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Is there a derivation of the Einstein–Smoluchowski relation without the assumption of the Boltzmann distribution? Every time I see a derivation, it always assumes the Boltzmann distribution, such as the derivation on wikipedia.

My problem is that the causality is upside down. For me, the Boltzmann distribution is caused by the dynamic motion of the atoms, not that the dynamic motion of the atoms is caused by the Boltzmann distribution. I have seen a claim here that it's a detailed balance, but even here they used the Boltzmann distribution assumption. I don't understand the Fluctuation-dissipation theorem, but there should be simple explanation for this.

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For me, the Boltzmann distribution is caused by the dynamic motion of the atoms, not that the dynamic motion of the atoms is caused by the Boltzmann distribution.

Caused is a dangerous word here. Rather the Boltzmann distribution tells you statistical properties of the motion. Therefore, nothing is wrong with deriving results about the statistics of the motion from the Boltzmann distribution: The Brownian particles will be in thermal equilibrium with the fluid, therefore they will follow a Boltzmann distribution and from that we get the above relation.

A more microscopical method to get the relations for Brownian motion are Langevin equations (statistical differential equations). There you model the influence of the thermal bath by a stochastic force. In this model you can observe how the systems reaches thermal equilibrium from some initial conditions. The relation in question is reached in the formalism by assuming a time scale large compared to the damping time (which is equivalent to the assumption of thermal equilibrium). Langevin's original derivation also assumed the equilibrium by using the virial theorem in the form $\frac 1 2 m\langle \ddot x \rangle^2 = \frac 1 2 \langle T \rangle$.

The bottom line is: The Einstein relation is a result about the kinematics of Brownian particles in equilibrium, therefore it is necessary to require the particles be in equilibrium to derive it.

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  • $\begingroup$ Thank you. Of course, you need to assume that it's in equilibrium, but don't need to use the Boltzmann distribution explicitly. I guess going through the Langevin equations all the way is much longer than just inserting Boltzmann distribution. $\endgroup$ – Eugen Hruska Oct 20 '15 at 16:36
  • $\begingroup$ It is not significantly longer (depending on where you start of course). The assumption of equilibrium is then hidden in the assumption of strong damping. But indeed, the Boltzmann distribution will not appear explicitly in your calculation with this method. (To determine the strength of the noise in the first place, however, one has to derive another fluctuation-dissipation theorem – again assuming the system is in equilibrium). $\endgroup$ – Sebastian Riese Oct 20 '15 at 16:45

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