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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) ...


4

There are several possible answers to this. One is to look at Shannon's definition of the entropy, $$ H = -\sum_i p_i \log p_i, $$ and note that it has the form of an expectation: $H$ is the expected value of $-\log p_i$, so it makes sense to give a name to that latter quantity. This is nice if you understand the value of the entropy. In Shannon's paper ('A ...


4

I think that you are really interested in the $q$-state clock model, which is similar to the Potts model, and is defined as follows. Fix an integer $q\geq2$. For each $i\in\mathbb{Z}^d$, let $$ \theta_i \in \bigl\{\frac{2\pi}{q} k\,:\, k\in\{0,1,\ldots,q-1\}\bigr\}, $$ and define the spin at site $i$ by $$ \mathbf{S}_i = (\cos\theta_i,\sin\theta_i) . $$ The ...


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 ...


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 ...


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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

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 ...


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

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

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 ...


1

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 ...


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

This is a very difficult question. In one dimensions a two body scattering is the minimum interaction you can have. In multiple dimensions you can have more bodies scattering as the minimum interaction. The Yang Baxter equation is formally, $$(R\otimes \mathbf{1})(\mathbf{1}\otimes R)(R\otimes \mathbf{1}) =(\mathbf{1}\otimes R)(R\otimes ...


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Yes, a high magnetic field can massively reduce the entropy of the dipoles for a given temperature. When the dipoles enter a field they become hotter (analogous to a gas when its volume is compressed) so they dissipate heat into their environment (into lattice vibrations and other degrees of freedom). Once the dipoles have returned to thermal equilibrium ...


1

Yes, the efficiency is then higher than 1 but that is fine because, although it can be described by a negative temperature, it does not per se correspond to a heat reservoir but rather to a very peculiar (statistically) excited state of a macroscopic system that is just waiting to release this energy. Now, this is not conceptually so problematic overall, ...


1

Here is a proof following Ojima, "Lorentz Invariance vs. Temperature in QFT", Letters in Mathematical Physics (1986) Vol. 11, Issue 1 (1986) 73-80. The first two pages of the paper are available for free here, but the website wants money for more of the paper. (Click the orange "Look Inside" button if the paper doesn't open automatically.) Fortunately, the ...



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