# Could a solid body have arbitrarily low entropy but arbitrarily high temperature?

Could a solid, isolated body have arbitrarily low entropy but arbitrarily high temperature?

I'm trying to build an intuition: if according to statistical mechanics, entropy is a measure of disorder and temperature is a measure of energy, could all particles (at least in theory) have high kinetic energy and be moving in a highly-predictable manner, leading to both high temperature and low entropy? Or would this only be possible for a short amount of time due to inherent randomness in particle movement, which would lead to entropy increasing despite the body being isolated?

Or is there a particular equation that gives a lower bound on entropy given its temperature / thermal energy? (I'm assuming fixed volume and isolated body throughout)

• I think the idea of a solid at arbitrarily high temperature is in contradiction with experimental facts. If we adhere to the known Physics, there will be a finite melting temperature. Difficult to say anything meaningful under impossible conditions. Jun 21, 2020 at 22:18
• By the way, the entropy you are referring to is the extensive entropy or the entropy per particle? As additional comments, I would add that temperature is not a measure of energy and entropy is not a measure of disorder without specifying which kind of disorder one is speaking about. Jun 21, 2020 at 22:22
• @GiorgioP Ah I didn't realize there were multiple definitions -- what I had in mind when I wrote was the Gibbs entropy formula $S=-k_B \sum_i p_i \log p_i$. I thought that the melting point would depend on temperature AND pressure, but if there is an upper bound on temperature for solids, then I'd reformulate saying: given some temperature below that upper bound, do we have a lower bound on $S$ as presented above, and if so, what is that lower bound on $S$.
– Gabi
Jun 24, 2020 at 9:53

## 3 Answers

Consider a thermal ensemble comprised of $$N$$ degrees of freedom at a temperature $$T$$. Let us look at this system from the perspective of the microcanonical ensemble. In the microcanonical ensemble, such a system with $$N$$ degrees of freedom is constructed such that its entropy is $$\Delta S = \log{N}$$, and such that its energies are spreaded in an interval $$E \pm \frac{\Delta E}{2}$$ about the average energy $$E$$. We demand that $$\Delta E \ll E$$, and that $$N$$ is a large number, properties which a system must possess to admit a thermodynamic description. The notion of temperature is meaningful if and only if there exists a thermodynamic description.

Now in this ensemble, the entropy, range of energies and temperature obey the following relation:

$$\frac{1}{T} = \frac{\Delta S}{\Delta E}.$$

As you can see, if we fix the range of energies $$\Delta E$$, then we have a simple relation between the entropy and the temperature. However you cannot have an arbitrary high temperature and an arbitrarily low entropy, because you will no longer remain in the thermodynamic regime, i.e. $$N = \exp{\Delta S}$$ will become a small number. You might demand a larger $$\Delta E$$, but this also has a domain of validity, because $$\Delta E \ll E$$. Besides you will need to take into account all the new degrees of freedom in the larger $$\Delta E$$, which will inevitably increase $$\Delta S$$.

EDIT: In response to the comment below, choosing your very specific configuration means that you already know which microstate the system is in, and of course, that means the entropy is zero. This amounts to a "fine-graining" of the system, and what you are calculating is the "fine-grained entropy" of your system where you have set $$N=1$$, and consequently you are not in the thermodynamic limit. The notion of temperature outside the thermodynamic limit is useless. Whereas note that the thermodynamic entropy is a "coarse-grained" observable. Here you do not know which of the $$N$$-th microstate you are in, where $$N$$ is a large number, and you only know the range of microstates accessible to the system and the range of energies. The existence of the thermodynamic limit is crucial to define the notion of temperature.

• Can I ask you to clarify: is $\Delta S$ the entropy, or difference in entropy? My understanding was that $S$ was entropy. Also, the $\log N$ part of things -- does that still apply if say we could somehow set a starting position/momentum for every single particle involved?
– Gabi
Jun 22, 2020 at 9:48
• @Gabi Sorry for the bad notation, I am denoting $\Delta S$ as the entropy here, as should be clear from the context. The above situation I considered is true for statistical systems in general. If you fix a starting positions/momenta within the above limits, then it should be true. Jun 22, 2020 at 11:10
• @Gabi I edited the answer to address your other query. Somehow I had missed answering this part of the question twice. Jun 22, 2020 at 15:02
• @Gabi Let's say you are using the Gibbs entropy formula, and have chosen a particular distribution. Then as you said the entropy is zero. However, in this case, you already know every possible detail about the system, i.e. you know the fine-grained description. You define thermodynamic quantities like temperature only (which is redundant in the fine-grained case) when you don't know the detailed description of the system, e.g. you have one mole of a gas where you don't know all $10^{23}$ positions and momenta. So statistically you write down the system in terms of a small number of variables. Jun 24, 2020 at 13:20
• @Gabi This is where the notion of temperature makes sense. Since you are coming from a statistical viewpoint, a perfect book to follow is Kardar, especially chapter 4, which you should study very carefully. All these things are mentioned here, but it is a bit terse. arxiv.org/abs/1307.0378 is a good resource too, here you can read the statistical description, and ignore the black hole parts. Jun 24, 2020 at 13:25

I think the validity of thermodynamic requiring large number of degrees of freedom, pointed by Bruce Lee, restricts entropy to be rather a large number. However, if you consider the two-state system (spin $$\uparrow, \downarrow$$), you may get such a situation. Let the $$p$$-probability for spin to point upwards, and this state will have energy = $$\varepsilon$$, whereas the spin down state we take to have energy = $$0$$. Then the entropy $$S$$ and energy are: $$S = p \ln p + (1 - p) \ln (1 - p)$$ $$E = p\varepsilon \Rightarrow\frac{1}{T} = \frac{\ln p - \ln(1-p)}{\varepsilon}$$ At the vicinity of $$p \rightarrow 1/2$$ the last expression approaches zero, therefore, the temperature is infinite. However, the definition for temperature, as pointed above, is meaningful only for large systems.

• Right, but if we could somehow magically set the momentum and position of every particle, then wouldn't we be able to have $p = 1$ at least for an instant, in which case $S$ would be $0$?
– Gabi
Jun 22, 2020 at 9:50
• +1, I think this is a good example which naturally follows from my answer. Jun 22, 2020 at 14:57

When you have increased the temperature and left the system alone The system will eventually try all the possible microstates. And the ordered arrangement is one of the possible various microstates.

But one of the specific well-behaved order is probably very less likely to happen among large number of microstate.

Entropy is a logarithm of number of microstate. Especially the ordered well behaved microstate is consistent with the high entropy. The system will also try this possibility. But at last, it will choose configuration in with energy is most spread out which is the condition of thermal equilibrium.

Order is not the same thing as low entropy And the second law isn't always a tendency towards disorder in thermodynamic entropy, the only special arrangements of particles that change entropy is the ones that change thermodynamic property and not the ones that spell out cuss words like mess up your room.

According to your idea of entropy, a black hole should have no entropy. But they have.

When you are talking about randomness as entropy then one should ask randomness of what?