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Note that I am working in unit-mass bases (with lower-case symbols). Since, We have, So that, Notes: The internal energy is a function of the constant volume specific heat, and the latter is a function of temperature. The enthalpy is a function of the constant pressure specific heat, and the latter is a function of temperature. Remember ...


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I assume that your question is: "How do I compute the helmholtz free energy given the dependence of the entropy on the intensive parameters $S(E,N,V)$?" In this case you start by inverting $S(E,N,V)$ to get a relation for $E(S,N,V)$. Then you can perform the Legendre transform, $$ F(T,N,V) = E(S(T,N,V),N,V) - S(T,N,V) \, T \, , $$ where the function ...


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This type of calculation can be done in Mathematica using the xTensor package (its free). There is a bit of a learning curve, but the documentation is great and they have a very active google-group. Typically you will need to write the Lagrangian explicitly as a polynomial in the Riemann Tensor. Once you do this, the VarD command (xTensor) can handle the ...


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Calorimetry is a nasty business. Whenever is possible entropy is measured by measuring mechanical parameters by taking advantage the Gibbs form of 2nd law. For example, if a thin rod is elastic then $dU = TdS+\sigma d\epsilon$ where $\sigma, \epsilon$ are the stress and strain. From the equality of derivatives you get $\frac {\partial S}{\partial \sigma}|_T ...


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For a reversible addition of heat, the entropy change is $\int \frac{dQ} {T}$, in other words the area under the graph of $\frac{1}{T}$ against $Q$ (heat added to system). And yes, when a small amount of heat $\Delta Q$ is added, temperature T changes only a little, so $\frac{\Delta Q} {T}$ is well-defined. When added up this gives the integral.


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Just to give an account of some of the most popular approaches that I have met so far about out of equilibrium thermodynamics and corresponding generalized definitions of entropy and thermodynamic potentials. On one end of the spectrum, on can follow a statistical inference approach to statistical mechanics in its very foundation (as it has been proposed ...


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$$dQ = T \ dS \tag1$$ $$dQ = C \ dT \tag2$$ Interesting, right? In $(1)$, the whole $T$ multiplies the infinitesimal $\frac{\text{J}}{\text{K}}$. In $(2)$ it's the opposite: the whole $\frac{\text{J}}{\text{K}}$ multiplies the infinitesimal $T$. But you hinted that you knew that yourself already. Let's cut to the chase: both are different beasts entirely, ...


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If you consider a constant volume transformation, the corresponding specific heat will be defined as: $C_v(T) \equiv \left( \frac{\partial U}{\partial T}\right)_{N,V}$ Now, it is not forbidden to use Leibniz rule for the decomposition of partial derivatives and for instance: $\left( \frac{\partial U}{\partial T}\right)_{N,V} = \left( \frac{\partial ...


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Taking the first law of thermodynamics you have $$dU = \delta Q + dE_{work}$$ I.e. the change of internal energy is the change of energy expended by work of the macroscopic parameters plus heat. It is a non-trivial result in thermodynamics that there exists a function $S$, entropy, and absolute temperature $T$ which can be used to write the first law as ...


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As a partial answer (a complete answer would require more time and more knowledge about classical thermodynamics on my part): As you thought, these are two entirely different objects. The heat capacity is a material-dependent object that - as you say - measures the difference in temperature when energy is absorbed by the material. It's object and process ...


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Newtonian mechanics isn't quite as predictable as it's made out to be, both in theory and in practice. In theory, it's an easy task to set up a Newtonian system of particles that will eventually violate the Lipschitz continuity condition. You can toss reversibility and predictability out the window when that happens. In practice, we can't know state ...


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Why "Surely every increase of say 1K increases the number of available microstates by the same multiplicative factor". It's not obvious to me why you say that. Generally, the more energy a system has, the more states become available for the addition of small amounts of additional energy. Consider an ideal gas where each particle contributes $p^2/2m$ to ...


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This question has bothered me for a long time, asked some real EM/optics experts but never got a satisfying answer for it, so do not expect one from me. At best I heard that it was not a stupid but an interesting question. Despite what I was ever taught I for one believe that diffraction is a manifestation of irreversibility of EM. If you start with a hole ...


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We define entropy in the electromagnetic case the same way we define it everywhere else. Entropy is the cause that (in absence of other physical changes) heat flows from hot to cold. Consider two temperature baths of different temperature facing each other. The surfaces of both temperature baths are being treated as black body radiators. Using the ...


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Is there a minimum energy content of information? Carl Witthoft's answer gives us a key hint: in terms of energy efficiency one can't really do better than using photons to encode bits. A photon residing in a container of linear size $R$ has minimum energy $\Delta E \approx \hbar c / R$. By increasing the size $R$ one can reduce the energy per bit below ...


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This may not be the best way to look at the problem, but: Suppose you send one photon for a "1" and no photon for a "0" according to some prearranged clock. What's the lowest possible photon energy? The answer, of course, is "asymptotically approaching zero." Not that I'd like to build an antenna capable of detecting a photon with $\lambda > 1$ ...


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See e.g. Landauer's principle http://en.wikipedia.org/wiki/Landauer's_principle and capacity of noisy channels http://en.wikipedia.org/wiki/Channel_capacity. Not everybody agrees with these limits, but to me they seem fairly reasonable based on relatively straight forward noise arguments. In an ideal world (i.e. temperature T=0), there is no lower limit ...


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[As requested, I convert my comment into an answer, as it might also be useful for other people.] There is a very interesting series of works by Lieb and Yngvason on entropy and the second law of thermodynamics, based on the kind of axiomatic approach you seem to be interested in. You can start with this introductory paper, or this, this or this more ...


