Consider a bath of Brownian particles at temperature $T$. If we sprinkle some larger particles in this (eg: pollen grains in water or dust motes in air), they'll diffuse with diffusion constant $D$ due to bombardments by the Brownian particles. For the same bombardments, any acceleration of these bigger particles due to an external force will die down to a terminal velocity $v_t=F/\gamma$, where $\gamma$ is a damping coefficient. The relation between their fluctuation and dissipation is given by a fluctuation-dissipation equation:

$\gamma D=k_BT$ (Einstein-Smoluchowski relation)

I now have a basic question regarding the behaviour of the individual terms on the left side. Suppose I were to slowly change just the temperature of the bath. That would change the product $\gamma D$. But how would $\gamma$ and $D$ separately change?

Drawing an analogy from the ideal gas state equation $PV = k_BT$, their individual behaviour might depend on the particular process in which I change $T$. So assume that my system (say a bath of water with pollen grains) remains at atmospheric pressure and at the same volume as I just notch up the temperature of the heat bath. How would $\gamma$ and $D$ change then?

  • 1
    $\begingroup$ You never told us what $\gamma$ is. $\endgroup$
    – N. Virgo
    Mar 26, 2013 at 1:49
  • $\begingroup$ I did, actually, in the 4th line: $v_t=F/\gamma$ (it's the damping coefficient). $\endgroup$ Mar 26, 2013 at 4:51
  • $\begingroup$ Yes, but you didn't tell us what it was. I fixed it for you. $\endgroup$
    – N. Virgo
    Mar 26, 2013 at 5:20
  • $\begingroup$ I did, because I wrote the definition of $\gamma$, which is a lot more than just the name. $\endgroup$ Mar 27, 2013 at 12:51
  • $\begingroup$ Writing down an expression that happens to contain a symbol isn't the same as saying what that symbol stands for. Why would you say it is? $\endgroup$
    – N. Virgo
    Mar 27, 2013 at 13:19

1 Answer 1


The temperature-evolution of $D$ and $\gamma$ are still based on models. However, for some fluids these are standard and accurate ones.

For Low Reynold's number, $\gamma$ is proportional to $\eta$, the viscosity of the fluid, by the Stoke's law $\gamma=6\pi\eta r$.

In the Arrhenius fluid model, $\eta$ drops with temperature if the fluid flow obeys the Arrhenius equation for molecular kinetics:

$\gamma\propto e^{E_a/RT}$ (See Temperature-Dependence of Liquid Viscosity.)

$D$ is the final observable that emerges from the other values via the Einstein-Smoluchowski equation $D=k_B T/\gamma$.

For Low Reynold's number, this thus becomes the Stokes-Einstein equation (because Stoke's law can be used): $D=k_B T/6\pi\eta r$, and therefore $D$ depends thus on temperature:

$D \propto Te^{-E_a/RT}$ (See Temperature-Dependence of Diffusion Coefficient.)


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