# Internal Energy of Degenerate Fermi ideal gas to the 4th power

• Othin
In summary, Homework Equations states that the energy spectrum of an ideal Fermi degenerate gas is given by: $$\varepsilon_k = \left( \frac{\hbar^2 k^2}{2m} \right) , k=\left( 2n\pi /L \right)$$ where ##N= \gamma V \int_0^\infty D(\varepsilon)f(\varepsilon) d\varepsilon## and: $$U= \gamma V \int_0^\infty \varepsilon D(\varepsilon)f(\vare Othin ## Homework Statement We are asked to derive the expression for the internal energy of an ideal Fermi degenerate gas using Sommerfeld expansions, writing out terms up to the fourth order in ##(\frac{T}{T_F} )## , that is, we must determine ## \alpha ## in the following expression:$$ U= \frac{3}{5}N\varepsilon_F \left\{ 1 + \frac{5\pi ^2}{12} {\left( \frac{T}{T_F}\right)}^2 + \alpha {\left( \frac{T}{T_F}\right)}^4 \right\}$$The method is basically the one used in the text of Kubo on Statistical Mechanics. ## Homework Equations We're treating the 3-D "particle in a box" model with periodic boundary conditions, so that the energy spectrum is given by:$$\varepsilon_k = \left( \frac{\hbar^2 k^2}{2m} \right) , k=\left( 2n\pi /L \right)$$We are also supposing that the number of particles is sufficiently large for we to approximate the summations characteristic of a discrete spectrum by integrals. We then have:$$
N= \gamma V \int_0^\infty D(\varepsilon)f(\varepsilon) d\varepsilon, \\
U= \gamma V \int_0^\infty \varepsilon D(\varepsilon)f(\varepsilon) d\varepsilon \\
$$where ## \gamma ## takes the spin degeneracy into account and:$$
D( \varepsilon ) = \frac{1}{4 \pi^2} {\left( \frac{2m}{\hbar ^2} \right)}^{3/2} \varepsilon ^{1/2} \equiv C \varepsilon ^{1/2} ,\ \ f(\varepsilon) = \frac{1}{exp[\frac{\varepsilon - \mu}{k_bT}] + 1}.
$$The Sommerfeld expansion is also given, whereby we can solve integrals of the form ## I = \int_0^\infty f(\varepsilon) \psi(\varepsilon) d{\varepsilon}## :$$
I = \int_0^{\mu}\psi(\varepsilon) d\varepsilon + \frac{1}{2!}I_2 {\left( \frac{d \psi}{ d \varepsilon} \right)}_{\varepsilon= \mu} + \frac{1}{4!}I_4 {\left( \frac{d^3 \psi}{ d\varepsilon^3} \right)}_{\varepsilon= \mu} + ...
$$Within approximations, the integral ##I_2## and ##I_4## are taken to equal, respectively, ## \frac{\pi^2 (k_bT)^2}{3} ## and ## \frac{7 \pi^4}{15}(k_bT)^4## ## The Attempt at a Solution [/B] Since U must be given in powers of ##T/T_F ##, we must first use the equation for N in order to write #/mu# in terms of T and #T_F#. Therefore, for N:$$
\psi(\varepsilon) = \varepsilon^{1/2} \implies N = \gamma VC \left \{ \frac{2}{3}\mu^{3/2} + \frac{\pi ^2}{12} (k_b T)^2 \mu^{1/2} - \frac{3}{8} \frac{1}{4!} \mu^{\frac{-5}{2}}I_4 + ... \right \} \\

\frac{N}{\gamma VC}= \frac{2}{3} \mu^{3/2} \left[ 1 + \frac{\pi^2}{8}\left (\frac{k_b T}{\mu} \right)^2 + \left( \frac{7 \pi^4}{640} \right) \left (\frac{k_bT}{\mu} \right)^4 \right]
$$Now, after this expression Kubo (and other books I've checked) make some kind of expansion around ## \mu_o ## and arrive at the following expression:$$
\mu = \mu_o \left[1- \frac{\pi^2}{12} \left( \frac{k_b T}{\mu_o}\right)^2 - \frac{\pi^4}{80} \left( \frac{k_b T}{\mu_o}\right)^4 \right]

and , no matter how I try, I can't get anything like that last expression (I managed to get the first term by truncating the series and using a binomial expansion to the first order, but when I include the next power I can't make it work)

I have a copy of Kubo. Are you OK with his derivation up through equation (6) on page 241?

TSny said:
I have a copy of Kubo. Are you OK with his derivation up through equation (6) on page 241?
View attachment 109883
Hi! Well, no, I don't get the approximation that leads to equation (6), though I'm ok with the one above it. Don't know how to solve that equation.

