Energy Fluctuations and Heat Capacity
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You should be comfortable calculating the canonical partition function \(Z\) for simple systems and understanding that the average energy is given by \(\langle E \rangle = -\frac{\partial \ln Z}{\partial \beta}\).
Why this matters
Heat capacity is not just a measure of how much energy a system absorbs; it is a direct measure of how much its energy fluctuates.
In the canonical ensemble, a system exchanges energy with a heat bath, meaning its instantaneous energy is not fixed but varies around a mean value. A common misconception is that thermodynamics describes only average, static properties. However, the connection between the variance of energy and the heat capacity reveals that macroscopic thermodynamic quantities are fundamentally rooted in microscopic statistical fluctuations. Understanding this link allows us to diagnose the stability of a system and understand why heat capacity diverges at phase transitions, where fluctuations become macroscopic.
The Intuitive Link
Imagine a small cup of water sitting in a large room. The room acts as a heat bath at a constant temperature. The water molecules are constantly jostling, and the total kinetic energy of the water molecules is not a single, rigid number. Instead, it wiggles up and down around an average value. If the water is very "stiff" in its energy response—meaning it takes a lot of heat to raise its temperature slightly—then those wiggles are small. Conversely, if the system is very "soft" or easily excited, the energy wiggles are large. The heat capacity \(C_V\) quantifies this "softness." Therefore, a high heat capacity implies large energy fluctuations, and a low heat capacity implies small fluctuations. This relationship is exact, not just an approximation, and it holds for any system in thermal equilibrium.
Mathematical Derivation of the Variance
To connect the fluctuation \(\langle (\Delta E)^2 \rangle\) to the partition function, we start with the definition of the average energy in the canonical ensemble. The probability of a state with energy \(E_i\) is \(p_i = \frac{e^{-\beta E_i}}{Z}\). The average energy is \(\langle E \rangle = \sum_i E_i p_i\). We can express this using the partition function \(Z = \sum_i e^{-\beta E_i}\) as \(\langle E \rangle = -\frac{1}{Z} \frac{\partial Z}{\partial \beta} = -\frac{\partial \ln Z}{\partial \beta}\).
Now, consider the second derivative of \(\ln Z\) with respect to \(\beta\). First, differentiate \(\langle E \rangle\): \[ \frac{\partial \langle E \rangle}{\partial \beta} = -\frac{\partial^2 \ln Z}{\partial \beta^2} \] Let us compute \(\frac{\partial \langle E \rangle}{\partial \beta}\) directly from the sum definition. \[ \frac{\partial \langle E \rangle}{\partial \beta} = \frac{\partial}{\partial \beta} \left( \frac{\sum_i E_i e^{-\beta E_i}}{\sum_j e^{-\beta E_j}} \right) \] Using the quotient rule, this expands to: \[ \frac{\partial \langle E \rangle}{\partial \beta} = \frac{ \left( \sum_i -E_i^2 e^{-\beta E_i} \right) Z - \left( \sum_i E_i e^{-\beta E_i} \right) \left( \sum_j -E_j e^{-\beta E_j} \right) }{Z^2} \] Simplifying the numerator by dividing by \(Z^2\): \[ \frac{\partial \langle E \rangle}{\partial \beta} = \frac{\sum_i -E_i^2 e^{-\beta E_i}}{Z} - \frac{\left( \sum_i E_i e^{-\beta E_i} \right) \left( \sum_j E_j e^{-\beta E_j} \right)}{Z^2} \] Recognising that \(\frac{\sum_i E_i^2 e^{-\beta E_i}}{Z} = \langle E^2 \rangle\) and \(\frac{\sum_i E_i e^{-\beta E_i}}{Z} = \langle E \rangle\), we get: \[ \frac{\partial \langle E \rangle}{\partial \beta} = -\langle E^2 \rangle + \langle E \rangle^2 = -\left( \langle E^2 \rangle - \langle E \rangle^2 \right) \] The term in the parentheses is precisely the variance of the energy, \(\langle (\Delta E)^2 \rangle\). Therefore: \[ \frac{\partial \langle E \rangle}{\partial \beta} = -\langle (\Delta E)^2 \rangle \] Combining this with our earlier result \(\frac{\partial \langle E \rangle}{\partial \beta} = -\frac{\partial^2 \ln Z}{\partial \beta^2}\), we arrive at the fundamental identity: \[ \langle (\Delta E)^2 \rangle = \frac{\partial^2 \ln Z}{\partial \beta^2} \]
Connecting to Heat Capacity
We now relate this variance to the constant-volume heat capacity \(C_V\). By definition, \(C_V = \left( \frac{\partial \langle E \rangle}{\partial T} \right)_V\). We need to change the derivative from being with respect to \(\beta\) to being with respect to \(T\). Recall that \(\beta = \frac{1}{k_B T}\). Differentiating \(\beta\) with respect to \(T\): \[ \frac{d \beta}{d T} = -\frac{1}{k_B T^2} \] Using the chain rule: \[ \frac{\partial \langle E \rangle}{\partial T} = \frac{\partial \langle E \rangle}{\partial \beta} \frac{d \beta}{d T} \] Substitute the expressions we have derived: \[ C_V = \left( -\langle (\Delta E)^2 \rangle \right) \left( -\frac{1}{k_B T^2} \right) \] \[ C_V = \frac{\langle (\Delta E)^2 \rangle}{k_B T^2} \] Rearranging this gives the final, crucial result: \[ \langle (\Delta E)^2 \rangle = k_B T^2 C_V \] This equation tells us that the mean square fluctuation of energy is directly proportional to the heat capacity. If \(C_V\) is large, the energy fluctuations are large. If \(C_V\) is zero (as in a ground state at \(T=0\) for many systems), the fluctuations vanish.
