Re: [問題] Quantum Physics by Eisberg 書上的一괠…
※ 引述《boyzone66 (打籃球ㄟ)》之銘言：
: 這可能不是近物的問題 是我微積分有問題 囧rz.....
: 但我真的自己導不出來...
: 請問 Eisberg 第2版藍皮書 第114-115頁
: 4-10 Sommerfeld's Model
: 我可以自己導出∮ Px dx = n(x) ‧ h = πab ( )內表示下標
: 但如果把原本的 Px和dx的積分
: 改成極座標形式 Pr和dr的積分
: 會得到115頁最上方的一個等式
: 用來說明角動量L和橢圓a/b的關係 (a/b=半長軸除以半短軸)
: L(a/b-1)=nh/2π
: 請問到底怎麼來的呢?
: 如果這個等式成立
: 那下方第二個等式必須等於2πL(a/b-1)
: ∮ L dθ = n(θ)‧ h
: ∮ Pr dr = n(r) ‧ h = 2πL(a/b-1)
: 極座標dr我不知道該怎麼定義積分範圍 也不知道Pr代表什麼?
: 而且照道理不管用什麼形式的座標積分 橢圓面積應該也還是πab
: 到底是怎麼和角動量L有關係???
In classical mechanics, the equation of motion governed by central force motion
satisfies a simple relation between r and theta.
alpha/r = 1 + epsilon*Cos(theta)
where aplpha and epsilon are constants related to energy and angular momentum.
Using this relation, we can rewrite ∮Prdr
= ∮epsilon^2 * (sin theta)^2 / (1 + epsilon Cso theta)^2 d(theta)
After doing the integration(with the help of Mathematica XD), we shall get a
nasty result of indefinite interal. However the loop integral should be quite
simple. And the result is just 2πL(a/b-1).
I encountered some difficulty obtaining the result when I simply used 2Pi and
0 as the upper and lower limits. It seems to result from the discontinuity at
Pi in the integral. By inspecting the symmetry of the integrand, we can
take the result to be the limit of the integral(approaching from below) at Pi
minus the value of integral at 0, and time the difference by a factor of 2.
Then we can happily get the final result.
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