# Why is blackbody radiation continuous?

I will preface this by saying I am not a physicist and have not taken a course on spectroscopy but I had some basic questions as I work with IR spectroscopy.

Is it accurate that photons are only emitted during electron transitions? I ask as during IR spectroscopy IR radiation is absorbed by specific molecular vibrations not electronic excitation. Also why is the spectrum from blackbody radiation continuous? Isn't it also due to electrons moving between energy levels? What is the mechanism for blackbody radiation? I understand that temperature can affect molecular vibration but what about translational, rotational and electronic excitation?

Also in the picture I understand that the dissociation energy can be obtained by the height of the peak essentially going from E0 Einf but what is the mechanism to jump vibrational energy levels? can temperature cause a jump or just IR radiation?

• Aren't molecular vibrations caused-indirectly by electronic transitions, that are then transferred to nuclear distance vibrations? As in - photon hits electron which recoils and pushes the nucleus (graphically/semi-classically). Blackbody radiation is an ideal calculation - however vibrations of free electrons on the surface of a metal box can lead to a continuous spectrum. And since temperature equilibration is molecules hitting each other, I'd imagine transitions of all sorts are possible if they are fast enough. Nov 9, 2019 at 23:31
• I didnt know that about molecular vibrations. I was still wondering about vibrational energy levels, can temperature cause a jump between those levels? Also is it still the electrons responsible for the radiation? Nov 10, 2019 at 0:45
• Many, many duplicates. e.g. physics.stackexchange.com/questions/59213/… physics.stackexchange.com/questions/508452/… Nov 10, 2019 at 7:42
• So youre saying its the way phonons interact with electrons that creates a continuous spectrum Nov 10, 2019 at 18:40

why is the spectrum from blackbody radiation continuous?

Blackbody radiation is continuous because the frequencies photons can have form a continuous spectrum.

The way the blackbody formula is obtained is by assuming that the energy in light is given in terms of its frequency $$E=\hbar\omega$$ (at the time Planck proposed it, he didn't think about light being made up of particles at all, it was Einstein who later interpreted this formula as saying light is made up of packets).

The core idea that makes this work is that Planck then assumes that, for some reason, light can be in thermal equilibrium with matter (at a given temperature $$T$$). The probability to find a wave (photon) with a given energy is then the same as it would be for any particle, proportional to a Boltzmann weight $$\exp(-\text{energy}/k_B T)$$ where $$k_B$$ is a constant (Boltzmann's constant).

With a bit of math you get to expectation value of the energy for any frequency (in a box with periodic boundary conditions):

$$\langle E_\omega\rangle=\frac{\sum_{j=0}^\infty(jE_\omega)e^{-jE_\omega/k_BT}}{\sum_{j=0}^\infty e^{-jE_\omega/k_BT}}=\frac{\hbar\omega}{e^{\hbar\omega/k_BT}-1}$$

After an integral over the space of wavenumbers you arrive at the final formula for blackbody radiation,

$$I(\omega)=\frac{\hbar}{\pi^2}\frac{\omega^3}{e^{\hbar\omega/k_BT}-1}$$

this gives you the energy radiated by a hot (black)body at a given temperature $$T$$, as a function of the frequency $$\omega$$. As you can see we needed to place no restriction in the values for $$\omega$$ (technically we did, but only for the expectation value, you take a limit of infinite size later and the restriction disappears), and that's why the distribution is continuous.

I said as much, but also notice that what we used is the expectation value, that is the average energy in that frequency. In reality both atomic (electron) and vibrational (phonon) states have discrete energy spectra, so that the energy absorbed and emitted comes in packets with the value of the gaps between energy levels (very similar to what is shown in your image).

Is it accurate that photons are only emitted during electron transitions?

I would say yes. Both the bound and free electrons emit photons when they make a transition (since the temperature is above absolute zero, fluctuations will make it so that you always have some free electrons). The bound electrons have a discrete energy spectrum but the free ones don't, theirs is continuous. But indirectly vibrations in the material can also affect the emission of photons via electron-phonon interactions (phonons are the quanta of vibrations).

what is the mechanism to jump vibrational energy levels? an temperature cause a jump or just IR radiation?

In this case it is also the interaction with the electrons that adds quanta to the vibration. On the other hand temperature doesn't exactly fit the picture here. Temperature is a statistical average quantity, it's (proportional to) the average the kinetic energy of the particles in any body (that's where the expression $$E=k_BT$$ comes from). But a higher temperature means more energy and it would imply a higher vibrational energy levels as well.

• Thanks this makes some sense. You are saying that blackbody radiation is an idealized case where we assume light and matter are in thermal equilibrium so the thermal kinetic energy is in equilibrium with the IR radiation. Also it is continuous because it is the small rotational and vibrational energy level jumps in reality Nov 10, 2019 at 1:37
• @ChemEng It is idealized in the sense that is a thermodynamic average, macroscopically you can take it to be true. It's also continuous because of the free electrons (they have a continuous spectrum of energy, unlike the bound electrons that have discrete jumps).
– user137661
Nov 10, 2019 at 2:01