Unitary Thermodynamics from Thermodynamic Geometry

Abstract

Degenerate Fermi gases of atoms near a Feshbach resonance show universal thermodynamic properties, which are here calculated with the geometry of thermodynamics, and the thermodynamic curvature R. Unitary thermodynamics is expressed as the solution to a pair of ordinary differential equations, a “superfluid” one valid for small entropy per atom z≡S/Nk _B , and a “normal” one valid for high z. These two solutions are joined at a second-order phase transition at z=z _c . Define the internal energy per atom, in units of the Fermi energy, as Y=Y(z). For small z, Y(z)=y ₀+y ₁ z ^α+y ₂ z ^2α+⋯, where α is a constant exponent, y ₀ and y ₁ are scaling factors, and the series coefficients y _i (i≥2) are determined uniquely in terms of (α,y ₀,y ₁). For large z the solution follows if we also specify z _c , with Y(z) diverging as z ^5/3 for high z. The four undetermined parameters (α,y ₀,y ₁,z _c ) were determined by fitting the theory to experimental data taken by a Duke University group on ⁶Li in an optical trap with a Gaussian potential. The very best fit of this theory to the data had α=2.1, z _c =4.7, y ₀=0.277, and y ₁=0.0735, with χ ²=0.95. The corresponding Bertsch parameter is ξ _B =0.462(40).

Appendices

Appendix 1: Geometric Equation

In this Appendix, I justify the uniqueness of the geometric equation

$$ R=-\frac{\kappa}{\phi}, $$

(41)

with κ a dimensionless constant of order unity, and the thermodynamic potential per volume, in units of k _B ,

$$ \phi=\frac{p}{k_B T}. $$

(42)

ϕ is always employed in conjunction with an appropriate background subtraction, depending on the singular point around which the solution is generated. There are two types of singular points: (1) singular points with |R|→∞, where the background subtraction is made from ϕ, and (2) singular points with R→0, where the background subtraction is made from 1/ϕ.

Singular points with |R|→∞ are encountered at critical points, where intermolecular interactions strongly organize the atoms. Singular points with R→0 are encountered in near ideal gases where intermolecular interactions have little effect. Considerations of simplicity, units, mathematical viability, and universality of κ turn out to give us little choice about the correct form of ϕ and in the subtraction of the nonsingular part. Although I state my arguments in the context of the specific physical problem here, such arguments have been used in quite different physical settings: the simple critical point [24], galaxy clustering [28], corrections to scaling [29], the paramagnetic ideal gas [30], and gases with power law interactions [31]. In all these cases, the geometric equation takes the same form as that presented here, and with the same prescription for the background subtraction.

Consider a singular point with |R|→∞, and start the discussion with hyperscaling. Widom [22] argued that, near a critical point, average fluctuations in the free energy Φ in a volume of size ξ ³ should be ∼k _B T. If we further take these fluctuations to be the singular part of the free energy itself, we get the hyperscaling assumption:

$$ \frac{\varPhi-\varPhi_0}{V}\sim\frac{k_B T}{\xi^3}, $$

(43)

where Φ ₀ denotes the value of Φ at the critical point, which we must subtract to get the singular properties corresponding to ξ→∞.

To connect ξ to thermodynamic properties in hyperscaling, it is necessary to express Φ in thermodynamic terms. Widom [22] was not explicit in this regard, but he nevertheless used Eq. (43) to write the hyperscaling exponent scaling relation 2−α=d  ν between the heat capacity (at constant volume) exponent α and the correlation length exponent ν, in spatial dimension d. Goodstein [23] made an argument similar to Widom’s in spirit, and picked Φ as the Gibbs free energy. He stated, however, that “it will turn out not to make any difference which energy function we choose.”

Such looseness with the precise definition of Φ is sufficient if we are interested only in the hyperscaling exponent scaling relation, but in the context of the geometric equation we must be more precise. Bringing in the thermodynamic curvature R forces a sharpening of the argument. In the critical regime, R connects to ξ via |R|∼ξ ³, a proportionality resulting both from direct calculations in a number of cases (see [5, 15, 19] for review), and from a covariant theory of thermodynamic fluctuations [16–18].

Replacing ξ ³ with R in Eq. (43) leads to

$$ R=-\kappa \biggl(\frac{k_B T V}{\varPhi-\varPhi_0} \biggr), $$

(44)

with κ a dimensionless constant of order unity and “∼” replaced by “=”. R has units of volume per molecule, so Φ must have units of energy. I will try all possible Φ’s constructed from the four free energy building blocks {E,−TS,pV,−μN}, each extensive and each with units of energy. I find that of the 14 nontrivial possibilities, only Φ=pV, and cases with Φ∝pV, are viable.

