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Derivation for black body radiation. What is the 2 bout?

  1. Dec 7, 2013 #1
    $${\rho}({\lambda}) d{\lambda}=E({\lambda})*f({E(\lambda}))*D({\lambda})d{\lambda}$$
    $${\rho}({\lambda}) d{\lambda}$$ is density of radiative energy,
    $$E({\lambda})= k_BT$$ is the energy of an atom vibrate in 3D,
    $$f({E(\lambda}))=1$$ is the probability distribution. Equals to 1 because we follow equipartition theorem,
    $$D({\lambda})d{\lambda}=2 *{\frac{4V\pi}{\lambda^4}}d{\lambda}$$ ,$$D({\lambda})d{\lambda}$$ is the number of state. And 2 there is because "Another technical problem is that you can have waves polarized in two perpendicular planes, so we must multiply by two to account for that. " from hyperphysics.

    My question is : do I define every terms above correctly?
    And I don't understand why wave polarize then we have to time two in the number of state? There will be double states because I the polarize wave are boson. But boson is not in classical.how do they so sure have to times two there?
  2. jcsd
  3. Dec 7, 2013 #2

    Claude Bile

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    Science Advisor

    It sounds like you answered your own question: The factor of two comes about because there a two polarisation modes through which atoms can shed their energy. Like having two taps to empty a water tank.

  4. Dec 7, 2013 #3
    Can you please explain more? How two taps to empty a water tank? Water go from taps to tank then?
    How should I prove the two polarization modes? Please tell me. Thanks.
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