Most probable values of the Maxwell Boltzman distribution

Trivial Man

To preface my problem, you should know what I'm deriving. When I try to find the most probable speed of a classical, nonrelativistic particle as described by the Maxwell speed distribution I find that it is v_mp=(2*k*T/m)^(1/2). The kinetic energy associated with this particle would then be E_K=(m*v_mp^2)/2=k*T. Next, when I use the Boltzman distribution to try and find the most probable kinetic energy of a particle I find that E_K=(k*t)/2.

Both of these values I've verified as correct with my book and even in another topic in the forum (http://www.physicsforums.com/showthread.php?t=120947). Multiple sources I've found, including my professor, make it a point to emphasize that the most probable kinetic energy is not the same as the kinetic energy of the most probable speed. However, one would intuitively expect these values to be the same since the speed and kinetic energies of a particle are related. Obviously the math dictates that they are, but from a more physical perspective why are these values different?

ehild

The probability densities mean probability only together with the associated interval . If you ask what is the probability that a molecule has 100 m/s speed, the answer is zero. Even among a lot of molecules in a vessel, the speed of no one will be exactly 100 m/s. You can ask what is the probability that a speed of a molecule is between 10 m/s and 11 m/s, and the answer is : The probability that the speed of a molecule is between v1 and v2 is ∫f(v)dv (integral from v1 and v2). If the interval is very narrow, you can approximate the probability that the speed is between v and v+dv with f(v)dv. If you want the probability that the speed of a molecule is between 100m/s and 100.1 m/s, you can say that it is f(100)*0.1.
The probability density itself is f(v)=dP/dv.

The probability that the speed of a molecule is between v and v+dv is f(v)dv. The probability that the KE of a molecule is between E and E+dE is g(E)dE, where g is the MB distribution function in terms of KE. It can be written in terms of speed: g(E)dE= g(E(v))(dE/dv)dv=f(v)(dE/dv)dv

If you want the most probable speed, it means you need to find v where f(v) is maximum. At the most probable energy, g(E) is maximum, but it is maximum at that speed where f(v)dE/dv = f(v)(mv) is maximum. The intervals dE and dv are not simply proportional.

ehild