In: Chemistry
In thermodynamics, we measure things
like pressure, temperature, and volume. These are high-level
descriptions. Together, these state variables form a macrostate.
The low-level description is the position and momentum of each
particle. This information forms a microstate.
In the card example, I assume we take the order of the cards to be
a microstate. However, it is unspecified what characteristic of the
cards make up the macrostate.
If the macrostate is, "the cards are in a pile", then you're right;
shuffling doesn't change the entropy. The entropy would always be
log52!, because any microstate would correspond to
that macrostate.
A more interesting macrostate might be the number of times a higher
card comes before a lower card. If we label the cards 1 - 52, we
could then look at all pairs of cards.
Whenever the higher card comes before the lower card, we add one to
our tally.
In the fresh deck, this number would be zero (depending on how we
number the cards) because the cards are in exactly the right order
and go from lowest to highest.
When we shuffle the deck, there would be some higher cards before
lower ones, and as we continued shuffling the number would increase
until, with a perfectly random order, it would be likely that very
nearly half the pairs of cards appear in the wrong order.
If NN pairs of cards appear in the wrong order, we could then solve
a combinatorics problem to figure out how many possible orderings
there are for that N. The result would be Ω(N). The entropy would
be
S=logΩ
and this entropy would be maximum for N
This choice of macrostate is arbitrary; I picked it as an example.
If you made some other choice, such as the number of cards touching
only cards of the same color, you would get a different result for
the entropy. However, for most choices you would find that the
original deck had low entropy, and for all choices you would find
that a randomly-shuffled deck would be most likely to have nearly
the highest possible entropy.