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I have an implementation of over-damped Brownian dynamics, with particles that follow the version of the Newtons law where the inertia is absent. This is a common thing to do at micrometer scale.

$m x''(t) = \Sigma F(t) - \gamma x'(t) + \xi (t)$

with no inertia, the equation is reduced to

$ x'(t) = (\Sigma F(t) + \xi (t)) / \gamma$

This equation I solve with a numerical method.

Of course in classical molecular dynamics there is a conserved quantity that allows us to validate the code

But here I lose access to velocity so that I cant calculate the kinetic energy.

Is there any other conserved quantity that can be used in this case instead of energy?

Qmechanic
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user13831679
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  • "Of course in classical molecular dynamics there is a conserved quantity that allows us to validate the code" -> which one? – Quillo Apr 27 '23 at 12:11
  • The energy between particles, $E = 0.5 \Sigma m v^2 + U$ – user13831679 Apr 27 '23 at 12:15
  • Concerning Hamiltonian formulations with dissipation, see e.g. this Phys.SE post. – Qmechanic Apr 27 '23 at 12:16
  • @user13831679 why do you think that $E$ should be conserved for the Langevin equation (overdamped or not)? $m x''(t) = \Sigma F(t) - \gamma x'(t) + \xi (t)$ contains a force, $\xi (t)$, that, as a matter of fact, is an external force not associated to $U$. Therefore, $E$ can not be conserved (even without friction). Am I missing something important? – Quillo Apr 27 '23 at 12:46
  • Well, it seems to me that you also have to switch off the friction $\gamma$! Unfortunately, overdamped means that inertia is negligible wrt friction, so you can not have the overdamped regime when you are supposed to switch off $\gamma$. – Quillo Apr 27 '23 at 12:58
  • Huh, you might be right, to validate langevin dynamics I would have to turn off both friction and noise – user13831679 Apr 27 '23 at 13:10

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