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\begin{document}
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\heading{E3021: Bosons with Spin in magnetic field}
\auname{Chen Giladi}
{\bf The problem:}
$N$ Bosons that have mass $m$ and spin1 are placed
in a box that has volume $V$. A magnetic field B is
applied, such that the interaction is $-\gamma B S_z$,
where $S_z=1,0,-1$ and $\gamma$ is the gyromagnetic ratio.
In items (c-f) assume the Boltzmann approximation
for the occupation of the $S_z\ne1$ states.
\Dn
(a) Find an equation for the condensation temperature $T_c$.
\Dn
(b) Find the condensation temperature $T_c(B)$ for $B=0$
and for $B\rightarrow \infty$.
\Dn
(c) Find the critical $B$ for condensation if $T$ is set
in the range of temperatures that has been defined in item(b).
\Dn
(d) Describe how $T_c(B)$ depends of $B$ in a qualitatively manner.
Find approximate expressions for moderate and large fields.
\Dn
(e) Find the condensate fraction as a function of $T$ and $B$.
\Dn
(f) Find the heat capacity of the gas assuming large but finite field.
\Dn
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}
\Dn
(a)
Before we find the condensation temperature $T_c$, we shold understand that we can treat the particles of the system like three different gasses, therfore the Hammiltonian is $\mathcal{H}=\frac{p^2}{2m}-\gamma B S_z$
where $s_z=-1,0,1$.
\Dn
\[
n=\frac{N}{V}=\underbrace{\sum_p\frac{1}{\frac{1}{\zeta}e^{\frac{\beta p^2}{2m}-\beta\gamma B}-1}}_{s_z=+1}
+\underbrace{\sum_p\frac{1}{\frac{1}{\zeta}e^{\frac{\beta p^2}{2m}}-1}}_{s_z=0}
+\underbrace{\sum_p\frac{1}{\frac{1}{\zeta}e^{\frac{\beta p^2}{2m}+\beta\gamma B}-1}}_{s_z=-1}
\]
So n equals to
\[
n=\frac{1}{\lambda ^{3}}[g_{3/2}(\zeta e^{\beta \gamma B})+g_{3/2}(\zeta )+g_{3/2}(\zeta e^{-\beta \gamma B})]
\]
The lowest energy state for $s_z=+1$ particles, all terms must be $\geq 0$. At the condensation we get $\zeta e^{\beta \gamma B}=1\Rightarrow \zeta =e^{-\beta \gamma B}$.
\Dn
At $T
\]
(b)
For $B=0$, n is
\[
n=\frac{3}{\lambda ^{3}}g_{3/2}(1)=3g_{3/2}(1)\cdot (\frac{2m\pi kT_c}{h^2})^{3/2}
\]
So the condensation temperature is
\[
kT_c=\left (\frac{n}{3\cdot 2.612} \right )^{2/3}\cdot \frac{2\pi \hbar^2}{m}
\]
*Note that $g_{3/2}(1)\approx 2.612$.
