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\begin{document}
\heading{E8481: Mass on a spring } \auname{Michaele Epshtein}

\textbf{The problem:}\\


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 ${x}$
direction. A small mass ${m}$ 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 $A(t)$.

\Dn 
\begin{description}
\item [{{(a)}}] Find the partition function $Z.$ 
\item [{{(b)}}] Find the average $\langle x(t)\rangle$ and $\langle x^{2}(t)\rangle.$ 
\item [{{(c)}}] Find the fluctuation $\langle\delta x^{2}\rangle=\langle(x-\langle x\rangle)^{2}\rangle$,
what is the minimal m which can be meaningfully measured? 
\item [{{(d)}}] Write the Langevin equation for $x(t)$ with friction
$\gamma$ and a random force $f(t)$. 
\item [{{(e)}}] Assuming $\langle f(t)f(0)\rangle=C\delta(t)$, Find
$\langle\tilde{x}^{2}(t)\rangle$ and the intensity of the random
force $f(t)$ that acts on the mass , from (b) find the coefficient
C. 
\item [{{(f)}}] Describe the external force by a scalar potential and
demonstrate FDT. 
\item [{{(g)}}] Describe the external force by a vector potential and
demonstrate FDT.
\end{description}
\Dn

\textbf{The solution:}\\


\noindent \begin{flushleft}
(a) The Hamiltonian of the system is: 
\par\end{flushleft}

\[
H=\frac{p^{2}}{2m}+\frac{1}{2}Kx^{2}-mgx\]


\[
Z=\frac{1}{2\pi\hbar}\underset{-\infty}{\overset{+\infty}{\int}}e^{-\frac{\beta p^{2}}{2m}}\, dp\underset{-\infty}{\overset{+\infty}{\int}}e^{-(\frac{\beta Kx^{2}}{2}-mg\beta x)}\, dx=\frac{1}{2\pi\hbar}\sqrt{\frac{2m\pi}{\beta}}\underset{-\infty}{\overset{+\infty}{\int}}e^{-(y^{2}-\sqrt{\frac{2\beta}{K}}mgy)}\, dy={\{y=\sqrt{\frac{\beta K}{2}}x\}}\]


\[
Z=\frac{1}{2\pi\hbar}\underset{-\infty}{\overset{+\infty}{\int}}e^{-\frac{\beta p^{2}}{2m}}\, dp\underset{-\infty}{\overset{+\infty}{\int}}e^{-(\frac{\beta Kx^{2}}{2}-mg\beta x)}\, dx=\frac{1}{2\pi\hbar}\sqrt{\frac{2m\pi}{\beta}}\underset{-\infty}{\overset{+\infty}{\int}}e^{-(y^{2}-\sqrt{\frac{2\beta}{K}}mgy)}\, dy={\{y=\sqrt{\frac{\beta K}{2}}x\}}\]


\[
=\frac{1}{\hbar\beta}\sqrt{\frac{m}{K\pi}}\underset{-\infty}{\overset{+\infty}{\int}}e^{-(y-\sqrt{\frac{\beta}{2K}}mg)^{2}}\, e{}^{\frac{\beta m^{2}g^{2}}{2K}}\, dy=\frac{1}{\hbar\beta}\sqrt{\frac{m}{K}}\, e^{\frac{\beta m^{2}g^{2}}{2K}}\]


\noindent (b) By definition of $\langle x\rangle$

\[
\langle x(t)\rangle=\frac{1}{Z}\frac{1}{2\pi\hbar}\underset{-\infty}{\overset{+\infty}{\int}}e^{-\frac{\beta p^{2}}{2m}}\, dp\underset{-\infty}{\overset{+\infty}{\int}x\,}e^{-(\frac{\beta Kx^{2}}{2}-mg\beta x)}\, dx\]


\[
=\sqrt{\frac{\beta K}{2\pi}}\, e^{-\frac{\beta m^{2}g^{2}}{2K}}\{-\frac{1}{\beta K}e^{-(\frac{\beta Kx^{2}}{2}-mg\beta x)}\underset{-\infty}{\overset{+\infty}{|}}+\frac{mg}{K}\underset{-\infty}{\overset{+\infty}{\int}}e^{-(\frac{\beta Kx^{2}}{2}-mg\beta x)}dx\}\]


The first part of the integral goes to zero and we get

\[
\langle x\rangle=\cfrac{mg}{k}\]