Relative Fluctuations and the Thermodynamic Limit
While the absolute variance \(\langle (\Delta E)^2 \rangle\) grows with the size of the system, it is more physically meaningful to look at the relative fluctuation, defined as the ratio of the standard deviation of energy to the average energy: \[ \frac{\sqrt{\langle (\Delta E)^2 \rangle}}{\langle E \rangle} \] For a system with \(N\) particles, the average energy \(\langle E \rangle\) is an extensive quantity, scaling linearly with \(N\) (i.e., \(\langle E \rangle \propto N\)). The variance \(\langle (\Delta E)^2 \rangle\) is also extensive, scaling linearly with \(N\) (i.e., \(\langle (\Delta E)^2 \rangle \propto N\)). Therefore, the standard deviation \(\sqrt{\langle (\Delta E)^2 \rangle}\) scales as \(\sqrt{N}\). \[ \frac{\sqrt{\langle (\Delta E)^2 \rangle}}{\langle E \rangle} \propto \frac{\sqrt{N}}{N} = \frac{1}{\sqrt{N}} \] As \(N \to \infty\) (the thermodynamic limit), this relative fluctuation goes to zero. This explains why, for macroscopic objects, we perceive energy as a fixed, deterministic quantity rather than a fluctuating one. The "noise" of the fluctuations becomes negligible compared to the "signal" of the average energy. However, for small systems (like nanoparticles or single molecules), \(N\) is small, and these relative fluctuations can be significant, leading to observable deviations from classical thermodynamic predictions.
The derivation
Energy Fluctuations and Heat Capacity
For a system in the canonical ensemble at temperature T, the second derivative of ln Z with respect to β = 1/kT equals the variance of the energy, and this variance is fixed by the constant-volume heat capacity: ⟨(ΔE)²⟩ = ⟨E²⟩ − ⟨E⟩² = ∂²ln Z/∂β² = kT²CV.
A common pitfall is assuming that because the relative fluctuations vanish in the thermodynamic limit, the absolute fluctuations also vanish. In fact, the absolute variance \(\langle (\Delta E)^2 \rangle\) grows in proportion to \(N\); it is only the relative fluctuation that vanishes. This distinction is critical when discussing phase transitions, where the heat capacity \(C_V\) itself diverges, causing the absolute fluctuations to become macroscopic and signalling the breakdown of the standard thermodynamic limit assumptions. C
Worked example
Problem. A system of \(N = 1000\) independent classical one-dimensional harmonic oscillators is in a heat bath at 300 K, so \(C_V = Nk_B\) and \(\langle E \rangle = Nk_BT\). Find the standard deviation of the energy and the relative fluctuation. Use \(k_B = 1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}\).
- Heat capacity: \(C_V = 1000 \times 1.38 \times 10^{-23} = 1.38 \times 10^{-20}\ \mathrm{J\,K^{-1}}\).
- Variance: \(\langle(\Delta E)^2\rangle = k_BT^2C_V = 1.38 \times 10^{-23} \times (300)^2 \times 1.38 \times 10^{-20} = 1.71 \times 10^{-38}\ \mathrm{J^2}\).
- Standard deviation: \(\sigma_E = \sqrt{1.71 \times 10^{-38}} = 1.31 \times 10^{-19}\ \mathrm{J}\).
- Mean energy: \(\langle E \rangle = Nk_BT = 1000 \times 1.38 \times 10^{-23} \times 300 = 4.14 \times 10^{-18}\ \mathrm{J}\), so \(\sigma_E/\langle E \rangle = 0.0316\).
- Check by scaling: \(\sigma_E/\langle E \rangle = \sqrt{k_BT^2 \cdot Nk_B}/(Nk_BT) = 1/\sqrt{N} = 1/\sqrt{1000} = 0.0316\). For a mole of oscillators it would be about \(1.3 \times 10^{-12}\).
Units check. \(\mathrm{J\,K^{-1}} \times \mathrm{K^2} \times \mathrm{J\,K^{-1}} = \mathrm{J^2}\), whose square root is in J; the relative fluctuation is dimensionless.
Three quick tests
- Does the variance of energy depend on the specific form of the Hamiltonian?
Answer
Only through the heat capacity: once \(C_V\) is known, the variance follows as \(k_BT^2C_V\), whatever the microscopic details. - If the heat capacity \(C_V\) is zero, what happens to the energy fluctuations?
Answer
The variance \(\langle (\Delta E)^2 \rangle\) becomes zero, meaning the energy is fixed. - Why do we observe deterministic energy in macroscopic objects?
Answer
Because the relative fluctuations scale as \(1/\sqrt{N}\), which becomes negligible for large \(N\).
Problem set for PU-203 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.