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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
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\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}