Thermodynamics of the independent harmonic oscillators with different frequencies in the Tsallis statistics in the high physical temperature approximation

Abstract

We study the thermodynamic quantities in the system of the N independent harmonic oscillators with different frequencies in the Tsallis statistics of the entropic parameter q ($1<q<2$) with escort average. The norm equations are derived, and the physical quantities are calculated with the physical temperature. It is found that the number of oscillators is restricted below $1/(q-1)$. The energy, the Rényi entropy $S_q^{(R)}$, and the Tsallis entropy $S_q^{(T)}$ are obtained by solving the norm equations approximately at high physical temperature and/or for small deviation $q-1$. The energy is q-independent at high physical temperature when the physical temperature is adopted, and the energy is proportional to the number of oscillators and physical temperature at high physical temperature. The form of the Rényi entropy is similar to that of the von-Neumann entropy, and the Tsallis entropy is given through the Rényi entropy. The physical temperature dependence of the Tsallis entropy is different from that of the Rényi entropy. The Tsallis entropy is bounded from the above, while the Rényi entropy increases with the physical temperature. The ratio of the Tsallis entropy to the Rényi entropy is small at high physical temperature. The relation between the physical temperature $T_{\mathrm {ph}}$ and the temperature T (the inverse of the Lagrange multiplier) is obtained, and the quantity as a function of T and q can be obtained through $T_{\mathrm {ph}}$. We calculate the free energy $F_q^{(R)}$ which is defined with $T_{\mathrm {ph}}$ and $S_q^{(R)}$ and the free energy $F_q^{(T)}$ which is defined with T and $S_q^{(T)}$. The relation between $\partial F_q^{(R)}/\partial T_{\mathrm {ph}}$ and $S_q^{(R)}$ and the relation between $\partial F_q^{(T)}/\partial T$ and $S_q^{(T)}$ are shown.

Graphical abstract

Appendices

Appendix A: Approximate expression of Hurwitz zeta function

The Hurwitz zeta function [26,27,28] is defined by

$$\begin{aligned}&\zeta _{\mathrm {H}}(s,\alpha ) := \sum _{n=0}^{\infty } \frac{1}{(\alpha + n)^s} . \end{aligned}$$

(A.1)

We treat the case of $s>1$ and $\alpha > 0$ in this appendix.

Let $B_n(x)$ be Bernoulli polynomials which are defined by

$$\begin{aligned} \frac{te^{xt}}{e^t-1} = \sum _{n=0}^{\infty } B_n(x) \frac{t^n}{n!} . \end{aligned}$$

(A.2)

The Bernoulli number $B_n$ in this paper is defined¹ by

$$\begin{aligned} B_n := B_n(x=1). \end{aligned}$$

(A.3)

We use the Euler–Maclaurin formula. Let a and b be integer with $a < b$ and let f(x) be continuously differentiable for M-times. The Euler–Maclaurin formula is

$$\begin{aligned} \sum _{n=a}^b f(n) =&\int _a^b dx f(x) + \frac{1}{2} (f(b)+f(a))\nonumber \\&+ \sum _{k=1}^{M-1} \frac{B_{k+1}}{(k+1)!} (f^{(k)}(b) - f^{(k)} (a)) \nonumber \\&- \frac{(-1)^M}{M!} \int _a^b dx B_M(x-[x]) f^{(M)}(x) , \end{aligned}$$

(A.4)

where $f^{(k)}$ is k-th derivative and [x] is the Gauss symbol (the floor function).

We attempt to find the expression of $\zeta _{\mathrm {H}}(1+z, \alpha )$ for $z>0$ using the Euler–Maclaurin formula. The right-hand side of Eq. (A.1) is an infinite series. Therefore, we first consider the following finite series:

$$\begin{aligned} \zeta _{H,m}(s,\alpha ) = \sum _{n=0}^m \frac{1}{(\alpha + n)^s} . \end{aligned}$$

(A.5)

We set f(x) as $1/(\alpha + x)^{1+z}$ and apply the Euler–Maclaurin formula. By taking the limit $m \rightarrow \infty$, we have the expression of $\zeta _{\mathrm {H}}(1+z,\alpha )$. The integral part converges when $\alpha$ is positive. We finally obtain

$$\begin{aligned}&\zeta _{\mathrm {H}}(1+z, \alpha ) \nonumber \\&\quad = \frac{1}{z\alpha ^z} + \frac{1}{2\alpha ^{1+z}}\nonumber \\&\qquad + \sum _{k=1}^{M-1} \frac{(-1)^{k+1} B_{k+1}}{(k+1)!} \frac{\varGamma (z+k+1)}{\varGamma (z+1)} \frac{1}{\alpha ^{z+k+1}} \nonumber \\&\qquad - \frac{(-1)^M}{M!} \int _0^{\infty } dx B_M(x-[x]) f^{(M)}(x) \quad (z>0, \alpha > 0) . \end{aligned}$$

(A.6)

The function $\zeta _{\mathrm {H}}(1+z, \alpha )$ can be rewritten [29]. For example, $\zeta _{\mathrm {H}}(1+z, \alpha )$ is given by

$$\begin{aligned}&\zeta _{\mathrm {H}}(1+z, \alpha ) \nonumber \\&\quad = \frac{1}{z\alpha ^z} + \frac{1}{2\alpha ^{1+z}} + \frac{1}{z} \sum _{k=2}^{M} \frac{B_k}{k!} \frac{\varGamma (z+k)}{\varGamma (z)} \frac{1}{\alpha ^{z+k}} \nonumber \\&\qquad - \frac{(-1)^M}{M!} \int _0^{\infty } dx B_M(x-[x]) f^{(M)}(x) \quad (z>0, \alpha > 0), \end{aligned}$$

(A.7)

because $B_{2n+1}$ is zero for $n \ge 1$ and $\varGamma (z+1) = z \varGamma (z)$.