From the virial theorem$\langle x\frac{\partial H}{\partial x}\rangle=k_{B}T$

\[
\langle x(Kx-mg)\rangle=K\langle x^{2}\rangle-mg\langle x\rangle=K\langle x^{2}\rangle-mg(\frac{mg}{K})=k_{B}T\]


\noindent \[
\langle x^{2}\rangle=\frac{k_{B}T}{K}+(\frac{mg}{K})^{2}\]


\noindent \begin{flushleft}
(c) $\langle\delta x^{2}\rangle=\langle(x-\langle x\rangle)^{2}\rangle=\langle x^{2}\rangle-\langle x{}^{2}\rangle=\cfrac{k_{B}T}{K}$ 
\par\end{flushleft}

The Fluctuation of the spring should be much smaller than the tension
itself, Therefor

\[
\langle x\rangle^{2}\gg\langle\delta x^{2}\rangle\;\;\rightarrow\;\;\cfrac{k_{B}T}{K}\ll\left(\cfrac{mg}{k}\right)^{2}\]


And, finally, we get following condition: $m\gg\sqrt{\frac{Kk_{B}T}{g^{2}}}$

\noindent \begin{flushleft}
(d) We use Hamilton equations: 
\par\end{flushleft}
\begin{enumerate}
\item $\frac{\partial H}{\partial p}=\frac{p}{m}=\dot{x}$ 
\item $\frac{\partial H}{\partial x}=Kx-mg=-\dot{p}=-m\ddot{x}$ 
\end{enumerate}
from the two equations, friction $\gamma$ that interact with velocity
and a random force $A(t)$ we find a Langevin's equation (the system
is in equilibrium)

\[
m\ddot{x}+m\gamma\dot{x}+Kx-mg=mf(t)\]


\noindent \begin{flushleft}
(e) By set $\tilde{x}(t)=x(t)+\langle x(t)\rangle$into the Langevin's
equation that we got and using Fourier transform functional relationships
$\frac{\partial^{n}}{\partial t^{n}}x(t)=(i\omega)^{n}X(\omega)$
we get 
\par\end{flushleft}
\begin{lyxlist}{00.00.0000}
\item [{{$(-m\omega^{2}+i\omega m\gamma+K)\tilde{X}(\omega)=mf(\omega)$}}] we
rearrange a little to have 
\end{lyxlist}
\[
\cfrac{|\tilde{X}(\omega)|^{2}}{|f(\omega)|^{2}}=\cfrac{1}{\omega^{2}}\cfrac{|\tilde{V}(\omega)|^{2}}{|f(\omega)|^{2}}=\cfrac{m^{2}}{(m\omega^{2}-k)^{2}+m^{2}\omega^{2}\gamma^{2}}\]


The intensity of the random force $A(t)$ that acts on the mass

\[
\nu_{A}(\omega)=\underset{-\infty}{\overset{+\infty}{\int}}C_{A}(t)\, e^{i\omega t}dt=\underset{-\infty}{\overset{+\infty}{\int}}C\delta(t)e^{i\omega t}dt=C\;\; where\;\; C_{A}(t)=\langle f(t)f(0)\rangle\]


From the following relation $\cfrac{\nu_{x}(\omega)}{\nu_{A}(\omega)}=\cfrac{\tilde{|X}(\omega)|^{2}}{|f(\omega)|^{2}}\;\;\rightarrow\;\;\nu_{x}(\omega)=\cfrac{C}{(\omega^{2}-k/m)^{2}+\omega^{2}\gamma^{2}}$ 
\begin{lyxlist}{00.00.0000}
\item [{{$\langle\tilde{x}(t)\tilde{x}(t+\tau)\rangle=\underset{-\infty}{\overset{+\infty}{\int}}\nu_{x}(\omega)e^{-i\omega\tau}d\omega\;\;\rightarrow\;\;\langle\tilde{x}^{2}(t)\rangle=\underset{-\infty}{\overset{+\infty}{\int}}\nu_{x}(\omega)d\omega=\frac{1}{2\pi}\int\frac{d\omega}{(\omega^{2}-K/m)^{2}+\gamma^{2}\omega^{2}}=\dfrac{m\pi C}{2\pi\gamma K}$}}] ~ 
\end{lyxlist}
By comparing this result with (b) we can find $C$

\[
\langle\tilde{x}^{2}(t)\rangle=\langle\left(x-\langle x\rangle\right)^{2}\rangle=\langle x^{2}\rangle-\langle x\rangle^{2}=\cfrac{k_{B}T}{K}\]


\[
\langle\tilde{x}^{2}(t)\rangle=\dfrac{m\pi C}{2\pi\gamma K}=\cfrac{k_{B}T}{K}\;\;\rightarrow\;\; C=\cfrac[l]{2k_{B}T\gamma}{m}\]


and finelly

\[
\nu_{x}(\omega)=\cfrac{2k_{B}T\gamma m}{(\omega^{2}-k/m)^{2}+\omega^{2}\gamma^{2}}\]


\[
\nu_{v}(\omega)=\cfrac{2k_{B}T\gamma m\omega^{2}}{(\omega^{2}-k/m)^{2}+\omega^{2}\gamma^{2}}\]


