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\exnumber{8484}
\heading{Galvanometer}
\auname{Tomer Dollberg}
{\bf The problem:}
\Dn
A galvanometer can be regarded as a spring-held pointer
that has mass $M$, natural oscillation frequency $\omega_0$,
and a damping coefficient ${\gamma}$.
The position $x$ of the spring indicates the current $I$.
It obeys the equation
%
\[\ddot{x}+\omega_0^2x = -\gamma \dot{x} + A(t) + \alpha I\]
%
where $A(t)$ represents an environmentally induced
white noise that has a spectral intensity $\nu$,
and $\alpha$ is a coupling constant.
\begin {itemize}
\item[(1)] On the basis of the above Langevin equation
write a $d\omega$ integral for the variance $\langle x^2 \rangle$
in the absence of current.
\item[(2)] Based on canonical FDT considerations
deduce what is the result of the integral
that you wrote in the previous item.
\item[(3)] For a constant $I$, what is the average position $\langle x \rangle$ of the pointer?
\item[(4)] Regarding $I$ as a driving source, write what is the conjugate
variable, what is the interaction term $\mathcal{H}_{int}$ in the Hamiltonian,
and what is the associate susceptibility $\chi(\omega)$.
\item[(5)] Write an expression for the average rate of energy absorption $\dot{W}$,
given that the current source has a frequency $\omega$ and RMS amplitude $I_{0}$.
\item[(6)] The expression for $\dot{W}$ is formally the same
as for a current source that is connected to a parallel RLC circuit.
Write expressions for the effective values of $R$ and $L$ and $C$.
\end {itemize}
{\bf Tip:} The equation of a parallel RLC circuit can
be written as ${G(\omega)V_{\omega}=I_{\omega}}$
where ${G(\omega)}$ is a sum of three terms.
Capacitors and inductors are described
by $I=C\dot{V}$ and by $V=L\dot{I}$ respectively.
\Dn
\Dn\Dn
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{\bf The solution:}
\Dn
(1)
In the absence of current the given equation becomes
\begin{equation}
\ddot{x}+\omega_0^2x = -\gamma \dot{x} + A(t)
\end{equation}
We take the Fourier transform of this equation and get
\begin{equation}
-\omega^2 x_{\omega} + \omega_0^2 x_{\omega} = \gamma i \omega x_{\omega} + A_{\omega}
\end{equation}
which gives the response of $x_{\omega}$ to the white noise $A_{\omega}$
\begin{equation}
x_{\omega} = \frac{A_{\omega}}{\omega_0^2 - \omega^2 - i \omega \gamma}
\end{equation}
Using the Wiener-Khinchin theorem this becomes
\begin{equation}\label{W-K}
\Tilde{C}_{x}(\omega) = \frac{\Tilde{C}_{A}(\omega)}{(\omega_0^2-\omega^2)^2 + \omega^2 \gamma^2}
\end{equation}
$A(t)$ is white noise with spectral intensity $\nu$ so its correlation function is
\begin{equation}
C_A(\tau)=\left< A(t_2)A(t_1) \right> = \nu \delta(t_2-t_1)
\end{equation}
Meaning its power spectrum is
\begin{equation}\label{A-power-spectrum}
\Tilde{C}_A(\omega) = \int_{-\infty}^{\infty} dt' \left< A(t')A(0) \right> e^{i \omega t'} = \nu
\end{equation}
plugging \eqref{A-power-spectrum} into \eqref{W-K} and taking the inverse Fourier transform with $t=0$ we get
\begin{equation}
\left< x^2 \right> = C_{x}(t=0)=\int_{-\infty}^{\infty} \Tilde{C}_x(\omega) d\omega = \int_{-\infty}^{\infty} \frac{\nu}{(\omega_0^2-\omega^2)^2 + \omega^2 \gamma^2} d\omega
\end{equation}
\Dn
(2)
Based on canonical considerations, we simply use the law of equipartition
\begin{equation}
\frac{1}{2} m \omega_0 \left< x^2 \right> = \frac{T}{2}
\end{equation}
to get
\begin{equation}
\left< x^2 \right> = \frac{T}{m \omega_0}
\end{equation}
\Dn
(3)
At equilibrium $\ddot{\left< x \right>}=\dot{\left< x \right>}=0$, and $\left< A \right>=0$ by definition, so $\left< x \right>$ can be extracted from the Langevin equation
\begin{equation}
\left< x \right> = \frac{\alpha I }{\omega_0^2}
\end{equation}
\Dn
(4)
From the Langevin equation we can deduce the interaction term to be
\begin{equation}
\mathcal{H} _{int} = -\alpha m x I
\end{equation}
meaning $\alpha m x$ is the conjugate variable of $I$.
