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\heading{3336: Condesation for general dispersion} 
\auname{Boris Shif}


{\bf The problem:}

\Dn
An ideal Bose gas whose single particle spectrum is given by ${\epsilon =C\cdot|p|^{s}, s>0}$ is contained in volume $V$ of dimention d. The gas is in uniform temperature $T$.
\Dn

(1)  Calculate the one-particle density of states.\\

(2) Find the condition on $s$ and $d$ for the existence of Bose-Einstein condensation.  In particular relate to relativistic (s=1)
and nonrelativistic (s=2) particles in two dimensions (d=2).\\

(3) Find the dependence of the number of particles $N$ on the chemical potential $\mu$.\\

(4) Find the dependence of the total energy $E$ on the chemical potential, and show how the pressure $P$ is obtained from this result.\\

(5) Find an expression for the heat capacity $C_v$ as a function of $N$ in the limit of infinite temperature.\\  

(6) Repeat item (1) for relativistic gas whose particles have finite mass such that their dispersion relation is $\epsilon =\sqrt{m^2c^4+c^2p^2}$.\\

(7) Consider a relativistic gas in $2D$. Find expressions for $N$ and $E$ and $P$. Should one expect Bose-Einstein condensation?

\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}
\Dn

(1) 
\\The particle spectrum is given by $\epsilon =C\cdot|p|^{s}\Leftrightarrow |p|^2=(\epsilon / C)^{2/s}$

\[
\nu(\epsilon)= \int\int_{H<E}{\frac{d^dxd^dp}{(2\pi)^d}}=\frac{V}{(2\pi)^d}\int\int_{k^2_1+\ldots+k^2_d<(\epsilon/C)^{2/s}}{d^dk}
=\frac{V}{(2\pi)^d}\cdot\Omega_d\int^{k}_{0}{k^{d-1}dk}=\frac{V}{(2\pi)^d}\frac{\Omega_d}{d}k^d
\]
\[
k^d=(k^{2})^{d/2}=((\epsilon / C)^{2/s})^{d/2}=(\epsilon / C)^{d/s}
\]
and therefore
\[
\nu(\epsilon)=\frac{s}{d}\frac{V}{s\cdot(2\pi)^d}\frac{\Omega_d}{C^{d/s}}\epsilon^{d/s}
\]
\[
g(\epsilon)=\frac{\partial\nu}{\partial\epsilon}=\frac{V}{s\cdot(2\pi)^d}\frac{\Omega_d}{C^{d/s}}\epsilon^{d/s-1}
\]
%%%%%%%%%%%%%%%
\Dn
\Dn

(2) 
\\We define $\alpha=\frac{d}{s}$ and $c=\frac{1}{s\cdot(2\pi)^d}\frac{\Omega_d}{C^{d/s}}$

\[
N(\epsilon)=V\cdot c\cdot\int^\infty_0{d\epsilon\frac{\epsilon^{\alpha-1}}{e^{\beta(\epsilon-\mu)}-1}}
\]

When we have BEC we get $\mu\rightarrow0$.\\
In order to exhibit  BEC the integral should converge and so for $\mu\rightarrow0$ the integrant should fulfill a condition that $\alpha>1\Leftrightarrow d>s$.\\
For 2D nonrelativistic particle $d=s=2$. From above the system will not exhibit BEC.\\
For 2D relativistic particle (with rest mass $m=0$) we have $d=2$ and $s=1$. From the above the particles can exhibit BEC.\\

%%%%%%%%%%%%%%%
\Dn

(3)
\\We define $z=e^{\beta\mu}$
\[
N(\epsilon)=V\cdot c\cdot\int^\infty_0{d\epsilon\frac{\epsilon^{\alpha-1}}{\frac{1}{z}e^{\beta\epsilon}-1}}
\underbrace{=}_{x=\frac{\epsilon}{T}}V\cdot c\cdot T^\alpha\cdot\int^\infty_0{dx\frac{x^{\alpha-1}}{\frac{1}{z}e^{x}-1}}=V\cdot c\cdot T^\alpha \cdot\Gamma(\alpha)\cdot F_{\alpha}(z)
\]
\\where
\[
F_{\alpha}(z)=\sum^\infty_{l=1}\frac{\pm(\pm1)^l}{l^\alpha}z^l=\sum^\infty_{l=1}\frac{\pm(\pm1)^l}{l^\alpha}e^{l\beta\mu}
\]
\[ 
\Rightarrow N(\mu)=V\cdot c\cdot T^\alpha \cdot\Gamma(\alpha)\cdot\sum^\infty_{l=1}\frac{\pm(\pm1)^l}{l^\alpha}e^{l\beta\mu}
 \]
%%%%%%%%%%%%%%
\Dn

