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\exnumber{8483}
\heading{Millikan experiment}
\auname{Noam Wunch}
{\bf The problem:}
\Dn
Consider a Millikan-type experiment whose purpose is to measure the charge~$e$ of
a particle with mass ~$\mass$. The particle is located beteen plates of capacitor,
where the electric field~$\mathcal{E}$ is in the "up" direction,
while the gravitation~$g$ is in the "down" direction.
The distance between the plates is~$L$,
and the temperature of the system is~$T$.
Due to the poor vacuum the particle executes a Brownian motion
that is described by a Langevin equation with friction force~$-\eta v$.
The charge of the electron is estimated
via ${\delta F = e\mathcal{E}-\mass g = 0}$.
In item~(1) the system is prepared with a single particle
in the middle. In item~(3) assume a uniform gas of~$N$ particles.
In both cases the current is integrated during a time interval~$t$,
and the charge $Q=\int I(t')dt'$ is inspected as "readout".
\begin{enumerate}
\item[(1)]
Assuming that $\delta F =0$, determine the time $t_d$ such that
for $t\ll t_d$ it is not likely to get charge readout.
\item[(2)]
What is the $\delta F$ for which the condition $t \ll t_d$
is no longer valid. We shall regard this value, call it $\delta_1$,
as the resolution of the measurement.
\item[(3)]
Assuming that $\delta F =0$, determine the
power spectrum $C(\omega)$ of the current $I(t)$.
\item[(4)]
Assume that the time of the measurement is~$t$.
What is the $\delta F$ for which the condition ${ \sqrt{\text{var}(Q)}
\ll \langle Q \rangle}$
is no longer valid. We shall regard this value, call it $\delta_N$,
as the resolution of the measurement.
\item[(5)]
Express the ratio $\delta_N/\delta_1$ as a function
of $N$ and $t/t_d$.
\end{enumerate}
{\bf Tips:}
In the absence of fluctuations ${\delta F =0}$ is indicated by having zero readout.
In item (3) the ``readout" is a current versus voltage (``IV") measurement, and ${\delta F =0}$ is indicated
by zero current. Due to the fluctuations there is some blurring which determines
the resolution $\delta_N$. In order to calculate the fluctuations in item (3) define
the one-particle current as the velocity (up to a prefactor).
\Dn\Dn
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}
\Dn
(1)
The Langevin equation is a stochastic force equation describing the motion of a particle executing Brownian motion:
\begin{equation}
\dot{v} = -\eta v +f(t)
\end{equation}
Where $\eta$ encompasses all drag forces acting on the particle (in this case collisions) and $f(t)$ is a random force with $ = 0$. Solving this equation for the spreading of the particle yields $\left<(x(0)-x(t))^2\right> = 2Dt$ and, for timescales longer than decorrelation time, D is constant and given by the Einstein relation $D = \frac{T}{\eta}$. It follows, that it would be unlikely to get a charge readout for:
\begin{equation}
\frac{T}{\eta} \cdot t \ll L^2 \longrightarrow t_d = \frac{\eta L^2}{T}
\end{equation}
%%%%%%%%%%%%%%%
\Dn
(2)
When $e \mathcal{E} - mg \neq 0$ a "drift" term is to be added to the Langevin equation:
\begin{equation}
\dot{v} = -\eta v + f(t) + \delta F
\end{equation}
Where $\delta F = e \mathcal{E} -mg$. The average velocity is no longer zero, and is given by $ = \frac {\delta F}{\eta}$. In this case a minimum measurment time $t > \frac{L}{}$ is required to get a reading. But we would also want this time to be shorter than the spreading time $t_d$ we found in the previous item. This leads to the condition:
\begin{equation}
\delta F > \frac{T}{L} \equiv \delta_1
\end{equation}
%%%%%%%%%%%%%%%
\Dn
(3)
The current of a single particle is $I^1 = \frac{e}{L}v$. The power spectrum can be expressed as:
\begin{equation}
<{|I^1_\omega|}^2> = \left(\frac{e}{L}\right)^2 <{|v_\omega|}^2>
\end{equation}
In frequency space, Eq(1) can be rewritten as:
\begin{equation}
v_\omega = (-i\omega +\gamma)^{-1}\frac{f_\omega}{m} \; \left(\gamma = \frac{\eta}{m}\right)
\end{equation}
From which the velocity power spectrum can be easily found:
\begin{equation}
<{|v_\omega|}^2> = (\omega^2 +\gamma^2)^{-1}\frac{C_\omega}{m^2}
\end{equation}
$C_\omega$ in the equation above is the, ensemble averaged, power of $f$ (which is also the correlation function). For white noise, the Einstein relation gives us $C_\omega = 2 m \gamma T$ (refered to as $\nu$ in D.C. notes). The total current is a sum over single particle currents and so the power of the total current will be N times the power from a single particle:
\begin{equation}
<{|I_\omega|}^2> = N\left(\frac{e}{L}\right)^2 \frac{T}{m} \frac{2\gamma}{\omega^2 +\gamma^2}
\end{equation}
%%%%%%%%%%%%%%%
\Dn
(4)
The readout is the total charge $Q = \int_0^{t}I(t')dt'$. For a significant readout we require ${\sqrt{\text{var}(Q)}
\ll \langle Q \rangle}$. $\delta F$ contributes to the RHS:
\begin{equation}
= * t = N\frac{e}{L} = N\frac{e}{L} \frac{\delta F}{\eta} t
\end{equation}
and the current variance contributes to the LHS (it can be calculated assuming $\delta F = 0$):
\begin{equation}
Var(Q) = * = N \left(\frac{e}{L}\right)^2\int_0^{t} \int_0^{t} dt' dt'' = N \left(\frac{e}{L}\right)^2 \frac{2T}{\eta} t
\end{equation}
The condition on $\delta F$ is then:
\begin{equation}
\delta F > \sqrt{\frac{T\eta}{Nt}} \equiv \delta_N
\end{equation}
%%%%%%%%%%%%%%%
\Dn
(5)
The ratio $\frac{\delta_N}{\delta_1}$ can be expressed as a function of $N$ and $t/t_d$:
\begin{equation}
\frac{\delta_N}{\delta_1} = \frac{1}{\sqrt{N\frac{t}{t_d}}}
\end{equation}
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