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Average speed of molecules in a Fermi gas

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}##
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.
 

DrClaude

Mentor
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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}##,
$$
\langle \sqrt{x} \rangle \neq \sqrt{\langle x \rangle}
$$

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?
 
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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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