(4)
\[
E(\epsilon)=V\cdot c\cdot\int^\infty_0{d\epsilon\frac{\epsilon^{\alpha-1}\epsilon}{\frac{1}{z}e^{\beta\epsilon}-1}}
\underbrace{=}_{x=\frac{\epsilon}{T}}V\cdot c\cdot T^{\alpha+1}\cdot\int^\infty_0{dx\frac{x^{\alpha}}{\frac{1}{z}e^{x}-1}}
\underbrace{=}_{\nu=\alpha+1}V\cdot c\cdot T^{\alpha+1}\cdot\int^\infty_0{dx\frac{x^{\nu-1}}{\frac{1}{z}e^{x}-1}}
\]
And we get
\[
E(\epsilon)=V\cdot c\cdot T^{\alpha+1} \cdot\Gamma(\nu)\cdot F_{\nu}(z)
=V\cdot c\cdot T^{\alpha+1} \cdot\Gamma(\alpha+1)\cdot F_{\alpha+1}(z)
\]
we should also notice that 
\[
E(\epsilon)=\frac{\alpha}{\beta}lnZ
\]
\\We know that

\[ P=\frac{1}{\beta}\frac{lnZ}{V}=\frac{1}{\alpha}\frac{E}{V}=\frac{c}{\alpha}\cdot T^{\alpha+1} \cdot\Gamma(\alpha+1)\cdot F_{\alpha+1}(z)\underbrace{=}_{\alpha=\frac{d}{s}}\frac{s}{d}\frac{E}{V} \]


%%%%%%%%%%%%%
\Dn

(5)
\\The limit of infinite temperature is Boltzman approximation.
\\For Boltzman approximation we get:
\[
E(\epsilon)=V\cdot c\cdot\int^\infty_0{d\epsilon\frac{\epsilon^{\alpha-1}\epsilon}{\frac{1}{z}e^{\beta\epsilon}}}
=V\cdot c\cdot T^{\alpha+1}\cdot\Gamma(\alpha+1)\cdot z=V\cdot c\cdot T^{\alpha+1}\cdot\alpha\cdot\Gamma(\alpha)\cdot z
\]
\\and
\[
N(\epsilon)=V\cdot c\cdot\int^\infty_0{d\epsilon\frac{\epsilon^{\alpha-1}}{\frac{1}{z}e^{\beta\epsilon}}}
=V\cdot c\cdot T^{\alpha}\cdot\Gamma(\alpha)\cdot z
\]
\[
\Rightarrow E=\alpha\cdot N\cdot T
\]
\[ C_v=\frac{\partial E}{\partial T}|_{V,N}=\alpha\frac{\partial (NT)}{\partial T}|_{V,N}=\alpha N=\frac{d}{s}N
\]
%%%%%%%%%%%%%
\Dn

(6)
\[\epsilon =\sqrt{m^2c^4+c^2|p|^2}\Rightarrow |p|^2=(\frac{\epsilon}{c})^2-m^2c^2\]
as we did in (1)
\[
\nu(\epsilon)= \int\int_{H<E}{\frac{d^dxd^dp}{(2\pi)^d}}=\frac{V}{(2\pi)^d}\int\int_{k^2_1+\ldots+k^2_d<(\frac{\epsilon}{c})^2-m^2c^2}{d^dk}
=\frac{V}{(2\pi)^d}\cdot\Omega_d\int^{k}_{0}{k^{d-1}dk}=\frac{V}{(2\pi)^d}\frac{\Omega_d}{d}k^d
\]
\[
k^d=(k^{2})^{d/2}=((\epsilon / c)^2-m^2c^2)^{d/2}
\]
and therefore
\[
\nu(\epsilon)=\frac{1}{d}\frac{V}{(2\pi)^d}\Omega_d((\frac{\epsilon}{c})^2-m^2c^2)^{d/2}
\]
\[
g(\epsilon)=\frac{\partial\nu}{\partial\epsilon}=\frac{V}{(2\pi)^d}\Omega_d((\frac{\epsilon}{c})^2-m^2c^2)^{d/2-1}\cdot\frac{\epsilon}{c^2}
\]
%%%%%%%%%%%%%
\Dn

