As I understand it a blackbody spectrum is continuous – every possible frequency between the upper and lower bounds for that specific spectrum is there. When we look at the sun (mainly hydrogen) or a tungsten filament (comprised of one element only) we observe a continous spectrum between the upper and lower frequency range for that object. However, each element has its own unique discrete emission spectrum which I cannot see accounting for the continous range of frequencies seen in the black body curves. What is happening at the atomic/subatomic level to cause generate the continuous observed spectra?

The only explanation I can think of is that deexcitation of electrons is not the source of the light but rather it is the vibrations of the whole atom (which increase in frequency with temperature), and hence the oscillating electrons and protons within, that ‘generate’ the time-varying field which is the EM waves.

I have already looked at all previous threads to related blackbody questions and this answer in below in quotations is the closest to what I am after but if this is correct more detail on how a collection of atoms have an increased range of energy levels would be appreciated!

"In a hand waving sense, the black body radiation is also due to transitions between energy levels in the lattices of the bodies, but these are dense enough to become a continuum, as the approximation of the black body radiation shows, fitting the black body spectrum."

  • $\begingroup$ There are many duplicates. In the case of the Sun, the continuum radiation in the visible part of the spectrum is provided by free-free and free-bound transitions associated with H$^{-}$ ions. Objects that ONLY emit and absorb discrete line spectra cannot be blackbodies. $\endgroup$
    – ProfRob
    Aug 5, 2020 at 12:43
  • $\begingroup$ Or this one physics.stackexchange.com/questions/3563/… The Sun is a dense plasma, opaque for visible light. There are no distinct H$^-$ ions. $\endgroup$
    – user137289
    Aug 5, 2020 at 14:33
  • $\begingroup$ thanks for the replies, i had read the links above before posting but the answers unfortunately hadn't helped. the answer below by Pablo is clearer to me $\endgroup$
    – tomd7824
    Aug 5, 2020 at 18:15

1 Answer 1


In the case you're looking at the spectrum of an ideal gas for example, particles don't interact among themselves, and then you only have transitions "inside" each atom, so if you have only one type of atom, the photons are generated by the same transition repeated a lot of times, so they have (more or less) the same frequency, and you get spectra with lines.

When atoms start interacting among themselves, things get complicated very quickly, because the number of avilable transitions grow exponentialy with the number of particles. These transitions are not as simple as the electron ones, for example, a lattice in a solid material can be modeled as point masses connected by springs, so moving one of these masses affect not only the immediate neighbours, but virtually all the lattice. Each one of these discrete transitions have different characteristic frequencies, and then the sum of all of them generate the continous spectrum,

  • $\begingroup$ thanks, this is much clearer! so am i right in saying that the increased range of vibrational frequencies increases the number of transitions available? does the vibration of the positive nucleus of the atoms contribute to the EM spectrum emitted? $\endgroup$
    – tomd7824
    Aug 5, 2020 at 18:18
  • 1
    $\begingroup$ The example I used before is only valid for a solid material arranged in a lattice, in that case it's not the vibration of individual atoms, but the vibration of the lattice, ie, the variation in distance between different atoms. However, this is not the case in stars for example, because it's formed by plasma, where atoms also interact with each other. However, the explanation is still basically the same: the interactions among different atoms emit photons with discrete frequencies, but there are so many of them that the sum results in a continous emission $\endgroup$ Aug 5, 2020 at 20:09

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