## Relativisitic Fermi gas

Posted by peeterjoot on April 20, 2013

## Question: Relativisitic Fermi gas ([1], pr 9.3)

Consider a relativisitic gas of particles of spin obeying Fermi statistics, enclosed in volume , at absolute zero. The energy-momentum relation is

where , and is the rest mass.

Find the Fermi energy at density .

With the pressure defined as the average force per unit area exerted on a perfectly-reflecting wall of the container.

Set up expressions for this in the form of an integral.

Define the internal energy as the average .

Set up expressions for this in the form of an integral.

Show that at low densities, and at high densities. State the criteria for low and high densities.

There may exist a gas of neutrinos (and/or antineutrinos) in the cosmos. (Neutrinos are massless Fermions of spin .) Calculate the Fermi energy (in eV) of such a gas, assuming a density of one particle per .

Attempt exact evaluation of the various integrals.

## Answer

We’ve found [3] that the density of states associated with a 3D relativisitic system is

For a given density , we can find the Fermi energy in the same way as we did for the non-relativisitic energies, with the exception that we have to integrate from a lowest energy of instead of (the energy at ). That is

Solving for we have

We’ll see the constant factor above a number of times below and designate it

so that the Fermi energy is

For the pressure calculation, let’s suppose that we have a configuration with a plane in the orientation as in fig. 1.1.

It’s argued in [4] section 6.4 that the pressure for such a configuration is

where is the number density and is a normalized distribution function for the velocities. The velocity and momentum components are related by the Hamiltonian equations. From the Hamiltonian eq. 1.1 we find \footnote{ Observe that by squaring and summing one can show that this is equivalent to the standard relativisitic momentum .} (for the x-component which is representative)

For we can summarize these velocity-momentum relationships as

Should we attempt to calculate the pressure with this parameterization of the velocity space we end up with convergence problems, and can’t express the results in terms of . Let’s try instead with a distribution over momentum space

Here the momenta have been scaled to have units of energy since we want to express this integral in terms of energy in the end. Our normalized distribution function is

but before evaluating anything, we first want to change our integration variable from momentum to energy. In spherical coordinates our volume element takes the form

Implicit derivatives of

gives us

Our momentum volume element becomes

For our distribution function, we can now write

where is determined by the requirement

The z component of our momentum can be written in spherical coordinates as

Noting that

all the bits come together as

Letting , this is

We could conceivable expand the numerators of each of these integrals in power series, which could then be evaluated as a sum of terms.

Note that above the Fermi energy also has an integral representation

or

Observe that we can use this result to remove the dependence of pressure on this constant

Now for the average energy difference from the rest energy

So the average energy density difference from the rest energy, relative to the rest energy, is

From eq. 1.0.24 and eq. 1.0.26 we have

or

This ratio of integrals is supposed to resolve to 1 and 2 in the low and high density limits. To consider this let’s perform one final non-dimensionalization, writing

The density, pressure, and energy take the form

We can rewrite the square roots in the number density and energy density expressions by expanding out the completion of the square

Expanding the distribution about , we have

allowing us to write, in the low density limit with respect to

**Low density result**

An exact integration of the various integrals above is possible in terms of special functions. However, that attempt (included below) introduced an erroneous extra factor of . Given that this end result was obtained by tossing all but the lowest order terms in and , let’s try that right from the get go.

For the pressure we have an integrand containing a factor

Our pressure, to lowest order in and is then

Our energy density to lowest order in and from eq. 1.0.33c is

Comparing these, we have

or in this low density limit

**High density limit**

For the high density limit write , so that the distribution takes the form

This can be approximated by a step function, so that

With a change of variables , we have

Comparing both we have

or

Wow. That’s pretty low!

**Pressure integral**

Of these the pressure integral is yields directly to Mathematica

where is a modified Bessel function [5] of the second kind as plotted in fig. 1.2.

Plugging this into the series for the pressure, we have

Plotting the summands for in fig. 1.4 shows that this mix of exponential Bessel and quadratic terms decreases with .

Plotting this sum in fig. 1.3 numerically to 10 terms, shows that we have a function that appears roughly polynomial in and .

For small it can be seen graphically that there is very little contribution from anything but the term of this sum. An expansion in series for a few terms in and gives us

This allows a and approximation of the pressure

**Number density integral**

For the number density, it appears that we can evaluate the integral using integration from parts applied to eq. 1.0.30.30

Expanding in series, gives us

Here the binomial coefficient has the meaning given in the definitions of \statmechchapcite{nonIntegralBinomialSeries}, where for negative integral values of we have

Expanding in series to a couple of orders in and we have

To first order in and this is

which allows a relation to pressure

It’s kind of odd seeming that this is quadratic in temperature. Is there an error?

**Energy integral**

Starting from eq. 1.0.30c and integrating by parts we have

The integral with the factor of doesn’t have a nice closed form as before (if you consider the a nice closed form), but instead evaluates to a confluent hypergeometric function [6]. That integral is

and looks like fig. 1.5. Series expansion shows that this hypergeometricU function has a singularity at the origin

so our multiplication by brings us to zero as seen in the plot. Evaluating the complete integral yields the unholy mess

to first order in and this is

Comparing pressure and energy we have for low densities (where )

or

It appears that I’ve picked up an extra factor of somewhere, but at least I’ve got the low density expression. Given that I’ve Taylor expanded everything anyways around and this could likely have been done right from the get go, instead of dragging along the messy geometric integrals. Reworking this part of this problem like that was done above.

# References

[1] Kerson Huang. *Introduction to statistical physics*. CRC Press, 2001.

[2] Peeter Joot. *Basic statistical mechanics.*, chapter {Non integral binomial coefficient}. \natexlab{a}. URL http://sites.google.com/site/peeterjoot2/math2013/phy452.pdf.

[3] Peeter Joot. *Basic statistical mechanics.*, chapter {Relativisitic density of states}. \natexlab{b}. URL http://sites.google.com/site/peeterjoot2/math2013/phy452.pdf.

[4] RK Pathria. *Statistical mechanics*. Butterworth Heinemann, Oxford, UK, 1996.

[5] Wolfram. *BesselK*, \natexlab{a}. URL http://reference.wolfram.com/mathematica/ref/BesselK.html. [Online; accessed 11-April-2013].

[6] Wolfram. *HyperGeometricU*, \natexlab{b}. URL http://reference.wolfram.com/mathematica/ref/HypergeometricU.html. [Online; accessed 17-April-2013].

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