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

  • Thread starter Outrageous
  • Start date
  • #1
374
0

Main Question or Discussion Point

$${\rho}({\lambda}) d{\lambda}=E({\lambda})*f({E(\lambda}))*D({\lambda})d{\lambda}$$
$${\rho}({\lambda}) d{\lambda}$$ 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.
 

Answers and Replies

  • #2
Claude Bile
Science Advisor
1,471
19
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.
 
  • #3
374
0
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.
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.
 

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

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
4
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
20
Views
5K
Replies
2
Views
1K
Top