# In what manner is the spectrum of a black body really a continuum?

Usually they say the spectrum of a black body is a continuum and emits radiation at every frequency. Is it really a continuum like a line or are there (small) intervals?

For example, can a body emit at 700nm and also at 400nm and everything in between, or at very small intervals like at 400.000001nm and 400.000002nm etc.? So is it really like a wave or line which implies a kind of infinity of wavelengths or is it more or less like discreet lengths although very close to each other? Or is it not really known because measuring is in a certain way always with discreet numbers (like 400.0001)?

As I understand it, it should always be at certain discreet numbers, firstly because the electrons from which the radiation comes only have a discreet values, and secondly because collisions between atoms/protons/electrons create energy at certain set wavelengths. But that is just my interpretation.

So what is the reality of black body radiation?

• By definition a black body is a perfect emitter at all wavelengths/frequencies, however there are no perfect black bodies in nature. Stars come very close but they have many emission and absorption lines depending on the composition of their atmosphere. – Dean Feb 23 '16 at 15:42

A blackbody spectrum is continuous.

As Dean comments, there are no true blackbodies. But this is not really the issue here:

It's true that most emitted photons from, say, a star, which is quite close to a blackbody, is the result of some discrete process. E.g. an electron is brought to an excited state by a collision, the atom de-excites, and a photon with the wavelength corresponding to the given transition is emitted.

But two mechanisms make the total spectrum of the star truly continuous:

1. Firstly, ionized atoms recombine with free electrons. The electron will go to any state, ground or excited, and with various probabilities. The energy of the photon emitted is equal to the energy it would take to remove the electron from the atom (i.e. ionize it again), plus the kinetic energy it had before recombination. And this kinetic energy can have any value (although if it's too large, the probability of recombination decrease).

2. Secondly, even without recombinations, you will not observe only the discrete lines. The reason is that the atoms of the blackbody are not stationary, but move with a ("thermal") velocity $v=\sqrt{k_\mathrm{B}T/m}$, where $k_\mathrm{B}$ is Boltzmann's constant, $T$ is the temperature, and $m$ is the mass of the atom. This motion causes a Doppler broadening of the emitted lines, resulting in a continuous, and Gaussian$^\dagger$, probability distribution.

$^\dagger$Actually the probability distribution is a convolution of the Gaussian random motion, and the Lorentzian natural line profile, resulting in a so-called Voigt profile. This profile is dominated by the Gaussian in the line center, and only becomes Lorentzian in the wings.