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\begin{document}
\heading{E7040: Nyquist theory for a ring RL-circuit} \auname{ Nimrod Cohen}

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\Dn

Question


Derive the Nyquist expression for the current-current correlation
function in a closed ring, taking into account its inductance.
\\

\Dn

Use the following procedure:


\begin{enumerate}

\item Cite an expression for the Inductance $L$ of the ring given its radius $R$ and its section radius $ r $ when $ R>r$.
\item Write the R-L circuit equation for the current I, where the flux $\Phi(t)$ through the ring is the driving parameter.
\item Identify the generalized susceptibility, and observe that it is formally the same expression as in the problem of Brownian motion.
\item Calculate the current-current correlation function $ \left\langle I(0)I(\tau)\right\rangle $  taking the classical / high temperature limit.
\item Calculate the current fluctuation $ \left\langle I^2\right\rangle $,Verify that $ \left\langle I^2\right\rangle $ agree with the canonical result.

\end{enumerate}
%%%%%%%%%%%%%
\Dn
Solution\\

\begin{enumerate}

\item $ L=\mu_{0}R \left[ ln\left( \frac{8R}{r} \right) -1.75\right]  $.
\item Let us write equation for the force by Using $ V_{EMF}= \dot{\Phi} $ we get
\\
\[ RI+L\dot{I}=\dot{\Phi} \]
By rearrange the equation and taking Transform Fourier we get 
\[I_{w}=\frac{i \omega}{R-i \omega L} \Phi_{\omega} \]
\item From the theory of liner respond we Identify the generalized susceptibility

\[\chi_{\omega}=\frac{i \omega}{R-i \omega L} \] 
For Brownian motion the susceptibility is $\chi_{\omega}=\frac{i \omega}{\eta-i \omega M}$ and we get the same expression.
\item  First we calculate $\Im(\chi_{\omega})$ 
\[\chi_{\omega}= \frac{-\omega^2L}{R^2+\omega^2L^2}+ i\frac{\omega R}{R^2+\omega^2L^2} \]
\[\Im(\chi_{\omega})=\frac{\omega R}{R^2+\omega^2L^2}\]
\[C_{II}^{Q}(\omega)=\hslash \cdot coth(\frac{\hslash \omega}{2K_{B}T})\Im(\chi_{\omega})\]
\[C_{II}^{classic}(\omega)=\underset{\hbar\rightarrow0}{lim}\:\: C_{II}^{Q}(\omega)\longrightarrow\frac{2K_{B}TR}{R^2+\omega^2L^2}\]
\[\left\langle I(0)I(\tau)\right\rangle=\int_{-\infty}^{\infty} e^{-i\omega\tau}C_{II}^{classic}(\omega)\, \frac{d\omega}{2\pi}\]    
\[\left\langle I(0)I(\tau)\right\rangle = \frac{K_{B}T}{L}e^{-\frac{R\tau}{L}}\]
In the case of Brownian motion we obtain the same expression $ \frac{K_{B}T}{M}e^{-\frac{\eta\tau}{M}} $
\item Now we calculate the current fluctuation by taking $ \tau\rightarrow0 $ to obtain
\[\left\langle I^2\right\rangle = \frac{K_{B}T}{L}\]
The Hamiltonian is $\mathcal{H}=\frac{1}{2}LI^2 $  by using equal division rule for each quadratic term in the Hamiltonian we get the same result
\[\left\langle \frac{1}{2}LI^2\right\rangle=\frac{K_{B}T}{2}\] and therefore \[\left\langle I^2\right\rangle= \frac{K_{B}T}{L}\]
 
\end{enumerate}
\end{document}