(f) Because of the external force that couples to $x$ the system
isn't in equilibrium, namely there is dissipation

\[
H=\frac{p^{2}}{2m}+\frac{1}{2}Kx^{2}-mgx+F(t)x\]


\noindent \begin{flushleft}
We use Hamilton equations: 
\par\end{flushleft}
\begin{enumerate}
\item $\frac{\partial H}{\partial p}=\frac{p}{m}=\dot{x}$ 
\item $\frac{\partial H}{\partial x}=Kx-mg+F=-\dot{p}=-m\ddot{x}$ 
\end{enumerate}
from the two equations, friction $\gamma$ that interact with velocity
and a random force $f(t)$ like in (d) we find a Langevin's equation

\[
m\ddot{x}+m\gamma\dot{x}+Kx-mg=-F+mf(t)\]


By same way as in (e) we get $\;(m\omega^{2}-i\omega m\gamma-K)\tilde{X}(\omega)=F(\omega)\;\;$
we rearrange a little to get the susceptibility $\chi_{x}$ that describes
the linear response of $x$ to $F$

\[
\cfrac{\langle\tilde{X}(\omega)\rangle}{F(\omega)}=\cfrac{1}{m\omega^{2}-K-i\omega m\gamma}=\chi_{x}\]


\[
Im\,\chi_{x}=\cfrac{m\omega\gamma}{(m\omega^{2}-k)^{2}+m^{2}\omega^{2}\gamma^{2}}\]


By compering with $\nu_{x}(\omega)$ we show that FDT holds. 

\[
\nu_{x}(\omega)=\cfrac{2k_{B}T}{\omega}Im\,\chi_{x}=\cfrac{2k_{B}Tm\gamma}{(m\omega^{2}-k)^{2}+m^{2}\omega^{2}\gamma^{2}}\]


(g) Because of the external vector potential A(t) that couples to
$v$ the system isn't in equilibrium, namely there is dissipation

\[
H=\frac{p^{2}}{2m}+\frac{1}{2}Kx^{2}-mgx+\cfrac{p}{m}A(t)\]


\noindent \begin{flushleft}
We use Hamilton equations: 
\par\end{flushleft}
\begin{enumerate}
\item $\frac{\partial H}{\partial p}=\frac{p}{m}+\frac{A}{m}=\dot{x}$ 
\item $\frac{\partial H}{\partial x}=Kx-mg=-\dot{p}=-m\ddot{x}+\dot{A}$ 
\end{enumerate}
from the two equations, friction $\gamma$ that interact with velocity
and a random force $f(t)$ like in (d) we find a Langevin's equation

\[
m\ddot{x}+m\gamma\dot{x}+Kx-mg=\dot{A}+mf(t)\]


By same way as in (e) we get $\;(-m\omega^{2}-i\omega m\gamma+K)\tilde{X}(\omega)=i\omega A(\omega)\;\;$
we rearrange a little to get the susceptibility $\chi_{v}$ that describes
the linear response of $v$ to $A$

\[
\cfrac{\langle\tilde{V}(\omega)\rangle}{A(\omega)}=i\omega\cfrac{\langle\tilde{X}(\omega)\rangle}{A(\omega)}=\cfrac{\omega^{2}}{m\omega^{2}-K-i\omega m\gamma}=\chi_{v}\]


\[
Im\,\chi_{v}=\cfrac{m\omega^{3}\gamma}{(m\omega^{2}-k)^{2}+m^{2}\omega^{2}\gamma^{2}}\]


By compering with $\nu_{v}(\omega)$ we show that FDT holds. 

\[
\nu_{x}(\omega)=\cfrac{2k_{B}T}{\omega}Im\,\chi_{x}=\cfrac{2k_{B}Tm\gamma\omega^{2}}{(m\omega^{2}-k)^{2}+m^{2}\omega^{2}\gamma^{2}}\]

\end{document}