The specific procedure for testing Φ’s is: (1) Pick a singular point where, on physical grounds, we expect |R|→∞, and guess some physically motivated series yielding the thermodynamics. Such a series should contain undetermined coefficients, to be evaluated by series solution of the geometric equation. (2) Expand R in terms of this series. (3) Construct a candidate Φ from the four building blocks {E,−TS,pV,−μN}, and evaluate it at the singular point to get Φ ₀. (4) Construct the series for −κk _B TV/(Φ−Φ ₀), with κ undetermined. (5) Equate this series to the one for R and see whether the resulting κ is universal. By “universal,” I mean independent of the specific constants found in the solution. (6) Repeat this procedure until all candidate Φ’s have been tried.

For unitary thermodynamics, we expect R→+∞ at z=0, characteristic of the noninteracting Fermi gas [20, 21]. Try a Puisseux series

$$ Y(z)=y_0+y_1 z^{\alpha}+y_2 z^{2 \alpha}+\cdots $$

(45)

to calculate the energy,

$$ E=N\epsilon_F(\rho) Y(z). $$

(46)

Equations (6), (45), and (46) lead to

$$ R=\frac{k_B\alpha V}{2S} \biggl[1- \biggl(\frac{(3 \alpha -5) y_1^2+10 y_0 y_2}{5 (\alpha-1) y_0 y_1} \biggr)x+O \bigl(x^2 \bigr) \biggr], $$

(47)

where x≡z ^α.

As the first free energy candidate, try Φ=pV, the case featured in this paper. Since {T,p}={E _,S ,−E _,V }, Eqs. (45) and (46) lead to

$$ p_0 = \frac{2}{3} c y_0 \rho^{5/3}, $$

(48)

with

$$ c = \frac{3^{2/3}\pi^{4/3}\hbar^2}{2m}, $$

(49)

and

$$ \phi_{\mathit{sing}}\equiv\frac{p-p_{0}}{k_B T} = \frac{2S}{k_B \alpha V} \biggl[ \frac{1}{3}- \biggl(\frac{y_2}{3 y_1} \biggr)x+O \bigl(x^2 \bigr) \biggr]. $$

(50)

Thus

$$ -\kappa=R \phi_{\mathit{sing}}=\frac{1}{3}- \biggl[\frac{(3 \alpha-5) y_1^2+5 (\alpha+1) y_0 y_2}{15 (\alpha-1) y_0 y_1)} \biggr]x + O \bigl(x^2 \bigr). $$

(51)

Clearly, κ must take the universal value κ=−1/3, regardless the values of α and the series coefficients. Setting the first-order term on the right-hand side of Eq. (51) to zero requires $y_{2}=-(3 \alpha-5) y_{1}^{2}/[5 (\alpha+1) y_{0}]$, and setting higher-order terms to zero uniquely determines all the series coefficients in terms of {α,y ₀,y ₁}.

All 14 possible nonzero ways of creating a free energy Φ are presented in Table 1. The form of the fundamental equation in Eq. (46) leads to E=3pV/2. Generally, we also have the Gibbs-Duhem equation E=TS−pV+μN. Hence, a number of the candidate Φ’s in Table 1 have Φ∝pV, and they all lead to the same form of the geometric equation as Φ=pV. Cases with Φ∝pV are the only ones with universal κ. All other cases have κ depending on the solution dependent α, and are hence unacceptable.

Table 1 All 14 nontrivial free energies Φ constructed from {E,−TS,pV,−μN}. Also shown are Φ ₀ and Rϕ _sing for small z, where ϕ _sing ≡(Φ−Φ ₀)/k _B TV. A number of cases have Φ∝pV, of which all have values of κ independent of solution. The other cases have κ depend on α, and are hence nonuniversal

Finally, turn attention briefly to singular points with R→0. For unitary thermodynamics, such singular points correspond to z→∞. For singular points with R→0, we must subtract 1/ϕ ₀ from 1/ϕ to get R=0, where ϕ ₀ is Φ evaluated at the singular point. I will not present an explicit analysis of this case, since it is clear that we must have Φ=pV to be consistent with the case above with z→0. Likewise, it turns out that there is little choice about the appropriate background subtraction. The geometric equation for singular points with R→0 is thus

$$ R = -\kappa \biggl[\frac{k_B T}{p}- \biggl(\frac{k_B T}{p} \biggr)_0 \biggr], $$

(52)

were the quantity in parentheses on the right-hand side is evaluated at the singular point. As shown in Sect. 3.2, this equation is entirely solvable, with κ=−1/3, the same as for the small z solution.

