Heat Capacity of Ideal Bose Gas: Temperature Dependence and the Cusp at T_c

1. Internal Energy and the Heat Capacity

Our starting point is the total internal energy, $E$, of the gas. As you've seen in previous derivations, for a non-relativistic gas in 3D, the energy is related to pressure via $E=\frac{3}{2}PV$. We previously found an integral expression for pressure, which leads to an expression for the energy in terms of the Bose function $g_{5/2}(z)$ and the thermal de Broglie wavelength $\lambda$:

where $z=e^{\beta \mu }$ is the fugacity and $\lambda =h/\sqrt{2\pi m{k}_{B}T}$.

The heat capacity at constant volume is defined as $C_V=(\partial E/\partial T{)}_{N,V}$. When we take this derivative, we must be careful. Both $\lambda$ and, in general, the fugacity $z$ are functions of temperature. Applying the chain rule gives:

This expression is the foundation for our analysis. Its evaluation depends critically on whether we are above or below $T_c$, because the behavior of $z(T)$ is fundamentally different in the two regimes.

2. Heat Capacity Below $T_c$

Let's first consider the low-temperature phase, $T<T_c$. As we established, in this regime, a macroscopic number of particles occupy the ground state. To accommodate this, the chemical potential is effectively pinned at its maximum value, $\mu \approx 0$, which means the fugacity $z=e^{\beta \mu }\approx 1$. Since $z$ is constant, its derivative with respect to temperature is zero, $dz/dT=0$. This significantly simplifies the calculation of $C_V$.

With $z=1$, the Bose function $g_{5/2}(1)$ is just the Riemann zeta function $\zeta (5/2)\approx 1.341$. The energy expression becomes:

Collecting all the constants, we see that the energy is purely a function of temperature with the dependence $E\propto T^{5/2}$.

The heat capacity is now straightforward to compute:

This shows that for $T<T_c$, the heat capacity follows a $T^{3/2}$ power law.

To get a more informative expression, we can normalize it by the value at $T_c$. The result is a simple and elegant scaling relation:

The following reading provides the detailed derivation of this result and calculates the value of $C_V$ at the critical point.

The value $C_V/(N{k}_B)\approx 1.925$ at $T_c$ is notably larger than the classical value of $1.5$ for a monatomic ideal gas, indicating that the quantum nature of the bosons leads to an enhanced ability to store energy near the transition.

3. Heat Capacity Above $T_c$

Now, let's turn to the normal phase, $T>T_c$. Here, there is no condensate, and the chemical potential must adjust with temperature ($\mu <0$, so $z<1$) to keep the total particle number $N$ fixed. The constraint is given by:

Since $N$ and $V$ are constant, we have $g_{3/2}(z)/{\lambda }^3=\text{const}$. Because $\lambda$ depends on $T$, $z$ must also depend on $T$. This means the term with $dz/dT$ in our general expression for $C_V$ is now non-zero.

The calculation proceeds in two steps:

1. Find an expression for $dz/dT$ by differentiating the particle number constraint equation with respect to $T$.
2. Substitute this expression into the full derivative for $C_V$.

This procedure is mathematically more involved than the case for $T<T_c$.

The final expression for the heat capacity per particle for $T>T_c$ is:

This expression is a function of temperature through the implicit temperature dependence of $z(T)$.

4. The Cusp at the Critical Temperature

We now have expressions for $C_V$ on both sides of $T_c$. The crucial question is: what happens at $T=T_c$?

Continuity of $C_V$:
First, let's check if the function $C_V(T)$ is continuous at $T_c$.

• From below, we found $C_V(T_c^{-})/(N{k}_B)=\frac{15}{4}\frac{\zeta (5/2)}{\zeta (3/2)}\approx 1.925$.
• From above, we must take the limit of our second expression as $T\to T_c^{+}$, which means $z\to 1^{-}$. As $z\to 1$, the first term becomes $\frac{15}{4}\frac{\zeta (5/2)}{\zeta (3/2)}$. For the second term, we need the behavior of $g_{1/2}(z)$ as $z\to 1$. The series definition is $g_{1/2}(1)={\sum }_{l=1}^{\infty }{l}^{-1/2}$, which is a divergent p-series. Thus, $g_{1/2}(1)=\infty$. The second term vanishes in the limit.

So, $C_V(T_c^{+})=C_V(T_c^{-})$. The heat capacity is continuous at the critical temperature.

Discontinuity of the Derivative $d{C}_V/dT$:
While $C_V$ is continuous, its slope is not. This is the origin of the "cusp".

• For $T<T_c$, we have $C_V\propto T^{3/2}$, so $d{C}_V/dT\propto T^{1/2}$. At $T_c$, this derivative is finite and positive.
• For $T>T_c$, one must differentiate the more complex expression for $C_V$. This is a rather tedious exercise in calculus, involving derivatives of the Bose functions.

The following resource shows the result of this calculation and demonstrates the discontinuity. You don't need to reproduce every algebraic step, but you should follow the logic to appreciate the result.

The discontinuity in the derivative of the heat capacity gives rise to the characteristic shape shown in the plot from the Rochester notes.

This cusp is a classic example of what is known as a lambda transition, named for the $\lambda$-shape of the specific heat curve. It is a hallmark of a continuous (or second-order) phase transition.

Conclusion

In this lesson, we have derived the temperature dependence of the heat capacity of an ideal Bose gas and uncovered one of the most important thermodynamic signatures of Bose-Einstein condensation.

Key Takeaways:

• The behavior of the chemical potential (or fugacity) dictates the calculation of $C_V$. Below $T_c$, $\mu \approx 0$ and is constant, while above $T_c$, $\mu$ varies with temperature to conserve particle number.
• For $T<T_c$, the heat capacity grows as $C_V\propto T^{3/2}$. This dependence arises directly from the energy of the thermally excited bosons, as the condensate itself has zero entropy and does not contribute to the heat capacity.
• For $T>T_c$, the expression for $C_V$ is more complex, but at high temperatures ($T\gg T_c$), it correctly approaches the classical ideal gas value of $\frac{3}{2}N{k}_B$.
• At the critical temperature $T_c$, the heat capacity function is continuous, but its derivative is discontinuous. This creates a distinctive cusp, a tell-tale sign of a continuous phase transition.

So far, our entire discussion of BEC has been for a gas of non-interacting, non-relativistic particles in three dimensions. In our next lesson, we will explore a crucial question: Why does Bose-Einstein condensation not occur in 1D or 2D for such a gas? This will deepen your understanding of how dimensionality plays a fundamental role in the physics of phase transitions.