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\begin{document}
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\exnumber{8484}
\heading{Galvanometer}
\auname{Tamar Simhon}
{\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 $\dot{\mathcal{W}}$,
given that the current source has a frequency $\omega$ and RMS amplitude $I_{0}$.
\item[(6)] The expression for $\dot{\mathcal{W}}$ 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
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{\bf The solution:}
\Dn
(1) In the absence of current the galvanometer equation becomes
\begin{equation}
\ddot{x}+\omega^2_0x=-\gamma\dot{x}+A(t),
\end{equation}
taking the Fourier transform we get
\begin{equation}
-\omega^2x_\omega+\omega^2_0x_\omega=i\gamma\omega x_\omega+A_\omega.
\end{equation}
Solving for $x_\omega$ we get its response to the white noise $A_\omega$
\begin{equation}
x_\omega=\frac{A_\omega}{\omega^2_0-\omega^2-i\gamma\omega}.
\end{equation}
Using the Wiener-Khinchin theorem
\begin{equation}
\tilde{C}_x(\omega)=\frac{\tilde{C}_A(\omega)}{(\omega^2_0-\omega^2)^2+(\gamma\omega)^2}.
\end{equation}
We have defined the correlation function of the stationary stochastic variable as:
\begin{equation}
C(\tau)\equiv C(t_2-t_1)=\langle A(t_2)A(t_1)\rangle=\nu\delta(t_2-t_1)\equiv \nu\delta(\tau),
\end{equation}
thus,
\begin{equation}
\tilde{C}_A(\omega)=\int_{-\infty}^{\infty}C(\tau)e^{i\omega\tau}d\tau=\nu.
\end{equation}
In that case we know that $\langle x\rangle=0$ thus,
\begin{equation}
\text{Var}(x)=\langle x^2\rangle=\int_{-\infty}^{\infty}\tilde{C}_x(\omega)d\omega=\int_{-\infty}^{\infty}\frac{\nu}{(\omega^2_0-\omega^2)^2+(\gamma\omega)^2}d\omega
\end{equation}
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(2) The correlation function $\langle x(t_2)x(t_1)\rangle$ for $t_1=t_2=t$ should be consistent with law of equipartition
\begin{equation}
\frac{1}{2}M\omega^2_0\langle x^2\rangle=\frac{T}{2},
\end{equation}
thus,
\begin{equation}
\langle x^2\rangle=\frac{T}{M\omega^2_0}.
\end{equation}
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(3) Using the galvanometer equation, the fact that $\langle A(t)\rangle=0$ and $\langle \ddot{x}\rangle=\langle \dot{x}\rangle=0$ at equilibrium
\begin{equation}
\langle x\rangle=\frac{I\alpha}{\omega^2_0}.
\end{equation}
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(4) Regarding $I$ as a driving source, we can deduce its interaction term in the Hamiltonian
\begin{equation}
\mathcal{H}_{int}=-\alpha MxI,
\end{equation}
thus $\alpha Mx$ is the conjugate variable of $I$.
Doing the same procedure as in section (1) for $I$ we get
\begin{equation}
x_\omega=\frac{\alpha I_\omega}{\omega^2_0-\omega^2-i\gamma\omega},
\end{equation}
but, the generalized force is $\mathcal{F}=-\frac{\partial\mathcal{H}}{\partial I}=\alpha Mx$, thus
\begin{equation}
\chi(\omega)=\frac{M\alpha^2}{\omega^2_0-\omega^2-i\gamma\omega}.
\end{equation}
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(5) Given $I(t)=Re[I_0e^{-i\omega t}]$, we know that $\dot{\mathcal{W}}=\eta(\omega)\frac{(I_0\omega)^2}{2}$, where $\eta(\omega)\equiv\frac{\text{Im}[\chi(\omega)]}{\omega}$, thus using (13) we get
\begin{equation}
\dot{\mathcal{W}}=\langle -\dot{I}\alpha Mx\rangle=\frac{I^2_0\omega^2}{2}\frac{M\alpha^2\gamma}{(\omega^2_0-\omega^2)^2+(\gamma\omega)^2}.
\end{equation}
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(6) Since this is a parallel RLC circuit $V_R=V_C=V_L$, thus we can write the equation for the current
\begin{equation}
I_R+I_C+I_L=I(t),
\end{equation}
where $I(t)$ is the current source.
Using the tip we can rewrite (15)
\begin{equation}
\frac{1}{R}V(t)+C\dot{V}+\int\frac{V(t)}{L}dt=I(t).
\end{equation}
Taking the Fourier transform we get
\begin{equation}
I_\omega=(\frac{1}{R}-i\omega C-\frac{1}{i\omega L})V_\omega\equiv G(\omega)V_\omega,
\end{equation}
we can use the relation $I=\dot{Q}$ in order to write
\begin{equation}
V_\omega=\frac{-i\omega}{\frac{1}{R}-i\omega C-\frac{1}{i\omega L}}Q_\omega,
\end{equation}
thus the susceptibility is
\begin{equation}
\chi(\omega)=\frac{\frac{\omega^2}{C}(\omega^2-\frac{1}{LC})-i\frac{\omega^3}{RC^2}}{(\frac{1}{LC}-\omega^2)^2+\frac{\omega^2}{R^2C^2}}.
\end{equation}
We can calculate the average rate of energy absorption where $V$ is the generalized force and $Q$ is the parameter
\begin{equation}
\dot{\mathcal{W}}=\langle-\dot{Q}V\rangle=\langle -IV\rangle=\frac{I^2_0}{2}\frac{\frac{\omega^2}{RC^2}}{(\frac{1}{LC}-\omega^2)^2+\frac{\omega^2}{R^2C^2}}.
\end{equation}
With analogue to (14) we can find the effective $R$,$L$ and $C$
\begin{equation}
L=\frac{M\alpha^2}{\omega^2_0},
\end{equation}
\begin{equation}
R=\frac{M\alpha^2}{\gamma},
\end{equation}
\begin{equation}
C=\frac{1}{M\alpha^2}.
\end{equation}
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\end{document}