I have recently been studying some statistical thermodynamics and I am currently trying to understand all the different concepts of the course. I was wondering about differences between MB statistics, MB distribution, and Boltzmann distribution, and how all three are related to each other.
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4$\begingroup$ I am sure you have read this link below, it's got links to your three areas of interest, but I think you would need to be more specific in your question, best of luck with it : en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution $\endgroup$– user108787Jun 25, 2016 at 9:43
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The Maxwell-Boltzmann distribution and the Boltzmann distributions are probability distributions, i.e. functions $\rho(\vec x,\vec v)$ of velocity and position of a particle, that say what is the probability density that the velocity and position belong to the small cube around the given value of them.
The Boltzmann distribution is the more general one, $\rho \sim \exp(-E/kT)$, saying that the probabilities exponentially decrease with the energy. The energy $E=E(\vec x,\vec v)$ is a function of position and velocity. When the particular dependence on the position (potential energy) is substituted for $E$, we usually talk about the Boltzmann part (factor) of the distribution.
When we include the kinetic energy $E_k=mv^2/2$, we get the Maxwell part of the distribution. Sometimes, $\rho\sim \exp(-mv^2/2kT)$ is known as the Maxwell-Boltzmann distribution even if no potential energy $E_P$ is included in $E$. So Maxwell and Boltzmann differ as "kinetic" and "potential" part of the general Boltzmann distribution.
On the other hand, Maxwell-Boltzmann statistics is a rule how to statistically treat the information about many particles of the same "species". The rule, the only one known in classical (pre-quantum) physics, says that even though they may have the same properties, they are distinguishable. There is no distribution here – it's a rule saying that the positions $(\vec x_1,\vec x_2)$ and $(\vec x_2,\vec x_1)$ of two particles are distinct.
The relationship between the statistics and the distribution (for one particle) is that the distribution may be derived from the statistics if we look for the most likely way to divide the energy among many particles of the same kind. The "most likely" means to maximize the volume of the phase space.
The Maxwell-Boltzmann statistics and distributions are being talked about especially in contrast with their quantum counterparts, the Bose-Einstein and Fermi-Dirac statistics/distributions. In those statistics, the particles are indistinguishable. Moreover, in the Fermi-Dirac statistics, at most 1 particle may have the same list of properties.
The BE and FD distributions are derived and look a bit different. While the Boltzmann one is based on $\exp(-E/kT) = 1/\exp(E/kT)$, the BE and FD distributions are $$ \sim \frac{1}{\exp(-E/kT) \mp 1}$$
answered Jun 25, 2016 at 13:31
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1$\begingroup$ If I understand it right, MB Statistics specify the statistics rules from which we can derive equations/distributions such as the Boltzmann distribution. And the Maxwell distribution would somehow be a specific case (which takes speed/velocity or kinetic energy into consideration) of the more general Boltzmann distribution (which considers any form of energy)? $\endgroup$– VoidtJun 25, 2016 at 14:24
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1$\begingroup$ Yes. The distributions are derived from the rules how to count the states of many particles and what states are allowed and distinct - the statistics. And Maxwell or Maxwell-Boltzmann distribution is a specific version of the Boltzmann distribution once the exponent contains a term $-mv^2/2kT$ from the kinetic energy. $\endgroup$ Jun 25, 2016 at 14:32
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1$\begingroup$ Alright, thank you! And one last thing, I have seen some plots of the Maxwell distribution function which, as expected, is represented as an inverse exponential, but I have also seen some which are some sort of bell-shaped curves. How can we explain that? $\endgroup$– VoidtJun 25, 2016 at 14:42
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2$\begingroup$ The word Maxwell does mean that there is the factor $\exp(-mv^2/2kT)$ as I wrote about thrice already, and this is a Gaussian (bell curve) function of $v$. So it has to be a Bell curve if you put $v$ on the $x$-axis. If you place $E$ or $v^2$ on the $x$-axis, it's just a decreasing exponential. $\endgroup$ Jun 25, 2016 at 14:44