# Chemistry - Why is the Maxwell–Boltzmann formula inapplicable at low temperatures?

## Solution 1:

Since the explanation was a little more complicated than I initially thought, I figured it would be worth it to combine my comments (and info from Physics SE) into an answer.

Quantum particles satisfy Fermi–Dirac or Bose–Einstein statistics depending on whether they are fermions or bosons. These distributions have the form $$\langle n_i\rangle=\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]\pm1}$$ where the plus/minus is for fermions/bosons.

To consider the high temperature limit, we need to note that not only is there a direct temperature dependence, but $$\mu$$ is also dependent on temperature. Specifically, in the high temperature limit, $$\mu<0$$ and $$|\mu|>k_bT$$. Combining this information, we can make the (very accurate) approximation $$\exp[(\epsilon_i-\mu)/k_bT]\pm1\approx\exp[(\epsilon_i-\mu)/k_bT]$$ as the exponential function will be much larger than $$1$$. So at high temperatures $$\langle n_i\rangle\approx\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]}$$ which matches the form of the Maxwell–Boltzmann distribution. We can see that at low temperatures these two distributions would not agree, as the additional $$\pm1$$ term in the denominator would become more significant as the exponential got smaller.

It's important to remember that the particles are always indistinguishable; all we have done in the high temperature limit is made an approximation that simplified the functional form. We should not take this coincidental agreement with the MB distribution (for which the particles are assumed to be distinguishable) to imply that the quantum particles have become distinguishable at high temperatures.

## Solution 2:

The Maxwell–Boltzmann distribution (MBD) is inextricably tied to ideal gas behavior, so where ideal gas behavior fails so will the MBD. It isn't just at low temperatures that the MBD fails, since the MBD will also fail at "very high" temperatures because the gases ionize. The gist is that to consider quantum effects you have to go to a more complicated model than the MBD since the MBD explicitly ignores quantum mechanics.

The situation is somewhat like comparing Newton's theory of gravity and Einstein's theory of relativity. So:

• The MBD was known before quantum mechanics was figured out, so the classical derivation doesn't use quantum mechanics.
• The MBD is a continuous function. The "real" distribution would have to be quantized because of the implications of the Plank constant, so it be a histogram rather than a continuous function.
• The MBD doesn't ignores gravity thus doesn't predict that a massive enough gas cloud would collapse due to gravitational attraction.
• Fermi–Dirac statistics and Bose–Einstein statistics are two quantum distributions that are opposites in the sense that Fermi–Dirac statistics applies to fermions and Bose–Einstein statistics applies to bosons. However the MBD ignores this bifurcation of particle types.
• Fermi–Dirac statistics and Bose–Einstein statistics are discrete distributions, but in the limit (ie lots of particles, energies great enough for implications of the Plank constant not to apply) reduce to the MBD which is a "continuous" distribution. (Think of the binomial distribution approximation of the Gaussian distribution in statistics.)
• A Bose-Einstein condensate, where Bose-Einstein statistics apply, isn't really a gas, but rather it is another state of matter.
• The MBD requires particles to be distinguishable, and that many particles can have the same energy state.
• Bose-Einstein Statistics requires particles to be indistinguishable, and that many particles can have the same energy state.
• Fermi–Dirac statistics requires particles to be indistinguishable, and that no two particles can occupy the same state.
• The MBD, as well as Fermi–Dirac statistics and Bose–Einstein statistics, ignore relativity in which the mass of a particle seems to increase as the velocity of the particle increases.
• The MBD, as well as Bose–Einstein statistics, fail at very high temperatures because the gas starts to ionize.