# Behavior of quantum gases at low temperature

I am taking an introductory stastistical mechanics course, and one question that was posed during lectures, was to graph the behavior of the average energy per particle ($$\overline{E}/N)$$, of a Bose-Einstein gas, a Fermi-Dirac gas, and a gas governed by equipartition from $$T = 0$$ to the semi-classical regime. My classmates and I have given significant thought to this, but have been unable to reach consensus, and have no solutions available. Regardless, the question is a good microcosm for my lack of understanding of how these gases behave.

Here is my best understanding of the things we do know, and don't know.

Things we know

• Equipartition gas is a straight line, passing through the origin, of slope $$3/2 k_B$$ (equipartition)

• Bose-Einstein line passes through 0

• Fermi-Dirac line passes through $$y = 3/5 \epsilon_f$$, at $$T = 0$$

• Fermi-Dirac, Bose-Einstein, and classical should all limit to the same curve as $$T \rightarrow \infty$$.

Things we don't know

• What is the slope of the Fermi-Diract curve at zero?

• What is the slope of the Bose-Einstein curve at zero?

• In our notes, the Stefan-Boltzmann law gives the functional form of $$E_{\mathrm{Bose-Einstein}}(T) \propto T^4$$ (as far as I can tell, for ALL $$T$$)- this contradicts the fact that it should limit to the classical result? ($$\propto T$$ at high temperatures?)

• Does the Fermi-Dirac curve or Bose-Einstein curves ever cross the classical line?

• Should there be some kind of discontinuity as a result of Bose-Einstein condensation?

Putting all of this together, and assuming that Fermi-Dirac and Bose-Einstein gases start with a slope of zero, and there is no Bose-Einstein condensation discontinuity our best guess for the behavior of the function is:

This really feels like the kind of graph that should be present in most textbooks, but my search there has turned up empty. Baierlin pg. 204 includes a plot of $$C_V(T)$$ for a Bose gas (so we could integrate that), (and it does have a discontinuity of the first derivative at Bose-Einstein condensation, but that will go away when we integrate):

As per the comment. This is a plot of Fig. 4. 8 from Coleman's book. The new question is, qualitatively why do we see the form (particularily of the Bose gas) that we do. Why is the slope larger for $$T < T_{\text{Bose-Einstein - condensation}}$$? than for $$T$$ larger?

• you have a similar graph in Coleman's Introduction to Many-Body Physics (Fig.4.8), but for pressure (which you can view as the grand canonical potential)
– LPZ
Commented Apr 24 at 14:59
• Thank you very much, unfortuantely I have been unable to find a copy of Coleman's book online that includes this figure (the google books copy omits pg. 89 and 90). Might you be able to post a photograph? (Edit: after some more digging, I realized my institution gives me access to some online materials so I've gotten that, and added it to the post)
– Jack
Commented Apr 27 at 11:41

I will focus on the internal energy. I will make everything dimensionless, and set $$s=D/2$$ for a nonrelativsitic gas in a box and $$s=D$$ for a gas in a harmonic trap or a massless (ultra relativistic) gas with $$D$$ the dimension. In dimensionless units, the density of states is: $$D(E) = \frac N{\Gamma(s)}E^{s-1}$$ Let $$z$$ be the fugacity, I'll only consider the grandcanonical ensemble in the thermodynamic limit $$N\to\infty$$.

Maxwell-Boltzmann

In the classical case: $$\frac UN = sT$$ according to equipartition. Formally, to unify the approach with the quantum gases, you first calculate the fugacity by inverting: \begin{align} N &= \int D(E)ze^{-\beta E}dE \\ &= T^s\frac N{\Gamma(s)}\int_0^{+\infty} ze^{-x}x^{s-1}dx \\ &= Nz T^s \end{align} so $$z=T^{-s}$$. You can then compute the energy: \begin{align} U &= T^{s+1}\frac N{\Gamma(s)}\int_0^{+\infty} ze^{-x}x^sdx \\ &= sNT^{s+1}z \\ &= sTN \end{align}

