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\exnumber{7041}
\heading{FDT for RLC circuit}
\auname{Afik Shachar}


{\bf The problem:}

\Dn


%Baruch's  D23.

An electrical circuit has in series components with capacitance $C$,
inductance $L$, resistance $R$ and a voltage
source $V_0\cos \omega t$ with frequency $\omega$.
\begin{itemize}
\item[(a)] Identify the responsefunction $\alpha_Q(\omega)=\langle Q(\omega)\rangle /(\half V_0)$  . Use this to write the energy dissipation rate.
\item[(b)] Use the fluctuation dissipation relation to identify the
Fourier transform $\Phi_Q(\omega)$ of the charge correlation function.
Evaluate $\langle Q^2(t)\rangle$ and compare with the result from equipartition.
\item[(c)] Evaluate the current fluctuations
$\langle I^2(t)\rangle$ and compare with the result from equipartition. Under what conditions does one get Nyquist's result $\langle I^2\rangle_{\omega_1 \leftrightarrow \omega_2 }  = \frac{{2k_BT}}{{\pi R}}\left( {\omega_2  - \omega_1 } \right)$ ?

Hint: $\int_{-\infty}^{\infty} \frac{d\omega
/2\pi}{(\omega^2-\omega_0^2)^2+\gamma^2\omega^2}=\frac{1}{2\gamma
\omega_0^2}\,,\qquad\int_{-\infty}^{\infty} \frac{\omega^2 d\omega
/2\pi}{(\omega^2-\omega_0^2)^2+\gamma^2\omega^2}=\frac{1}{2\gamma}\,.$\\
\end{itemize}

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{\bf The solution:}

\Dn

(a) 

\Dn

this RLC circuit is discribed by the equation:
\[ L\langle\ddot{Q}\rangle+R\langle\dot{Q}\rangle+\frac{1}{C}\langle Q\rangle=V_0\cos\left(\omega t\right) \] 
applying Fourier transform:
\[ -L\Omega^2 \langle Q(\Omega)\rangle-i\Omega\langle Q(\Omega)\rangle+\frac{1}{C}\langle Q(\Omega)\rangle=V_0\sqrt{\frac{\pi}{2}}\bigg[\delta(\Omega-\omega)+\delta(\Omega+\omega)\bigg] \]
\[ \langle Q(\Omega)\rangle=\chi(\Omega) FT\left[V(t)\right]=\frac{V_0\sqrt{\frac{\pi}{2}}\big[\delta(\Omega-\omega)+\delta(\Omega+\omega)\big]}{\left(\frac{1}{C}-L\Omega^2\right)-i\Omega R} \]
\[\langle Q(\omega) \rangle=V_0\sqrt{\frac{\pi}{2}}\frac{1}{\left(\frac{1}{C}-L\omega^2\right)-i\omega R}=V_0\sqrt{\frac{\pi}{2}}\frac{\left(\frac{1}{C}-L\omega^2\right)+i\omega R}{\left(\frac{1}{C}-L\omega^2\right)^2+\omega^2 R^2} \]
\[ \alpha_Q(\omega)=\frac{\langle Q(\omega)\rangle}{\half V_0}=\sqrt{2\pi}\frac{\left(\frac{1}{C}-L\omega^2\right)+i\omega R}{\left(\frac{1}{C}-L\omega^2\right)^2+\omega^2 R^2} \]
\[ \chi(\omega)=\frac{\alpha_Q(\omega)}{\sqrt{2\pi}}=\frac{\left(\frac{1}{C}-L\omega^2\right)+i\omega R}{\left(\frac{1}{C}-L\omega^2\right)^2+\omega^2 R^2} \]
and the energy dissipation rate (from lecture notes, sec. 12.2, eq. 619):
\[ \dot{\mathcal{W}}=\half\eta(\omega)V_0^2\omega^2=\half\omega^2 V_0^2 \frac{Im\left[\chi(\omega)\right]}{\omega}=\half V_0^2\frac{\omega^2 R}{\left(\frac{1}{C}-L\omega^2\right)^2+\omega^2 R^2} \]
we denote $ \gamma = \frac{R}{L} $ and $\omega_0^2=\frac{1}{CL} $
\[ \dot{\mathcal{W}}=\frac{1}{2L}\frac{\gamma V_0^2 \omega^2}{\left(\omega-\omega_0^2\right)^2+\gamma^2 \omega^2 } \]