Appendix 2: LDA → TRAP

In this Appendix, I present the basics of connecting the uniform thermodynamics (LDA) to the measured overall properties of a fluid in a trap. This topic was discussed by Haussmann and Zwerger [26], and I add to their discussion in this Appendix a simple ad hoc power law toy example, which I fit to experimental data from the Duke University group [6]. This fit is not expected to be particularly good, but it does raise some useful points for discussion.

Consider a thermodynamic system in a trap where an atom at position $\vec{r}$ experiences a known external potential energy per atom $U(\vec {r})$ in addition to the net potential energy contributed by the other atoms in the system. Let $U(\vec{r})$ have a minimum U(0)=0, and increase monotonically with $r=|\vec{r}|$ in all directions. Assume that the LDA is also known and is given by the fundamental equation E(S,N,V)=Ve(s,ρ), where e(s,ρ)=E/V is the internal energy per volume, s=S/V, and ρ=N/V. The temperature and the thermodynamic chemical potential are given by {T,μ}={e, _s ,e, _ρ }. Logically, in going from an LDA to a trap thermodynamics, the volume variable V gets replaced by the potential energy per atom $U(\vec{r})$ [32].

T and the total chemical potential $\mu_{0} = \mu+ U(\vec{r})$ are both constant throughout the trap. The basic question is: for a known LDA and given T, μ ₀, and $U(\vec{r})$, what are the energy E _t , entropy S _t , and number of atoms N _t in the trap? To determine this transformation $\{T, \mu_{0}, U(\vec{r})\}\to\{E_{t}, S_{t}, N_{t}\}$, proceed as follows: (1) Determine the local $\mu=\mu_{0} - U(\vec {r})$ for all $\vec{r}$. (2) Algebraically solve (T=e, _s ,μ=e, _ρ ) for $(s(\vec{r}), \rho(\vec{r}))$ in terms of (T,μ ₀) for all $\vec{r}$. (3) Identify the boundary (or edge) of the trap by finding the surface over which ρ=0. (4) Integrate over the volume of the trap out to the edge:

$$ E_t=\int \bigl[e\bigl(s(\vec{r}),\rho(\vec{r})\bigr) + \rho(\vec {r})U(\vec{r}) \bigr]\, d^3r, $$

(53)

$$ S_t=\int s(\vec{r})\,d^3r, $$

(54)

and

$$ N_t=\int\rho(\vec{r})\,d^3r. $$

(55)

Several questions come up at the edge of the trap. In some direction, is there a finite distance r where ρ→0, or does ρ instead slowly peter out only as r→∞? Need $U(\vec{r})$ diverge to infinity to contain the atoms in the trap? Does ρ→0 imply z→∞? Do the integrals above for (E _t ,S _t ,N _t ) converge?

To illuminate these issues, consider a simple toy example based on the scaled equation of state in Eq. (15):

$$ e(s, \rho) = \rho \epsilon_F(\rho) Y(z), $$

(56)

with z=s/k _B ρ. Take a power law,

$$ Y(z) = y_0 + y_1 z^{\alpha}, $$

(57)

with constants y ₀>0, y ₁>0, and α>1. T=e, _s yields

$$ \rho(T,z) = \frac{2\sqrt{2}}{3 \pi^2} \biggl(\frac{m k_B T}{\alpha y_1z^{\alpha-1} \hbar^2} \biggr)^{3/2}. $$

(58)

The trap edge ρ→0 clearly corresponds to z→∞, physically reasonable since the space available to an atom, and hence the entropy per atom, grows without limit as ρ→0. The condition $e_{,\rho}=\mu= \mu_{0} - U(\vec{r})$ and Eq. (58) yields

$$ -k_B T z + \frac{5 k_B T z}{3 \alpha} + \frac{5 y_0 k_B T }{3\alpha y_1 z^{\alpha-1}} = \mu_0 - U(\vec{r}). $$

(59)

