Maxwellian velocity distribution vs. speed distribution

[weezy]

Maxwellian velocity distribution is obtained by $$g(v_x)\propto e^{-mv_x^2/2k_B T}$$ and when extended to 3 dimensions the distribution becomes: $$\propto e^{-mv_x^2/2k_B T}e^{-mv_y^2/2k_B T}e^{-mv_z^2/2k_B T} = e^{-mv^2/2k_B T}$$
Now looking at the speed distribution we take a spherical shell in phase space between $v , v+dv$ and obtain: $f(v)dv \propto 4\pi v^2 dv e^{-mv^2/2k_B T}$

My question is that are the two distributions equivalent since $f(v)$ has a extra $v^2$ term but logically the two seem equivalent?

 

[Charles Link]

Maxwellian velocity distribution is obtained by $$g(v_x)\propto e^{-mv_x^2/2k_B T}$$ and when extended to 3 dimensions the distribution becomes: $$\propto e^{-mv_x^2/2k_B T}e^{-mv_y^2/2k_B T}e^{-mv_z^2/2k_B T} = e^{-mv^2/2k_B T}$$
Now looking at the speed distribution we take a spherical shell in phase space between $v , v+dv$ and obtain: $f(v)dv \propto 4\pi v^2 dv e^{-mv^2/2k_B T}$

My question is that are the two distributions equivalent since $f(v)$ has a extra $v^2$ term but logically the two seem equivalent?

The velocity distribution goes from minus infinity to plus infinity for each direction and peaks and has mean at v=0. The speed distribution goes from zero to plus infinity and has value zero for speed less than or equal to zero. (It is really undefined for negative speeds.) The mean value of the speed for the Maxwellian is $v_{mean}=\sqrt{(8/ \pi) k T/m}$. For the speed, $v_{rms}=\sqrt{3 k T/m}$ and most probable speed is $v_{mp}=\sqrt{2 k T/m}$. Perhaps the rms speed is the place where your statement that the two seem to be equivalent is most applicable. It is necessary to compute the precise form of the speed distribution, (with the spherical shell), in order to compute $v_{mean}$ and $v_{mp}$.

 

[mathman]

The difference reflects the fact that there are two different coordinate systems for the integration. In the first case the integral is over three dimensions with $dv_xdv_ydv_z$. The the second case the integral is based on spherical coordinates $v^2dvsin\phi d\phi d\theta$ and then the angular variables integrate out to $4\pi$.

 

[vanhees71]

Well the name says it all: You have distribution! So to transform from one distribution to the distribution of a function of the independent variables you have to take into account the corresponding Jacobian.

In your case you have given the velocity distribution, i.e., the number of particles in an infinitesimal cube $\mathrm{d}^3 \vec{v}$ in velocity space around $\vec{v}$ is
$$\mathrm{d} N = \mathrm{d}^3 \vec{v} f(\vec{v}).$$
Now you have $v=|\vec{v}|$. It's clear that here the case is pretty simple, because you can just use spherical coordinates, in which
$$\mathrm{d} N = \mathrm{d} v \mathrm{d} \vartheta \mathrm{d} \varphi v^2 \sin \vartheta f(\vec{v}).$$
Now you want to "forget" about the angular distribution. So you have to take the integral over $\vartheta$ and $\varphi$, i.e.,
$$\tilde{f}(v)=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi v^2 \sin \vartheta f(\vec{v}).$$
Since in your case $f(\vec{v})=f(v)$, you can do the angle integrals
$$\tilde{f}(v)=v^2 f(v) \int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \sin \vartheta=4 \pi v^2 f(v).$$