In mathematics, the family of Debye functions is defined by D n ( x ) = n x n ∫ 0 x t n e t − 1 d t . {\displaystyle D_{n}(x)={\frac {n}{x^{n}}}\int _{0}^{x}{\frac {t^{n}}{e^{t}-1}}\,dt.}
The functions are named in honor of Peter Debye, who came across this function (with n = 3) in 1912 when he analytically computed the heat capacity of what is now called the Debye model.
The Debye functions are closely related to the polylogarithm.
They have the series expansion[1] D n ( x ) = 1 − n 2 ( n + 1 ) x + n ∑ k = 1 ∞ B 2 k ( 2 k + n ) ( 2 k ) ! x 2 k , | x | < 2 π , n ≥ 1 , {\displaystyle D_{n}(x)=1-{\frac {n}{2(n+1)}}x+n\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k+n)(2k)!}}x^{2k},\quad |x|<2\pi ,\ n\geq 1,} where B n {\displaystyle B_{n}} is the n-th Bernoulli number.
lim x → 0 D n ( x ) = 1. {\displaystyle \lim _{x\to 0}D_{n}(x)=1.} If Γ {\displaystyle \Gamma } is the gamma function and ζ {\displaystyle \zeta } is the Riemann zeta function, then, for x ≫ 0 {\displaystyle x\gg 0} ,[2] D n ( x ) = n x n ∫ 0 x t n d t e t − 1 ∼ n x n Γ ( n + 1 ) ζ ( n + 1 ) , Re n > 0 , {\displaystyle D_{n}(x)={\frac {n}{x^{n}}}\int _{0}^{x}{\frac {t^{n}\,dt}{e^{t}-1}}\sim {\frac {n}{x^{n}}}\Gamma (n+1)\zeta (n+1),\qquad \operatorname {Re} n>0,}
The derivative obeys the relation x D n ′ ( x ) = n ( B ( x ) − D n ( x ) ) , {\displaystyle xD_{n}^{\prime }(x)=n\left(B(x)-D_{n}(x)\right),} where B ( x ) = x / ( e x − 1 ) {\displaystyle B(x)=x/(e^{x}-1)} is the Bernoulli function.
The Debye model has a density of vibrational states g D ( ω ) = 9 ω 2 ω D 3 , 0 ≤ ω ≤ ω D {\displaystyle g_{\text{D}}(\omega )={\frac {9\omega ^{2}}{\omega _{\text{D}}^{3}}}\,,\qquad 0\leq \omega \leq \omega _{\text{D}}} with the Debye frequency ωD.
Inserting g into the internal energy U = ∫ 0 ∞ d ω g ( ω ) ℏ ω n ( ω ) {\displaystyle U=\int _{0}^{\infty }d\omega \,g(\omega )\,\hbar \omega \,n(\omega )} with the Bose–Einstein distribution n ( ω ) = 1 exp ( ℏ ω / k B T ) − 1 . {\displaystyle n(\omega )={\frac {1}{\exp(\hbar \omega /k_{\text{B}}T)-1}}.} one obtains U = 3 k B T D 3 ( ℏ ω D / k B T ) . {\displaystyle U=3k_{\text{B}}T\,D_{3}(\hbar \omega _{\text{D}}/k_{\text{B}}T).} The heat capacity is the derivative thereof.
The intensity of X-ray diffraction or neutron diffraction at wavenumber q is given by the Debye-Waller factor or the Lamb-Mössbauer factor. For isotropic systems it takes the form exp ( − 2 W ( q ) ) = exp ( − q 2 ⟨ u x 2 ⟩ ) . {\displaystyle \exp(-2W(q))=\exp \left(-q^{2}\langle u_{x}^{2}\rangle \right).} In this expression, the mean squared displacement refers to just once Cartesian component ux of the vector u that describes the displacement of atoms from their equilibrium positions. Assuming harmonicity and developing into normal modes,[3] one obtains 2 W ( q ) = ℏ 2 q 2 6 M k B T ∫ 0 ∞ d ω k B T ℏ ω g ( ω ) coth ℏ ω 2 k B T = ℏ 2 q 2 6 M k B T ∫ 0 ∞ d ω k B T ℏ ω g ( ω ) [ 2 exp ( ℏ ω / k B T ) − 1 + 1 ] . {\displaystyle 2W(q)={\frac {\hbar ^{2}q^{2}}{6Mk_{\text{B}}T}}\int _{0}^{\infty }d\omega {\frac {k_{\text{B}}T}{\hbar \omega }}g(\omega )\coth {\frac {\hbar \omega }{2k_{\text{B}}T}}={\frac {\hbar ^{2}q^{2}}{6Mk_{\text{B}}T}}\int _{0}^{\infty }d\omega {\frac {k_{\text{B}}T}{\hbar \omega }}g(\omega )\left[{\frac {2}{\exp(\hbar \omega /k_{\text{B}}T)-1}}+1\right].} Inserting the density of states from the Debye model, one obtains 2 W ( q ) = 3 2 ℏ 2 q 2 M ℏ ω D [ 2 ( k B T ℏ ω D ) D 1 ( ℏ ω D k B T ) + 1 2 ] . {\displaystyle 2W(q)={\frac {3}{2}}{\frac {\hbar ^{2}q^{2}}{M\hbar \omega _{\text{D}}}}\left[2\left({\frac {k_{\text{B}}T}{\hbar \omega _{\text{D}}}}\right)D_{1}{\left({\frac {\hbar \omega _{\text{D}}}{k_{\text{B}}T}}\right)}+{\frac {1}{2}}\right].} From the above power series expansion of D 1 {\displaystyle D_{1}} follows that the mean square displacement at high temperatures is linear in temperature 2 W ( q ) = 3 k B T q 2 M ω D 2 . {\displaystyle 2W(q)={\frac {3k_{\text{B}}Tq^{2}}{M\omega _{\text{D}}^{2}}}.} The absence of ℏ {\displaystyle \hbar } indicates that this is a classical result. Because D 1 ( x ) {\displaystyle D_{1}(x)} goes to zero for x → ∞ {\displaystyle x\to \infty } it follows that for T = 0 {\displaystyle T=0} 2 W ( q ) = 3 4 ℏ 2 q 2 M ℏ ω D {\displaystyle 2W(q)={\frac {3}{4}}{\frac {\hbar ^{2}q^{2}}{M\hbar \omega _{\text{D}}}}} (zero-point motion).