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\exnumber{3710}
\heading{Fermions in gravitational field of a star} 
\auname{Ofek Asban}


{\bf The problem:}

\Dn

Consider an artificial model of neutron star where the gas of ${N}$  neutrons is held together by a gravitational potential ${U=-\frac{mMG}{r}}$ generated by the solid core of the star. 
($M$ and ${r_{0}}$ are the mass and radius of the core, ${G}$ - gravity constant, ${m}$ - mass of the neutrons and ${r}$ - distance from the center).

\Dn

\begin{itemize}

\item [(1)]
Assume the neutron gas as Fermi gas. Find the density ${n(r)}$ at ${r>r_0}$ for $T=0$  and the fermi energy ${\epsilon_{f}(N)}$.

\Dn

\item [(2)]
Find the flux of particles that escapes the gravitational field of the core for low temperature, (${\mu \approx \epsilon_f}$).

\Dn

\item [(3)] 
Write a differential equation for ${N(t)}$ as a function of time.

\Dn

\item [(4)] 
Find the the condition in which Boltzmann approximation is valid for all the particles in the system. 


\end{itemize}

\Dn

\textbf{The solution:}

\begin{itemize}

\item [(1)]


 The number of states

\[N=2\int_r\int_p\frac{{d^3}r{d^3}p}{(2\pi)^3} \quad \Rightarrow \quad n(r)=2\int_p\frac{{d^3}p}{(2\pi)^3}\]

 After integrating over the angular degrees of freedom we get-

\[n(r)=2\int_0^{p_f}\frac{4\pi p^2}{(2\pi)^3}dp=2\frac{4\pi}{3(2\pi)^{3}}{p_f}^3\]

Where ${p_f}$ is the fermi momentum.

After substituting ${\epsilon=\frac{p^2}{2m}+U \quad \Rightarrow \quad p_f=[2m(\epsilon_f-U)]^{\frac{1}{2}}}$ we get

\[n(r)=\frac{(2m)^{\frac{3}{2}}}{3\pi^2}(\epsilon_f+\frac{mMG}{r})^{\frac{3}{2}}\]

\Dn

The fermi energy can be found from calculating the total number of particles ${N}$

\[N=\int_r n(r)d^3r=4\pi\int_{r_0}^{r_{m}}n(r)r^2dr=\frac{4}{3\pi}(2m)^{\frac{3}{2}}\int_{r_0}^{r_{m}}(\epsilon_f+\frac{mMG}{r})^{\frac{3}{2}}r^2dr \quad \Rightarrow \quad \epsilon_f(N)\]

Where ${r_{m}}$ is the maximal radius of which there are neutrons, and can be found from the condition ${n(r)>0}$

\[(\epsilon_f+\frac{mMG}{r})^{\frac{3}{2}}>0  \quad \Rightarrow \quad \epsilon_f>-\frac{GMm}{r} \quad \Rightarrow \quad r_{m}=\frac{GMm}{|\epsilon_f|}\]


\Dn\Dn

\item [(2)]

For ${T>0}$ there is emission of particles (see appendix).
In case of low temperature ($\mu\approx\epsilon_f$) the particles that escapes obey Boltzmann statistics, it can be seen from the fugacity

\[z=e^{\beta(\mu+U)}=e^{-\beta(|\epsilon_f|+\frac{GMm}{r})}<<1\]

According to the detailed balance principle the number of particles emitted from the star is the same as the number of particles incident to the star 
  
\[J_{e}=J_{i}=J\]

And therefore we can avoid the details of the potential and calculate the incoming free particles that are not influenced by it.

\Dn

The number of particles at speed $v=(\frac{2\epsilon_k}{m})^\frac{1}{2}$ out side the potential $U$  

\[N(v)=V (\frac{m}{2\pi})^3 e^{-\beta(\frac{1}{2}mv^2-\mu)}4\pi v^2 dv\]

Which is Boltzmann velocity distribution. Accordingly the incident flux

\[J=\frac{1}{4}\int\frac{N(v)}{V}vdv=\frac{m^3}{(2\pi)^2}e^{\beta\mu}\int_0^\infty e^{-\beta \frac{1}{2}mv^2}v^3dv =\frac{m}{2(\pi)^{\frac{3}{2}}}T^2e^{\mu/T}= \frac{m}{2(\pi)^{\frac{3}{2}}}T^2e^{-|\epsilon_f(N)|/T}\]

We got Richardson law, Where the fermi energy ($\epsilon_f$) is the work function $W$.

