This Physics question

[Corribus]

LOL, I didn't realize the second set had any order. No matter; fundamentally it changes nothing.

First, it is not really useful to assign either case an absolute value of entropy (zero, as you've done), because when dealing with thermoDYNAMICS we're really interested in change between states rather than states themselves.

Second, thermodynamics is a statistical concept which deals with large numbers of either particles or observances, so again picking out two states and asking which one has a larger entropy again doesn't really serve much purpose.

That said:

Let's assume for a second that the second sequence is random. This will make things easier. Also, at present we are speaking of just abstract combinations of numbers, but without those numbers being related to anything, it makes little sense to talk about entropy - an empirical concept. So, we will change the problem by speaking of balls.

Suppose I have ten balls, all identical in shape and color. I then label each of the balls with a number ranging from 0 to 9 and arrange them in a straight line along a single dimension. Now it is at least appropriate to consider two possible scenarios, where they are arranged
0, 1, 2, 3, 4, 5, 6, 7, 8, 9

and

0, 2, 3, 6, 7, 1, 9, 4, 5, 8

Given that entropy is often interpreted as being related to "order" (or "disorder"), it's tempting to conclude that the first sequence is lower entropy than the second. This is because we quickly recognize a pattern in the first sequence but do not in the second (remember, we're assuming the second one is "random" at the moment).

This however would be an erroneous conclusion because the labels only exist for bookkeeping purposes and have no functional meaning. A system in which the balls are ordered 0,1,2,3,4,5,6,7,8,9 is functionally equivalent to a system in which the balls are ordered SECOND ORDER. That is - if I were to randomly determine the ordering of the balls, the probability of arriving at the first order is exactly the same as arriving at the second order. Because the labels don't really mean anything (energetically, say), just because the balls are in numerical order in the former case doesn't mean it's any more "rare" or "ordered" than the apparently random arrangement in the latter case. The fact that we recognize an artificial source of order in the former case tricks us into thinking the former case is somehow "special". But it's really not. Imagine, for example, if you weren't familiar with arabic numerals - in this case both arrangements would appear "random". The fact that your second arrangement actually has a hidden code, of sorts, should clue you into the fact that, lacking some sort of external meaning, the ball labels don't really mean anything.

Now - suppose we ascribe some other meaning to the labels. In this way, we make the combinations functionally inequivalent. Suppose, for instance, we define a characteristic observable Q, such that each ball in the system contributes to the total Q-value a value equal to the number of balls locally arranged in increasing number. For instance, because your first arrangement has 10 balls in increasing order, each ball is worth 10 points, for a total Q-value of 100. Your second system (0,2,3,6,7,1,9,4,5,8) would have a Q-value of 16 (2,3 are in order and 6,7 are in order and 4,5 are in order, so each of those is worth 2 points, and the remaining numbers are worth 1 point each). So Q-values for all possible systems range from 10 (no balls in order) to 100 (all balls in order), with many values in between). [In essence, what we've done here is make an artificial probabilistic energy value of the system.]

Now you see that because your first and second arrangements have different Q-values, they no longer are functionally equivalent. The first arrangement, for example, is the only possible arrangement to have a Q-value of 100. It is quite unique. On the other hand, your second arrangement, which has a Q-value of 16, is not so unique. There are numerous possible arrangements that could have this Q-value. [(1,2,3,6,7,0,8,4,5,9), for instance, would also have this Q-value, and would thus be functionally equivalent to your second arrangement.)

So, if we were to invent some machine that would randomly rearrange the balls once a minute, and we were to observe the system once each minute, we would see that, on average, we would be more likely to see the system with a Q-value of 16 than with a Q-value of 100. Meaning that the first arrangement would have a lower entropy than the second arrangement, because there are less stastical ways to observe an arrangement with this Q-value than one with 100. Not that we would never observe the rare case of Q = 100 - just we would be more likely to observe one with a lower Q-value. Even in this case, however, we must be careful to stipulate that the second arrangement is a microstate - one possible member that makes up the state defined by Q-value of 16, all of which are functionally equivalent. Thus it would still be inappropriate to ask whether 0,1,2,3,4,5,6,7,8,9 or 0,2,3,6,7,1,9,4,5,8 has a higher entropy; it would be more appropriate to ask whether a state with Q-value of 100 or a state with Q-value of 16 has a higher entropy, because this is what differentiates the microstates, not the specific labels.

So the answer is that you can't compare the entropies of your two arrangements, no matter what apparent or hidden codes there are, because the sequences of numbers aren't associated with any sort of observable value.

[Banedon]

I think that's a very cogent explanation Corribus. Thanks.