twofish-quant said:
This doesn't make sense to me. I have a cup of ice. There is an entropy associated with that cup of ice. (41 J/ (mol K)) Now if I watch a cup of ice melt into water and the entropy is now 70 J(mol K), are you trying to tell me that because of some weird quantum entanglement effect the entropy of the universe is constant?
Yes
What you seem to be saying is that anytime ice melts, then there is some weird quantum mechanical effect that causes something weird to happen in some other part of universe.
Not that something weird happens on the other part of the universe, and in fact the rest of the universe excluding the melting ice also has it's entropy go up. But as the ice melts, it is interacting with everything else and there are correlations between the ice's state and the "rest-of-the-universe"'s state. These correlations when included in the calculation of the entropy of the universe-as-a-whole will reduce the sum of the entropes if you just add the calculated entropy of each part.
I'm just watching ice melt. Are you saying that I can't understand ice melting without quantum entanglements?
Of course not. Likewise we don't need QM to track the Moon's orbit. And as long as one is talking about some piece of the universe (which necessarily excludes the mechanisms used to observe that piece) then you needn't worry about this.
Now you find this idea perplexing and counter-intuitive. Well it is, as is so much in QM. But you can "do the math". Take a system of two particles which are maximally entangled, say two spin-1/2 particles in a sharp spin-0 composite state. Now let one of the particles reside in your "system" and shoot the other one into space, or better yet into a Black Hole.
Since the composite is sharply defined it has zero entropy.
To describe one of the particles alone and you must do a partial trace over the other, you get a maximum entropy density matrix diag(1/2,1/2). Its entropy = k ln(2). (k = Boltzmann's constant.)
Likewise if you were to determine the entropy of the other half of the entangled pair you'd get k ln(2). Add and you get 2kln(2) but that's not the entropy of the composite. It is rather 0. Entropy doesn't "add up" in QM.
Finally let me mention that when you have an entangled pair, it typically undergoes decoherence due to interaction with the environment. But that's just the entanglement "swapping" to other systems. The particle in hand soon becomes entangled with photons (which have interacted with the entangled partner) shooting off into space at speed c, never to be recovered. That's where the irreversibility comes into play.
Similarly if you consider a high entropy system and wish to "refrigerate" it to lower entropy you are basically swapping the entanglements between your lab system and far flung photons to entanglement between your heat sink and far flung photons. Your system's entropy goes down... and remarkably the entropy of all but your system goes down since there are now more correlations within the exosystem.
Ultimately according to this view of entropy, at any given time the entropy of a system is equal to the entropy of all the universe minus that system which is to say both represent the same quantity. It is the amount of entanglement across the boundary between them.
Weird indeed! No?