出自http://li.mit.edu/Stuff/PM/main.pdfp144頁
更新1:
出自Notes on Physical Metallurgy 第144頁
該如何泰勒展開Lennard-Jones Potential?
2021-03-09 6:01 pm
有人知道該如何用泰勒展開Lennard-Jones Potential成如下圖黃色框的方程式嗎
出自http://li.mit.edu/Stuff/PM/main.pdfp144頁
出自http://li.mit.edu/Stuff/PM/main.pdfp144頁
回答 (1)
2021-03-10 8:17 am
簡化符號:
u(r) = 4ε[(σ/r)^12 - (σ/r)^6]
令 d = r - 2^(1/6)σ= r - kσ.
(σ/r)^n = (k+d/σ)^(-n)
≒ k^(-n){1-nd/(kσ)+[n(n+1)/2]d^2/(kσ)^2}
當 |d/(kσ)| << 1
所以 (k^(-6) = 1/2)
(σ/r)^12 - (σ/r)^6
≒ 1/4{1-12d/(kσ)+78d^2/(kσ)^2}
- 1/2{1-6d/(kσ)+21d^2/(kσ)^2}
= -1/4 + 9d^2/[2^(1/6)σ]^2
當 |d/(2^(1/6)σ)| << 1, 得
u(r) ≒ 4ε{-1/4 + 9d^2/[2^(1/6)σ]^2}
= -ε + {36ε/[2^(1/6)σ]^2}(r - 2^(1/6)σ)^2
u(r) = 4ε[(σ/r)^12 - (σ/r)^6]
令 d = r - 2^(1/6)σ= r - kσ.
(σ/r)^n = (k+d/σ)^(-n)
≒ k^(-n){1-nd/(kσ)+[n(n+1)/2]d^2/(kσ)^2}
當 |d/(kσ)| << 1
所以 (k^(-6) = 1/2)
(σ/r)^12 - (σ/r)^6
≒ 1/4{1-12d/(kσ)+78d^2/(kσ)^2}
- 1/2{1-6d/(kσ)+21d^2/(kσ)^2}
= -1/4 + 9d^2/[2^(1/6)σ]^2
當 |d/(2^(1/6)σ)| << 1, 得
u(r) ≒ 4ε{-1/4 + 9d^2/[2^(1/6)σ]^2}
= -ε + {36ε/[2^(1/6)σ]^2}(r - 2^(1/6)σ)^2
收錄日期: 2021-04-24 08:40:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20210309100143AAXbbVu