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\exnumber{8481}
\heading{Mass on a spring} 
\auname{Gabi Socolovsky}

{\bf The problem:}

\Dn

A balance for measuring weight consists of a sensitive spring which
hangs from a fixed point. The spring constant is ${K}$. The balance
is at temperature ${T}$ and gravity acceleration is ${g}$ in the ${x}$
direction. A small mass ${\mass}$ hangs at the end of the spring. There
is an option to apply an external force $F(t)$, to which $x$ is
conjugate or apply an external vector potential $A(t)$.

\begin{itemize}

\item [{{(a)}}] Find the partition function $Z$.
 
\item [{{(b)}}] Find $\langle x \rangle$ and $\langle x^{2}\rangle$ and $\mbox{Var}(x)$.

\item [{{(c)}}] Write a Langevin equation for $x(t)$, with friction $\gamma$, and a random force $f(t)$. 

\item [{{(d)}}] Assuming $\langle f(t)f(0)\rangle=C\delta(t)$, find $\mbox{Var}(x)$, 
and deduce what is $C$ by comparing with the canonical result.  

\item [{{(e)}}] 
Assuming $x$ is measured in the lab by averaging over time period $t_0$, 
what is the minimal mass that can be meaningfully measured?

\item [{{(f)}}] Describe the external force $F(t)$ by a scalar potential and demonstrate FDT. 

\item [{{(g)}}] Describe the external force $F(t)$ by a vector potential and demonstrate FDT.

\end{itemize}

\Dn

Note: $\int \frac{d\omega}{(\omega^2-\omega_0^2)^2+\gamma^2\omega^2}=\frac{\pi}{\gamma \omega_0^2}$.

\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn


(a) The Hamiltonian of the system is:
\begin{equation}
H=\frac{p^{2}}{2m}+\frac{1}{2}Kx^{2}-mgx
\end{equation}
We rewrite the Hamiltonian in the following way \\
\begin{equation}
H=\frac{p^{2}}{2m}+\frac{1}{2}K(x-x_{0})^{2}+const
\end{equation}
where $x_{0}=\frac{mg}{K}$.
Therefore, the partition function is
\begin{equation}
Z=\frac{1}{\lambda_{T}}\int_{-\infty}^{\infty}{e^{-\beta\frac{K}{2}(x-x_{0})^2}\times const}
\end{equation}
(b) The Gaussian in the partition function is centered around $x_{0}$, therefore we deduce that\\
\begin{equation}
\langle x \rangle=x_{0}
\end{equation}
In order to find $\Var(x)$ we use the equipartition theorem -  $\langle x\frac{\partial H}{\partial x} \rangle=T$, \\
\begin{equation}
\langle xK(x-x_{0})\rangle=T
\end{equation}
\begin{equation}
\langle x^2\rangle=\frac{T}{K}+x_{0}^2
\end{equation}
The variance of $x$ is
\begin{equation}
\Var (x)=\langle x^2 \rangle - \langle x \rangle^2=\frac{T}{K}
\end{equation}

(c) The Langevin equation for this system is:\\
\begin{equation}
m\ddot{x}+\gamma\dot{x}+Kx-mg=f(t)
\end{equation}\\
(d) Redefining $x$ around the equilibrium, $x\rightarrow x-x_0$, so that in equilibrium $\langle x \rangle=0$, Eq. (8) becomes\\
\begin{equation}
m\ddot{x}+\gamma\dot{x}+Kx=f(t)
\end{equation}
Fourier transforming Eq. (9) 
\begin{equation}
(-m\omega^2-i\gamma\omega+K)x_{\omega}=f_{\omega}
\end{equation}
Using the Weiner-Khinchin theorem, we get\\
\begin{equation}
\tilde{C}_{xx}(\omega)=\frac{\tilde{C}_{ff}(\omega)}{(-\omega
^2 m+K)^2+(\gamma\omega)^2}
\end{equation}\\ 
where the Fourier transform of the force-correlation is $\tilde{C}_{ff}(\omega)=C$. \\
Using Eq. (11) we calculate $\Var(x)$ -
\begin{equation}
\Var(x)=C_{xx}(t=0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\tilde{C}_{xx}(\omega)d\omega}=\frac{1}{2\pi}\frac{C}{m^2}\int_{-\infty}^{\infty}{\frac{d\omega}{(\omega ^2-\frac{K}{m})^2+(\frac{\gamma\omega}{m})^2}}=\frac{C}{2\gamma K}
\end{equation}
Finally, comparing the result with with Eq (7), we get 
\begin{equation}
C=2\gamma T
\end{equation}
Hence, Eq. (11) becomes\\
\begin{equation}
\tilde{C}_{xx}(\omega)=\frac{2\gamma T}{m^2[(\omega
^2 -\frac{K}{m})^2+(\frac{\gamma\omega}{m})^2]}
\end{equation}