Bose Einstein

You have BEC transition when $$s>1$$. The critical temperature is at: $$T_c = \zeta(s)^{-1/s}$$ In the normal phase $$T>T_c$$: \begin{align} N &= T^s\frac N{\Gamma(s)}\int_0^{+\infty} \frac1{e^x/z-1}x^{s-1}dx \\ &= NT^s\text{Li}_s(z)\\ U &= T^{s+1}\frac N{\Gamma(s)}\int_0^{+\infty} \frac1{e^x/z-1}x^sdx \\ &= sNT^{s+1}\text{Li}_{s+1}(z) \end{align} so internal energy differs from equipartition by the factor: $$\frac UN = sT\frac{\text{Li}_{s+1}(z)}{\text{Li}_s(z)}$$ The extra factor is always less than $$1$$. You can check this from the power series expansion of the polylogarithm: $$\text{Li}_s(z) = \sum_{n=1}^\infty\frac{z^n}{n^s}$$ so by comparing each term, $$\text{Li}_{s+1}(z)>\text{Li}_s(z)$$.

In the BEC phase $$T, $$z=1$$ so: $$U = sNT^{s+1}\text{Li}_{s+1}(1)\\ \frac UN = s\zeta(s+1)T^{s+1}$$ The functions do not agree between the two phases, so you expect some kind of discontinuity. From the heat capacity graph, you expect the discontinuity to be of second order, so the curve should still be continuously differentiable. You can check this by asymptotic expansion $$\text{Li}$$ at $$z=1$$. The hierarchy of the terms will depend on $$s$$, you need to distinguish different cases depending on where is $$s$$ compared to $$2$$.

You need to be careful with the application of the Stefan-Boltzmann law. It is for photons (which are ultra relativistic) in 3D, so correspond to $$s=3$$. You do not recover the usual $$U\propto T$$ because you are not thinking in terms of fixed number of photons but rather fixed fugacity $$z=1$$ (i.e. chemical potential $$\mu=0$$ at the brink of condensation). Every thing is still consistent though, because $$N\propto T^3$$ as $$T\to\infty$$ so the Stefan-Boltzmann law is equivalent to equipartition.

Fermi-Dirac

You now have: \begin{align} N &= T^s\frac N{\Gamma(s)}\int_0^{+\infty} \frac1{e^x/z+1}x^{s-1}dx \\ &= NT^s(-\text{Li}_s(-z))\\ U &= T^{s+1}\frac N{\Gamma(s)}\int_0^{+\infty} \frac1{e^x/z+1}x^sdx \\ &= sNT^{s+1}(-\text{Li}_{s+1}(-z)) \end{align} You have a similar discrepancy with Maxwell-Boltzmann: $$\frac UN = sT\frac{\text{Li}_{s+1}(-z)}{\text{Li}_s(-z)}$$ This time the prefactor is always greater than $$1$$.

At very low temperature, $$z\to+\infty$$ or more precisely, using the chemical potential defined as: $$\mu = T\ln z$$ You have: $$\mu\to\epsilon_F$$ with: $$\epsilon_F = \Gamma(s+1)^{1/s}$$ As you've claimed, the energy converges to finite positive value as $$T\to0$$. The slope is $$0$$ thanks to the Sommerfield expansion which tells you that the next leading order is quadratic.

This should answer most of your questions. More importantly, you now have the analytic expressions of $$U/N$$ in terms of the standard polylogarithms, so you should be able to plot these curves yourself.

Hope this helps.

This answer is - as desired - qualitative and encompasses only the new question:

"The new question is, qualitatively why do we see the form (particularily of the Bose gas) that we do. Why is the slope larger for $$T < T_{\text{Bose-Einstein - condensation}}$$? than for $$T$$ larger?"

The real reason why the slope changes at this point is that the gaslaw-gas (ideal gas) is gradually transformed into a gas without quantum mechanics, which has less pressure at the same temperature because from $$T_{Bose}$$ downward gradually no longer all thermal levels are occupied, since enough thermal energy is no longer available for this.