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

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.

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.