# 德拜模型

## 推导

${\displaystyle \lambda _{n}={2L \over n}\,,}$

${\displaystyle E_{n}\ =h\nu _{n}\,,}$

${\displaystyle E_{n}=h\nu _{n}={hc_{s} \over \lambda _{n}}={hc_{s}n \over 2L}\,,}$

${\displaystyle E_{n}^{2}={p_{n}}^{2}{c_{s}}^{2}=\left({hc_{s} \over 2L}\right)^{2}\left(n_{x}^{2}+n_{y}^{2}+n_{z}^{2}\right)\,.}$

${\displaystyle U=\sum _{n}E_{n}\,{\bar {N}}(E_{n})\,,}$

${\displaystyle U=\sum _{n_{x}}\sum _{n_{y}}\sum _{n_{z}}E_{n}\,{\bar {N}}(E_{n})\,.}$

${\displaystyle \lambda _{\rm {min}}={2L \over {\sqrt[{3}]{N}}}\,,}$

${\displaystyle n_{\rm {max}}={\sqrt[{3}]{N}}\,.}$

${\displaystyle U=\sum _{n_{x}}^{\sqrt[{3}]{N}}\sum _{n_{y}}^{\sqrt[{3}]{N}}\sum _{n_{z}}^{\sqrt[{3}]{N}}E_{n}\,{\bar {N}}(E_{n})\,.}$

${\displaystyle U\approx \int _{0}^{\sqrt[{3}]{N}}\int _{0}^{\sqrt[{3}]{N}}\int _{0}^{\sqrt[{3}]{N}}E(n)\,{\bar {N}}\left(E(n)\right)\,dn_{x}\,dn_{y}\,dn_{z}\,.}$

${\displaystyle \langle N\rangle _{BE}={1 \over e^{E/kT}-1}\,.}$

${\displaystyle {\bar {N}}(E)={3 \over e^{E/kT}-1}\,.}$

${\displaystyle U=\int _{0}^{\sqrt[{3}]{N}}\int _{0}^{\sqrt[{3}]{N}}\int _{0}^{\sqrt[{3}]{N}}E(n)\,{3 \over e^{E(n)/kT}-1}\,dn_{x}\,dn_{y}\,dn_{z}\,.}$

${\displaystyle \ (n_{x},n_{y},n_{z})=(n\cos \theta \cos \phi ,n\cos \theta \sin \phi ,n\sin \theta )}$

${\displaystyle U\approx \int _{0}^{\pi /2}\int _{0}^{\pi /2}\int _{0}^{R}E(n)\,{3 \over e^{E(n)/kT}-1}n^{2}\sin \theta \,dn\,d\theta \,d\phi \,,}$

${\displaystyle N={1 \over 8}{4 \over 3}\pi R^{3}\,,}$

${\displaystyle R={\sqrt[{3}]{6N \over \pi }}\,.}$

${\displaystyle U={3\pi \over 2}\int _{0}^{R}\,{hc_{s}n \over 2L}{n^{2} \over e^{hc_{s}n/2LkT}-1}\,dn}$

${\displaystyle U={3\pi \over 2}kT\left({2LkT \over hc_{s}}\right)^{3}\int _{0}^{hc_{s}R/2LkT}{x^{3} \over e^{x}-1}\,dx}$

${\displaystyle T_{D}\ {\stackrel {\mathrm {def} }{=}}\ {hc_{s}R \over 2Lk}={hc_{s} \over 2Lk}{\sqrt[{3}]{6N \over \pi }}={hc_{s} \over 2k}{\sqrt[{3}]{{6 \over \pi }{N \over V}}}}$

${\displaystyle {\frac {U}{Nk}}=9T\left({T \over T_{D}}\right)^{3}\int _{0}^{T_{D}/T}{x^{3} \over e^{x}-1}\,dx=3TD_{3}\left({T_{D} \over T}\right)\,,}$

${\displaystyle T}$微分，我们便得到无量纲热容：

${\displaystyle {\frac {C_{V}}{Nk}}=9\left({T \over T_{D}}\right)^{3}\int _{0}^{T_{D}/T}{x^{4}e^{x} \over \left(e^{x}-1\right)^{2}}\,dx\,.}$

## 德拜的推导

${\displaystyle n\sim {1 \over 3}\nu ^{3}VF\,,}$

${\displaystyle U=\int _{0}^{\infty }\,{h\nu ^{3}VF \over e^{h\nu /kT}-1}\,d\nu \,,}$

${\displaystyle 3N={1 \over 3}\nu _{m}^{3}VF\,.}$

${\displaystyle U=\int _{0}^{\nu _{m}}\,{h\nu ^{3}VF \over e^{h\nu /kT}-1}\,d\nu \,,}$
${\displaystyle =VFkT(kT/h)^{3}\int _{0}^{T_{D}/T}\,{x^{3} \over e^{x}-1}\,dx\,,}$

${\displaystyle =9NkT(T/T_{D})^{3}\int _{0}^{T_{D}/T}\,{x^{3} \over e^{x}-1}\,dx\,,}$
${\displaystyle =3NkTD_{3}(T_{D}/T)\,,}$

## 低温极限

${\displaystyle {\frac {C_{V}}{Nk}}\sim 9\left({T \over T_{D}}\right)^{3}\int _{0}^{\infty }{x^{4}e^{x} \over \left(e^{x}-1\right)^{2}}\,dx}$

${\displaystyle {\frac {C_{V}}{Nk}}\sim {12\pi ^{4} \over 5}\left({T \over T_{D}}\right)^{3}}$

## 高温极限

${\displaystyle {\frac {C_{V}}{Nk}}\sim 9\left({T \over T_{D}}\right)^{3}\int _{0}^{T_{D}/T}{x^{4} \over x^{2}}\,dx}$
${\displaystyle {\frac {C_{V}}{Nk}}\sim 3\,.}$

## 参考文献

1. ^ Debye, Peter. Zur Theorie der spezifischen Waerme. Annalen der Physik. 1912, 39 (4): 789–839. Bibcode:1912AnP...344..789D. doi:10.1002/andp.19123441404 （德语）.