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The Maxwell-Boltzmann probability density function for the energy is $$ f_{MB}\left(E\right)=2\left(\frac{1}{\text{K}_{B}T}\right)^{3/2}\sqrt{\frac{E}{\pi}}\exp\left(-\frac{E}{\text{K}_{B}T}\right) $$ In this expression it is usually implied that the energy levels are continuous. However, in statistical physics it is shown that in a discrete system of energy levels one has $$ P_i = A \exp\left(-\frac{E_i}{\text{K}_{B}T}\right) $$ Where $E_i$ is the energy at $i^{th}$ level and $A$ is a normalization constant (the inverse of the partition function).

My question is whether it is possible, by passing to the limit $\Delta E \rightarrow 0$, to demonstrate the first formula above using the second formula?

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    $\begingroup$ No. This is not continuum versus discrete. The first equation takes into account the density of states. $\endgroup$
    – user137289
    Commented Nov 11, 2020 at 16:36
  • $\begingroup$ @Pieter - although I did not fully understood your comment, I edited my question to remove the continuum vs discrete. When you say No, do you mean it is not possible? $\endgroup$
    – Blue
    Commented Nov 11, 2020 at 16:46
  • $\begingroup$ I mean that it is not possible to go from the Boltzmann factor to the Maxwell-Boltzmann distribution without taking into account the density of states. $\endgroup$
    – user137289
    Commented Nov 11, 2020 at 16:52

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  • It is indeed possible to derive a continuous version of the Boltzmann distribution from its discrete version, by using a suitable continuum limit.

  • The Maxwell-Boltzmann distribution is not the same thing as the Boltzmann distribution. The latter is much more general and applies to arbitrary systems, whereas the former is specific to a gas in three dimensions.

    The difference is that in the Boltzmann distribution you're describing the probability of being in a state $i$, which happens to have energy $E_i$, but you leave open the possibility of there being other microstates with the same energy. In the Maxwell-Boltzmann case, you are interested in the probability of having a pre-defined energy $E$, and as such you have to take into account that some energies have more microstates than others. (This is because of the angular degree of freedom, and results in the factor of $\sqrt{E}$.)

All of this is standard material and it is discussed in depth in any suitable (late-undergraduate) textbook on statistical mechanics.

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  • $\begingroup$ So to obtain the Maxwell-Boltzmann distribution above (for a 3D gas) I have to take into account that there are several configurations (microstates) with differente velocity components that are not considered in the "usual" derivation of the Boltzmann Distribution? $\endgroup$
    – Blue
    Commented Nov 11, 2020 at 18:17
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    $\begingroup$ That is inaccurate. The usual derivation of the Boltzmann distribution does take into account the existence of multiple microstates with the same energy. The discrepancy is that the two calculations you're trying to match are calculating different quantities (specifically, the occupation probability of an individual microstate that happens to have energy $E$ versus the occupation probability of all the microstates at that energy). The usual calculation accounts for this difference using the 'density of states', which is the key search term if this is still confusing. $\endgroup$ Commented Nov 11, 2020 at 18:30
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    $\begingroup$ Yes, this is still a little confusing to me. But I think you and Pieter above have pointed me in the direction. I was studying using introductory physics books. I will study the density of states as suggested. Anyway, just wanted to thank you both! $\endgroup$
    – Blue
    Commented Nov 11, 2020 at 19:02

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