Most probable energy and speed for Maxwell-Boltzmann distribution

[addaF]

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 $E = mv^2 / 2$

 

[mjc123]

Physically, there is no such thing as the most probable energy or speed, because these are continuous not discrete variables. The probability of having exactly a specified speed, e.g. exactly 1000 m/s, is vanishingly small. Probability must be evaluated over a finite range dv or dE. Now if you make all your speed slices of the same width dv, the maximum probability is at the maximum of the probability density function P(v). Similarly for equal widths dE, the maximum probability is at the maximum of the function P(E). But since dE and dv are not linearly related, you can't simultaneously have your slices of equal dv and equal dE. If they are of equal dv, then dE increases with E, and P(E)dE is not proportional to P(E), so its maximum doesn't occur at the maximum of P(E).

 

[addaF]

Physically, there is no such thing as the most probable energy or speed, because these are continuous not discrete variables. The probability of having exactly a specified speed, e.g. exactly 1000 m/s, is vanishingly small. Probability must be evaluated over a finite range dv or dE. Now if you make all your speed slices of the same width dv, the maximum probability is at the maximum of the probability density function P(v). Similarly for equal widths dE, the maximum probability is at the maximum of the function P(E). But since dE and dv are not linearly related, you can't simultaneously have your slices of equal dv and equal dE. If they are of equal dv, then dE increases with E, and P(E)dE is not proportional to P(E), so its maximum doesn't occur at the maximum of P(E).

For simplicity I'm thinking to an unidimensional gas with a temperature $T$. For what we said i expect that, on average, the particles moves at $v = \sqrt{2 kT /M}$ which is also called thermal velocity. But from equipartition theorem, since i have only one dimension, i also expect that the average energy is $E = kT/2$, which is precisely the average energy given by the MB distribution.

Your answer makes perfectly sense, but it strictly refers to the change of variables. But if i physically measure the velocity and energy of such particles, supposing to make a large amount of measurements, why should i find such "discrepancy"?

 

[mjc123]

The discrepancy isn't physical, it's mathematical. It's about how you treat your measurements. Do you put them in "bins" of equal dv width, or of equal dE width? You will get a different answer in each case for what is "most probable".

 

[rbd]

As I did not fully understand the explanation with the "bins" I did some further research and eventually understood the concept. As such I thought to come back and try to explain it in other words. I have 2 methods how one could look at it.

The bins that are mentioned in previous answers refer to dE and dv. As E=mv^2/2, one can easily see that a constant splitting up an energy range in equal dE's and then converting it to a speed range would not result in equal dv's because a infinitisimal range dE corresponds to a smaller infinitisimal range dv if it is located at a higher E (eg E(0-1J)->v(0-1m/s) while E(100-101)->v(10-10,04m/s) for 2kg). As such, the 'bins' for both cases have different sizes depending on where they are located.

The difference between the 'most probable' points is because the scale of the x-axis is different. This concept can be explained with the following thought experiment:
Assume that there is an uniform distribution between 0 and 100 for a certain variable t. By plotting this probability density distribution on an x1-axis where x1 = t, a constant line between 0 and 100 is found. However when one would plot the probability density distribution on an x2-axis where x2=t^2, a rising quadratic function between x2=0 and x2=10 is found. As such, looking at these probability density distributions,the most probable point for x2=10 and one would assume that the most probable point for t is thus at t = 100. This is however not correct for the variable t as the probabiltiy for t is uniform. This thought experiment indicates how the choice of variable can seemingly influence the most probable point.