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From a thermodynamical point of view, living beings are able to reduce their entropy by exporting entropy to the external world. This does not contradict the 2nd principle, since living beings are open systems. For this reason, in a thermodynamically homogeneous universe (heat death), no change in the entropy can occur, and consequently no living beings (nor ...


4

I would say it looks like nothing. The heat death requires that the whole universe is thermodynamically homogeneous, and that the universe has reached its maximum entropy. This means that every thing becomes a disordered lump of very sparse matter, without anything to see whatsoever. It's as if the universe is in a state akin to the "chaotic nothingness" ...


3

Since state $\sigma$ is not in thermal equilibrium I don't think one can use your definition of "thermodynamical" entropy. In fact, one should instead use Von Neumann entropy, which is a correct measure of statistical (so not quantum!) uncertainty. There is no other "classical" or "thermodynamical" entropy in quantum systems. As you mentioned, for a thermal ...


2

Your mistake is that by "state" you mean following: particle either in section A or section B. In reality what you use for computing entropy is number of ways for particle to be in some point of phase space, that is having particle coordinate and impulse (all 3D). But you can omit these details. Consider box with single particle and 2 sections divided by ...


2

I think you have a fundamental misunderstanding of what the heat death really is. Any observer, whether they are a time traveler, observer from another universe, or whatever, would just see a lot of empty space. The first thing to know is that the heat death is not a single event. The universe, after heat death, is dead in the sense that nothing is ...


1

Answering your questions (1) As long as the irreversibility arises solely as a consequence of the fact that the system and the enviornment are at different temperatures, the methods outlined below work. You can calculate $\Delta S_{\textrm{sys}}$ as normal, and calculating $\Delta S_{\textrm{env}}$ is also straight-forward if we treat its temperature as ...


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Here is a reference for the continuous versions: http://arxiv.org/abs/1402.2966


1

Rubber consists of many long-chain polymers. In an unstressed sample, these are randomly arranged. As a mental model think of them as anchor chains, where the angle between each link is entirely random - the overall polymer is in essence a random walk through the medium. Now, you pull on it. The net result is to better align the backbone of the polymer ...


1

Normally $H(X|Y)$ means conditional entropy. In this case I don't think there is any generally accepted definition of Renyi counterpart. There is, however, a recent Master thesis which lists some possibilities including a new propositions by the author, which seems to be quite reasonable: http://web.math.leidenuniv.nl/scripties/MasterBerens.pdf If you ...


1

As Joannes says, the two principles belong to two different theories: the least action principle is a principle about the (conservative) laws of motion and a proposition about the paths actually followed by the degrees of freedom of mechanical bodies the maximum entropy principle refers either to thermodynamics to figure out in which direction will a ...


1

Your mistake is in the last line. You assert: \begin{equation*} \langle f \rangle=\sum_i f_i e^{\lambda f_i} \end{equation*} but that isn't true, rather: \begin{equation*} \langle f \rangle=\sum_i f_i e^{\lambda f_i-1+\mu} \end{equation*} To convert correctly between your two expressions for $\langle f \rangle$, you must figure out a way to express $\mu$ ...


1

You're almost there. The thing to realise about the partition function is that it is the normalisation. If statistical mechanics were being developed today I'm sure its name would be "normalisation factor", but "partition function" has stuck historically. Now, the partition function is in fact defined as $Z = \sum_i e^{\lambda f_i}$, so that $$ p_i = ...


1

The formula $$ dS = dQ/T $$ only applies to thermodynamic processes that can be described by a path in thermodynamic state space (representing a quasistatic process, where the system is in thermodynamic equilibrium at all stages). Only for such processes it is meaningful to talk about change of thermodynamic entropy. Stirring a fluid is not such a simple ...



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