Consider a system of $N$ particles, subject to some interaction potential $U$ (e.g. Lennard-Jones) and to thermal noise. The equation of motion is given using the Langevin equation:

$$m_i \ddot{\bf r}_i = -\nabla U({\bf r}_i) -\lambda\dot{\bf r}_i + {\boldsymbol \eta}_i,$$

where $\lambda$ is the friction coefficient and ${\boldsymbol \eta} = [\eta_{i,x} ~ \eta_{i,y} ~\eta_{i,z}]$ is a vector of random gaussian forces, with $0$ mean and

$$\langle \eta_{i,\alpha}(t), \eta_{j,\beta}(t') \rangle = 2\lambda k_BT \delta_{i,j}\delta_{\alpha, \beta}\delta(t-t').$$

It seems (and maybe I'm wrong!!!) that $T$ is the temperature of the system. Well, I have difficulties to understand this point, since we are assuming that $T$ is a constant parameter (i.e. a thermal bath), but the system evolves over time, gaining (or loosing) kinetic energy, and therefore the temperature is changing over time as well!!!


$$T = \frac{1}{3Nk_B}\sum_{i=1}^N m_i \dot{\bf r}_i^2$$

changes over time according to the evolution of the velocity $\dot{\bf r}_i$.

What am I missing about the definition of this system and the role of the temperature?

  • $\begingroup$ The δ-function form of the correlations means that the force at a time t is assumed to be completely uncorrelated with it at any other time. This is an approximation; the actual random force has a nonzero correlation time corresponding to the collision time of the molecules. $\endgroup$ – drvrm Aug 23 '18 at 20:02
  • $\begingroup$ @drvrm thanks for your comment! Anyway, how does this relate to the temperature problem? $\endgroup$ – the_candyman Aug 23 '18 at 20:21
  • $\begingroup$ @the_candyman-<but the system evolves over time, gaining (or losing) kinetic energy, and therefore the temperature is changing over time as well!!!> your statement drove me to look for reasons-see- en.wikipedia.org/wiki/Langevin_equation $\endgroup$ – drvrm Aug 23 '18 at 20:26
  • $\begingroup$ @drvrm I've already read that page a lot of time. I still don't get the role of the parameter $T$ $\endgroup$ – the_candyman Aug 23 '18 at 20:29
  • $\begingroup$ @drvrm I mean, I don't understand what you are trying to say to me! $\endgroup$ – the_candyman Aug 23 '18 at 20:32

You are completely correct, that $T$ is a fixed parameter that defines the temperature of the system. The mistake is identifying this with the instantaneous "temperature" defined via your equation involving velocities. This equation is only true if you ensemble-average the right hand side, the appropriate ensemble being the canonical ensemble. Incidentally, the Langevin equation can be proven to generate states (coordinates and momenta) sampled from the canonical ensemble at temperature $T$. So, $$T = \left\langle \frac{1}{3N}\sum_{i=1}^N m_i \dot{\bf r}_i^2\right\rangle$$ You should think of this equation not as a definition of $T$, but as a statement of what the ensemble average on the right turns out to be, given that the system is at equilibrium at temperature $T$. The same equation applies to each individual atom $i$, if we omit the summation over $i$ and the normalising factor $N$.

Alternatively, if you consider a single realisation of the Langevin equation, you can consider time-averaging the right hand side of that equation and (subject to some reasonable assumptions about ensemble equivalence, ergodicity and so on) the result will also be $T$.

The time dependent analogue of $T$, which you can call $T(t)$ if you like, although I prefer to emphasise the difference by calling it $\mathcal{T}(t)$ or something similar, is a mechanical variable: a function of coordinates and momenta in general (in this case, just momenta, or velocities). It fluctuates by definition, and is only equal to $T$ after averaging, and in an equilibrium ensemble. You could, with equal validity, define an instantaneous "temperature" for each atom $i$, using a similar formula, omitting the sum and the $1/N$, all fluctuating and all different. However they would all give the same ensemble average.

This distinction is not unique to the Langevin equation. For example, in molecular dynamics using the completely deterministic Nosé-Hoover thermostat, $T$ is again a fixed parameter, and it is possible to define an instantaneous "temperature" from the kinetic energy, with the same equation (more or less) as you have given, whose average is $T$.

  • $\begingroup$ Thanks a lot for your answer (+1)! Then, you are saying that under suitable hypotheses: $$T = \lim_{M \to +\infty}\frac{1}{M}\int_{0}^{M}\mathcal{T}(t)dt,$$ where $$\mathcal{T}(t) = \frac{k_B}{3N}\sum_{i=1}^N m_i \dot{\bf r}_i(t)^2,$$right? $\endgroup$ – the_candyman Aug 24 '18 at 10:59
  • $\begingroup$ Sorry, I am creating confusion with the definitions. The right one is: $$\mathcal{T}(t) = \frac{1}{3Nk_B}\sum_{i=1}^N m_i \dot{\bf r}_i(t)^2.$$ $\endgroup$ – the_candyman Aug 24 '18 at 20:53
  • 1
    $\begingroup$ Yes, that's what I'm trying to say. Implicit in that time-average expression, though, has to be an average over the statistical properties of the random force term. When you solve a stochastic differential equation, they need to be taken into account. This gets a bit circular, though, because the random force term is defined to satisfy the fluctuation-dissipation theorem, and hence to generate the correct average kinetic "temperature". This is why $T$ appears in the expression for $\langle\eta_{i\alpha}(t)\eta_{j\beta}(t')\rangle$. This is all discussed in standard texts, by the way. $\endgroup$ – user197851 Aug 24 '18 at 21:54

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