Boltzmann vs Maxwell distribution?

[Toby_phys]

So I worked through the Boltzmann distribution and got:
$$
P\propto e^{\frac{-E}{k_BT}}
$$
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?

 

[Charles Link]

The states are counted in momentum space, and that is why the results follow how they do. e.g. the energy is $E=\frac{p_x^2+p_y^2+p_z^2 }{2 m}$, but the number of states in an energy interval depends upon the block $\Delta N=(\frac{\Delta p_x \, \Delta p_y \, \Delta p_z}{h^3})V$. ($V$ is the volume of the sample). When converting to speeds, a spherical symmetry in the p-space is assumed and gives $\Delta N=(\frac{4 \pi p^2 \, m \, dv}{h^3})(V)$, ( where $v$ is the speed). In addition to the Boltzmann factor $e^{-\frac{E}{kT}}$ for the relative occupancy, the number of states(density of states) needs to be included to find the entire distribution function.

 

[Toby_phys]

Why are the states counted in momentum space?? I haven't seen a derivation where this is necessary

 

[Charles Link]

Why are the states counted in momentum space?? I haven't seen a derivation where this is necessary

The Statistical Physics book by Reif gives a pretty good description. It may seem like a handwaving approach at first, but this is the easiest way to demonstrate it (this derivation is found in Reif's book): $\\$ In a containter of volume $V$, with length $L_x$, $L_y$, and $L_z$, you set up periodic boundary conditions: $e^{ik_x x}=e^{ik_x(x+L_x)}$ and similarly for y and z. This means $k_x L_x=n_x (2 \pi)$, for integer $n_x$, and similarly for y and z. Thereby $\Delta N=\Delta n_x \Delta n_y \Delta n_z =\frac{\Delta^3 \vec{k}}{(2 \pi)^3} V$. (Note $V=L_x L_y L_z$). With the substitution $\Delta^3 \vec{p}=\hbar^3 \Delta^3 \vec{k}$, the equation I have in post #2 should follow.

 

[vanhees71]

The Maxwell-Boltzmann distribution is derived by maximizing the entropy of a $N$-body state in the (grand-)canonical ensemble of non-interacting particles in a box of volume $V$. A convenient single-particle basis is chosen in the previous posting, namely momentum space. Let's consider scalar particles for convenience. For the many-body system we use the corresponding Fock (occupation-number) basis of the common eigenvectors of the number occupying each single-particle state labeled by the complete set of discrete momenta $\vec{p} \in 2 \pi/L \mathbb{Z}^3$ (I make the simplification to set $L_1=L_2=L_3=L$, i.e., use a cube as the "quantization volume").

Instead of evaluating the simple partition sum of the grand-canonical ensemble, we rather calculate the functional
$$Z[\beta(\vec{p})]=\mathrm{Tr} \exp\left \{-\sum_{\vec{p}} [\beta(\vec{p}) E(\vec{p}) -\alpha] \hat{N}(\vec{p}) \right \}.$$
At the end we get the usual partition sum by setting $\beta=\frac{1}{k_{\text{B}} T}$ and $\alpha(\vec{\beta})=\beta \mu$, where $T$ is the absolute temperature, $\mu$ the chemical potential for the conserved particle number and $E(\vec{p})=\vec{p}^2/(2m)$. Since the $\hat{N}(\vec{p})$ all commute with each other we have
$$Z[\alpha(\vec{p})]=\prod_{\vec{p}} \sum_{N(\vec{p})=0}^{\infty} \exp\{[-\beta E(\vec{p})+\alpha(\vec{p})]N(\vec{p}) \} = \prod_{\vec{p}} \frac{1}{1-\exp\{[-\beta E(\vec{p})+\alpha(\vec{p})]\}}.$$
We have made use of the fact that spin-0 particles are bosons, and thus each $N(\vec{p}) \in \mathbb{N}_0$.

Now it's easier to use
$$W[\alpha(\vec{p})]=\ln Z[\alpha(\vec{p})]=-\sum_{\vec{p}} \ln \{1-\exp\{[-\beta E(\vec{p})+\alpha(\vec{p})]\}.$$
The mean number of particles in the state $\vec{p}$ is given by
$$n(\vec{p}) = \langle \hat{N}(\vec{p}) \rangle=\left [\frac{\partial W}{\partial \alpha(\vec{p})} \right]_{\alpha(\vec{p})=\beta \mu}=\frac{\exp [-\beta E(\vec{p})+\beta \mu]}{1-\exp [-\beta E(\vec{p})+\beta \mu]}=\frac{1}{\exp [\beta E(\vec{p})-\beta \mu]-1}.$$
Now making $L$ very large we can take $\vec{p} \in \mathbb{R}^3$ making use of the above derived single-particle density we find
$$\mathrm{d} N=f(\vec{p}) \mathrm{d}^3 \vec{p}= f_{\text{B}}[E(\vec{p})] V \frac{\mathrm{d}^3 \vec{p}}{(2 \pi \hbar)^3}.$$
For a very diluted gas, for which $\beta \mu$ is a very small negative number, we can neglect the 1 in the denominator of the Bose-Einstein distribution, and we get the classical limit,
$$f(\vec{p})=\frac{V}{(2 \pi \hbar)^3} exp[-\beta E(\vec{p})+\beta \mu].$$
This is the phase-space distribution function of the ideal gas.

Now to get the distribution function for $v=|\vec{p}|/m=p/m$ we note that $\mathrm{d}^3 \vec{p}=\mathrm{d} p \mathrm{d}^2 \Omega p^2=\mathrm{d} v \mathrm{d}^2 \Omega m^3 v^2$ and thus
$$\mathrm{d} N=f(\vec{p}) \mathrm{d}^3 \vec{p} =\frac{V}{(2 \pi \hbar)^3} \exp(-\beta m v^2/2+\alpha) m^3 v^2 \mathrm{d} p \mathrm{d}^2 \Omega$$
Integrating over the whole solid angle and using that the integrand doesn't depend on the angles we simply get
$$\frac{\mathrm{d} N}{\mathrm{d} v} = \frac{V}{(2 \pi \hbar)^3} 4 \pi m^3 v^2 \exp(-\beta m v^2/2+\alpha).$$
You have to be aware that if you change the variable of interest for a distribution, you always get the original distribution multiplied with the appropriate Jacobian changing the "volume element" from one set of variables to another.

You can also check that the energy-distribution function is
$$\frac{\mathrm{d} N}{\mathrm{d} E} \propto \sqrt{E} \exp(-\beta E+\alpha)$$
using
$$E=m v^2/2 \; \Rightarrow \; v=\sqrt{2 m E} \; \Rightarrow \; \mathrm{d} v=\mathrm{d} E \sqrt{2 m}\frac{1}{2 \sqrt{E}}=\mathrm{d} E \sqrt{\frac{m}{E}}.$$