If α=5/3, the terms linear in z in Eq. (59) cancel, and Eqs. (58) and (59) lead to

$$ s(T,\mu_0,\vec{r}) = \frac{2}{5}\sqrt{\frac{6}{5}} \frac {k_B}{\pi^2} \biggl[\frac{m k_B T}{\hbar^2 y_1} \biggr]^{3/2}, $$

(60)

with density profile

$$ \rho(T,\mu_0,\vec{r}) = \frac{2}{5}\sqrt{\frac {6}{5}} \frac{1}{\pi^2} \biggl\{ \frac{m[\mu_0-U(\vec{r})]}{\hbar^2 y_0} \biggr\} ^{3/2}. $$

(61)

This $\rho(T,\mu_{0},\vec{r})$ is independent of T and follows the Thomas-Fermi density profile [33]. $e(T,\mu _{0},\vec{r})$ now follows from e(s,ρ) since s and ρ are known at this point in terms of $(T,\mu_{0},\vec{r})$. Clearly, a real valued ρ at $\vec{r}$ requires $\mu_{0}>U(\vec{r})$, which is consistent with Eq. (57). As r increases from zero in some direction, $U(\vec{r})$ increases until $U(\vec{r}) = \mu_{0}$, assuming that μ ₀ is not too big. When $U(\vec{r})\to\mu_{0}$, we get ρ→0 and z→∞, corresponding to the trap edge. Assuming that we have such a trap edge in every direction, the integrals for E _t , S _t , and N _t will all converge, since the volume of integration is finite.

With α>1, but α≠5/3, the linear z terms dominate in Eq. (59) as z→∞. But diverging z now requires $|\mu _{0}-U(\vec{r})|\to\infty$, which clearly cannot happen since $0\le U(\vec {r})\le\mu_{0}$, and μ ₀ has been set to some fixed value characteristic of the entire system. α=5/3 is thus the only exponent leading to a clear trap edge, and with no need for any infinity in $U(\vec{r})$ to confine the atoms.

As a practical exercise, let me compare the toy LDA in Eq. (57), with α=5/3, with the experimental trap data collected by the Duke group [6]. The Duke group used an optical trap with a Gaussian potential

$$ U(\vec{r})=U_0 \bigl[1-\mbox{exp} \bigl(-2 \tilde {r}^2 \bigr) \bigr], $$

(62)

where U ₀=10 μK k _B , $\tilde{r}^{2} = (x_{1}/a_{1})^{2} + (x_{2}/a_{2})^{2} + (x_{3}/a_{3})^{2}$, (x ₁,x ₂,x ₃) are the spatial coordinates, and {a ₁,a ₂,a ₃}={52.20,45.44,1153.2} μm. Define also the trap Fermi energy (for a harmonic trap)

$$ E_F(N_t)= \hbar(\omega_1 \omega_2\omega_3)^{1/3}(3 N_t)^{1/3}, $$

(63)

used to scale the experimental energy data. The two transverse and the axial trap frequencies are {ω ₁,ω ₂,ω ₃}=2π{665,764,30.1} Hz, respectively, with $\omega _{i}=\sqrt{4U_{0}/ma_{i}^{2}}$ (i=1,2,3), and m=6.015 amu is the mass of a ⁶Li atom. The trap edge has $\mu_{0}=U(\vec{r})$, corresponding to

$$ \tilde{r} = \sqrt{\frac{1}{2} \ln \biggl(\frac{U_0}{U_0-\mu_0} \biggr)}. $$

(64)

Clearly, $\tilde{r}$ increases as μ ₀ increases from 0, and $\tilde{r}\to\infty$ as μ ₀→U ₀.

An essential quantity in the data analysis of a function y depending on x is

$$ \chi^2\equiv\frac{1}{n}\sum_i\biggl[\frac {(y-y_i)}{\sigma_i} \biggr]^2, $$

(65)

where y and y _i denote theoretical and experimental values, respectively, for the i’th of the n data points, and σ _i is the standard deviation for y _i . If there are error bars on both the x and the y axes, we take [34]

$$ \sigma_i^2\rightarrow\sigma_y^2+ \biggl(\frac{dy}{dx} \biggr)^2 \sigma_x^2, $$

(66)

where σ _x and σ _y are the errors in x and y, respectively.

