# Why is Black body radiation continuous?

1. Oct 5, 2015

My name is Bradley and I am a first year university student attending Intro to Quantum mechanics lectures but didn't understand...

Why the black body radiation curve (unlike the quantized emission seen from atomic spectra), is continuous over all frequencies. I am wondering what exactly gives rise to a basically continuous black-body radiation curve in real objects? Since atomic energy states are quantized, it seems real life black-body curves would have some degree of measurable quantization to them (as themal radiation comes from electrons moving to a lower energy state).

If you could explain to me why the spectrum is continuous for all frequencies and not just the wavelengths corresponding to difference in energy levels in the atoms that emit the radiation, I would be very grateful.

2. Oct 5, 2015

### Orodruin

Staff Emeritus
This is where your misconception lies. Thermal radiation is not due to energy transitions in atoms or molecules.

3. Oct 5, 2015

Thank you Orodruin. Then what is the process which gives rise to thermal radiation?

4. Oct 5, 2015

### Orodruin

Staff Emeritus
This depends on your black body. There are very few things in real life which are actual black bodies, although it is a good first approximation for many things.

5. Oct 6, 2015

### Demystifier

For a hot gas or liquid, the thermal radiation is mainly created by the collisions of atoms and molecules. For a hot solid we don't have a true collision, but something which is effectively the same. The atoms vibrate in random directions, so they effectively "collide" by coming close to each other without really "touching each other".

6. Oct 6, 2015

### vanhees71

This is a very misleading statement. Of course, radiation in thermal equilibrium is due to the quantized emission and absorption of radiation energy (aka photons). Equilibrium occurs due to this continuous absorption and emission processes due to the principle of detailed balance, i.e., it is a state of maximum entropy and thus described by the Bose-Einstein distribution for photons. The mean photon number is determined completely by temperature.

In a cavity the black-body spectrum is in principle discrete, because the possible wave modes are quantized due to the boundary conditions for the electromagnetic field in the cavity, however the possible frequencies are dense since the possible photon momenta are separted by about $2 \pi \hbar/L$, where $L$ is a typical extension of the cavity, and this is a very small number. So you can usually take the thermodynamic limit (formally volume to infinity keeping the photon density constant). In this limit the black-body spectrum becomes continuous. The ensity of states is $\mathrm{d}^3 \vec{k}/(2 \pi \hbar)^3$ and thus the mean density of photons at temperature ##T#
$$\langle N \rangle=2 \int_{\mathbb{R}^3}} \mathrm{d}^3 \vec{k} \frac{1}{(2 \pi \hbar)^3} \frac{1}{\exp(c |\vec{k}|/k_B T)-1},$$
where the factor 2 comes from the two polarization states for each photon mode.

The best black-body spectrum in Nature is the cosmic microwave background in the universe. Here the details are a bit different. Here there was thermal equilibrium between ions and radiation up to a time about 400000 years after the big bang. Then the same mechanism as described above for the cavity black-body radiation was at work: The photons (i.e., quantized radiation energy) was absorbed and emitted all the time by collisions among the photons with the ions and among the atoms. Due to the principle of detailed balance (which is, btw, following from the unitarity of the S matrix and thus a very fundamental principle) you have radiation and ions in thermal equilibrium as long as the emission and absorption rates are small compared to the expansion rate of the universe. Thus, at some time when the universe cooled down to the Mott transition temperature, where the ions (i.e., electrons and protons, and the light primordial nuclei) combine (funnily one talks about "recombination" although there were no atoms before this time of the universe's evolution) to electrically neutral atoms, the emission and absorption rates went down drastically (practically to 0). From this moment on the photons were decoupled from the heatbath ("freeze-out"). Now photons are massless and thus there is no intrinsic scale in the black-body spectrum (the only intrinsic energy scale is the temperature of the universe). Thus the only thing happening to the photons is gravitational redshift multiplying all photon momenta and frequencies by a common red-shift factor. Thus the CMBR looks still like a black-body spectrum but with a much smaller temperature by the same redshift factor as the photon frequencies. This becomes also clear quantitatively: The Mott-transition temperature is around 3000K and the red-shift factor from the time of "recombination" is around 1000, giving about 3K for the background radiation, which is not bad compared to the very precise value (around 2.75 K) from COBE, WMAP an PLANCK.

7. Oct 6, 2015

### Orodruin

Staff Emeritus
I do not see how this contradicts my statement. It is one thing to be a quantised emission of photons and another to be quantised emission based on the emission lines of a single atom, which is what the OP was referring to.

8. Oct 6, 2015

### vanhees71

Ok, then I've misunderstood you. Sorry for that.

9. Oct 6, 2015

### lightarrow

I'm not completely sure the OP has understood that the blackbody radiation from a solid object is, also, due to "electrons moving to a lower energy state"(*), but that these states are not of single atoms or molecules, but of the entire solid.

(*) Of course it's not exactly a "movement" but a de-excitation from higher to lower energy states.

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lightarrow

Last edited: Oct 6, 2015
10. Oct 6, 2015

### vanhees71

The point is that it doesn't matter what's the material and how you describe it. Harmonic oscillators (even a single one) can do this job. Thermal equilibrium is the "most boring" state, i.e., the state of least information. From black body radiation you learn nothing about the material except its temperature. That's it.

11. Oct 6, 2015

### lightarrow

Certainly. But there must be physical electromagnetic oscillators in order for bb radiation to exist. Which are the physical oscillators in a solid material?

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lightarrow

12. Oct 7, 2015

### vanhees71

I guess that depends on the temperature. There are various "oscillators" in a solid, i.e., the electrons bound to the atoms and the "crystal lattice" (phonons) itself. Everything carrying an electric charge can exchange energy with the radiation field.

13. Oct 7, 2015

### lightarrow

Exactly :-)

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lightarrow

14. Oct 12, 2015

### BillyT

As two isolated atoms come closer to each other, their energy levels are perturbed. Very strongly in solids. Thus there is some modified energy level that can absorb any discrete energy photons (A continuum, except it is more complex process for higher than UV photons, like X-rays).
If the solid is a conductor, some of the energy of absorbed photons can "heat" the free electron distribution, but any such intense brief "flash" of photon energy will quickly be transferred to the solid's lattice too so both have the same temperature (in sub nano seconds, I think)

If any wave length can be absorbed then it also must be emitted too. - If that were not true, one could make net energy form an pair of same temperature sources, but showing that violation of thermodynamics is more than I care to do now.