Repeating the procedure done in (1) with $I\ne0$, we get
\begin{equation}
\left_{\omega} = \frac{\alpha I_{\omega}}{\omega_0^2 - \omega^2 - i \omega \gamma}
\end{equation}
which implies
\begin{equation}\label{chi}
\chi (\omega) = \frac{m \alpha^2}{\omega_0^2 - \omega^2 - i \omega \gamma}
\end{equation}
\Dn
(5)
For $I(t)=I_0 \cos(\omega t)$, the average rate of energy absorption $\dot{W}$ is given by
\begin{equation}
\dot{W} = \left< -\dot{X} \mathcal{F} \right>_t= \left< -\dot{I} \alpha m x \right>_t = \frac{I_0^2 \omega^2}{2}\eta(\omega) = \frac{I_0^2 \omega}{2} \im[\chi(\omega)]
\end{equation}
and plugging \eqref{chi} in we get
\begin{equation}\label{energy-absorption}
\dot{W} = \frac{I_0^2}{2} \frac{m \alpha^2 \omega^2 \gamma}{(\omega_0^2 - \omega^2)^2 + \omega^2 \gamma^2}
\end{equation}
\Dn
(6)
For each of the components in the parallel RLC circuit we have
\begin{subequations}
\begin{align}
&V_L = L\dot{I}_L \implies I_L(\omega) = \frac{V_L(\omega)}{- i \omega L} \\
&V_R = I_R R \implies I_R(\omega) = \frac{V_R(\omega)}{R}\\
&V_C = \frac{Q_C}{C} \implies I_C(\omega) = - i \omega C V_C(\omega)
\end{align}
\end{subequations}
The sum of all currents must equal that of the source, and since all components are connected in parallel, $V_L=V_C=V_R$, leading to
\begin{equation}
I_{\omega} = \left(- \frac{1}{i \omega L} + \frac{1}{R} - i \omega C \right) V_{\omega}
\end{equation}
Using the relation between charge and current, $\dot{Q}=I$, we rewrite this equation as
\begin{equation}
V_{\omega} = \frac{-i \omega}{- \frac{1}{i \omega L} + \frac{1}{R} - i \omega C} Q_{\omega}
\end{equation}
where $V$ is the generalized force and $Q$ the parameter. Hence $\chi(\omega)$ is
\begin{equation}
\chi(\omega) = \frac{\frac{\omega^2}{C} \left(\omega^2 - \frac{1}{LC}\right) -i\frac{\omega^3}{R C^2}}{\left( \frac{1}{LC} -\omega^2\right)^2 + \frac{\omega^2}{R^2 C^2}}
\end{equation}
The energy absorption is then given by
\begin{equation}
\dot{W} = \left< -\dot{Q} V \right> = \left< -I V \right> = \frac{I_0^2}{2 \omega} \im[\chi(\omega)] = \frac{I_0^2}{2} \frac{\frac{\omega^2}{R C^2}}{\left( \frac{1}{L C} - \omega^2 \right)^2 + \frac{\omega^2}{R^2 C^2}}
\end{equation}
and by comparison with \eqref{energy-absorption} we get the following effective values
\begin{subequations}
\begin{eqnarray}
L = \frac{m \alpha^2}{\omega_0^2} \\
R = \frac{m \alpha^2}{\gamma} \\
C = \frac{1}{m \alpha^2}
\end{eqnarray}
\end{subequations}
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\end{document}