- #1
Outrageous
- 373
- 0
$${\rho}({\lambda}) d{\lambda}=E({\lambda})*f({E(\lambda}))*D({\lambda})d{\lambda}$$
$${\rho}({\lambda}) d{\lambda}$$ 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.
$${\rho}({\lambda}) d{\lambda}$$ 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.
