Boltzmann vs Maxwell distribution?

In summary, the Boltzmann distribution is derived by maximizing the entropy of a non-interacting gas in a box using the grand-canonical ensemble. The states are counted in momentum space and the resulting distribution is for the momentum components. This is because the number of states in an energy interval is dependent on the momentum space volume and not the energy space volume. Therefore, when converting to speeds, a Jacobian is introduced and the resulting distribution is for the speed components, resulting in the Maxwell distribution. This approach may seem handwaving, but it is the easiest way to demonstrate the relation between the two distributions.
  • #1
Toby_phys
26
0
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?
 
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  • #2
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.
 
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  • #3
Why are the states counted in momentum space?? I haven't seen a derivation where this is necessary
 
  • #4
Toby_phys said:
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.
 
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  • #5
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
$$E=m v^2/2 \; \Rightarrow \; v=\sqrt{2 m E} \; \Rightarrow \; \mathrm{d} v=\mathrm{d} E \sqrt{2 m}\frac{1}{2 \sqrt{E}}=\mathrm{d} E \sqrt{\frac{m}{E}}.$$
 
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Related to Boltzmann vs Maxwell distribution?

1. What is the difference between Boltzmann and Maxwell distribution?

The Boltzmann distribution describes the distribution of particles in thermal equilibrium, taking into account both energy and entropy. The Maxwell distribution, on the other hand, describes the distribution of particles based solely on their kinetic energy at a given temperature.

2. Which distribution is more commonly used in scientific research?

The Boltzmann distribution is more commonly used in scientific research, as it provides a more accurate description of the distribution of particles in thermal equilibrium.

3. How are Boltzmann and Maxwell distributions related?

The Maxwell distribution can be derived from the Boltzmann distribution by considering only the kinetic energy term and assuming a perfect gas with no interactions between particles.

4. What are some real-world applications of Boltzmann and Maxwell distributions?

The Boltzmann distribution is used in fields such as thermodynamics, statistical mechanics, and astrophysics to describe the distribution of particles in thermal equilibrium. The Maxwell distribution is applied in gas dynamics, plasma physics, and atmospheric science.

5. Can Boltzmann and Maxwell distributions be applied to non-equilibrium systems?

While the Boltzmann and Maxwell distributions are based on the assumption of thermal equilibrium, they can also be applied to non-equilibrium systems by considering the system as a series of equilibrium states. However, the accuracy of these distributions may be limited in non-equilibrium systems.

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