Open Thread, Feb. 2 - Feb 8, 2015

Gondolinian

If it's worth saying, but not worth its own post (even in Discussion), then it goes here.

Notes for future OT posters:

1. Please add the 'open_thread' tag.

2. Check if there is an active Open Thread before posting a new one. (Immediately before; refresh the list-of-threads page before posting.)

3. Open Threads should be posted in Discussion, and not Main.

4. Open Threads should start on Monday, and end on Sunday.

If it's worth saying, but not worth its own post (even in Discussion), then it goes here.

Previous Open Thread

Notes for future OT posters:

1. Please add the 'open_thread' tag.

2. Check if there is an active Open Thread before posting a new one. (Immediately before; refresh the list-of-threads page before posting.)

3. Open Threads should be posted in Discussion, and not Main.

4. Open Threads should start on Monday, and end on Sunday.

Nope, sorry.

Also, I still don't buy the claim about the temperature. You said in the linked comment that putting a known-microstate cup of tea in contact with an unknown-microstate cup of tea wouldn't really be thermal equilibrium because it would be "not using all the information at your disposal. And if you don't use the information it's as if you didn't have it."

If I know the exact state of a cup of tea, and am able to predict how that state will evolve in the future, the cup of tea has zero entropy.

Then suppose I take a glass of water that is Boltzmann-distributed. It has some spread over possible microstates - the bigger the spread, the higher entropy (And also temperature, for Boltzmann-distributed things).

Then you put the tea and the water in thermal contact. Now, for every possible microstate of the glass of water, the combined system evolves to a single final microstate (only one, because you know the exact state of the tea). The combined sytem is no longer Boltzmann in either subsytem, and has the same entropy as the original glass of water, just moved into different microstates.

Note that it didn't matter what the water's temperature was - all that mattered was that the tea's distribution had zero entropy. The fact that there has been no increase in entropy is the proof that all the information has been used. If the water had the same average energy as the tea, so that no macroscopic amount of energy was exchanged, then these thing would be in thermal equilibrium by your standards.

Then you put the tea and the water in thermal contact. Now, for every possible microstate of the glass of water, the combined system evolves to a single final microstate (only one, because you know the exact state of the tea).

After you put the glass of water in contact with the cup of tea, you will quickly become uncertain about the state of the tea. In order to still know the microstate, you need to be fed more information.

-1passive_fist11y

That's wrong on both counts. Firstly, even if you actually had a block of ice at 0 K and put it in thermal contact with a warm glass of water, the total system entropy would increase over time. It is completely false that the number of initial and final microstates are the same. Entropy depends on volume as well as temperature. (To see why this is the case, consider that you're dealing with a continuous phase space, not a discrete one). Additionally, your example doesn't apply to what I'm talking about, because nowhere are you using the information about the cup of tea. Again, as I said, if you don't use the information it's as if you didn't have it. I am fully aware that saying it in this way is clumsy and hard to understand (and not 100% convincing, even though it really is true). That's why I'm looking for a more abstract, theoretical way of saying it.

7

Open Thread, Feb. 2 - Feb 8, 2015

7

7

7

Open Thread, Feb. 2 - Feb 8, 2015

7

7