Average speed of molecules in a Fermi gas

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SUMMARY

The average speed of molecules in a Fermi gas can be calculated using the relation for mean speed, specifically for a gas of N spin-1/2 particles at absolute zero temperature (T = 0). The established formula is = (3/4)u_F, where u_F represents the Fermi velocity. The Fermi statistic f(ε) is defined as f(ε) = 2 for all ε ≤ ε_F, and it is crucial to note that velocity scales as √ε, leading to the conclusion that ⟨√x⟩ ≠ √⟨x⟩. This distinction is essential for accurate calculations in quantum statistical mechanics.

PREREQUISITES
  • Understanding of Fermi gas principles
  • Familiarity with Fermi statistics and the equation f(ε) = 2
  • Knowledge of kinetic energy relations in quantum mechanics
  • Basic grasp of statistical mechanics at absolute zero temperature
NEXT STEPS
  • Study the derivation of the Fermi velocity (u_F) in quantum gases
  • Learn about the implications of Fermi-Dirac statistics at T = 0
  • Explore integration techniques for non-simple probability distributions
  • Investigate the relationship between energy and velocity in quantum systems
USEFUL FOR

Physicists, particularly those specializing in quantum mechanics and statistical physics, as well as students studying Fermi gases and their properties.

Silicon-Based
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Homework Statement
Prove that the mean speed ##<u>## in a gas of ##N## spin-1/2 particles at ##T = 0## is ##<u> = \frac{3}{4}u_F##, where ##u_F## is the Fermi velocity
Relevant Equations
##f(\epsilon) = \frac{1}{e^{\beta(\epsilon-\mu)}-1}##
My first most obvious attempt was to use the relation ##<\epsilon> = \frac{3}{5}\epsilon_F## and the formula for kinetic energy, but this doesn't give the right answer and I'm frankly not sure why that's the case. My other idea was to use the Fermi statistic ##f(\epsilon)## which in this case should be ##f(\epsilon)=2## for all ##\epsilon \leq \epsilon_F##, and then integrate somehow to find the mean velocity, but it isn't obvious to me what to integrate as ##f(\epsilon)## isn't a simple probability distribution.
 
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Silicon-Based said:
My first most obvious attempt was to use the relation ##<\epsilon> = \frac{3}{5}\epsilon_F## and the formula for kinetic energy, but this doesn't give the right answer and I'm frankly not sure why that's the case.
It doesn't work because of the fact that velocity scales as ##\sqrt{\epsilon}##,
$$
\langle \sqrt{x} \rangle \neq \sqrt{\langle x \rangle}
$$

Silicon-Based said:
My other idea was to use the Fermi statistic ##f(\epsilon)## which in this case should be ##f(\epsilon)=2## for all ##\epsilon \leq \epsilon_F##, and then integrate somehow to find the mean velocity, but it isn't obvious to me what to integrate as ##f(\epsilon)## isn't a simple probability distribution.
Generally speaking, if you wanted to calculate ##\langle x \rangle## for a Fermi gas, what equation would you set up?
 
Silicon-Based said:
Problem Statement: Prove that the mean speed ##<u>## in a gas of ##N## spin-1/2 particles at ##T = 0## is ##<u> = \frac{3}{4}u_F##, where ##u_F## is the Fermi velocity
Relevant Equations: ##f(\epsilon) = \frac{1}{e^{\beta(\epsilon-\mu)}-1}##

f(ϵ)f(ϵ)f(\epsilon) isn't a simple probability distribution.

Are you sure about this? What happens to f at T=0?
 

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