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I understand the formula for the Boltzmann distribution to be

$P(E_i) = e^{-E_i/(kT)}/Z$

When the energy levels vary continuously illustrations of pdf for either the energy or the velocity at a fixed temperature show a skewed bell shaped distribution over the non-negative reals somewhat like a smooth version of the Poisson distribution. but the above formula is a decaying exponential wrt the energy level and everything else is constant at fixed temperature, so I would expect the graph of the pdf of energy to be monotonic decreasing & convex. The pdf of the speed would then to be the the right half of the Gaussian distribution which is still monotonic decreasing. Either way I do not see how the pdf could be increasing near 0. Clearly I am missing something really obvious, but I cannot figure out what it is.

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What you are missing is the density of energy levels. Your $P(E_i)$ is the probability for a single state of energy $E_i$ (in a discrete situation). But the number of such states in an interval $(E, E + \Delta E)$ will not be constant. For example, for a single particle (with no internal degrees of freedom, just kinetic energy) in a box, this number is approximately proportional to $E^{1/2}\ \Delta E$.

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  • $\begingroup$ How does this work in the continuous case like the ideal gas? Here the energy levels seem evenly distributed. $\endgroup$ Commented Jan 13, 2014 at 6:34
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    $\begingroup$ There you look at volume in phase space. In three dimensions, the volume of the ball $ m(v_x^2 + v_y^2 + v_z^2)/2 < E $ is $V = 4/3 \pi (2E/m)^{3/2}$, so the density is $dV/dE = 4 \pi (2E/m)^{1/2}/m$. $\endgroup$ Commented Jan 13, 2014 at 16:14

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