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1

Christoph gives a good answer from the point of view of statistical physics, but the first context we actually encounter entropy is in classical thermodynamics. It is useful even without the interpretation of being "the measure of disorder". This got slightly out of hand and is now quite technical, but I think it is still digestible and provides a complete ...


2

What is entropy, more than disorder? Don't look at entropy as disorder. Thinking of it as disorder has long been a source of confusion. Many texts are moving away from using the disorder description. Macroscopically, it's better to think of entropy as a measure of energy dispersion rather than as a measure of disorder. Microscopically, it's better to think ...


3

What is entropy, more than disorder. Mathematically, entropy is just a measure of spread of a probability distribution: The lower the entropy, the more spiked the distribution. In statistical mechanics, a state is generally only partially defined via some macroscopic constraints, and entropy is a measure of microscopic indeterminacy. Maximizing entropy ...


1

This is a very difficult problem. I try to explain why. If we want to obtain the mathematical proof of the second law, we must consider the mathematical proof of the entropy first, as a state function. Clausius’ definition $dS=δQ/T$ cannot be proven in mathematics, as an exact differential, so the definition $dS=δQ/T$ must depend on imaginary reversible ...


0

There is a new statement on the second law: "irreversibility root in a fundamental principle: the gradients of the four thermodynamic forces spontaneously tend to zero". Please see http://arxiv.org/abs/1201.4284


3

I believe it was Boltzmann who first made the connection between entropy and micro states. chapter 12 of "Classical and Statistical Thermodynamics" by Ashley H. Carter discusses Boltzmann's arguments. To summarize from that book: Entropy ($S$) corresponds to a particular configuration of an ensemble of particles called a macro state. A macro state can be ...


0

I'm not sure what our question is exactly, so I'll try to answer what I guess your question is. In general, the entropy of a fluid is a function of both $V$ and $T$. During an isentropic compression, the decrease in entropy from the reduction of volume is compensated by an increase due to the temperature rise. The net effect is zero. For an ideal gas we ...


1

By definition, $dS =\frac{dQ}{T}$, so an adiabatic process doesn't change entropy. But you can find more details at http://en.wikipedia.org/wiki/Adiabatic


0

I'd have used $\Omega$ rather than $W$, and I'd say either way that it's dimensionless. Additonally the Boltzmann constant serves only to define the temperature unit (in the case of SI that's kelvins). So your presmise is incorrect and that's the origin of your problem. Read a textbook, a good one is Schreoeder.


0

In the video lecture mentioned in the question, the guy always use $T_{sur}$ in the inequality TdS>dU+PdV. In deriving the equation the "surrounding" (also see "The Principles of Chemical Equilibrium" by Denbigh, p.82, the term "thermostat" is used instead) is included as a bigger isolated system. Let's say the entropy of these bigger system is Si. ...


3

We have a perfectly unambiguous definition of temperature for canonical ensembles, and this temperature may be negative in bounded-energy systems. This kind of negative temperature is indisputable, and some would argue it has been realized in spin-inversion experiments. The problem is that there are two decent but imperfect definitions for the entropy of a ...


3

What I am asking, then, is whether someone on StackExchange might be able to shed some light on the matter as to how there can be a disagreement about something that seems should be a mathematical fact. The main disagreement seems to be about which definition of the word "entropy" in the context of statistical physics is "correct". Definition is an ...


2

Simply the thermodynamical quantities used in the original paper were not suitable for that problem. They in particular calculated $T=\frac{\partial U}{\partial S}$ where $ U$ is the internal energy and $S$ the entropy. However a wrong definition of entropy has been used. Mathematicians has proved that the use of that specific entropy was wrong and that ...


5

First question: Is this equation applicable for irreversible processes? From the first law, we have: $$ \mathrm{d} U = \mathrm{d}Q + \mathrm{d}W $$ where $\mathrm{d} U$ is an exact differential, and $\mathrm{d}Q$ and $\mathrm{d}W$ are inexact differentials. It is thus remarkable to see that the sum of of two inexact differentials makes an exact ...


1

As I understand it, the solar system evolved from a massive molecular cloud. To me, this seems to break the second law of thermodynamics, as I think it suggests order from disorder. There are two problems here. One is the concept of entropy as disorder. A number of thermodynamics texts have now discarded this old concept. For one thing, it doesn't help ...


2

Combining the first and second law of thermodynamics we can get the following equation $$TdS=dU-P_{ext}dV$$ There is little reason to use $P_{ext}$ here. Also the sign is wrong. The correct way to write the differential relation between entropy, energy and volume is $$ TdS = dU + PdV $$ where $T, P$ are temperature and pressure of the system and ...


1

You have provided the von Neumann entropy definition which is derived from its density matrix. I would consider it an intrinsic rather than fundamental property, but this is just semantics. Some recent work by John Baez has investigated the dynamics of quantum entropy called quantropy.


0

The order of the events never changes in Special Relativity (as Jerry Schirmer stated above) to preserve causality. If you are thinking of Boltzmann entropy then I would guess its still increasing as the number of states after evolution (no matter how slow compared to a different reference frame) goes up just like in any other system.


8

First of all, unlike mass/energy, the entropy is not an observable. The word "observable" may be understood both as an adjective and as a noun ("an observable"). Entropy isn't an observable because it is not given by a linear operator acting on the Hilbert space (or on the space of density matrices), $$ L: |\psi\rangle \to L|\psi\rangle $$ Instead, the ...


2

If you think of the future as a probabilistic distribution of events, for the far future there are an infinite number of possible events. As you approach those events in time, past (and present) actions force the future to collapse to a single event (assuming two can't happen simultaneously). You could think (and even predict) one event would happen over ...


5

It is a reasonable question at the elementary particle physics level , since the mathematical formulae of all the models we have are reversible as to time. It is in the thermodynamic manifestation of the laws that an arrow of time appears, and in special relativity which separates observations in timelike and spacelike regions. So it is one of those ...


0

Christoph Schiller argues that the log term arises from the different orientations that a black hole can have in space. His proposal also fixes the value of the constant K, as he explains, to 3/2 times the Boltzmann constant k_B. I found this in the section "Entropy of horizons" on page 269 and 270 of "The strand model - a speculation on unification". I ...


0

The amount of 'randomness' is not the only definition of entropy. Entropy is also known as the amount of unusable energy that's present in a system. So if you have a lot of heat energy in a system, since all of that heat energy can be used up, its entropy would be low. If you have less heat in a system, only a small amount of heat in that system can be used ...


1

The usual argument is a proof of the existence of a function whose differential is $\delta Q/T$. The argument usually proceeds as follows: Fact. (Physics) $\int_\gamma \frac{\delta Q}{T} = 0$ for any closed path $\gamma$ in thermodynamic state space. Once you have this fact, you can prove that Claim 1. $\delta Q/T$ is conservative, namely ...


2

The "Mexican Hat Potential" (although now more politically correctly called the "Champagne Bottle Potential" after the punt at the base) is the potential energy curve for the vacuum expectation value (VEV) of the Higgs field. Think of the blue dot as being "the vacuum", and the radial direction as turning up the strength of a background field that permeates ...



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