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What is the quantum mechanics behind blackbody radiation?

  1. Apr 16, 2013 #1
    I can't seem to find a definitive answer to my questions surrounding this topic, every textbook I read is rather vague. When Max Planck scratched the surface of quantum mechanics he introduced the idea of quantisation of energy in blackbodies. Could someone please explain how exactly a blackbody works using quantum mechanics. Is it that the electrons in the atoms in the walls (and interior) of the blackbody absorb specific photons then emit them? Does it mean that in order to emit a photon of a specific wavelength it needs to absorb a photon of the same wavelength. For example, does a blackbody need to absorb an ultraviolet photon in order to emit one?

    The theory behind continuous spectra is also very vague and I can't find any detailed information anywhere. To produce a continuous spectra, do electrons in atoms still undergo energy levels changes? The answer I always get is that continuous spectra involves vibrating atoms at a macroscopic scale but surely there is a quantum explanation? At the subatomic level what specifically causes the spectra to be continuous rather than producing a line spectra.

    Thanks
     
  2. jcsd
  3. Apr 16, 2013 #2

    fzero

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    A black body is an idealization. In practice, the main example of a quantum system with a continuous energy spectrum is the free particle (in an infinite space). The energy is associated to the translational motion of the particle, and the momentum (and therefore the energy) can take on a continuous range of values. This is the case for an ideal gas in thermal equilibrium.

    If we were to consider an ideal gas in a finite box, then we would have discrete energy levels given by the usual formula for a particle in a box. But if the average energy (or temperature) of a particle is large compared to the ground state energy, then the spectrum is approximately continuous. We can see this by computing the energy spacing of highly excited states in one dimension:

    $$ \frac{E_{n+1} - E_n } {E_n} = \frac{ (n+1)^2 - n^2 }{n^2} \sim \frac{1}{n} \rightarrow 0 ~~\mathrm{as}~~ n\rightarrow \infty.$$

    The Planck distribution is typically derived by using the energy formula for particles in a box, but the continuous energy spectrum appears because the size of the box is taken to be large compared to the length scale set by the temperature.

    Rotational and vibrational spectra will also approach a continuous spectrum when the average energy is large compared to the spacing between energy levels. The calculation is analogous to the one above.
     
  4. Apr 17, 2013 #3
    Thank you. For your example of a free particle, I thought it would need to accelerate to emit photons/electromagnetic waves? So if I'm getting it right, can the free particle take on any values because it is not bound to an atom? Will you always get an emission spectra for an ideal gas in a finite box?
     
  5. Apr 17, 2013 #4

    vanhees71

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    The black-body spectrum is best derived with QED. I work in natural units ([itex]\hbar=c=1[/itex]. You start with an ideal photon gas in a cubic box of length [itex]L[/itex] and the corresponding partition sum
    [tex]Z=\mathrm{Tr} \exp(-\beta \hat{H})=\left [\prod_{\vec{p}} \sum_{n(\vec{p})=0}^{\infty} \exp \left (-\beta n[\omega(\vec{p})] \omega(\vec{p}) \right) \right ]^2.[/tex]
    The product runs over [itex]\vec{p}=\frac{2 \pi}{L} \vec{n}, \quad \vec{n} \in \mathbb{Z}^3[/itex]. We have taken the trace using the occupation-number basis of the bosonic Fock space for photons. The sum is a geometric series, leading to
    [tex]Z=\left (\prod_{\vec{p}} \frac{1}{1-\exp[-\beta \omega(\vec{p})]} \right )^2.[/tex]
    Taking the logarithm leads you to
    [tex]\Omega=\ln Z = -2\sum_{\vec{p}} \ln \left [1-\exp[-\beta \omega(\vec{p})] \right].[/tex]
    In the limit of a very large volume you can approximate the sum as an integral
    [tex]\Omega=-2 \int_{\mathbb{R}^3} \frac{L^3}{(2 \pi)^2} \ln \left [1-\exp[-\beta \omega(\vec{p})] \right].[/tex]
    The mean internal energy is
    [tex]U=-\frac{\partial \Omega}{\partial \beta}=2 \int_{\mathbb{R}^3} \frac{L^3}{(2 \pi)^2} \omega(\vec{p}) f_{\mathrm{B}}[\omega(\vec{p})].[/tex]
    This leads to the phase-space distribution funktion
    [tex]\mathrm{d} N=\mathrm{d}^3 \vec{p} \frac{2}{(2 \pi)^3} f_{\mathrm{B}}[\omega(\vec{p})].[/tex]
    That's the Planck spectrum.
     
  6. Apr 17, 2013 #5

    fzero

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    Well an ideal gas is also a bit too much of an idealization. A real gas of molecules would have small interactions that would precipitate exchange of light (and therefore heat).


    I was talking about the translational spectrum of the molecule, not the constituent electrons. For simplicity, let's assume that the temperature is not so hot that the electrons start to disassociate.
     
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