- #1

Toby_phys

- 26

- 0

$$

P\propto e^{\frac{-E}{k_BT}}

$$

Where $E$ is the energy. So surely this means the kinetic energy (and therefore speed) of particles is distributed over a Boltzmann distribution. Or in equation:

$$

P\propto e^{\frac{-mv^2}{2k_BT}}

$$

However, this is not what my book says. The books says velocity is distributed in this way. They appear to have split up the components of energy

Speed is distributed as follows, as a Maxwell distribution.

$$

P\propto v^2e^{\frac{-mv^2}{2k_BT}}

$$

I understand that the probability of the speed is the sum of all the different degrees of freedom and such.

However, why can you split up the components of the kinetic energy is this way? Note, I am not asking how to get from Boltzmann to Maxwell, I am asking why velocity, and not kinetic energy, is Boltzmann distributed?