On the Second Law of Thermodynamics one: Eliezer does something a bit naughty in this one which has the effect of making it look weirder than it is. He says: suppose you have a glass of water, and suppose somehow you magically know the positions and velocities of all the molecules in it. And then he says: “Does that make its thermodynamic entropy zero? Is the water colder, because we know more about it?”, and answers yes. But those are two questions, not one question; and while the answer to the first question may be yes (I am not an expert on thermodynamics and I haven’t given sufficient thought to what happens to the thermodynamic definition of entropy in this exotic situation), the answer to the second question is not.
The water contains the same amount of heat energy as if you didn’t know all that information. It is not colder.
What is different from if you didn’t know all that information is that you have, in some sense, the ability to turn the water into ice plus electricity. You can make it cold and get the extracted energy in usable form, whereas without the information you couldn’t do that and in fact would have to supply energy on net to make the water cold.
(“In some sense” because of course if you gave me a glass of water and a big printout specifying the states of all its molecules I couldn’t in fact use that to turn the water into ice plus electrical energy. But in principle, given fancy enough machinery, perhaps I could.)
Another way of looking at it: if I have a glass of room-temperature water and complete information about how its molecules are moving, it’s rather like having a mixture of ice and steam containing the same amount of thermal energy, just as if those molecules were physically separated into fast and slow rather than just by giving me a list of where they all are and how they’re moving.
Another difference from separated hot & cold reservoirs is that the time horizon for being able to make use of the information is on the order of nanoseconds before the information is useless. Even without quantum messiness and prescribing perfect billiard-ball molecules, just a few stray thermal photons from outside and a few collisions will scramble the speeds and angles hopelessly.
As far as temperature goes it is really undefined, since the energy in the water is not thermal from the point of view of the extremely well informed observer. It has essentially zero entropy, like the kinetic energy of a car or that of a static magnetic field. If you go ahead and try to define it using statistical mechanics anyway, you get a division by zero error: temperature is the marginal ratio of energy to entropy, and the entropy is an unchanging zero regardless of energy.
I think that last bit only applies if we suppose that you are equipped not only with a complete specification of the state of the molecules but with a constantly instantly updating such specification. Otherwise, if you put more energy in then the entropy will increase too and you can say T = dE/dS just fine even though the initial entropy is zero. (But you make a good point about the energy being not-thermal from our near-omniscient viewpoint.)
If you just have a snapshot state (even with an ideal model of internal interactions from that state) then any thermal contact with the outside will almost instantly raise entropy to near maximum regardless of whether energy is added or removed or on balance unchanged. I don’t think it makes sense to talk about temperature there either, since the entropy is not a function of energy and does not co-vary with it in any smooth way.
On the Second Law of Thermodynamics one: Eliezer does something a bit naughty in this one which has the effect of making it look weirder than it is. He says: suppose you have a glass of water, and suppose somehow you magically know the positions and velocities of all the molecules in it. And then he says: “Does that make its thermodynamic entropy zero? Is the water colder, because we know more about it?”, and answers yes. But those are two questions, not one question; and while the answer to the first question may be yes (I am not an expert on thermodynamics and I haven’t given sufficient thought to what happens to the thermodynamic definition of entropy in this exotic situation), the answer to the second question is not.
The water contains the same amount of heat energy as if you didn’t know all that information. It is not colder.
What is different from if you didn’t know all that information is that you have, in some sense, the ability to turn the water into ice plus electricity. You can make it cold and get the extracted energy in usable form, whereas without the information you couldn’t do that and in fact would have to supply energy on net to make the water cold.
(“In some sense” because of course if you gave me a glass of water and a big printout specifying the states of all its molecules I couldn’t in fact use that to turn the water into ice plus electrical energy. But in principle, given fancy enough machinery, perhaps I could.)
Another way of looking at it: if I have a glass of room-temperature water and complete information about how its molecules are moving, it’s rather like having a mixture of ice and steam containing the same amount of thermal energy, just as if those molecules were physically separated into fast and slow rather than just by giving me a list of where they all are and how they’re moving.
Another difference from separated hot & cold reservoirs is that the time horizon for being able to make use of the information is on the order of nanoseconds before the information is useless. Even without quantum messiness and prescribing perfect billiard-ball molecules, just a few stray thermal photons from outside and a few collisions will scramble the speeds and angles hopelessly.
As far as temperature goes it is really undefined, since the energy in the water is not thermal from the point of view of the extremely well informed observer. It has essentially zero entropy, like the kinetic energy of a car or that of a static magnetic field. If you go ahead and try to define it using statistical mechanics anyway, you get a division by zero error: temperature is the marginal ratio of energy to entropy, and the entropy is an unchanging zero regardless of energy.
I think that last bit only applies if we suppose that you are equipped not only with a complete specification of the state of the molecules but with a constantly instantly updating such specification. Otherwise, if you put more energy in then the entropy will increase too and you can say T = dE/dS just fine even though the initial entropy is zero. (But you make a good point about the energy being not-thermal from our near-omniscient viewpoint.)
If you just have a snapshot state (even with an ideal model of internal interactions from that state) then any thermal contact with the outside will almost instantly raise entropy to near maximum regardless of whether energy is added or removed or on balance unchanged. I don’t think it makes sense to talk about temperature there either, since the entropy is not a function of energy and does not co-vary with it in any smooth way.