The classic macroscopic variables one typically measures for an ideal gas are $P$, $V$, $T$, $n$, - pressure, volume, temperature, and amount, respectively. I am curious what the corresponding variables are for analogous system I'll call a 'bitgas', and the relationship between the infodynamics and thermodynamics.
A 'bitgas' for the purposes of this question is a string over the alphabet $\{0,1\}$ which is written into the state of a localized physical system. That is, there is a closed 3 dim'l boundary around the system, with finite volume.
As an example, let's take a solid state hard drive $H$ whose capacity $C = 15 \, mol*bit$, or ~1 yottabyte. Suppose the volume is $V= 66.8 cm^{3}$. If Moore's law holds, such a device may be commonplace by 2040. The hard drive is in a room at $300Kel$, and we may or may not hook up a cable $I/O$ (a USB cable, or SATA + power) which can transfer data and/or power.
Let $x$ be a bitstring representing the state of the hard drive. $I/O$ can act on $x$ in one of three ways i) $swap_{ij}$, apply a transposition $(i\,j)$ swapping the bits at position $i,j$ ii) $write_{i}(y)$ where $y\in\{0,1\}$ and the bit at position $i$ is $y$ after the operation iii) $read_i$ transfers the bit $y_i$ out of position $i$.
In this analogy, the hard drive atoms, mechanics, and enclosure represent the classical "container", and the $1's$ which are written in the hard drive are the "gas atoms". In the classical scenario, the gas is the thing with macroscopic thermodynamic properties like pressure and temperature. Here, the hard drive does of course have a temperature and take up volume, but it is a solid. Changing the temperature below a certain critical temperature $T_c$ at which the hard drive melts or burns shouldn't affect $x$.
Define the following variables for the bitgas $H$:
- $K$ = information content, the Kolmogorov complexity $K(x)$
- $C$ = capacity of $H$
- $T$ = temperature of $H$
- $n_1$ = number of ones in $x$
$n_0 = C - n_1$ is the number of zeros.
Suppose $n_0=n_1=r$, so that there are as many 0's as 1's, and restrict to the case where we only allow the operation $swap$.
For any compression algorithm, we know that some strings will be incompressible, and have large information content so that $K\approx C$. Other strings, such as $x_r=0^{r} 1^{r}$ have low information content.
$n_1/(n_0+n_1)=r/C=1/2$ is constant in this example. However, if we very slowly heat up the hard drive close to its failure temperature $T_c$, we expect errors to occur and bits to start flipping, which may change $n_1/C$.
If we initialize the hard drive to an initial state $x_r$, it would seem $K_0=K(x_r)$ is small and constant while $T<<T_c$, but as $T$ approaches the critical temperature $K$ begins to increase until it reaches $\approx C$.
When $K\approx C$, we will have put about $15*N_{A}*k_{B}*300Kel \approx 37kJ$ into the bitgas.
One could imagine holding a candle at, say, one corner of the hard drive. If that side is all zeros, it will begin to become corrupted. This 'corrupted' portion would contain a lot of information about where the flame was held.
A less extreme example would be to put the hard drive on a hot plate, and slowly increase the temperature until bits start flipping.
It seems that $K$ depends on $T$, and I am wondering what the relationship is, exactly. In other words, what is $\displaystyle \frac{\partial K}{\partial T}$ when $C$ is held constant?
I'll point out that $K$ does depend on $n_1$. When $n_1=C$, all the bits are 1, which is a highly compressible state, so that $K \approx \log(C)$.
