# Origin of black body spectra [duplicate]

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."

• 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. – Rob Jeffries Aug 5 at 12:43
• Or this one physics.stackexchange.com/questions/3563/… The Sun is a dense plasma, opaque for visible light. There are no distinct H$^-$ ions. – Pieter Aug 5 at 14:33
• 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 – tomd7824 Aug 5 at 18:15