(e) We measure $x(t)$ in the lab and average over time period $t_{0}$. This measurement introduces a new random variable $X$ -\\
\begin{equation}
X=\frac{1}{t_{0}}\int_{0}^{t_{0}}{x(t)dt}
\end{equation}
This variable has a mean value $\langle X \rangle$ and standard deviation $\sigma(X)$. In order that the measured value of $X$ will be within the resolution
of the measurement, it has to obey the condition
- $\langle X \rangle \gg \sigma(X)$. From this condition, the minimal mass $m_\text{min}$ will be found. The mean value of $X$ is  
\begin{equation}
\langle X\rangle=x_{0}
\end{equation}     
In order to find $\sigma(X)$ we subtract the mean value from $X$ 
\begin{equation}
X-x_{0}=\frac{1}{t_{0}}\int_{0}^{t_{0}}{\big(x(t)-x_{0}\big)dt}
\end{equation}
Redefining $x\rightarrow x-x_{0}$ yields
\begin{equation}
\sigma^2(X)=\frac{1}{t_{0}^2}\int_{0}^{t_{0}}{dt'}\int_{0}^{t_{0}}{dt''\langle x(t')x(t'')\rangle}=\frac{1}{2t_{0}}\tilde{C}_{xx}(\omega=0)=\frac{T\gamma}{t_{0}K^2}
\end{equation}
where we used $C_{xx}(t'-t'')=\langle x(t')x(t'')\rangle$.\\ 
Finally, the minimal mass is given by 
\begin{equation}
\langle X \rangle \gg \sigma(X)
\end{equation}
\begin{equation}
m \gg \frac{1}{g}\sqrt{\frac{T\gamma}{t_{0}}}=m_\text{min}
\end{equation}

(f) The force $F(t)$ is described by a scalar potential $\varepsilon$, so the conjugate variables are $x$ and $\varepsilon$.\\ Averaging the Langevin formula\\  
\begin{equation}
m\langle \ddot{x}\rangle +\gamma\langle \dot{x}\rangle+K\langle x\rangle=\varepsilon
\end{equation} 
and Fourier transforming Eq. (22), we get\\
\begin{equation}
x_{\omega}=\chi_{\omega} \varepsilon_{\omega}
\end{equation}
where $\chi_{\omega}=\frac{1}{(-m\omega^2+K)-i\omega\gamma}$.
Hence, we get the correlation function\\
\begin{equation}
\tilde{C}_{xx}(\omega)=\frac{2T}{\omega}\operatorname{Im}\{\chi_{\omega}\}=\frac{2\gamma T}{m^2[(\omega^2-\frac{K}{m})^2+(\frac{\gamma\omega}{m})^2]}
\end{equation}
This result is exactly the same as Eq. (14). 
\Dn

(g) The force $F(t)$ is described by a vector potential $A(t)$, so the conjugate variables are $v$ and $A$. The averaged Langevin formula becomes\\
\begin{equation}
m\langle \dot{v}\rangle +\gamma\langle v\rangle+K\langle x\rangle=-\dot{A}
\end{equation}
Fourier transforming Eq. (26), we get\\
\begin{equation}
v_{\omega}=\chi_{\omega}A_{\omega}
\end{equation}
where $\chi_{\omega}=\frac{\omega^2}{-m\omega^2-i\gamma\omega+K}$. The correlation function is
\begin{equation}
\tilde{C}_{vv}(\omega)=\frac{2T}{\omega}\operatorname{Im}\{\chi_{\omega}\}=\frac{2\gamma T \omega^2}{(K-m\omega^2)^2+(\gamma\omega)^2}
\end{equation}
Next, we calculate $\Var(v)$\\
\begin{equation}
\Var(v)=C_{vv}(t=0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\tilde{C}_{vv}(\omega)d\omega}=\frac{T}{m}
\end{equation}     
where we have used the identity $\int \limits _{-\infty}^{\infty}{\frac{x^2}{(x^2-a^2)^2+(bx)^2}dx}=\frac{\pi}{b}$. This result coincides with the equipartition theorem - $\frac{m\langle v^2\rangle}{2}=\frac{T}{2}$. 
\end{document}
