9

I'm wondering if the information lost by rearranging the letters of a book is measurable as a difference in its initial and final mass.

Choose a long, random string over an alphabet, say $\{0,1\}$, of length $N$. It should be random in the sense that it is incompressible. You might also choose a big book, at random, and compress it.

Once you have the book, or have written down the string in a book, measure the book's mass $m_0$.

Convert the letters into a standard alphabet by using, say, the ASCII encoding scheme. The letters should be more or less distributed uniformly, unlike English which has a rank-frequency distribution for the letters. Rearrange the letters into the complete works of Shakespeare, or as much literature as you possibly can. Then apply the encoding to get a bitstring. This process can be represented by a 0/1 permutation matrix $\sigma_1$ which acts on the bitstring.

Finally, move all the 0's to the left, and 1's to the right. This can be represented as another permutation matrix $\sigma_2$. Measure the mass of the book to get $m_2$.

It appears that the information content of the book at the beginning is $S_0=N$ bits. The information content $S_1$ of the complete works of Shakespeare is around 1.98MB (less than really, size of zipped text file). The last state is very compressible, and $S_2 \approx 2\log_2(N/2)$.

Suppose $N$ is large, say Avagadros' number $N=N_{A}=6.02214076*10^{23}$, more than a zetta and less than a yotta. Then $\triangle S = S_0 - S_2 \approx N_A$. If 1 bit represents about $10^{-23} J/K$, then at $300K$ the information lost corresponds to $~20.1$ picograms.

I suppose the lost information is carried away by the matrices $\sigma_1$, $\sigma_2$ if no one watches or records the rearrangement as it occurs. Is that correct?

Jackson Walters
  • 567

3 Answers3

10

Rearranging letters in a book destroys semantic information, the information the text gives a reader who can understand it. But it does not change the information in the Shannon or thermodynamic sense, as distinguishable states.

When I read a book I am exploiting pre-existing correlations between my brain and the text, so that reading a certain set of symbols triggers some mental representations. The meaning of the text resides in this mutual information rather than the text itself - a different set of symbols might tell me the "same" message in a different language (or even a permutation). But since the meaning is not in the book it has no effect on the mass or any other physical property.

The information that matters thermodynamically is how many book microstates corresponds to the same macrostate. Rearranging the ink a bit is a minuscule change compared to all the degrees of freedom in the paper molecules that do not matter for the message. Still, a very low-entropy state (all bits zero) would have a slightly different Gibbs free energy $E-TS$ from a high-entropy state (bits randomly zero or one). But it has nothing to do with how much meaning there is in the book. It also does not affect the mass: the stress-energy tensor $T_{ij}$ in general relativity is (as far as I can understand relativistic thermodynamics) independent of the entropy currents.

Anders Sandberg
  • 34,657
2

Summary: Your question is tantamount to asking whether mass-energy equivalence has been extended to mass-energy-information (or mass-energy-entropy). As far as I can see, essentially no one in the physics community (outside of a handful of outliers) accepts such an extension. So the answer to your question seems to be, as far it's known, no. I discuss this below, and also explain why entropy and energy are qualitatively different, even though they are both included in free energy expressions.

Anders Sandberg raises the objection that rearranging the letters destroys semantic but not Shannon information. This can be addressed by posing a somewhat different question:

Let's compare two otherwise identical books. Does a book in which the first half are all zeros, and the last half are all ones, have a different mass from an otherwise identical book in which the ones and zeros are randomly arranged?

The latter would have a higher Shannon entropy, since it requires more information to describe the sequence of numbers in the latter book than the former. Having said this, I believe Anders is correct: Even if the books did have different Shannon entropies, their masses would not differ because of this.

I'll make two related arguments, one based on thermodynamics, and another based on the nature of science generally.

The thermodynamic argument: Mass-energy equivalence applies to, well, energy. Energy is not the same as free energy. Free energies consist of an energy term minus an entropy term (of the form TS). One of the great utilities of a free energy is that it enables us to determine the ability of a system to do work. The more the entropy of the system can increase during a process, the more work (everything else being equal) can be obtained from the system.

So you might ask: Doesn't that mean a lower entropy system has more energy than a higher entropy system? The answer is no. Entropy does not contribute to a system's energy. It is, instead, a measure of the quality of a system's energy—specifically, of how useful (or useless) a system's energy is for doing work.

Sean Carroll has a nice discussion of this on his Preposterous Universe blog:
https://www.preposterousuniverse.com/blog/2010/11/22/using-information-to-extract-energy/

Thus energy and entropy are qualitatively different things. Hence it would require a significant expansion of the concept of mass-energy equivalence to include entropy as a form of energy.

This then leads to an argument against their equivalence having been established, based on how modern science is done: Mass-energy equivalence and entropy-information equivalence are very important in physics. Hence, if mass-entropy (or mass-information) equivalence (which is really what you're asking about) had also been established, this would be a well-known result (because it would connect mass-energy equivalence and entropy-information equivalence!).

Instead, in checking both Google and Google Scholar, I was only able to find a few papers about mass-entropy/mass-information equivalence, including this one:

https://ui.adsabs.harvard.edu/abs/2019AIPA....9i5206V/abstract

Vopson, M.M., 2019. The mass-energy-information equivalence principle. AIP Advances, 9(9), p.095206.

In reading the abstract, and examining the citations, it is clear that mass-energy-information equivalence is, currently, highly speculative.

Hence while is may be possible that the book with lower Shannon entropy (and thus higher free energy) would weigh more, there doesn't appear to be even a small school of physicists that currently accept this.

theorist
  • 599
0

Well if you compress the more ordered book into code and write the code in a new book. Yes, the new book will be lighter.

The original book will not be lighter because although you have ordered the letters according to a scheme that a given algorithm can compress well, how does the Universe know which algorithm you are thinking of? It will compress well according to one algorithm but badly according to almost every other.