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\heading{E8025: Thermal flow via a Brownian particle (Exam 2011)}
\auname{Harel Hecht}
{\bf The problem:}
\Dn
A Brownian particle in one dimension that has mass ${\mass=1}$,
is in contact with two baths:
A hot bath that has temperature $T_2$
that induces friction with coefficient $\gamma_2$,
and a cold bath that has temperature $T_1$ that
induces friction with coefficient $\gamma_1$.
Accordingly the motion of the particle is described
by a Langevin equation that includes two friction
terms and two independent white noise terms $f_1(t)$ and $f_2(t)$.
The purpose of this question is to calculate
the rate of heat flow $\dot{Q}$ from the hot to the cold bath.
\Dn
{\em Note:} Each bath exerts on the particle a force that
has two components: a systematic "friction" component
plus a fluctuating component. The rate of heat flow~$\dot{Q}$
equals the rate of work which is done by the force
that is exerted on the particle by the hot bath.
In steady state, on the average, it equals in absolute value
to the rate of work which is done by the force that is exerted
on the particle by the cold bath.
\Dn
(1) Write the Langevin equation for the velocity $v(t)$.
Specify the intensity of the noise terms.
\Dn
(2) Find the steady state value of $\langle v^2 \rangle$.
\Dn
(3) Express the instantaneous $\dot{Q}$
at time $t$, given $v(t)$ and $f_2(t)$.
\Dn
(4) Find an expression for $\langle\dot{Q}\rangle$ at steady state.
\Dn
\Dn\Dn
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{\bf The solution:}
\Dn
\Dn
(1) Langevin equation for the velocity is \[\dot{v}=\gamma_1v+\gamma_2v+f_1(t)+f_2(t)\] where the intensity for white noise \[\langle f_i (t') f_j (t'') \rangle = 2T\gamma_i \delta_ij \delta(t'-t'') \] we define \[ \nu=\nu_1+\nu_2=2\gamma_1T_1+2\gamma_2T_2\]
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\Dn
(2) For \[\dot{v}=\gamma v+f(t)\] we get \[v(t)= \int_{0}^{t}dt'f(t')e^{-\gamma(t-t')} = \int_{-\infty}^{t}dt'f(t')e^{-\gamma(t-t')} + c \rightarrow c=v(-\infty)=0\] and now
\[\langle v^2\rangle =\int_{-\infty}^{t}\int_{-\infty}^{t}dt'dt''e^{-\gamma(t-t')}e^{-\gamma(t-t'')} \underbrace{\langle f(t')f(t'') \rangle}_{\nu \delta(t'-t'')} = \lbrace t\rightarrow\infty \rbrace =\frac{\nu}{2\gamma}=\frac{\gamma_1T_1+\gamma_2T_2}{\gamma_1+\gamma_2}\]
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\Dn
(3) \[\dot{Q}=\dot{W}=Fv=\lbrace T_2 > T_1 \rbrace =F_2v(t) =(\gamma_2v(t)+f_2(t))v(t)\]
\Dn
(4)\[\langle v(t)f_2(t) \rangle =\int_{0}^{t}dt'e^{-\gamma(t-t')} (\underbrace{\langle f_2(t)f_1(t') \rangle}_{0} + \underbrace{\langle f_2(t)f_2(t') \rangle}_{\nu_2 \delta(t-t')}) =\nu_2*\frac{1}{2} \]
\[ \langle \dot{Q}\rangle = \langle (\gamma_2v(t)+f_2(t))v(t) \rangle = \gamma_2 \langle v(t)^2 \rangle + \langle v(t)f_2(t) \rangle = \frac{\gamma_1 \gamma_2}{\gamma_1 +\gamma_2} (T_2-T_1)\] for \[ T_2=T_1 \rightarrow \langle \dot{Q}\rangle =0\]
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