(7)
\\We consider $2D$ relativistic gas therefore
\[
N(\epsilon)=\Sigma_kf(\epsilon_k-\mu)= \frac{V}{(2\pi)^2}\cdot\int^\infty_0{2\pi kdk\frac{1}{e^{\beta(\sqrt{k^2c^2+m^2c^4}-\mu)}-1}}
\]
we take $\epsilon=\sqrt{k^2c^2+m^2c^4}\Rightarrow\frac{d\epsilon}{dk}=\frac{kc^2}{\sqrt{k^2c^2+m^2c^4}}\Rightarrow\epsilon d\epsilon/c^2=kdk$ and get
\[
N(\mu)=\frac{V}{2\pi c^2}\cdot\int^\infty_{mc^2}{\epsilon d\epsilon\frac{1}{e^{\beta(\epsilon-\mu)}-1}}
\underbrace{=}_{\epsilon^`=\epsilon-mc^2}\frac{V}{2\pi c^2}\cdot\int^\infty_0{ d\epsilon^`\frac{\epsilon^`+mc^2}{e^{\beta(\epsilon^`+mc^2-\mu)}-1}}
\]
\[
=\frac{V}{2\pi c^2}\cdot(\int^\infty_0{ d\epsilon^`\frac{\epsilon^`}{e^{\beta(\epsilon^`+mc^2-\mu)}-1}}+\int^\infty_0{ d\epsilon^`\frac{mc^2}{e^{\beta(\epsilon^`+mc^2-\mu)}-1}})=\frac{V}{2\pi c^2}\cdot(k_b^2T^2\Gamma(2)F_2(z^`)+mc^2k_bT\Gamma(1)F_1(z^`))
\]
where $z^`=e^{\beta(\mu-mc^2)}$ and $F_1(z^`)=-ln(1-z^`)$.
\\The same way
\[
E(\epsilon)=\Sigma_k\epsilon f(\epsilon_k-\mu)= \frac{V}{(2\pi)^2}\cdot\int^\infty_0{2\pi kdk\frac{\epsilon}{e^{\beta(\sqrt{k^2c^2+m^2c^4}-\mu)}-1}}
=\frac{V}{2\pi c^2}\cdot\int^\infty_0{ d\epsilon^`\frac{(\epsilon^`+mc^2)^2}{e^{\beta(\epsilon^`+mc^2-\mu)}-1}}
\]
\[
\Rightarrow E(\mu)=\frac{V}{2\pi c^2}\cdot(k_b^3T^3\Gamma(3)F_3(z^`)+2mc^2k_b^2T^2\Gamma(2)F_2(z^`)+(mc^2)^2k_bT\Gamma(1)F_1(z^`))
\]
We want to calculate $P=\frac{1}{\beta}\frac{lnZ}{V}$
\[
lnZ=-\Sigma_k(ln(1-e^{-\beta(\epsilon_k-\mu)}))=- \frac{V}{(2\pi)^2}\cdot\int^\infty_0{2\pi kdk ln(1-e^{-\beta(\sqrt{k^2c^2+m^2c^4}-\mu)})}
\]
doing integration by parts and taking $\epsilon=\sqrt{k^2c^2+m^2c^4}$ as before we get
\[
lnZ=\frac{V\beta}{2\pi}\cdot\int^\infty_{mc^2}{ d\epsilon((\frac{\epsilon}{c})^2-m^2c^2)\frac{1}{e^{\beta(\epsilon-\mu)}-1}}
=\frac{V\beta}{2\pi c^2}\cdot\int^\infty_0{ d\epsilon^`\frac{(\epsilon^`+mc^2)^2-m^2c^4}{e^{\beta(\epsilon^`+mc^2-\mu)}-1}}
\]
therefore
\[
P(\mu)=\frac{1}{\beta}\frac{lnZ}{V}
=\frac{1}{2\pi c^2}\cdot\int^\infty_0{ d\epsilon^`\frac{(\epsilon^`+mc^2)^2-m^2c^4}{e^{\beta(\epsilon^`+mc^2-\mu)}-1}}
=\frac{1}{2\pi c^2}\cdot(k_b^3T^3\Gamma(3)F_3(z^`)+2mc^2k_b^2T^2\Gamma(2)F_2(z^`))
\]
Now we want to consider weather the system can exhibit BEC.
\[
N(\mu)=\frac{V}{2\pi c^2}\cdot\int^\infty_0{ d\epsilon^`\frac{\epsilon^`+mc^2}{e^{\beta(\epsilon^`+mc^2-\mu)}-1}}
\underbrace{=}_{\mu^`=\mu-mc^2}\frac{V}{2\pi c^2}\cdot\int^\infty_0{ d\epsilon^`\frac{\epsilon^`+mc^2}{e^{\beta(\epsilon^`-\mu^`)}-1}}
\]
\\For $\mu^`=0\Leftrightarrow\mu=mc^2$ the integral does not converge and so the system  can`t get BEC.


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