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A Brownian particle of mass $m$ in a heat bath at temperature $T_{1}$ can be described by:

$$\frac{\mathrm dv}{\mathrm dt} = -\frac{\gamma}{m}v + \frac{1}{m}\xi .$$

However, if I assume that a Brownian particle of mass $m$ is experiencing the simultaneous action of two heat baths at temperatures $T_1$ and $T_2$, where this action may be characterized by friction coefficients $\gamma_{1}$ and $\gamma_{2}$ and white Gaussian noises $\xi_{1}(t)$ and $\xi_{2}(t)$, I am unable to find the expression for the probability distribution and write the correct motion equation.

I wrote something like: $$\frac{\mathrm dv}{\mathrm dt} = -\frac{\gamma_{1} + \gamma_{2}}{m}v + \frac{1}{m}\xi_{1} + \frac{1}{m}\xi_{2},$$ but with this equation of motion I can't reach to the velocity probability distribution that I'm supposed to:

$$p(v) \propto \exp\left[\frac{-mv^{2}}{2k_{B}T_\text{ef}} \right] ,$$ where

$$T_\text{ef} = \frac{\gamma_{1}T_{1} + \gamma_{2}T_{2}}{\gamma_{1} + \gamma_{2}} .$$

What am I missing?

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  • $\begingroup$ Oh sorry, I have some typos. Let me correct it. $\endgroup$
    – RKerr
    Aug 17, 2022 at 13:22
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    $\begingroup$ It should work similarly. How do you get p for the normal case? $\endgroup$
    – Kuhlambo
    Aug 17, 2022 at 13:34
  • $\begingroup$ That's where I'm shaky. I tried to find the velocity probabily distribution using the Focker-Planck equation, but I was unsuccessful. I think I corrected everything now. $\endgroup$
    – RKerr
    Aug 17, 2022 at 13:40
  • $\begingroup$ Do you still think that the noise terms are wrong? That's okay, thank you for helping! $\endgroup$
    – RKerr
    Aug 17, 2022 at 13:46
  • $\begingroup$ No i think that is better, sorry. Zwanzig s first chapter on that is confusing but it's not nothing look at that.. $\endgroup$
    – Kuhlambo
    Aug 17, 2022 at 13:50

3 Answers 3

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The Langevin Equation for this system can be written as follows (assuming $m = 1$):

\begin{align} \dot{v} &= - \left(\gamma_{1}v + \gamma_{2}v\right)+ \xi_{1}(t) + \xi_{2}(t) \\ &= - \gamma v + \xi(t) \tag{$\star$} \end{align} where the intensity of the white noise: $$\langle \xi_{i}(t) \xi_{j}(t^{\prime}) \rangle = 2k_{B}T\gamma_{i}\delta_{ij}\delta\left(t - t^{\prime} \right) \, .$$

For convenience, let us also define $\nu = k_{B}\left(2\gamma_{1}T_{1} + 2\gamma_{2}T_{2} \right) \, .$

Since the random force of the system is white noise, the probability distribution will follow a Gaussian one. Therefore,

$$p(v) \propto \exp\left[-\frac{1}{2}\left(\frac{v - \mu}{\sigma} \right)^{2}\right] \tag{$\square$}$$ where $\sigma^{2} = \langle v^{2} \rangle - \langle v \rangle ^{2}$ and $\mu = \langle v \rangle$.

Integrating $(\star)$, one can write:

\begin{align} v(t) &= \int_{0}^{t}dt^{\prime}\xi(t^{\prime}) \exp \left[-\gamma \left(t - t^{\prime} \right) \right] \\ &= \int_{-\infty}^{t}dt^{\prime}\xi(t^{\prime}) \exp \left[-\gamma \left(t - t^{\prime} \right) \right] + c \rightarrow c \\ &= v(-\infty) = 0 \end{align}

Therefore,

$$\langle v \rangle = \int_{-\infty}^{t}dt^{\prime}\langle\xi(t^{\prime})\rangle \exp \left[-\gamma \left(t - t^{\prime} \right) \right] = 0$$

\begin{align} \langle v^{2} \rangle &= \int_{-\infty}^{t}\int _{-\infty}^{t}dt^{\prime}dt^{\prime \prime}\langle\xi(t^{\prime})\xi(t^{\prime \prime})\rangle \exp \left[-\gamma \left(t - t^{\prime} \right) \right]\exp \left[-\gamma \left(t - t^{\prime \prime} \right) \right] \\ &= \int_{-\infty}^{t}\int_{-\infty}^{t}dt^{\prime}dt^{\prime \prime} \exp \left[-\gamma \left(t - t^{\prime} \right) \right]\exp \left[-\gamma \left(t - t^{\prime \prime} \right) \right] \nu \delta \left( t^{\prime} - t^{\prime \prime} \right) \\ &= \frac{\nu}{2\gamma} = k_{B}\frac{\gamma_{1}T_{1} + \gamma_{2}T_{2}}{\gamma_{1} + \gamma_{2}} \equiv k_{B}T_\text{eff} \end{align}

Finally, substituting in $(\square)$:

$$p(v) \propto \exp\left[-\frac{1}{2}\left(\frac{v - \mu}{\sigma} \right)^{2}\right] = \exp\left[\frac{-v^{2}}{2k_{B}T_\text{eff}} \right] \, ,$$ as we wanted.


Final note:

If we did not assume $m=1$, we would arrive at the same conclusion. However, the probability distribution would have the mass term:

$$p(v) \propto \exp\left[-\frac{mv^{2}}{2k_{B}T_\text{eff}} \right] $$

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  • $\begingroup$ I think you dropped a $k_B$ in your white noise autocorrelation. $\endgroup$
    – Kyle Kanos
    Aug 25, 2022 at 21:23
  • $\begingroup$ You're right, thank you for your answer! $\endgroup$
    – RKerr
    Aug 25, 2022 at 22:00
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For the two heat baths, you can write, $$m\frac{\mathrm{d}v}{\mathrm{d}t}=-\gamma_{12}v+\xi_{12}\tag{1}$$ where $\gamma_{12}=\gamma_1+\gamma_2$ and $\xi_{12}=\xi_1(t)+\xi_2(t)$. Note that the autocorrelation of the two noises gives, $$\langle\xi_i(t)\xi_j(t')\rangle=2k\gamma_iT_i\delta_{ij}\delta(t-t')=2k(\gamma_1T_1+\gamma_2T_2)\delta(t-t')=2kT_{12}\delta(t-t')$$ Equation (1) follows the same derivation as standard Langevin equation, leading to a mean square velocity of, $$\langle v^2\rangle=v^2(0)\mathrm{e}^{-2\gamma_{12}t/m}+\frac{kT_{12}}{\gamma_{12}m}\left(1-\mathrm{e}^{-2\gamma_{12}t/m}\right)$$ We expect equilibrium in the long time limit, so our exponential terms will drop to zero, leading us to, $$m\langle v^2\rangle=\frac{kT_{12}}{\gamma_{12}}=k\frac{\gamma_1T_1+\gamma_2T_2}{\gamma_1+\gamma_2}=kT_\text{eff}$$ And which we use in the probability distribution function to get the expected relation, $\log(p)\sim -mv^2/kT_\text{eff}$.

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If this hint helps solve the exercise tell me, otherwise ignore.

For long times without noise the velocity would decay to zero, but this kind of gaussian noise will eventually cause the velocity to have a determined mean and variance, which might tell you all you need to know.

That would be my ansatz, so to speak. Try Zwanzig chapter 1, for the case with one white noise term.

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