The total flow of emitted newtrons is the flux summed over the emitting surface $I=J4\pi {r_m^2}$.

\item[(3)] 

\[I=\frac{dN(t)}{dt}=\frac{2r_m^2}{\pi^{1/2}}mT^2e^{-|\epsilon_f(N(t))|/T}\]

We can see from this equation that N doesn't stay constant, therefore the system is out of equilibrium with the environment and Richardson law will be a good approximation only for a short time after the change of temperature from $T=0$ to $ T>0$.

\item[(4)]

The condition for Boltzmann approximation 

\[z=e^{\beta(\mu-U)} = e^{\beta(-|\mu|+\frac{GMm}{r})} \ll 1 \quad \Rightarrow \quad T \ll  T_f-\frac{GMm}{r} \quad \Rightarrow \quad T \ll T_f-\frac{GMm}{r_0}\]

The last equality says that if Boltzmann approximation is valid at the surface of the core ($r_0$), then it will be valid at any distance ($r>r_0$) as well.

\Dn
\Dn

Appendix - The density of particles at radius $r$ and $T>0$.

The total number of particles is given by the integral over the fermi distribution, with factor of 2 for spin ${\frac{1}{2}}$ degeneracy.

\[N = 2\int_r\int_p\frac{{d^3}r{d^3}p}{(2\pi)^3}\frac{1}{e^{\beta(\frac{p^2}{2m}+U-\mu)}+1} =\frac{2}{(2\pi)^3}\int_{r_0}^{\infty} d^3r\int_{-\infty}^{\infty} d^3p\frac{1}{\frac{1}{z}e^{\beta(\frac{p^2}{2m})}+1}\]

Where ${ z=e^{\beta(\mu-U)} }$ is the fugacity and ${U=-\frac{mMG}{r}}$.

\Dn

The number of particles in a volume element at radius $r$

\[n(r)=\frac{2}{(2\pi)^3} \int_{-\infty}^{\infty} d^3p\frac{1}{\frac{1}{z}e^{\beta(\frac{p^2}{2m})}+1}\]

After substituting ${p=(2m\epsilon_k)^{\frac{1}{2}} \quad \Rightarrow \quad p^{2}dp=\frac{1}{2}(2m)^{\frac{3}{2}}(\epsilon_k)^{\frac{1}{2}}}$, Where ${\epsilon_k}$ is the kinetic energy, and  integrating over the angular degrees of freedom we get

\[n(r)=\frac{(2m)^{\frac{3}{2}}}{2\pi^2} \int_{0}^{\infty} d\epsilon_k \frac{{\epsilon_k}^{\frac{1}{2}}}{\frac{1}{z}e^{\beta \epsilon_k}+1}\]

Switching to dimensionless  paramater ${x=\beta\epsilon \quad \Rightarrow \quad {\epsilon_k}^{\frac{1}{2}}d\epsilon_k=\frac{1}{\beta^{\frac{3}{2}}}x^{\frac{1}{2}}dx}$

\[n(r)=\frac{1}{2\pi^2}(\frac{2m}{\beta})^{\frac{3}{2}} \int_{0}^{\infty} dx \frac{{x}^{\frac{1}{2}}}{\frac{1}{z} e^x+1}=\frac{1}{2\pi^2}(\frac{2m}{\beta})^{\frac{3}{2}}F_{\frac{3}{2}}(z)\]

Observing the Last expression we see that $n(r\rightarrow\infty)>0$ which says there is emission of particles for $T>0$.
For $T=0$ we get same result as section [1] i.e. no emmition.

\end{itemize}
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\end{document} 
