I know the Bekestein bound is the upper bound for the information content of a region of space, but is it possible to actually calculate that information content (number of bits, not the bits themselves)?

For example, given 1mL of water at room temperature, I can say its energy is MC^2 (mass-energy equivalence) + HTM (heat capacity * temperature * mass, its thermal energy) / kTln2 (Landauer's principle, joules per bit of information entropy). Does this mean that 1mL of water contains 3.13 * 10^34 bits of information, or is this misguided of me to think?

1mL of water is only 3.34 * 10^22 molecules, based on the molarity, so the 3.13E34 number seems very high to me...

  • $\begingroup$ The mc^2 at the thermodynamic level is irrelevant since most of that energy is bound in the nucleus, (a tiny bit in the atoms and molecules). Thermodynamic entropy does not know about mc^2 in these terms. $\endgroup$ – anna v Mar 20 '15 at 4:56
  • $\begingroup$ Point taken, but without the MC^2 term, the result only describes the number of bits of information potentially able for doing thermodynamic work and not all of the information contained in the matter, correct? $\endgroup$ – haydnv Mar 23 '15 at 3:44

m includes all kinds of energies for the mass at rest, including the thermal energy, and those from the other degrees of freedom, such as the ones internal to the nucleus, as AnnaV mentioned. So you only have to compute mc^2, the HTM term is redundant.

  • $\begingroup$ Without the HTM term it seems like any material at any temperature would have exactly the same information content as any other material of the same mass, which seems wrong to me $\endgroup$ – haydnv Mar 21 '15 at 0:47
  • $\begingroup$ Given two balls who have the same mass at T=0K will weight different at different temperatures. That is, the hotter ball will have a larger rest mass than the colder ball. In the classical approximation they will have the same mass, of course $\endgroup$ – user66432 Mar 21 '15 at 1:16
  • $\begingroup$ But still, your suggested form of (MC^2) / kTln2 would give exactly the same result for, i.e., 1g of water at 300K and 1g of hydrogen gas at 300K and 1g of solid gold at 300K, yes? How can that be? Isn't the free energy very different in those forms and isn't that potentially an information-bearing degree of freedom? $\endgroup$ – haydnv Mar 23 '15 at 3:50
  • $\begingroup$ When you are thinking of total energy, water or nuclei or electrons are irrelevant in that the same states can be reached with creation and annihillation and radiation. What the existence of quantum numbers that identify individual states does is to pick up a subsystem of the total potential in number of degrees of freedom. $\endgroup$ – anna v Mar 23 '15 at 5:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.