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How about lifting a litre or kilogram of water? Lifting a kilogram of water $235.3\text{ m}$ (772 feet in sensible units) involves a potential energy change:$$PE=mgh=1\times9.8\times235=2.31 \text{ kJ}$$ Warming a kilogram of water, (specific heat $4.179\text{ kJ/kg/K}$), through a temperature change of $0.556\text{ K}$ (1 degree Farenheit) ...

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Here's something I believe is a simple proof. Unfortunately it uses a little bit of cohomology. Consider the canonical 2-form in extended phase space $T^*M \times \mathbb{R}$ $$\omega = \sum_{i=1}^N dq_i \wedge dp_i - dH(\vec{q},\vec{p},t) \wedge dt ,$$ where $N = dim(M)$. A function $f: M \to M$ is said to be a canonical transformation iff $f^* \omega = ... 4 This is a good question. I can only answer half of it: The eigenvalues of$M^{l}$are some times complex. Take a look at the RG flow of Einstein gravity close to its UV fixed point. The two relevant couplings (newton's constant and cosmological constant) spiral around the fixed point as they move away from it. The real part of the critical exponents ... 3 This relation is not true for general processes. For a closed system, the general relation is$\delta Q \leq TdS$, as is illustrated by the Clausius Theorem (http://en.wikipedia.org/wiki/Clausius_theorem). Another way of writing it is$dS=\delta Q/T + dS_{irr}$where$ dS_{irr}$is the entropy change due to irreversibilitiy in the closed system ... 3 To add to Whelp's Answer. Even though the $$\bar{d}Q=T\,dS$$ does not hold in an irreversible process, it still gives us something general. Consider a thermodynamic system linked to a system of reservoirs and which can only exchange heat and nothing else with the reservoirs. Let's call the thermodynamic system an engine for our convenience. Then the engine ... 2 I guess, one could start by considering the microcanonical and canonical ensembles as entirely different concepts. For they represent different statistical ensembles: in the former the energy of the system is fixed while in the second the energy can span all possible values allowed by the energy spectrum of the system with some penalty attributed to high ... 2 By the main theorem of connectedness in general topology, continuous maps preserve connectedness. Time evolution of Hamiltonian systems preserves connectedness because it is continuous. I think it is independent of from Liouville's theorem, it just requires the proving Hamiltonian time evolution is continuous. This is just a formal way of restating ... 2 Bose particles cannot be identified as different in a given state, whereas boltzmann particles can (even though both types can occupy a given energy state with more than one particle). Thus boltzmann statistics need to take into account the permutations ($n!$) of the$n$particles into a given state, in contrast to the bose particles (which are not ... 2 An order parameter distinguishes two different phases (or orders). In one phase the order parameter is zero and in another phase it is non-zero. It does not have to be macroscopic. For example, in the BCS theory of superconductivity the order parameter is called the gap$\Delta$. It can be interpreted as the binding energy of a Cooper pair, namely two ... 2 There are four state kets$|\uparrow_1\uparrow_2\rangle,|\uparrow_1\downarrow_2\rangle,|\downarrow_1\uparrow_2\rangle,|\downarrow_1\downarrow_2\rangle$representing four possible configuration of the two spins, the eigenvectors of the Hamiltonian is the linear combination of them. There will be four eigenvectors and four eigenenergy. You can also obtain the ... 2 No, if$S$is really the only thing you know about your system then there is no way to know its energy. There is no relationship between the energy and the entropy that doesn't involve some other quantity such as temperature. ...but surely you know something about your system, other than its entropy? I mean, you must know something about what it's made of, ... 2 The problem with the Boltzmann definition is, as you have neatly shown, that its usefulness depends on the assumption that your system is in equilibrium with its surroundings. Without first assuming equilibrium and subsequently setting the temperatures as equal, one cannot show that the Boltzmann entropy satisfies the First Law and hence meaningfully define ... 1 Several sources give a higher (experimental) value of the Debye temperature for diamond - about 2220K: http://www.sbfisica.org.br/bjp/download/v03/v03a03.pdf , http://www.cvd-diamond.com/properties_en.htm , http://www.chm.bris.ac.uk/motm/diamond/diamprop.htm . 1 In contrast to the other answers, I would like to mention that it is possible to compute rigorously the value of the critical temperature of the two-dimensional Ising (and Potts) model, without computing explicitly the free energy (which is in any case not possible for general Potts models). In the Ising case, this has been known for a long time, and there ... 1 Let's consider just one cycle of the Szilard engine. Aside from discussion of energy free state polling, one of the main points of Bennett paper (if you mean Charles Bennett, "The Thermodynamics of Computation: A Review", Int. J. Theo. Phys., 21, No. 12, 1982) here is that you must build a finite state machine (a very simple three-state machine) as a minimal ... 1 Suppose you have some radioactive material with a half life$\tau_{1/2}$. What that term "half life" means is that the amount of material$m(t)$you have left after a time$t$is $$m(t) = m(0) \exp[ -t / \tau_{1/2}] . \qquad (*)$$ However, the material is made up of discrete atoms and each one decays in a random way. Therefore, it's not 100% guaranteed ... 1 Actually, all of your references are correct. In a MD simulation, for homogeneous systems, you can calculate the ensemble average of a thermodynamic property in several different ways. You can simply average over all the particles in your system during one time step. You can average over a single particle over many time steps. You can average over all ... 1 Since you have the expression of$\Omega$your work is almost done. First remember that the entropy for a micro-canonical (fixed energy) system at thermal equilibrium is given the very famous Boltzmann's formula : $$S=k_B\,ln(\Omega)$$ Then, simply use the Stirling's approximation to evaluate$ln(N!)\approx Nln(N)-N$(because$N>>1\$, i.e. very long ...

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Interesting question. I would have thought that if you were aware of the exact number of energy states and the populations thereof, you could apply boltzmann statistics to each of the levels in order to fit an appropriate temperature to the population in each state. This temperature, if comparable amongst the included levels, would therefore require that the ...

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In string theory appears the so-called Hagedorn behavior, which is an exponential behavior of the density of states. As you point out, in this case, temperature does not vary with energy (so it means that the system has an infinite heat capacity!) This behavior appears in Little String Theory, e.g. http://arxiv.org/abs/hep-th/0010169.

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REMARK. Perhaps I wrongly interpreted the question. I interpreted it as if were referred to the total volume of phase space. The answer is negative if the question regards general changes in time of topology of the total space of phases and if you do not impose any generic restriction on the topology of the spaces, like compactness (see the final ...

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