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Blackbody radiation

  1. Jun 13, 2014 #1
    I have a quick question about blackbody radiation.

    Planck modelled a blackbody as a collection of harmonic oscillators. Then he assumed that each oscillator could only have a energy [itex]E[/itex] equal to [itex]nh\nu[/itex], where [itex]\nu[/itex] is the frequency of the oscillator.

    My question is, how can a collection of oscillators like this ever emit radiation which does not have frequency [itex]\nu[/itex] as well?
  2. jcsd
  3. Jun 13, 2014 #2


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    It's a collection of harmonic oscillators of different frequencies; all the frequencies are represented.
  4. Jun 13, 2014 #3

    I have a follow up question: it's easy to show that the average energy of one of these oscillators is [tex]<E>=\frac{h\nu}{e^{h\nu / kT}-1}.[/tex]

    How to you go from this formula to the energy density of the electromagnetic field in equilibrium?
  5. Jun 13, 2014 #4
    Imagine a cubic metal box of side length L. In this box only a discrete set of electromagnetic modes is allowed. Compute the expected energy in each mode and sum over the (infinite number of) modes. Divide by the box volume. Take ##L \to \infty## (the easy way to do this is to approximate the sum over discrete modes by an integral; this approximation becomes good in the large-L limit).
  6. Jun 13, 2014 #5


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    Black bodies are opaque. That means the photon does not come to your eye straight from being emitted. On the way it scatters off atoms, and each scattering results in a change of frequency. The photons and the atoms have frequent interactions, and are in thermal equilibrium with each other.
  7. Jun 13, 2014 #6
    How do you relate the expected energy in each mode to the average energy of an oscillator?
  8. Jun 13, 2014 #7
    It still seems like the photon's energy has to stay a multiple of ##h \nu##. If the photon's energy starts off as a multiple ##h \nu##, and if every atom's energy can only change by a multiple of ##h \nu##, then it seems like the photon's energy has to stay a multiple of ##h \nu##.
  9. Jun 13, 2014 #8


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    No, no, what makes you think the atom's energy has to change by ##h \nu##?? We are talking about collisions and recoils, not excitation to a higher energy level. Atoms move, they carry kinetic energy! :smile: That's what thermal equilibrium is all about. A collision will transfer some of the atom's kinetic energy to or from the photon.
    Last edited: Jun 13, 2014
  10. Jun 13, 2014 #9
    Right, but I'm not thinking about a real solid but rather Planck's model of a blackbody, made up of a large number of quantized oscillators. (I shouldn't have used the word atom.)
  11. Jun 14, 2014 #10
  12. Jun 14, 2014 #11
    The expected energy in a mode with frequency ##\nu## is given by your formula in post #3.
  13. Jun 14, 2014 #12

    Jano L.

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    There are more ways to do that.

    The easy way is to say that the mode with the same frequency as the resonator is, albeit abstract, just another harmonic oscillator, so the average energy at a given temperature should be the same.

    The hard way would be to calculate motion of the resonator under prescribed EM radiation and derive relation between average amplitudes of the EM field and that of the resonator. I think Planck did something along these lines in Part IV of his book The theory of heat radiation, Blakiston's son & Co., 1914 Philadelphia

  14. Jun 14, 2014 #13
    The container for the black body radiation does not have to be metal. The "standing waves" criterion comes from the requirement for thermal equilibrium. Hyperphysics and many other sites are not clear on this point. A good explanation is on then QuantumBaking channel of YouTube. There are 11 videos on black body radiation and the sixth one speaks to this point:
    Last edited by a moderator: Sep 25, 2014
  15. Jun 15, 2014 #14

    Jano L.

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    True, the formal operations in the derivation of the Rayleigh-Jeans formula seem to work even if the cuboid is made of wood or has entirely fictive boundary and thus no condition is imposed on the values of the field on the walls; the assumption that the average energy due to one Fourier component is the same as average energy of ordinary harmonic oscillator - ##k_B T## - is sufficient.

    However, perfectly reflecting cavity is considered in discussions of equilibrium radiation for a different, older reason: the Rayleigh-Jeans calculation addresses equilibrium radiation: existence of equilibrium requires that the system radiation + matter does not lose energy systematically through the (real of fictive) walls. Perfect conductor is able to prevent escape of EM energy, since (I believe) the Poynting vector has to have zero component normal to the walls. This way, matter can achieve thermodynamic equilibrium with radiation inside the cavity.

    In a sense, perfect conductor is like adiabatic wall in thermodynamics - non-existent, but very useful in theory.

    Of course, in experiments attempting to realize equilibrium radiation perfect conductors are not available, so people use next best thing - metals. Although metallic wall is not a perfect conductor, for long enough waves it is very close to being one. And even for high frequencies, it is best we can do.

    By the way, this means that part of the mismatch between the Rayleigh-Jeans formula and experiment is due to the fact that we cannot really prepare equilibrium radiation for ##all## frequencies - frequencies high enough go right through the walls, since they are not made of perfect conductor.

    Perhaps there are other reasons for why metallic cavity is assumed in the calculations of Rayleigh-Jeans type - if you know some, please let us know.
  16. Jun 15, 2014 #15
    Most experiments with black body radiation use glass or ceramic as the container. Common metals melt at typical temperatures used for black body experiments.

    There is no need to prevent long radio waves and short gamma rays from penetrating the walls. All that is needed is thermal equilibrium between the radiation field and the walls. The radiation field cannot pump energy from one side of the container to the other. There must be equal amounts of left and right going waves, which is roughly equivalent to saying standing waves.
  17. Jun 15, 2014 #16

    Jano L.

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    I do not know how measurements of intensity of equilibrium radiation are done today exactly, but I recall the experiments of Lummer and Pringsheim used blackened metallic cavity. This works well if temperatures measured are low enough. The reason metal was used is that the metal prevents the heat radiation energy from escaping away.

    Even if the radiation has temperature greater than the temperature of the metal, the metal does not have to melt, since it reflects most of the radiant energy. The little heat that does get absorbed by the metal can be removed by coolant applied from the outside so no melting occurs.

    I suppose other materials can be used too inside the cavity if they help to establish and maintain state of radiation close to equilibrium. But to isolate the cavity from radiating away its energy to the outer space, metal seems best.

    Establishment and maintaining thermal equilibrium requires that the system does not lose energy too quickly. If the walls are transparent at some frequency, I do not see how the intensity of radiation at this frequency inside the cavity could be claimed to have equilibrium value.

    Well, if the radiation inside was at equilibrium with the walls, the walls would be penetrated by the radiation and had to have the same temperature as the radiation. Thus they would radiate away to the outer space (outside the cavity). To maintain this equilibrium, such system would have to be enclosed in a still larger reflective (metallic) cavity.

    Energy can flow from one side to the other, as the thermal EM field is chaotic and has fluctuations. Only on average, there should not be any transfer of energy.

    ...in equilibrium, which happens because we isolated the system well, for example by metallic cavity.
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