energy level jumps, but I could be wrong.

I really try to develop mental models of physical events, and I need some verification on the initiating object in order to have a chance of being successful here.

2. Nov 10, 2009

### Staff: Mentor

Hot objects, whether a hot exhaust manifold, light bulb or the sun give of em radiation at a peak frequency dependent on their temperature. This is thermal or blackbody radiation: http://en.wikipedia.org/wiki/Thermal_radiation

3. Nov 10, 2009

### Infrasound

Yes, I know the name of the phenomenon. Sorry that I didn't use the proper terminology in my post (although Feynman would probably tell me not to worry as much about that).

I would like to know the physical events behind the observations. What object and what motion is causing the EM waves. Oscillations of electrons as they ride along with the atoms (seems too slow)? Oscillations of electrons as they interact with each other when atoms collide?

Im having trouble with the wiki explanation. I'm not very good at doing the math problems as a means to developing a mental picture.

4. Nov 11, 2009

### mikeph

I think it is based on quantum mechanics and statistical physics, and the derivation is fairly long and interesting. I am re-learning this right now, but my current understanding is that it is due to the population of quantum energy levels of a system. The higher the temperature, the higher the population of high-energy states. The rate of spontaneous emission is proportional to the population of the energy level, so as a temperature is higher, there is more emission from that energy level.

Now consider the system is divided into near-infinite energy levels representing a particle's translational energy, then in this limit, the emissions of the system as a whole becomes a distribution reflecting the distribution of energy state populations, and this black body curve is only dependant on temperature.

If you get a chance, have a look at Mandl on Statistical Physics, it's a very interesting subject.

5. Nov 11, 2009

### Born2bwire

It can sometimes be surprisingly difficult to find the answer. I think the reason is because the black body radiation problem was developed prior to quantum mechanics, is an idealization, and can be applied to different kinds of systems. So when Planck solved the black body radiator, he only assumed quantization of the energy in the radiator, but he did not know why or how. In addition, a true object does not have the density of states that a black body radiator has since the black body is an idealization and ignores the actual physics of a material. Finally, a black body is a system that absorbs and reradiates all energy. This idea can be applied to more abstract systems than a heated material, like a resonant cavity.

So a lot times, people look to the black body radiator for the answer but the descriptions given are for a more general case and do not explicitly say what is being quantized in the material.

The quantization of the energy in the material is due pretty much to the phonons, vibrational modes of the materials atoms and/or molecules. The density of states for the vibrational modes is very dense and similar to a black body radiator. This is opposed to say the electron excitation modes of the individual atoms or bulk material. The energy levels allowed in an atom are very discrete and you can only get very specific bands of radiation. Likewise, in a bulk material, you can have large bandgaps in the electron energy bands (like in an insulator or semiconductor) that would restrict the frequency range of emitted photons.

6. Nov 11, 2009

### MaxwellsDemon

Classically you view it as atomic charges oscillating due to thermal collisions within the object. Oscillation is acceleration, so they emit electromagnetic radiation. The problem with this view was that predictions were in conflict with the experimentally observed curve for light frequency vs. temperature. Of course the resolution of this conflict by assuming that the oscillators could only have certain discrete vibrational energies led to the birth of quantum mechanics.

7. Nov 11, 2009

### Infrasound

So which picture seems more accurate?

1. The whole atom is oscillating back and forth, with electrons just along for the ride and by default oscillating with the atom.

or

2. The electon itself is oscillating relative to the atom because of the collisions of atoms that causes the wave to be produced.
In other words, the atoms are moving too, but its the shaking of electrons themselves (at a different speed than the whole atom) from the collisions that really gives off the visible light.

The fact that only certain frequencies would be produced seems to suggest to me that the second explanation is more correct.

8. Nov 11, 2009

### mikeph

visible light is generally from electron transitions, but this isn't black body radiation. mostly at room temperature black body is in the infra-red. it's the atom vibrating. ALL frequencies are produced by black-body because there are theoretically near-infinite transitions possible due to vibrations, get the system hot enough and it will emit strongly in the visible.

but colour from say a sodium lamp is due to electron transitions. there are some great replies above ^.

9. Nov 11, 2009

### Infrasound

Hopefully I've got this right.

So visible light in chemical reactions and in atoms/molecules involved in the exchanging of electrons will only emit specific frequencies (i.e. electrons moving from one energy level to another). (*A side question might be, does an electron emit light when jumping from one atom to another, or just when moving within a single atom?)

While black body radiation can give off any color assuming sufficient energy/density.

And yes, as always, there ARE some great replies above. Including yours. Thanks.

10. Nov 11, 2009

### mikeph

Yes, that is my understanding. In fact black body does give off every colour at any temperature, since it is a distribution; in the ideal case the energy density of radiation should be positive across the entire spectrum. But the peak and majority of the radiation will only be in the visible spectrum, for sufficiently high temperature.

As for electrons jumping between atoms.. I don't actually know... in a flame you have a plasma, which I understand is that colour because of the energy of complete dissociation of the electron from the atom- perhaps this is an example of electrons jumping between different molecules. Chemical reactions are beyond what I have learnt though in QM, so I can't help you there.

11. Nov 11, 2009

### Born2bwire

The movement of the electrons throughout the bulk material, jumping between atoms as you say, can be described by the energy band structure for the bulk. This is something that is often discussed in terms of semiconductors or crystals. In a perfect conductor, the energy band for the electron is continuous, it has free movement about the bulk material and can achieve any energy state (for all intents and purposes we think of the band being continuous as opposed to being discrete because the energy differences between levels is smaller than the typical thermal energy). However, if we look at an insulator, there is a limit to the energy levels that the electrons can attain. And a semiconductor has large gaps in its bands. When we talk about LED's for example, the light is created by the energy given off from the electron dropping down from the top edge of a bandgap to the lower edge of the gap (from the conduction to the valence band).

I am not sure if you can describe radiation due to the change in the electron's state with in the continuous bulk energy bands. Generally we only talk about the electron transitioning between adjacent states, in which the photon energy would be very small. The exception is that a bandgap, like in a semiconductor, allows the energy difference between the adjacent states to be large enough for visible light to be emitted (or something similar). It would not surprise me that electrons can drop down multiple states at once since we see that in the atomic transitions, but I do not know if the rates for doing so would create a black body radiation spectrum.

Hmmm... Well the density of states would at least be defined using the Fermi-Dirac equation. This would give us the energy distribution of the electrons for a given temperature (without adjusting for bandgaps). Well, a good statistical mechanics book may be helpful here.

12. Nov 12, 2009

### @tomvy

wild guess...........antimatter?