Why is the macroscopic Heat capacity related to the standard deviation of microscopic energy fluctuations? In one of the problems in my textbook (Schroeder Intro to Thermal Physics), it is shown that within the canonical ensemble, the standard deviation of the energy fluctuations for a single microscopic system within the ensemble (denoted $\sigma_E$), is related to the macroscopic heat capacity of the entire system (All atoms/subsystems that make up the thermodynamics system) by the following formula
$$\sigma_E =kT\frac{\sqrt{C}}{\sqrt{k}}$$
where $C$ is the Heat capacity of the macroscopic system. This seems to be a very profound result. Is there any way to understand why this is the case? I can't seem to figure out why a greater amount of thermal fluctuations per particle ($\sigma_E$) should imply a larger heat capacity on the macroscopic level. Is there any way to visualize this?
 A: The question asks for intuition. The equation shown in the question can be derived from the fact that $\sigma_E^2$ and $C$ can both be expressed in terms of derivatives of the partition function with respect to $\beta=1/kT$, as shown here. We can extract some intuition from that derivation.
In the canonical ensemble, the probability of a state with energy $E$ is proportional to $e^{-E/kT}$. The partition function can be written either as a sum over states,
$$
 Z = \sum_n e^{-E_n/kT},
\tag{1}
$$
or as a sum over energies,
$$
 Z = \sum_E \rho(E) e^{-E/kT}
\tag{2}
$$
where $\rho(E)$ is the density of states with energy $E$. In a comment, you described some prior intuition about heat capacity:

atoms that have a large amount of degrees of freedom (translational+ rotational +perhaps a few vibrational etc) have a higher heat capacity than those that have few degrees of freedom. This is because the former can store more energy than the later.

To get that intuition from (2), use the fact that for each of those types of degree of freedom (translational, rotational, vibrational), the number of such states in an energy-shell of given width $dE$ is an increasing function of the energy $E$. Consequently, the more different types of degrees of freedom the system has, the more rapidly $\rho(E)$ grows as a function of $E$. That's the key. A system with more degrees of freedom has larger-magnitude derivatives of the partition function with respect to $\beta=1/kT$, for any given value of $T$. Since $\sigma_E^2$ and $C$ are both expressed in terms of the first- and second-derivatives of the partition function with respect to $\beta$, this explains intuitively why both of them have the same dependence on the number of degrees of freedom. In particular, it explains intuitively why both of them are larger (for a given $T$) in systems with more degrees of freedom.
Here's a little more detail about the intuition behind $\sigma_E^2$. Consider the summand in (2). We can think of $\sigma_E$ as the width of the peak in the graph of $\rho(E)e^{-E/kT}$ as a function of $E$. If we make $\rho(E)$ grow more rapidly, we shift the peak to a larger value of $E$, and that also makes the peak wider because $e^{-E/kT}$ doesn't decrease as rapidly for larger values of $E$ (it's derivative approaches zero as $E\to \infty$). That's why increasing the number of degrees of freedom makes $\sigma_E^2$ larger, for given value of $T$.
The intuition is tied to the mathematical form of the canonical ensemble (1)-(2). That's unavoidable, because we can contrive different ensembles in which $\sigma_E^2$ and $C$ are not be related to each other as shown in the question. Then again, since the canonical ensemble is essentially a consequence of the microcanonical ensemble (which is the least-presumptuous ensemble), we don't normally need to consider other ensembles, so the relationship shown in the question is relatively robust.
