At low temperatures, a mixture of ^3 He and ^4 He atoms form a liquid which separates into two

At low temperatures, a mixture of ^3 He and ^4 He atoms...

At low temperatures, a mixture of ${ }^3 \mathrm{He}$ and ${ }^4 \mathrm{He}$ atoms form a liquid which separates into two phases: a concentrated phase (nearly pure ${ }^3 \mathrm{He}$ ), and a dilute phase (roughly $6.5 \%{ }^3 \mathrm{He}$ for $T \leq 0.1 \mathrm{~K}$ ). The lighter ${ }^3 \mathrm{He}$ floats on top of the dilute phase, and ${ }^3 \mathrm{He}$ atoms can cross the phase boundary (see Fig. 2.23). The superfluid ${ }^4 \mathrm{He}$ has negligible excitation, and the thermodynamics of the dilute phase can be represented as an ideal degenerate Fermi gas of particles with density $n_{\mathrm{d}}$ and effective mass $m^*$ ( $m^*$ is larger than $m_3$, the mass of the bare ${ }^3 \mathrm{He}$ atom, due to the presence of the liquid ${ }^4 \mathrm{He}$, actually $m^*=2.4 m_3$ ). We can crudely represent the concentrated phase by an ideal degenerate Fermi gas of density $n_{\mathrm{c}}$ and particle mass $m_3$. (a) Calculate the Fermi energies for the two fluids. (b) Using simple physical arguments, make an estimate of the very low temperature specific heat of the concentrated phase $c_c\left(T, T_{\mathrm{F}_{\mathrm{c}}}\right)$ which explicitly shows its functional dependence on $T$ and $T_{\mathrm{Fc}}$ (where $T_{\mathrm{Fc}}$ is the Fermi temperature of the concentrated phase, and any constants independent of $T$ and $T_{F C}$ need not be determined). Compare the specific heats of the dilute and concentrated phases. (c) How much heat is required to warm each phase from $T=0 \mathrm{~K}$ to $T ?$ FIGURE CAN'T COPY. (d) Suppose the container in the figure is now connected to external plumbing so that ${ }^3 \mathrm{He}$ atoms can be transferred from the concentrated phase to the dilute phase at a rate of $N$, atoms per second (as in a dilution refrigerator). For fixed temperature $T$, how much power can this system absorb?

At low temperatures, a mixture of ${ }^3 \mathrm{He}$ and ${ }^4 \mathrm{He}$ atoms form a liquid which separates into two phases: a concentrated phase (nearly pure ${ }^3 \mathrm{He}$ ), and a dilute phase (roughly $6.5 \%{ }^3 \mathrm{He}$ for $T \leq 0.1 \mathrm{~K}$ ). The lighter ${ }^3 \mathrm{He}$ floats on top of the dilute phase, and ${ }^3 \mathrm{He}$ atoms can cross the phase boundary (see Fig. 2.23).

The superfluid ${ }^4 \mathrm{He}$ has negligible excitation, and the thermodynamics of the dilute phase can be represented as an ideal degenerate Fermi gas of particles with density $n_{\mathrm{d}}$ and effective mass $m^*$ ( $m^*$ is larger than $m_3$, the mass of the bare ${ }^3 \mathrm{He}$ atom, due to the presence of the liquid ${ }^4 \mathrm{He}$, actually $m^*=2.4 m_3$ ). We can crudely represent the concentrated phase by an ideal degenerate Fermi gas of density $n_{\mathrm{c}}$ and particle mass $m_3$.
(a) Calculate the Fermi energies for the two fluids.
(b) Using simple physical arguments, make an estimate of the very low temperature specific heat of the concentrated phase $c_c\left(T, T_{\mathrm{F}_{\mathrm{c}}}\right)$ which explicitly shows its functional dependence on $T$ and $T_{\mathrm{Fc}}$ (where $T_{\mathrm{Fc}}$ is the Fermi temperature of the concentrated phase, and any constants independent of $T$ and $T_{F C}$ need not be determined). Compare the specific heats of the dilute and concentrated phases.
(c) How much heat is required to warm each phase from $T=0 \mathrm{~K}$ to $T ?$
FIGURE CAN'T COPY.
(d) Suppose the container in the figure is now connected to external plumbing so that ${ }^3 \mathrm{He}$ atoms can be transferred from the concentrated phase to the dilute phase at a rate of $N$, atoms per second (as in a dilution refrigerator). For fixed temperature $T$, how much power can this system absorb?

Key Concepts

Degenerate Fermi Gas

A degenerate Fermi gas is a system of fermions at temperatures much lower than their Fermi temperature, where the majority of particles occupy states up to the Fermi level. In this regime, quantum effects dominate and the distribution of particles is governed by the Pauli exclusion principle, leading to a sharp cutoff in occupied energy states even at zero temperature. This concept is central to understanding the behavior of electrons in metals and quasiparticles in systems such as helium mixtures at very low temperatures.

