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To derive the Maxwell-Boltzmann distribution the first step requires the knowledge that the equation will take the form $$f(v)=Ke^{-\epsilon/ kT}$$ where $\epsilon$ is the sum of the kinetic energies in the directions of the x,y and z axes. I understand that there needs to be exponential decay and why there will need to a constant. Also, I understand that the eponent needs to be dimensionless but why is KT used? This seems to pop up everywhere without much of an explaination of why (as in this case). Can you help me understand how to get this first step? Thanks.

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    $\begingroup$ You could easily refer this in any text of statistical mechanics. Try helping yourself before asking. Hence I'm down voting this question. $\endgroup$ – shre_sudh_97 Mar 29 '16 at 17:14
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KT comes in from requiring agreement with the ideal gas law.

See "section 1.5.3 $\langle v^2\rangle$ should agree with the ideal gas law" and section 1.3 "The functional form of the dependence of T on $\langle\varepsilon\rangle$ cannot be determined solely from kinetic theory, since the temperature scale may be chosen in many possible ways. In fact, one way to define the temperature is through the ideal gas law: T = pV/(nR)" in Kinetic Theory of Gasses

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The Maxwell-Boltzmann distribution of speeds is derived from the Boltzmann distribution, which says that

$$\frac{N_i}{N_j}=e^{-(\epsilon_i-\epsilon_j)/kT}$$

This formula gives the ratio of the number of particles ($N_i$, $N_j$) in states with energy $\epsilon_i$ and $\epsilon_j$. From this it is easy to the that $f(v)=Ke^{-\epsilon/ kT}$ as you noted.

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  • $\begingroup$ how does the numerator of the exponent change to kinetic energy? Also why is KT in the denominator? Is this derived or found empirically? $\endgroup$ – RobChem Dec 24 '14 at 15:02

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