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From Boltzmann distribution: $P_{j}=\frac{e^{\frac{-mv^{2}}{2kT}}}{Z}$ with partition function in the form, $Z=\int e^{\frac{-mv^{2}}{2kT}} d\bar{v}^{3}$, where $ d\bar{v}^{3} =dv_{x}dv_{y}dv_{z}$ we get the Maxwell distribution for molecular velocities:

$P\left (v \right )=\left ( \frac{m}{2\pi kT} \right )^{\frac{3}{2}}e^{\frac{-mv^{2}}{2kT}}$

Now, I'am trying to understand how to pass from this distribution to the speed distribution. It looks that it is enought to multiply $P\left (v \right )$ for a volume $d\omega $ of a spherical layer of a proper thickness in space velocities to obtain a speed distribution for the interval $v$ e $v+dv$.

Considering the above, I have some problem to understand the following,

  1. Why exactly $P\left (v \right )=\left ( \frac{m}{2\pi kT} \right )^{\frac{3}{2}}e^{\frac{-mv^{2}}{2kT}}$ is considered a velocity distribution, the kinetic energy in the exponent is not a vector, so there are no velocity but speed in this formula.

  2. why a probability, $P\left (v \right )$ multiplied by a volume give another probability. In my understanding the probability is given by $P=\frac{dN}{N}=fd\omega$ ($f$ is a distribution function) so how $P\left ( v \right )d\omega $ can be another probability?

  3. Go from velocity to speed distribution, how it is really done?

franchino
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2 Answers2

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The key point is the word distribution.

The distribution of velocities is a function $P( \vec v)$ giving the probability of finding a molecule in a velocity space volume according to the formula: $$ {\mathrm {Probability~of~velocity~in~a~volume~d} }\vec v~{\mathrm {centered~around~}}\vec v= P( \vec v) {\mathrm d}\vec v. \tag{1} $$ Notice that $P( \vec v)$ depends on the vector $\vec v$, but in some cases like that of Maxwell distribution, it may be rotationally invariant and the it is a function of the modulus of the vector. Still, what qualifies it as a velocity distribution, is the fact that the link to probabilities is given by equation (1).

The distribution of speeds is a function $p(v)$ giving the probability of finding a molecule in a small interval of speeds ${\mathrm d}v$, around $v$, according to the formula $$ {\mathrm {Probability~of~speed~in~an~interval~d} } v~{\mathrm {centered~around~}}v= p( v) {\mathrm d}v. \tag{2}. $$

The relation between $P$ and $p$ in the case of the Maxwell distribution can be easily obtained by recalling that the latter corresponds to integration of the former over all the directions. Therefore, by using polar coordinates $$ {\mathrm d}\vec v=v^2{\mathrm d}v \sin \theta{\mathrm d}\theta{\mathrm d}\phi $$ and exploiting the dependency of $P$ on the modulus of the velocity, one can easily get the relation between $p(v)$ and $P({\vec v})$: $$ p(v) = 4 \pi v^2 P(|\vec v|) $$

GiorgioP-DoomsdayClockIsAt-90
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  1. p(v) is the velocity distribution: understand the speed. Velocity and speed are only for high school. In real Life, nobody cares about the distinction.

  2. P(v) is not a probability, but a distribution: i.e it is a density of probability. P(v) dv is a probability.

  3. idem than 1

Shaktyai
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