The Langevin equation for a Brownian particle without the friction term is: \begin{equation} m\dot{v}=F(t) \end{equation} Where $F(t)$ is the random force acting on the Brownian particle due to collisions with the heat bath particles. With the usual assumptions that: \begin{equation} \langle F(t)\rangle=0, \langle F(t)F(t')\rangle=2B\delta(t-t') \end{equation} One gets the following expression for the average of the velocity squared: \begin{equation} \langle v^2(t)\rangle=\frac{kT}{m}+\frac{2Bt}{m} \end{equation} This result is wrong, as it should be, for a Brownian particle in a heat bath. The average of the velocity squared increases with time. Even though the result is wrong I wonder if there is some intuition behind this still. Why would the absence of friction cause this behavior? Is there some intuition behind this or is this just what Mathematics tells us?
2 Answers
As stated in this question, the integral of a random force, as you described, gives Brownian motion. But in this case the Brownian motion is not performed by $x$, but by $v$. A variable that undergoes Brownian motion tends to drift away from the origin, which in captured by $\langle v^2\rangle\propto t$ for large $t$.
So this basically boils down to the question "if a particle undergoes a random force with mean 0, will the energy increase over time?". As you showed, it does. This is not unreasonable, because the force on a particle should depend on $v$. The larger its velocity with respect to the mean velocity, the more likely it is to have a collision that saps away energy. This is precisely captured by the friction term and leaving it out produces unphysical results.
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The drag force is a restoring force that opposes $v$ and returns it to zero mean. The velocity in this case fluctuates around 0.
In the absence of a restoring force the velocity does an unrestricted random walk. As your relationship shows, the mean squareed velocity increases linearly with time, just as $\langle x^2\rangle$ does in the presence of drag.
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