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addaF

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I just recall the two expression for the Maxwell-Boltzmann distribution:

$$ P(v)dv = \left( \dfrac{m}{2 \pi k T} \right)^{3/2} 4 \pi v^2 \exp \left(- \dfrac{mv^2}{kT} \right) dv \qquad P(E)dE = \left( \dfrac{4E}{\pi} \right)^{1/2} \dfrac{e^{-E/kT}}{\left( kT \right)^{3/2}} dE$$

The left expression gives the probability to have the particle with speed in the range ## [v,v+dv] ##, while the right one gives the probability to have an energy in the range ##[E,E+dE]##. I'm now struggling a bit to find a physical reason for the following fact:

the most probable speed is ## v = \sqrt{2kT/m}## and the corresponding energy is ##E = kT##; the most probable energy is ##E = kT/2##. The mathematical reason is clear: the change of variable is not linear. However (now) the difference between the two energies seems a bit illogical to me, and I'm trying to find a physical reason for this but i cannot figure out. Can anyone help me?

Thank you in advance

Edit: I forgot to say it, i assume ##E## as the classical kinetic energy, namely ##E = mv^2 / 2##

$$ P(v)dv = \left( \dfrac{m}{2 \pi k T} \right)^{3/2} 4 \pi v^2 \exp \left(- \dfrac{mv^2}{kT} \right) dv \qquad P(E)dE = \left( \dfrac{4E}{\pi} \right)^{1/2} \dfrac{e^{-E/kT}}{\left( kT \right)^{3/2}} dE$$

The left expression gives the probability to have the particle with speed in the range ## [v,v+dv] ##, while the right one gives the probability to have an energy in the range ##[E,E+dE]##. I'm now struggling a bit to find a physical reason for the following fact:

the most probable speed is ## v = \sqrt{2kT/m}## and the corresponding energy is ##E = kT##; the most probable energy is ##E = kT/2##. The mathematical reason is clear: the change of variable is not linear. However (now) the difference between the two energies seems a bit illogical to me, and I'm trying to find a physical reason for this but i cannot figure out. Can anyone help me?

Thank you in advance

Edit: I forgot to say it, i assume ##E## as the classical kinetic energy, namely ##E = mv^2 / 2##

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