Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
    Thanks.
     
  2. jcsd
  3. Dec 7, 2013 #2

    Claude Bile

    User Avatar
    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.

    Claude.
     
  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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Derivation for black body radiation. What is the 2 bout?
  1. Black Body Radiation (Replies: 2)

  2. Black Body Radiation (Replies: 1)

Loading...