Kote is absolutely right.
Andy, you seem not to realize that entropy depends on the observer. Entropy is a measure of our uncertainty about a system - i.e., the uncertainty of some particular observer. For two different observers, the entropy of a system can differ.
See for example the Wikipedia page on entropy:
Wikipedia said:
The interpretative model has a central role in determining entropy. The qualifier "for a given set of macroscopic variables" above has very deep implications: if two observers use different sets of macroscopic variables, then they will observe different entropies.
Andy Resnick said:
Sigh. If the entropy is different, the mass must also be different.
No. Because the amount of entropy fundamentally depends on the observer (see above), your statement implies that the mass of the system depends on the knowledge of the observer. That is clearly nonsensical.
Andy Resnick said:
If there is no difference in energy, there is still be energy associated with the entropy: kT ln(2) per bit of information, and again, there will be a change in mass.
No, that's a misunderstanding of Landauer's principle (again, look it up on Wikipedia). The energy kT ln 2 is not associated with
information, it is associated with the act of
erasing information - specifically with the entity doing the erasing, not the memory device itself.
If a bit is either 0 or 1, and you
don't know which, and you want to set it to a particular state regardless of its previous state, then you're erasing the information it previously contained. In terms of thermodynamical entropy, there were previously two possible and equiprobable microstates, but now there's only one, therefore the entropy has decreased by k ln 2 (for
you). But the second law prohibits the decrease of entropy in a closed system, so if we take the closed system of the bit and you (the bit alone is not a closed system as you manipulate it), this means that
your entropy must have increased, namely by spending energy kT ln 2 (transforming some of your energy from usable form into useless heat). It's important to note that total energy - thus mass - of you and the bit never changes and there needn't be any net energy transfer between you and the bit (unless the bit encodes 0 and 1 with different energy states).
As with anything about entropy, this whole thing depends on the particular observer: if you
do know the previous value of the bit, then the second law doesn't require you to spend kT ln 2 to change the bit, because the entropy of the bit doesn't change (for you). No information gets erased and the change is reversible (you know the previous state so you can change it back again).
To sum up: the energy is not associated with the memory device, but with the entity rewriting it, and depending on its knowledge. The energy - or mass - of a memory device doesn't depend on the "entropy" of the contents, as the amount of entropy fundamentally depends on the choice of observer. For practical devices, the mass of the device can differ only if different energy states are used to encode different values.
P.S.: Just to make it clear - the energy expense of kT ln 2 to erase a bit is the theoretical minimum imposed by the second law. Nothing prevents one from spending more energy than that to rewrite a bit, and inefficiencies inherent in practical devices of course vastly exceed that minimum.
. The theory is that the universe is inherently discrete and made of some physical equivalent of bits. We don't exactly have the technology to make flash drives that store information at the qubit level (if there is such a level). We can't just count how much mass it would take to get 8gb worth of qubits (is there even a straight conversion?). Even if we could, we would have to know the mechanism by which the 1s and 0s are encoded into qubits. Which particle property corresponds with a 1 and which corresponds with a 0? Does the particle mass change when we change between these expressions of properties? Is compression used?
