Re: [問題] 黑體輻射
ε(υ) = Σ(nhυ) exp(-nhυ/kT) / Σexp(-nhυ/kT)
set 1/kT = β
=> ε(υ) = Σ(nhυ) exp(-nhυβ) / Σexp(-nhυβ)
= d{ Σexp(-nhυβ) }/d(-β) / Σexp(-nhυβ)
Σexp(-nhυβ) = 1 / (1 - exp(-hυβ))
d(Σexp(-nhυβ)) / d(-β) = d (1 / (1 - exp(-hυβ))) / d(-β)
= -hυ/(1 - exp(-hυβ))^2
so
ε(υ) = -hυ/(1 - exp(-hυβ))^2 / { 1 / (1 - exp(-hυβ)) }
= -hυ / (1 - exp(-hυβ))
= hυ / (exp(-hυβ) - 1)
註: 這裡使用的d都是指偏微分
※ 引述《boyuuu (羅德曼)》之銘言：
: 普朗克黑體輻射的推導，
: 在算空腔壁上振子的平均能量時，
: 是由古典的波茲曼分佈得到的，
: 即平均ε(υ)=ΣεP / ΣP (其中P為波茲曼分佈且正比exp(-ε/kT))
: =Σ(nhυ) exp(-nhυ/kT) / Σexp(-nhυ/kT)
: =.....
: =hυ / exp(hυ/kT) -1
: =hυ×(bose分佈函數)
: 想問的是，為什麼用古典的波茲曼分佈來計算但最後會得到bose分佈函數?
: P為什麼不用光子遵守的bose分佈函數而用古典的波茲曼分佈呢?
: 雖然說普朗克黑體輻射公式是半古典的理論...
: 希望高手能指點一下 謝謝!
--
I am the bone of my paper. Pen is my body, and ink is my blood.
I have finished over a thousand assignments.
Unknown to pass. Nor known to fail.
Have withstood pain to take many midterms.
Yet, those hands will never write anything.
So as I pray, Unlimited homeworks.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.115.222.36
※ 編輯: nightkid 來自: 140.115.222.36 (08/22 23:46)
推
08/22 23:53, , 1F
08/22 23:53, 1F
→
08/22 23:53, , 2F
08/22 23:53, 2F
→
08/22 23:55, , 3F
08/22 23:55, 3F
→
08/22 23:56, , 4F
08/22 23:56, 4F
→
08/23 11:47, , 5F
08/23 11:47, 5F
→
08/23 11:50, , 6F
08/23 11:50, 6F
→
08/23 12:05, , 7F
08/23 12:05, 7F
→
08/23 12:10, , 8F
08/23 12:10, 8F
討論串 (同標題文章)