It is possible to estimate the integral of Eq. (A.6) by setting M. For example, the upper value of the integral with $M=2$ is estimated:

$$\begin{aligned} \left| \frac{1}{2!} \int _0^{\infty } B_2(x-[x]) f^{(2)}(x) \right| \le \frac{C_2}{2!} \int _0^{\infty } \left| f^{(2)}(x) \right| , \end{aligned}$$

(A.8)

where $C_2$ is the maximum value of $|B_2(x)|$ in the range of $0 \le x \le 1$.

From Eq. (A.6), we find that the $\zeta _{\mathrm {H}}(1+z, \alpha )$ for $\alpha \gg 1$ behaves

$$\begin{aligned} \zeta _{\mathrm {H}}(1+z,\alpha ) \sim \frac{1}{z \alpha ^{z}}. \end{aligned}$$

(A.9)

Appendix B: Approximate expression of Barnes zeta function

The Barnes zeta function [30, 31] is defined by

$$\begin{aligned}&\zeta _{\mathrm {B}}(s,\alpha |\mathbf {\omega }_N) \nonumber \\&\quad := \sum _{n_1,\cdots ,n_N=0}^{\infty } \frac{1}{(\alpha + \omega _1 n_1 + \cdots + \omega _N n_N)^s}, \nonumber \\&\qquad \qquad \mathbf {\omega }_N = (\omega _1, \omega _2, \cdots , \omega _N), \end{aligned}$$

(B.10)

where $s>N$, $\alpha > 0$, and $\omega _j > 0$.

We define $\varOmega _N$ as

$$\begin{aligned} \varOmega _N := \alpha + \omega _1 n_1 + \cdots + \omega _N n_N . \end{aligned}$$

(B.11)

The function $\zeta _{\mathrm {B}}$ is represented as

$$\begin{aligned} \zeta _{\mathrm {B}}(s,\alpha |\mathbf {\omega }_N)&= \sum _{n_1,\cdots ,n_N=0}^{\infty } \frac{1}{(\varOmega _N)^s} \nonumber \\&= \frac{1}{(\omega _N)^s} \sum _{n_1,\cdots ,n_{N-1}=0}^{\infty } \sum _{n_N=0}^{\infty } \frac{1}{((\varOmega _{N-1}/\omega _N)+n_N)^s} \nonumber \\&= \frac{1}{(\omega _N)^s} \sum _{n_1,\cdots ,n_{N-1}=0}^{\infty } \zeta _{\mathrm {H}}(s, \varOmega _{N-1}/\omega _N) . \end{aligned}$$

(B.12)

We have $\zeta _{\mathrm {H}}(1+z,\alpha ) \sim 1/(z \alpha ^z)$ for $\alpha \gg 1$. Therefore, for sufficiently large $\alpha$, we have

$$\begin{aligned} \zeta _{\mathrm {B}}(1+z,\alpha |\mathbf {\omega }_N)&\sim \frac{1}{(\omega _N)^{1+z}} \sum _{n_1,\cdots ,n_{N-1}=0}^{\infty } \frac{1}{z (\varOmega _{N-1}/\omega _N)^{z}}\nonumber \\&= \frac{1}{z \omega _N} \zeta _{\mathrm {B}}(z, \alpha |\mathbf {\omega }_{N-1}) . \end{aligned}$$

(B.13)

Using the recurrence relation, Eq. (B.13), we have the approximate expression of $\zeta _{\mathrm {B}}$ for $\alpha \gg 1$:

$$\begin{aligned}&\zeta _{\mathrm {B}}(1+z, \alpha |\mathbf {\omega }_N) \nonumber \\&\quad \sim \frac{1}{z (z-1) \cdots (z-(N-1))} \frac{1}{\omega _1 \omega _2 \cdots \omega _N} \frac{1}{\alpha ^{z-(N-1)}} \nonumber \\&\quad = \frac{1}{\Bigg (\displaystyle \prod \nolimits _{j=0}^{N-1} (z-j)\Bigg ) \Bigg (\displaystyle \prod \nolimits _{j=1}^{N} \omega _j \Bigg ) \alpha ^{z-(N-1)}} ~ (z {-} (N{-}1) {>} 0). \end{aligned}$$

(B.14)

The condition $z - (N-1) > 0$ is rewritten as $1+z - N >0$. This condition is equivalent to the condition $s>N$ with $s=1+z$ for $\zeta _{\mathrm {B}}(s, \alpha |\mathbf {\omega }_N)$.