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Now we have the second law of thermodynamics, that says that entropy always increases. Second law does not say exactly that. It has more formulations, some of which use the concept of entropy. One such formulation is When thermally insulated system changes its state from one equilibrium state to another, its entropy cannot decrease. This statement ...


1

Are not we simply saying that things more likely to occur, occur more times? Isn't it then, that the second law is simply an inmense tautology? No, this argument doesn't suffice to prove the second law. This argument only proves that thermal fluctuations away from equilibrum should be rare and short-lived. That's a statement that doesn't have anything ...


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Ultimate physical motivation Strictly in the sense of physics, the entropy is less free than it might seem. It always has to provide a measure of energy released from a system not graspable by macroscopic parameters. I.e. it has to be subject to the relation $${\rm d}U = {\rm d}E_{macro} + T {\rm d} S$$ It has to carry all the forms of energy that cannot be ...


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Equilibrium is defined in the original notions of thermodynamics as the asymptotic static state. I.e., by this definition no macroscopical quantity varies in equilibrium. Statistical physics however tells us that the system varies in a certain "random walk" around all the possible states and never stops. We just cannot distinguish most of these states ...


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Thermodynamically entropy is defined by \begin{equation} \mathrm{d}S = \frac{\mathrm{d}Q_{rev}}{T} \, ,\end{equation} where $\mathrm{d}Q_{rev}$ is the heat, transferred reversibly. As you point out it can be shown that this quantity is a function of state. This implies that the entropy of any thermodynamic system has, up to a constant, a well defined ...


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A certain volume of space with a uniform distribution of particles has maximum entropy. That is correct for non-interacting particles, but wrong for particles with the gravitational interaction. When gravity condenses these particles, it increases the entropy of the system, not decreases it, at least when the Jeans instability condition is satisfied. ...


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Based on some "google research" I get the impression that the popularity of the perfume thought experiment stems from a 1975 Scientific American article written by David Layzer called The Arrow of Time. The article featured this figure visualizing the thought experiment: Of course, the notion that the second law of thermodynamics implies an asymmetry ...


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Unitarity of quantum mechanics prohibits information destruction. On the other hand, the second law of thermodynamics claims entropy to be increasing. If entropy is to be thought of as a measure of information content, how can these two principles be compatible? I don't think there's anything inherently quantum-mechanical about this paradox. The same ...


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Prohibition of information destruction due to unitarity is a hypocrisy (sorry, I go to reiterate some stuff from the Where does deleted information go? posting), and the concept of entropy is foggy, especially in the quantum context: in spite of definitions mentioned in previous answers, there is no possibility to ever know the quantum state of a ...


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Henri Poincaré, in discovering limit cycles, used a thought experiment containing a box with a partition. One side had a gas, and the other didn't. When the partition was removed, the gas would diffuse through the opening and occupy both sides of the container. He first published works describing limit cycles somewhere in 1881-1882. I am unsure if he ...


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The Second Law of Thermodynamics states that the entropy of the universe always increases or stays constant. This means that we can reduce the entropy of the gas in the box (gas compressed from the whole box to half the box at constant temperature), only if we increase the entropy somewhere else. For example, we can compress the gas, doing work on it, and ...


1

Can the law of conduction be derived from the assumptions of classical thermodynamics? The answer should be no, because classical thermodynamics does not deal with description of irreversible processes in time; it only deals with equilibrium states. Second law of thermodynamics does not assert that entropy decreases in time, only that after irreversible ...


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One cannot "derive" any non-equilibrium rate law from thermodynamics, simply because they are beyond the scope of the theory. Thermodynamics simply does not deal with such phenomena and hence cannot tell you how such processes occur (in this case heat conduction). All that thermodynamics does is relate mean values of certain properties of systems amongst ...


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Short answer is no but even then it depends on what you mean by "the 2nd law of thermodynamics". In conventional treatments of so-called equilibrium thermodynamics Fourier's law of heat conduction is completely independent of the rest. In what is called "rational thermodynamic" where the 2nd law is formulated as the "Clausius-Duhem inequality" it is in fact ...


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Conservation of information in quantum mechanics is a hypocrisy as much as it is so in classical mechanics. It is not conserved in the same sense as energy, charge, or momentum. When sages like Hawking and Penrose discuss whether does “information” survive destruction in a spacetime singularity, they mean something utterly different from the concept familiar ...


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No, the reasoning is wrong. When you need a N-qubit state, no difference entangled or not, you just initialize N qubits and then apply a unitary transformation (reversible). So, entropy produced during the initialization is proportional to the amount of information, without exponents.


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A reference that demonstrates the fact:'When a is a quantity with units (and b is dimensionless), it's perfectly valid to write alnb, but it is not equal to ln(ba), because ba is undefined and so is its logarithm.' I want to get more information/details that asserts the information you provided through your useful answer.


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Observer, if you assume as someone watching the experiment or activity, is plainly wrong. Anything that can be detected and measured and thus, in principle, from infinitely hard calculation can tell us about the past or previous states, can be said to be information, and thus, entropy increased while the process was being carried out. Take for example, a box ...


0

To begin with, entropy is a classical thermodynamics concept. Different statistical frameworks , assuming some postulates, can define an entropy. The basic formulation of entropy defined by statistical mechanics where kB is the Boltzmann constant, equal to 1.38065×10^−23 J K−1. The summation is over all the possible microstates of the system, and p_i ...


0

In quantum statistical mechanics entropy is not defined via the probability density of a single state but through the density matrix which talks about the "non-quantum uncertainty" over the states. This is the "probability density" over which all the entropy theorems are proven in the quantum world. That is, if we know the system to be in a sharp quantum ...



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