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If we assume the energy of particles in an ideal gas follows a Boltzmann distribution, then the energy distribution function can be defined as below:

$g(E)\propto e^{-\frac{E}{k_BT}}$, where $k_B$ is the Boltzmann constant

Since the energy of particles in an ideal gas are assumed to only consist of translational kinetic energy (as they don't interact),

$E=\frac{1}{2}mv^2$, where v is the speed of the particle

However, for a certain level of kinetic energy, there is only one speed that can be associated with it, as speeds take positive value and hence there is a one-to-one relationship between the speed of a particle and its kinetic energy.

By this reasoning, the distribution function of energy should be proportional to the distribution function of velocity. In other words, the velocity distribution function is also proportional to a Boltzmann factor, with a constant factor of proportionality.

However, the Maxwell-Boltzmann speed distribution shows the velocity distribution having the following form:

$f(v)\propto v^2e^{-\frac{mv^2}{2k_BT}}$

Although I can follow the derivation of the Maxwell-Boltzmann distribution, I fail to see why the line of reasoning I described in previous paragraphs is wrong. Any help would be much appreciated. Thank you!

semisimpleton
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    Your starting point ( $g(E) \propto e^{-\frac{E}{k_BT}}$ ) is wrong. There should be a factor depending on energy to account for the degeneracy of each energy level. Even better, since you want to obtain the one-particle velocity distribution, is to start with the distribution probability of velocities. You may find useful this question on SE physics and the first answer. https://physics.stackexchange.com/questions/144763/derivation-of-the-maxwell-boltzmann-speed-distribution?rq=1 – GiorgioP-DoomsdayClockIsAt-90 Feb 09 '20 at 06:42
  • May I know why the total (kinetic) energy of each particle doesnt follow a Boltzmann distribution? Also, by degeneracy, are you referring to the degrees of freedom that can store energy (in this case, 3 translational degrees of freedom)? Thank you for the reply! – semisimpleton Feb 09 '20 at 07:21
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    By degeneracy I mean that, even for one particle, the same value of $v^2$ can be obtained by several choices of the independent variables $v_x,v_y,v_z$. – GiorgioP-DoomsdayClockIsAt-90 Feb 09 '20 at 09:27
  • Why doesnt the total kinetic energy follow a Boltzmann distribution, but its degenerate components do? – semisimpleton Feb 09 '20 at 11:17
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    The key point is that Boltzmann factor $e^{-\frac{E_i}{k_BT}}$ is the distribution probability of microscopic states, not energies. In order to get a distribution probability for energies, one has to sum over all the microscopic states with the same energy. It is that summation which introduces a factor, not depending on temperature, which makes the difference. – GiorgioP-DoomsdayClockIsAt-90 Feb 09 '20 at 15:06

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The Boltzmann distribution doesn't say: The probability to have energy $E$ is proportional to $e^{-E/k_B T}$.

It says: The probability to be in any one microstate which has the energy $E$ is proportional to $e^{-E/k_B T}$.

So for each energy $E$, you need to count all the states with energy $E$. The correspondence $E \leftrightarrow v$ is one-to-one, as you say. But $v$ is the magnitude of the velocity vector $\vec{v}$, which is what defines the state. The amount of velocity vectors with $|\vec{v}| = v$ is roughly proportional to $4\pi v^2$, the area of a sphere in $\vec{v}$-space.

Elias Riedel Gårding
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