\Dn
The term $g_{3/2}( e^{-2\beta \gamma B})$ is equal to zero, if $\gamma B\gg kT$, because we keep only lowest order correction for large B.\Dn
For $\gamma B\gg kT$ n is
\[
n=\frac{1}{\lambda ^{3}}\left ( g_{3/2}(1)+g_{3/2}(e^{-\beta \gamma B})\right )
\]
Note that $g_{v}(z)=z+\frac{z^2}{2^v}+o(z^3)$. So for $z\ll 1$ we get $g_v(z)\approx z$. Hence,
\[
n\approx \frac{1}{\lambda ^{3}}\left ( g_{3/2}(1)+e^{-\beta \gamma B}\right )
\]
So the condensation temperature is
\[
kT_{c}=\left ( \frac{n}{g_{3/2}(1)+e^{-\beta \gamma B}} \right )^{2/3}\cdot \frac{2\pi\hbar^2}{m}\approx \left ( \frac{n}{2.612} \right )^{2/3}\cdot \frac{2\pi\hbar^2}{m}
\]
\Dn
(c+d)
If we calculate the ratio between the condensation temperature $T_c(B)$ for $B=0$
and between for $B\rightarrow \infty$, we get the following number
\[
\frac{kT_{c}^{B=0}}{kT_{c}^{B\rightarrow \infty}}\approx \frac{1}{3^{2/3}}
\]
As $B$ is increased, $T_c$ rises until $B_c$ is reached. At $B=B_c$, $T=T_c$ and the condensation occurs. Then we have from the case of $\gamma B\gg kT$
\[
n\lambda ^{3}=g_{3/2}(1)+g_{3/2}(e^{-\beta \gamma B_c})\approx g_{3/2}(1)+e^{-\beta \gamma B_c}
\]
Hence the critical magnetic field is
\[
B_c=\frac{-kT}{\gamma }ln\left ( n\lambda ^{3}-g_{3/2}(1) \right )
\]
%graph from file ex3021_fig1
(e)
We should calculate $\frac{}{n}$. We have already showed that in
\[
n=\frac{1}{\lambda ^{3}}\left ( g_{3/2}(1)+e^{-\beta \gamma B} \right )+
\]
So the condensate fraction as function of $T$ and $B$ is
\[
\frac{}{n}=1-\frac{g_{3/2}(1)+e^{-\beta \gamma B}}{n\lambda ^{3}}
\]
%graph from file ex3021_fig2
(f)
Before we can calculate the heat capacity of the gas, we have to formulate an equation for the energy.
\[
E=-\left ( \frac{\partial }{\partial \beta}ln(Z) \right )=VkT^3\sum_{s_z=-1,0,1}\frac{\partial}{\partial T}\left ( \frac{g_{3/2}(\zeta e^{\beta \gamma B S_z})}{\lambda ^{3}} \right )_{\zeta ,V}
\]
At $T\leq T_c$ :
\[
\frac{E}{V}=kT^2\frac{\partial}{\partial T}\left [ \frac{g_{3/2}(1)}{\lambda ^{3}}+\frac{e^{-\beta \gamma B}}{\lambda ^{3}}+\frac{e^{-2\beta \gamma B}}{\lambda ^{3}} \right ]
\]
\[
\approx kT^2\frac{\partial}{\partial T}\left [\frac{g_{3/2}(1)}{\lambda ^{3}}+\frac{e^{-\beta \gamma B}}{\lambda ^{3}} \right ]=\frac{kT}{\lambda ^{3 }}\cdot\frac{3}{2}\left ( g_{3/2}(1)+e^{-\beta \gamma B} \right )+\frac{kT}{\lambda ^{3 }}\cdot \frac{\gamma B}{kT}e^{-\beta \gamma B}
\]
\[
\approx \frac{kT}{\lambda ^{3}}\cdot \frac{3}{2}g_{3/2}(1)+\frac{\gamma B}{\lambda ^{3}}e^{-\beta \gamma B}
\]
Note that for $\gamma B \gg kT$ the third term is more significant then the second one. Also, we have used the following results
\[
\frac{\partial}{\partial T}\left ( \frac{1}{\lambda ^{3}} \right )=\frac{3}{2T}\cdot \frac{1}{\lambda ^{3}}
\]
\[
\frac{\partial}{\partial T}\left ( \frac{e^{-\beta \gamma B}}{\lambda ^{3}} \right )=\frac{e^{-\beta \gamma B}}{\lambda ^{3}}\cdot \left ( \frac{3}{2T}+\frac{\gamma B}{kT^2} \right )
\]
For heat capacity of the gas assuming large but finite field ( $\gamma B \gg kT$ ) is
\[
\frac{c_v}{k_B}=\frac{9}{4}\cdot \frac{1}{\lambda ^{3}}g_{3/2}(1)+\frac{1}{\lambda ^{3}}\left ( \frac{\gamma B}{k_B T} \right )^2e^{-\beta \gamma B}
\]
\Dn
\Dn
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\end{document}