While reading Ken Dill's Molecular Driving Force, I noticed this paragraph when the author is trying to explain the Boltzmann distribution.

Particles don’t have an intrinsic preference for lower energy levels. Fundamentally, all energy levels are equivalent. Rather, there are more arrangements of the system that way. It is extremely unlikely that one particle would have such a high energy that it would leave all the others no energy.

However, I wonder if there are thermodynamic systems where there are more microstates for higher energy levels so that particles would be unable to reach the lowest energy level. For example (I hope this makes sense, correctly me please if I'm talking nonsense) if we expose some simple polypeptide polymers to strongly denaturing solutions than I guess they will unfold despite this corresponds to a high energy configuration for all atoms of these peptides.

P.S. I've read answers to another question very similar to mine here: The Preference for Low Energy States.

And I think I understand the ramifications of the first and second laws of thermodynamics.

  • $\begingroup$ Oh, by the way, I'm a biology student so please tell me about my wrong usage of technical language, if there is. $\endgroup$ – Macrophage Nov 27 '18 at 2:07

I wonder if there are thermodynamic systems where there are more microstates for higher energy levels...

The answer to this part is yes. In a typical system, the number $n(E)$ of microstates having total energy $E$ increases with increasing $E$. In fact, the absolute temperature $T$ is defined by $$ \frac{1}{T} \equiv \frac{\partial S}{\partial E} \tag{1} $$ where the entropy $S(E)$ is related to $n(E)$ by $S(E)=\log n(E)$. (I'm using units where Boltzmann's constant is equal to $1$.) The fact that the absolute temperature is normally positive is a manifestation of the fact that the number of microstates is typically an increasing function of the total energy.

The prototype of a "typical system" is the classical ideal gas with a fixed number $N$ of particles. In this case, the number $n(E,V)$ of microstates as a function of the total energy $E$ and total volume $V$ is $$ n(E,V)\propto \left(\frac{m}{\hbar^2}E\right)^{3N/2} V^N, \tag{2} $$ which is an increasing function of $E$, where $m$ is the single-particle mass and $\hbar$ is Planck's constant. (Even though we call this a "classical" system, it invokes some quantum physics via the assumption that the number of measurably-distinct microstates is finite in a finite volume.)

There are exceptions, such as spin systems with a finite number of spins in an external magnetic field, where the maximum number of microstates occurs at intermediate energy when half of the spins are "up" (higher individual energy) and half are "down" (lower individual energy). In these systems, the function $n(E)$ has a maximum at an intermediate value of $E$, so the definition (1) says that the absolute temperature is negative at higher values of $E$. Negative absolute temperatures are hotter, not colder, than positive absolute temperatures. The oxymoronic flavor of this statement is an artifact of using $T$ to quantify the temperature (instead of, say, $-1/T$).

The excerpt shown in the question might be referring to configurations with a fixed total energy $E$. In a typical system, most of the microstates with a given total energy $E$ are such that the energy is well-distributed among all of the particles, and relatively few of them have most of the energy assigned to just one of the particles. This is true even in a typical system where $n(E)$ is an increasing function of energy.

...so that particles would be unable to reach the lowest energy level.

One example of this could be a finite spin system as mentioned above. If we consider a situation in which the total energy of the system is very close to its maximum possible value (which is finite in such a system, by construction), then almost all of the spins will have to be in their higher-energy orientations in order to add up to the desired total energy.

I don't know enough about polymer physics to comment on possible examples in that area.

  • $\begingroup$ Thank you, sir. It's very well explained in your answer! $\endgroup$ – Macrophage Nov 27 '18 at 3:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.