Fermi Energy and Fermi Temperature

The Fermi energy is the highest occupied energy level in a degenerate Fermi gas at absolute zero temperature, and it serves as a fundamental energy scale of the system. The Fermi temperature is derived by relating the Fermi energy to temperature via Boltzmann’s constant. These quantities set limits on the thermal excitations and determine the behavior of the system’s thermodynamic properties. Calculating the Fermi energy is crucial for predicting phenomena like specific heat and thermal conductivity in quantum fluids.

Low Temperature Specific Heat in Fermi Systems

At very low temperatures, the specific heat of a degenerate Fermi gas is dominated by excitations near the Fermi surface, leading to a linear temperature dependence. This behavior arises because only a small number of particles within kBT of the Fermi energy can be thermally excited. The concept is essential in determining the thermodynamic response of systems like electrons in metals or quasiparticles in helium mixtures, where the microscopic energy scale provided by the Fermi energy controls the macroscopic heat capacity.

Phase Separation and Mass Renormalization

Phase separation refers to the phenomenon where a mixture segregates into distinct regions with different compositions, as seen in helium-3 and helium-4 systems at low temperatures. In such systems, the interaction with the surrounding medium can effectively increase the mass of the quasiparticles—a process known as mass renormalization. This renormalized effective mass has significant implications for the dynamics and thermodynamics of the concentrated and dilute phases, affecting quantities such as the Fermi energy and specific heat.

Heat Transfer in Quantum Liquids and Dilution Refrigeration

In systems where atoms or quasiparticles are transferred between phases, such as in dilution refrigerators, the rate of transfer and the associated change in energy distributions become crucial. The concept involves understanding how the energy linked with each quantum of matter contributes to the overall heat flow, thereby setting limits on the cooling power or heat absorption capacity of the system. This interplay between quantum statistics and energy exchange underlies the techniques used to achieve ultralow temperatures in experimental setups.

Transcript

0:00 Hi there.

00:01 So for this problem, we need to model the alien 3 as a non -interacting fermi gas.

00:09 Although alien 3 liquidifies at low temperatures, the liquid has an unusual low density and behaves in many waves like a gas because the forces between the atoms are so weak.

00:20 So helium -3 atoms are a spin -half, one -half fermions because of the impaired neutron in the nucleus.

00:30 So for part a of this problem, we need to, pretending that liquid ilium 3 is a non -interrupting fermi gas, we need to calculate the fermi energy and the fermi temperature.

00:43 The molar volume allowed temperatures, we know that it is 37 cubic centimeters.

00:54 So to start this problem, we know that the fermi energy of a gas of ilium 3 atoms with the given density is from the following equation.

01:08 We know that that is.

01:10 Plam's constant to the square divided by 8 times the mass and 3 times the number n divided by pi times the volume.

01:29 And this elevated to 2 over 3.

01:33 So in here, substituting the values, we have 6 .63 times 10 to the minus 34 joules times seconds and that elevated to the square and this divided by 8 times 3 times the mass for a proton that we know is 1 .66 times 10 to the minus 27 kilograms.

01:56 So we multiply this by three times the number n that we are going to assume is the abogartis number.

02:03 So it's 6 .02 times 10 to the 23.

02:12 And this divided by the volume that we are given.

02:15 So this is pi times 37 times 10 to the minus 6 because we convert this from centimeters to meters, so meters square.

02:27 And all of this elevated to 2 over 3.

02:33 So from this we obtain fermi energy.

02:37 That is equal to 6 .9 times 10 to the minus 23 joules.

02:45 That we can also write as 4 .3 times 10 to the minus 4 electron bulbs.

02:53 So the fermit temperature is therefore simply this value that we just have obtained divided by bulsman constant.

03:03 So we will have that this is 4 .3 times 10 to the minus 4 .4.

03:08 Electron balls divided by the balsman constant which we know is 8 .62 times 10 to the minus 5 jules per kelvin.

03:17 So from this we obtain a value of 5 kelvin.

03:28 So that's a solution for the temperature and for the fermi energy.

03:35 Now for part b of this problem we are asked to calculate the heat capacity for that temperature less than the fermi temperature and compared to the experimental result.

03:52 So for this part of the problem, we have as predicted by equation 7 .48, the heat capacity should be the heat capacity divided by n times pulsman constant times the temperature.

04:24 And this is equal to pi to the squared times bolzman constant divided by two times the fermi energy.

04:46 And from this, we can simplify this in terms of the fermi temperature.

04:53 So we obtain that this is pi to the square divided by two times the fermi temperature.

04:58 So from this we obtain 1 kelvin to the minus 1...

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