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The Planck Function and Flux

  1. Jul 4, 2011 #1
    For my own edification.

    The Planck Function when integrated over all wavelengths will give the radiated power per unit area for a black body. Would I be right in thinking, that if I had a known temperature for the black body source or indeed a kT value for that source that by using the function unintegrated, an emitted flux can be found for a single photon energy (frequency) being emitted by that source rather than a range? does Bv = emitted flux in the usual sense?
  2. jcsd
  3. Jul 7, 2011 #2

    Philip Wood

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    Gold Member

    No. The formulae for spectral energy density and spectral emissive power (per unit area) 'approximate' the quantised distributions to smooth distributions. I've put 'approximate' in quotes, because the approximation is, to all intents and purposes, undetectable.

    So when you use these distributions, the energy density or emitted energy per unit area is zero at any particular wavelength (or frequency), just in the same way as the probability of finding on a beach a pebble of mass 1.0000000000000000000000000000... gram (or 1.08172305825769346720638560... gram) is zero.

    That's why the formulae are often written as u[itex]\lambda[/itex] d[itex]\lambda[/itex] = [stuff]d[itex]\lambda[/itex] and E[itex]\lambda[/itex] d[itex]\lambda[/itex] = [similar stuff]d[itex]\lambda[/itex], including d[itex]\lambda[/itex] on each side.

    Note that you can also define spectral energy density and spectral emissive power as distributions in terms of frequency. The formulae are somewhat different, and it is easy to make mistakes.
    Last edited: Jul 7, 2011
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