# Example of both energy being/not being a monotonically increasing function of entropy and vice versa?

In (https://theoreticalminimum.com/courses/statistical-mechanics/2013/spring) Dr. Susskind says we can assume that in most of the cases energy is a monotonically increasing function of entropy and vise versa except unusual cases. Can someone please give me an example of both cases?

An example of the former is an ideal gas. However, as he says, pretty much anything you come across has this property. Another example I can think of is a ferromagnet. Here is a video of an Ising model with increasing temperature: https://m.youtube.com/watch?v=5l-6OITKi9I The black represents spin in one direction, say up, and white represents spin in the other direction. Clearly, as the temperature increases, the entropy also increases. Remember that the entropy is the logarithm of the number of microstates compatible with the macrostate, and in this case the only macroscopic parameters are number of spins and energy. You can see that at lower energies it is pretty stable because only “nice” microstates have the right energy (neighboring spins in opposite directions have higher energy in this model). As the energy increases, the number of compatible microstates (and as a result, the entropy) increases.

In an ideal gas, the reason entropy increases with temperature is that a larger amount of momentum space is accessible to the system, since there’s more energy to go around.

To have a system which has decreasing entropy with increasing energy, you would have to have fewer accessible states at higher energies. There’s nothing theoretical that says this is impossible I don’t think, but it’s what we would call “pathological”. In other words, it’s an edge case that breaks our assumptions, but doesn’t really matter because it’s pretty much invented to test the limits of the theory, not to describe something observed in the real world.

A subtle detail here is that in the infinite well (which is what we use for ideal gas), the energy level spacing increases as energy does, since $$E \sim n^2$$. However, when we discuss an ideal gas, the energy is being divided between many particles, and the number of microstates is the number of ways we can distribute this energy to all the particles. So $$E \sim \sum_i n_i^2$$, where $$n_i$$ are the energy quantum number of each particle. This amounts to counting the number of points that land on the surface of a hypersphere with the dimension of the number of particles. Really the number exactly on the surface is going to be very irregular and somewhat small, so it’s better to count the states in a thin shell which corresponds to a small range of energies. It should be clear that the larger the radius of the hypersphere (i.e. the higher the energy), the more microstates will be enclosed, and thus the entropy will be higher.

Pathra and Beale’s Statistical Mechanics has a really nice discussion about this, I recommend that if you can you should look at this. I would also happily clarify further.

• Thank you very much for your answer. I will let you know if I need further clarification :) – user650585 Nov 13 '18 at 23:18
• Would you please tell me the section or the page number at which this is discussed in Pathra and Beale’s Statistical Mechanics? – user650585 Nov 14 '18 at 1:06
• The discussion is really distributed throughout the beginning of the book, but the parts specifically relevant to what I have to say are section 1.4 and end of p31-beginning p32. (third edition) – Mr. HelloBye Nov 14 '18 at 2:01

The internal energy of an ideal gas is a monotonically increasing function of temperature, but is independent of volume. On the other hand, the entropy is a monotonically increasing function of both temperature and volume. So if the temperature increases, the internal energy must also increase; however, depending on whether the volume increases or decreases, the entropy can go either way. So, generally, increases in internal energy do not have to be accompanied by entropy increases.

Spin systems, or any system with a limited number of energy levels, are examples of entropy decreasing with energy. When the energy is at the maximum possible value, the entropy is just as low again as in the ground state. Somewhere in between there is a maximum of the entropy, where the thermodynamic beta is zero.

This is simplest for a system with non-interacting two-level states, for example spin 1/2 particles in a magnetic field. The internal energy is proportional to the number of particles in the high-energy state.

The multiplicity (the number of microstates at a given energy) is given by the binomial distribution. It is maximal when half the spins are up and half the spins are down. Here is an image for 400 spins.

The bottom panel shows the logarithm of the multiplicity, the entropy, both exakt and in the Gaussian approximation. The inset shows the derivative (the thermodynamic beta) and the temperature. The temperature diverges to plus or minus infinity at the maximum of the multiplicity.

Negative temperatures are hotter than infinity. In the real world, such systems always have some coupling to ordinary systems. The states with negative temperature cannot be in a real equilibrium. Eventually, energy will leak away to a "normal" reservoir.

• thank you for the answer. Would you please elaborate further on energy levels and entropy since I am not that familiar with these concepts in physics? Also what do you mean by "such systems always have some coupling to ordinary systems." ? – user650585 Nov 14 '18 at 0:49
• Entropy isn't about the number of energy levels, it's about the multiplicity of the energy levels. Even for a spin system, the higher energy states have higher multiplicity, see the Ising model. – Mr. HelloBye Nov 14 '18 at 1:59
• @Mr.HelloBye I have added a figure for non-interacting spins. But also the Ising system is symmetric for high or low energy, the multiplicity is the same when all states are up or all states are down. Usually in such simulations, one does not add energy but one raises the temperature, which is not the same thing. – Pieter Nov 14 '18 at 8:52
• @user650585 I elaborated with a figure. Spin systems at negative temperatures cannot be in real thermodynamic equilibrium. Energy will always leak away to a systems that behave normally. – Pieter Nov 14 '18 at 9:02
• I see what you mean for noninteracting spins in a magnetic field. But in the Ising model, the highest energy states are the ones with alternating spins, and the states with the spins all up or all down have the lowest energy. This is why above the critical temperature, since higher energy states are more accessible, domains dissolve since they aren't so statistically favored anymore. And the reverse is true, when you cool an Ising model, you force it into lower energy states, where most of the spins are aligned with their neighbors. – Mr. HelloBye Nov 15 '18 at 2:19