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?

  • $\begingroup$ My intuition tells me that the modelling of the environment is wrong. The molecules in a real gas or fluid have finite velocity and are not rigid, so the collision time is not infinitesimal and the max. momentum transfer is limited. The average velocity of the Brownian particle should not exceed the average velocity of the molecules and that is already an overestimate, if we go by the velocity distribution of ideal gases with different masses. $\endgroup$ Commented Apr 23, 2023 at 21:50
  • $\begingroup$ The term $2Bt / m$ is probably related to the time-integral of the power imparted to the particle from the environment. In other words, it may be the energy given from the environment to the particle, which in a real life situation is exactly offset by the energy given from the particle to the environment. $\endgroup$
    – DanielSank
    Commented Apr 23, 2023 at 21:54
  • $\begingroup$ You re applying an external force F, so it's normal that the kinetic energy varies in time. Friction dissipates this energy pumped in the system till equilibrium is reached (on average, the dissipated energy rate equals the pumped one). The fact that F is Browninan just gives the typical scaling with t (rather than, say, t^2 for a constant force). $\endgroup$
    – Quillo
    Commented Apr 23, 2023 at 22:24
  • $\begingroup$ Friction and flcutuations are related via fluctuation-dissipation theorem (of which "Einstein relations" is a particular case). Having one but not the other violates thermodynamic laws. Btw, it is not clear where you get the temperature from in the final equation. $\endgroup$
    – Roger V.
    Commented Apr 25, 2023 at 19:08

2 Answers 2


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.

  • $\begingroup$ Why does a larger velocity mean the particle is more likely to have a collision that saps away energy? $\endgroup$ Commented Apr 25, 2023 at 15:04
  • $\begingroup$ @ΜπαμπηςΠοζουκιδης In elastic collisions the particle with the largest momentum has a higher probability to lose energy. If two particles with equal masses collide, their new momenta will be rotated versions around the average momentum. To say this exactly: for the collision the particles have momenta $\vec p_1,\vec p_2$. Define $\vec P=\tfrac 1 2(\vec p_1+\vec p_2)$ and $\vec {\Delta p}=\tfrac 1 2(\vec p_1-\vec p_2)$ such that $\vec p_1=\vec P+\vec {\Delta p}$ and $\vec p_2=\vec P-\vec {\Delta p}$ $\endgroup$ Commented Apr 25, 2023 at 15:19
  • $\begingroup$ The new momenta will be given by $\vec q_1=\vec P+\vec{\Delta q}$ and $\vec q_2=\vec P-\vec{\Delta q}$, where $\vec{\Delta q}$ is just a rotated version of $\vec{\Delta p}$. $\endgroup$ Commented Apr 25, 2023 at 15:19

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.

  • $\begingroup$ This makes sense. But now I realize I don't have an intuition why $\langle x^2 \rangle$ grows linearly with time. $\endgroup$ Commented Apr 25, 2023 at 15:05
  • $\begingroup$ @Μπαμπη The variance of the sum $X_N=x_1+\cdots x_N$ of $N$ random variables drawn from the same distribution is $\sigma^2_N = N\sigma^2_1$. In a random walk $X_N$ is the sum of $N$ random steps drawn from the same distribution. $\endgroup$
    – Themis
    Commented Apr 25, 2023 at 17:28

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