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\begin{document} 
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\exnumber{3230}
\heading{Heat Capacity of He4 system, energy gap} 
\auname{Orr Be'er}

{\bf The problem:}

\Dn

The specific heat of ${He^{4}}$ at low temperatures has the form

\[ \ C_{v}= A(T) + B(T) e^{-\Delta/T}\]

This is explained by having a dispersion 
relation that give rise to long wavelength 
phonons, ${\omega = c|\bm{k}|}$ and short wavelength  
rotons ${\omega(k)= \Delta + b(|\bm{k}|-k_0)^2}$, 
where $k_0=1/a$ is comparable to the mean 
interparticle separation.

\begin{itemize}
\item [(a)] Find explcit expressions for 
the coefficients $A(T)$ and $B(T)$ 

\item [(b)]
What would be the power in the $T$ dependence
of the coefficients if the system was two dimensional?

\Dn\Dn

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

\item [(a)]
For phonons: $\varepsilon={\omega}$
\[
k=\frac{\omega}{c}
\]
\[
f(\omega)= \frac{1}{e^{\frac{\omega}{T}}-1}
\]
\[
E(x,k)=\int {\frac{d^3x d^3k}{(2\pi)^3}\varepsilon (k)f(k) }
\]
\[
E(\omega)=\frac{4\pi V}{(2\pi)^3}\int{\omega f(\omega)k^2(\omega)dk(\omega)}=\frac{V}{2\pi^2}\int_0^\infty{\frac{\omega^2}{c^2}\frac{1}{e^{\frac{\omega}{T}}-1}\omega d\left(\frac{\omega}{c}\right)}
\]
\[
E(\omega)_{phonons}= \frac{VT^4}{2\pi^2c^3} \int_0^\infty \left( \frac{\omega}{T}\right)^3 \frac{1}{e^{\frac{\omega}{T}}-1}d\left(\frac{\omega}{T}\right)=\frac{VT^4}{2\pi^2c^3}\left(\frac{\pi^4}{15}\right)
\]




The dispersion relation for rotons: $\varepsilon=\omega(k)= \Delta + b(|\bm{k}|-k_0)^2$ 
\[
E(k)=\frac{4\pi V}{(2\pi)^3}\int{\omega(k) f(k)k^2dk}
\]
For short wavelength rotons at low temperatures, we take $\frac{\omega}{T}>>1$ and so $\frac{1}{e^{\frac{\omega}{T}}-1}\approx e^{-\frac{\omega}{T}}$
\[
E(k)=\frac{V}{2\pi^2}\int_0^\infty  (\Delta + b(|\bm{k}|-k_0)^2){e^{-\frac{\Delta + b(|\bm{k}|-k_0)^2}{T}}k^2dk}
\]
Denote $x=\sqrt{\frac{b}{T}}(|\bm{k}|-k_0)$ 
\[
E(T)=\frac{V}{2\pi^2}\int_{0}^\infty\left(\Delta+Tx^2\right)e^{-(\frac{\Delta}{T}+x^2)}\left(\sqrt{\frac{T}{b}}x+k_0 \right)^2 \sqrt{\frac{T}{b}}dx
\]
\[
E(T)=\frac{Vk_0^2}{2\pi^2}\sqrt{\frac{T}{b}}e^{-\frac{\Delta}{T}} \int_{0}^\infty\left(\frac{1}{k_0} \sqrt{\frac{T}{b}}x+1 \right)^2\left(\Delta+Tx^2\right)e^{-x^2}dx
\]
We are interested in low temperature and short mean interparticle separation: $\frac{T}{k_0^2b}<<1$ The contributing parts of the integrand are symmetric, by that the energy takes the form:
\[
E(T)=\frac{Vk_0^2}{2\pi^2}\sqrt{\frac{T}{b}}e^{-\frac{\Delta}{T}} \int_{0}^\infty\left(\Delta+Tx^2\right)e^{-x^2}dx
\]
\[
E(T)=\frac{Vk_0^2}{2\pi^2}\sqrt{\frac{T}{b}}\left(\Delta\frac{\sqrt{\pi}}{2}+T\frac{\sqrt{\pi}}{4}\right)e^{-\frac{\Delta}{T}}
\]
\[
E_{rotons}=\frac{ V k_0^2}{4\pi^{\frac{3}{2}}\sqrt{b}}\left(\Delta\sqrt{T}+ \frac{T^{\frac{3}{2}}}{2}\right)e^{-\frac{\Delta}{T}}
\]
Thus the heat capacity is:
\[
\frac{\partial}{\partial T} \left (E_{phonos}+E_{rotons}\right)=\frac{2V\pi^2}{15c^3}T^3+\frac{ V k_0^2}{4\pi^{\frac{3}{2}}\sqrt{b}}\left(\frac{3}{4}\sqrt{T}+\frac{\Delta}{\sqrt{T}}+\frac{\Delta^2}{T^{\frac{3}{2}}}\right)e^{-\frac{\Delta}{T}}
\]
And
\[A(T) =\frac{2V\pi^2}{15c^3}T^3\]
\[B(T)=\frac{ V k_0^2}{4\pi^{\frac{3}{2}}}\left(\frac{3}{4}\sqrt{T}+\frac{\Delta}{\sqrt{T}}+\frac{\Delta^2}{T^{\frac{3}{2}}}\right)\]
%%%%%%%%%%%%%%%
\Dn

\item[(b)]
For two dimentional phonons system:
\[
N(\omega)_{phonons}=\frac{A}{2\pi c^2}\int_0^\infty\omega d\omega
\]
\[
E(T)_{phonons}= \frac{AT^3}{2 \pi c^2} \int_0^\infty { \frac{1}{e^{\frac{\omega}{T}}-1}\left(\frac{\omega}{T}\right)^2d\left( \frac{\omega}{T}\right)}= \frac{AT^3}{2  c^2\sqrt{\pi} }
\]
For two dimentional rotons system:
\[
E(k)_{rotons}=\frac{A}{2\pi c^2}\int_0^\infty{e^{-\frac{\Delta + b(|\bm{k}|-k_0)^2}{T}} (\Delta + b(|\bm{k}|-k_0)^2)kdk}
\]
\[
E(T)_{rotons}=\frac{Ak_0}{2\pi c^2}\sqrt{\frac{T}{b}}e^{-\frac{\Delta}{T}} \int_{0}^\infty\left(\frac{1}{k_0} \sqrt{\frac{T}{b}}x+1 \right)\left(\Delta+Tx^2\right){e^{-x^2}}=\frac{Ak_0}{4 c^2}\sqrt{\frac{T}{b}}\left(\Delta\sqrt{\pi}+T\frac{\sqrt{\pi}}{2}\right)e^{-\frac{\Delta}{T}}
\]
Without further assumpstion, we can deduce:
$A(T)_{2D}\propto\frac{1}{T}A(T)_{3D}$ and  $ B(T)_{2D}\propto B(T)_{3D}$

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


\end{itemize}


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