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\Dn

(b)
\[ \Phi_Q(\omega)=FT\left[\langle Q(0)Q(\tau)\rangle\right] \]
from FDT:
\[ \frac{Im\left[\chi(\omega)\right]}{\omega}=\frac{1}{\hbar\omega}\tanh\left(\frac{\hbar\omega}{2k_B T}\right) \Phi_Q(\omega) \]
for $ \hbar\omega <<k_B T$ :
\[ \frac{Im\left[\chi(\omega)\right]}{\omega}=\frac{1}{2k_B T} \Phi_Q(\omega) \]
\[ \Phi_Q(\omega)=\frac{1}{L}\frac{2k_B T \frac{R}{L}}{\left(\frac{1}{CL}-\omega^2\right)^2+\omega^2\frac{R^2}{L^2}} \]
again $ \gamma = \frac{R}{L} $ and $\omega_0^2=\frac{1}{CL} $
\[ \Phi_Q(\omega)=\frac{1}{L}\frac{2k_B T \gamma}{\left(\omega^2-\omega_0^2\right)^2+\omega^2\gamma^2} \]
and we recall that:
\[ \langle Q(0)Q(\tau)\rangle=\int_{-\infty}^\infty \Phi_Q(\omega) \exp^{-i\omega\tau}\frac{d\omega}{2\pi} \]
therefore:
\[ \langle Q^2 (t)\rangle = \int_{-\infty}^\infty \Phi_Q(\omega)\frac{d\omega}{2\pi}=\frac{2k_B T \gamma}{L} \int_{-\infty}^\infty\frac{d\omega/2\pi}{\left(\omega^2-\omega_0^2\right)^2+\omega^2\gamma^2}=\frac{2k_B T \gamma}{2\gamma\omega_0^2 L} \]
\[ \langle Q^2 (t)\rangle =C k_B T\]
from equipartition we have
\[ \left\langle \half CV^2\right\rangle=\left\langle\half \frac{Q^2}{C}\right\rangle=\half k_B T \]
we then get the same result:
\[ \left\langle Q^2 (t)\right\rangle=C k_B T \]

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

(c)
\[ I=\dot{Q} \]
\[ I(\omega)=-i\omega Q(\omega) \]
therefore
\[ \Phi_I(\omega)=\omega^2\Phi_Q(\omega)=\frac{1}{L}\frac{2k_B T \gamma\omega^2}{\left(\omega^2-\omega_0^2\right)^2+\omega^2\gamma^2} \]
\[ \langle I^2 (t)\rangle = \int_{-\infty}^\infty \Phi_I(\omega)\frac{d\omega}{2\pi}=\frac{2k_B T \gamma}{L} \int_{-\infty}^\infty\frac{\omega^2 d\omega/2\pi}{\left(\omega^2-\omega_0^2\right)^2+\omega^2\gamma^2}=\frac{2k_B T \gamma}{2\gamma L} \]
\[ \langle I^2 (t)\rangle = \frac{k_B T}{L} \]
from equipartition we have
\[ \left\langle \half LI^2 \right\rangle=\half k_B T \]
we then get the same result:
\[ \left\langle I^2 (t)\right\rangle=\frac{k_B T}{L} \]
to get Nyquist's result, we look again at $\Phi_I(\omega)$
\[\Phi_I(\omega)=\frac{1}{L}\frac{2k_B T \gamma}{\omega^2\left(1-\frac{\omega_0^2}{\omega^2}\right)^2+\gamma^2}  \]
it can be seen that for $ \omega\left(1-\frac{\omega_0^2}{\omega^2}\right) << \frac{R}{L} $, this becomes
\[ \Phi_I(\omega)=\frac{2k_B T}{R} \]
which in turn gives
\[ \left\langle I^2\right\rangle_{\omega_1 \leftrightarrow \omega_2}=2\int_{\omega_1}^{\omega_2}\Phi_I(\omega)\frac{d\omega}{2\pi}=\frac{2k_B T}{\pi R}\int_{\omega_1}^{\omega_2}d\omega\]
\[ \left\langle I^2\right\rangle_{\omega_1 \leftrightarrow \omega_2}=\frac{2k_B T}{\pi R}\left(\omega_2 - \omega_1\right)\]

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