To get the second equation of (6) from the first equation, write the bracket part of the top equation as ##f(x) = \left(1 + a x + b x^2 \right )^{-2/3} ## where ##x = \left( \frac{kT}{\mu} \right)^2## and do a Taylor expansion about ##x = 0## to second order in x.

Or, are you having difficulty getting to the first equation of (6)?

Othin
TSny said:
To get the second equation of (6) from the first equation, write the bracket part of the top equation as ##f(x) = \left(1 + a x + b x^2 \right )^{-2/3} ## where ##x = \left( \frac{kT}{\mu} \right)^2## and do a Taylor expansion about ##x = 0## to second order in x.

Or, are you having difficulty getting to the first equation of (6)?
Wow, thanks a lot! I was carrying the expansion around the ## \mu_o##. So, let me see if I got it right: I make a Taylor expansion around x=0, but that implies T=0, and, for that reason and the definition of the fermi energy, I wrtite ##\mu_o ## insead of ## \mu##?

You are just treating ##x = \left(\frac{kT}{\mu} \right)^2## as small compared to 1. In the Taylor expansion, you will still write ##x## with ##\mu## rather than ##\mu_0##. Note how the bracketed expression in the second equation in (6) is still in terms of ##\mu##.

You will still have some work to do to get from (6) to (7), where the right-hand side is in terms of ##\mu_0##.

Othin
TSny said:
You are just treating ##x = \left(\frac{kT}{\mu} \right)^2## as small compared to 1. In the Taylor expansion, you will still write ##x## with ##\mu## rather than ##\mu_0##. Note how the bracketed expression in the second equation in (6) is still in terms of ##\mu##.

You will still have some work to do to get from (6) to (7), where the right-hand side is in terms of ##\mu_0##.
From (6) to (7) must I do another expansion (around ##\mu_o##)? There is no equation relating ## \mu_o \text{and } \mu## that I remember, but neither my expansions around ##\mu_o## nor T=0 lead to the desired expression :(.

To fist order in ##x =\left( \frac{kT}{\mu} \right)^2##, (6) tells you that ##\mu = \mu_0 \left\{1-\frac{\pi^2}{12} \left( \frac{kT}{\mu} \right)^2 \right\}##.

Convince yourself that this is still accurate to first order if you replace ##\mu## by ##\mu_0## in the last term on the right. So, you get the first-order approximation given in Kubo below equation (6).

To get the second-order approximation in ##\left( \frac{kT}{\mu} \right)^2##, you need to go back to (6). In the second term inside the { }, you need to substitute for ##\mu## the first-order expression given just below (6). This term can then be approximated to order ##\left( \frac{kT}{\mu} \right)^4##. The third term in the { } in (6) is easy to approximate to order ##\left( \frac{kT}{\mu_0} \right)^4## by just letting ##\mu = \mu_0##.

Othin

## What is the Internal Energy of a Degenerate Fermi Ideal Gas to the 4th power?

The Internal Energy of a Degenerate Fermi Ideal Gas to the 4th power is a measure of the total energy of a system of fermions at absolute zero temperature. It takes into account the energy contributions from both the kinetic energy and the potential energy of the particles.

## How is the Internal Energy of a Degenerate Fermi Ideal Gas to the 4th power calculated?

The Internal Energy of a Degenerate Fermi Ideal Gas to the 4th power is calculated using the Fermi-Dirac distribution function, which takes into account the energy levels of the fermions and their probability of occupation. This function is then integrated over all possible energy states to obtain the total Internal Energy.

## What is the significance of the Internal Energy of a Degenerate Fermi Ideal Gas to the 4th power?

The Internal Energy of a Degenerate Fermi Ideal Gas to the 4th power is a fundamental quantity in understanding the behavior of degenerate fermionic systems, such as electrons in a metal. It is also important in determining the properties of white dwarf stars and other highly dense objects in astrophysics.

## How does the Internal Energy of a Degenerate Fermi Ideal Gas to the 4th power change with temperature?

At absolute zero temperature, the Internal Energy of a Degenerate Fermi Ideal Gas to the 4th power is at its minimum value, since all fermions are in their lowest energy state. As temperature increases, more energy levels become occupied and the Internal Energy also increases. However, at high temperatures, the relationship between temperature and Internal Energy becomes more complex due to the effects of thermal excitations.

## Are there any real-world applications of the Internal Energy of a Degenerate Fermi Ideal Gas to the 4th power?

Yes, the Internal Energy of a Degenerate Fermi Ideal Gas to the 4th power has many practical applications in fields such as condensed matter physics, astrophysics, and materials science. It is used to understand the behavior of metals and other materials at low temperatures, as well as the thermodynamic properties of highly dense objects in outer space.

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