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i have given a simplyfied one-dimensional Fokker-Planck equation : $\psi(p,t)=\frac{1}{\sqrt{2\pi vt}}\exp(-\frac{p^2}{2vt})$

My thoughts :
ok, this looks pretty similar to the gaussian distribution : $f(x)=\frac{1}{\sqrt{2\pi \sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2})$

obviously there are parallels ...
so is $\sigma^2 = vt$ and $\mu = 0 $

How do i derive this mathematically ?

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You are right. Your solution $\psi(p,t)$ takes the same form as that of a Gaussian pdf where the Gaussian has mean $\mu = 0$ and variance $\sigma^2 = vt$. Reaching this conclusion by comparing the Gaussian pdf to your equation is perfectly valid and rigorous mathematically and there is nothing more you need to "derive" to justify this. – Dinesh Nov 16 '11 at 11:13
For what it's worth, this is the distribution of particle momentums $p$ in a one-dimensional ideal gas with temperature $t$ and particle masses $m=v/k$, where $k$ is the Boltzmann constant $k=1.38×10^{−23}$ Joules/Kelvin. – Chris Taylor Nov 16 '11 at 12:15
Beyond saying $\sigma^2 = vt$ and $\mu = 0 $, what is there to derive? – Michael Hardy Nov 16 '11 at 14:29

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