The Duke experiment measured (E _t ,S _t ,N _t ) directly, with the experiment done at constant N _t ≃1.3×10⁵. There was no use of a heat bath or an atom bath, so (T,μ ₀) for any particular data run were not known a priori. I determine (T,μ ₀) as needed by data fitting in the context of some theoretical LDA. To connect some theoretical LDA to experimental data spanning a range of T, I proceed as follows: (1) Set some small T. (2) Set some μ ₀, and adjust its value until N _t in Eq. (55) matches the experimental value. (3) Integrate over the trap with these values of (T,μ ₀) to find the theoretical E _t and S _t . This step requires the transformation method $\{T, \mu_{0}, U(\vec{r})\}\to\{E_{t}, S_{t}, N_{t}\}$ described above in this Appendix. (4) Increment T to a higher value and repeat with step 2 until we have a theoretical curve of E _t /E _F (N _t )N _t versus S _t /k _B N _t spanning the full experimental data curve.² (5) Calculate χ ² for the data consisting of n pairs of (S _t /k _B N _t ,E _t /E _F (N _t )N _t ). (6) Repeat this entire procedure with incremented y ₀ and y ₁ to minimize χ ² for the best fit between experiment and theory.

Results are shown in Fig. 7 for the toy LDA in Eq. (57), with the best fit on varying the two parameters y ₀ and y ₁ having χ ²=3.63. Clearly, this toy model, with just two fit parameters and no phase transition, does not produce a particularly good fit. The results in Sect. 4, with the two-piece LDA constructed from the geometric equation, are much superior.

Fig. 7

The Duke data [6] (E ₈₄₀/E _F versus S ₁₂₀₀/k _B ) and the trap integrated toy model Y(z)=0.2401+0.1392z ^5/3, which was found to minimize χ ²=3.63 (Color figure online)

The Duke experiment operated at constant N _t , which in the context of the toy LDA Eq. (57) translates to constant μ ₀. With the best fit in Fig. 4, we have μ ₀/U ₀=0.0546, a value which clearly has the atoms down near the bottom of the trap.

Appendix 3: The Virial Theorem

In this Appendix 1 discuss the viral theorem

$$ E_t = 2 \int\rho(\vec{r})U(\vec{r})\,d^3 r, $$

(67)

which enables experimentalists to determine E _t just by measuring density profiles $\rho(\vec{r})$. Thomas et al. [25] argued that the virial theorem, valid for the ideal gas, holds as well in the strongly interacting Fermi fluid.

However, the Thomas derivation [25] assumes implicitly that the pressure goes to zero at the edge of the trap. This should be the case if the gas behaves like an ideal gas near the trap edge. But with a power law divergence, a surface term involving the pressure enters the picture, as I will now demonstrate. The Gibbs-Duhem equation at constant T yields ρ  dμ=dp. Since dμ=−dU, we get

$$ \nabla p(\vec{r}) + \rho(\vec{r})\nabla U(\vec{r}) = 0, $$

(68)

the condition of hydrostatic equilibrium. Assume now a harmonic potential, for which $\vec{r}\cdot\nabla U(\vec{r}) = 2 U(\vec {r})$, and assume p=2E/3V, valid for the scaled fundamental equation in Eq. (15). Taking the dot product of both sides of Eq. (68) with $\vec{r}$, and integrating over the trapped sample, leads to

$$ E_t = 2 \int\rho(\vec{r})U(\vec{r})\,d^3 r + \frac{1}{2}\oint p(\vec{r}) \vec{r} \cdot\hat{n}dA, $$

(69)

where $\hat{n}$ is a unit normal to the surface ρ=0.

The surface term in Eq. (69) is zero if we assume that the thermodynamics at the trap edge is that of the ideal gas, resulting in the virial theorem Eq. (67). For the monatomic ideal gas, we have the Sackur-Tetrode equation

$$ \frac{E}{N\epsilon_F} = e_0\exp \biggl( \frac{2S}{3k_B N} \biggr), $$

(70)

where e ₀ is a constant, and ϵ _F ∝ρ ^2/3. Since E=3Nk _B T/2, then for given T as ρ→0, we get z→∞. Also, p=ρk _B T, which leads immediately to p→0 as ρ→0, and the surface term in Eq. (69) is zero.

For the power law behavior, the series for Y _H (z) Eq. (26) yields a pressure

$$ p = \frac{4 \sqrt{\frac{6}{5}} m^{3/2}}{25 \pi^2 \hbar^3}\frac{(k_B T)^{5/2}}{ y_{-1}^{3/2}}, $$

(71)

which does not go to zero at fixed T as either ρ→0 or z→∞. The surface term in Eq. (69) will thus modify the virial theorem except at very small T. But, this should not affect the analysis in this paper.