# thermal energy while calculating Langevin Forces

I have a quick question from thermodynamics.

I remember that we take kT/2 as the kinetic energy per degree of freedom in kinetic theory of gases. But when we do langevin forces (for example in cavity dynamics), do we take the average thermal energy as kT/2 or kT ?

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## 1 Answer

The average thermal energy is $kT$ for each Harmonic oscillator, which is split equally between $kT/2$ kinetic and $kT/2$ potential. The average kinetic energy in a nonrelativistic system (or one with a quadratic kinetic energy) is always $kT/2$, for a quadratic potential, you get an equal potential contribution, while for a confining box-potential you get no potential energy contribution on average, because the potential acts in a negligible fraction of the total trajectory.

To find the Langevin dynamics appropriate to a given system coupled to a thermal bath, the coefficient of the thermal noise is determined by the condition that the Boltzmann distribution is stationary. Whether this is $1/2 kT$ or $kT$ depends on the Boltzmann distribution in question, but it's always known, since you know the energy as a function of the coordinates and velocities (or field values).

Once you find the coefficient of the Brownian noise, you make a Smulochowski approximation to find the pure Brownian limit of long times. See this answer for more detail on this limit: Cross-field diffusion from Smoluchowski approximation.

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