In Black Body radiation, Planck's law has a postulate saying

The wall of black body contains oscillators of all possible frequencies,ν.

There is one more postulate which says

The energy of these oscillators is not continuous but discrete valued.

Of course the second one is very well known but doesn't it contradict the first one? Shouldn't it mean that frequencies of vibration are quantised like maybe in case of standing waves? Please correct me if I am wrong in stating the postulates itself.


2 Answers 2


Your first statement can be understood if you take into account that a blackbody, is -by definition- an object or system which absorbs all radiation incident upon it and emits energy which is characteristic of this radiating system only, not dependent upon the frequencies which are incident upon it. So, it has to contain osscillators of all possible frequencies, since it absorbs (and re-emits) all frequencies.

The meaning of the second statement, is that since the re-emited energy is produced by standing waves or resonant modes of the cavity which is radiating, it has to be discrete.

In other words, although we have oscillators at all possible frequencies, each one of them has a discrete spectrum.

  • $\begingroup$ So what you are saying is that although oscillators in a particular cavity are having discrete energy, there are many cavities and each of these may have different set of allowed frequencies, and so together all possible frequencies can be attained. And the fact that all of them do add up to capture all possible frequencies is because it is a blackbody which must absorb all radiation. Have I understood it correctly? $\endgroup$
    – ModCon
    Commented Sep 11, 2017 at 17:45
  • $\begingroup$ well I meant that there is a single cavity containing oscillators at all possible frequencies. $\endgroup$
    – QCrypt
    Commented Sep 11, 2017 at 17:47
  • $\begingroup$ And it has to contain oscillators at all possible frequencies because it is a blackbody? Because intuitively it seems to me that the standing waves of a cavity may not capture all possible frequencies. Though I am using the classical picture of standing wave of strings to visualise this which may be wrong. $\endgroup$
    – ModCon
    Commented Sep 11, 2017 at 17:50
  • $\begingroup$ "it has to contain oscillators at all possible frequencies because it is a blackbody" -> yes this is what i mean. $\endgroup$
    – QCrypt
    Commented Sep 11, 2017 at 17:51
  • $\begingroup$ This clears my question. But I would like to ask how is it that oscillators of all frequencies are contained? Is it the shape of the body? And thanks! $\endgroup$
    – ModCon
    Commented Sep 11, 2017 at 17:59

I think the source of confusion is relative sizes of the cavity and typical wavelengths of standing waves in the cavity. In the Plank's law derivation the size of the cavity is considered to be much bigger than the wavelengths, that is why the oscillator frequency spectrum is considered to be continuous.


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