# Tag Info

## Hot answers tagged statistical-mechanics

6

Here's an intentionally more conceptual answer: Entropy is the smoothness of the energy distribution over some given region of space. To make that more precise, you must define the region, the type of energy (or mass-energy) considered sufficiently fluid within that region to be relevant, and the Fourier spectrum and phases of those energy types over that ...

5

Here's an elegant way to show that any linear combination of (anti)symmetric states is always (anti)symmetric. We use Dirac notation here for the states, and we assume, for simplicity, that we are dealing with a two-component system so that states of the system are linear combinations of products $|\psi\rangle = |\psi_1\rangle|\psi_2\rangle$. First, we ...

5

In terms of the temperature, the entropy can be defined as $$\Delta S=\int \frac{dQ}{T}\tag{1}$$ which, as you note, is really a change of entropy and not the entropy itself. Thus, we can write (1) as $$S(x,T)-S(x,T_0)=\int\frac{dQ(x,T)}{T}\tag{2}$$ But, we are free to set the zero-point of the entropy to anything we want (so as to make it convenient)1, ...

3

The entropy of a system is the amount of information needed to specify the exact physical state of a system given its incomplete macroscopic specification. So, if a system can be in $\Omega$ possible states with equal probability then the number of bits needed to specify in exactly which one of these $\Omega$ states the system really is in would be ...

3

You can set the entropy of your system under zero temperature to zero in compliance with the statistical definition $S=k_B\ln\Omega$. Then the S under other temperature should be $S=\int_0^T{\frac{dQ}{T}}$.

3

A superposition of (anti)symmetric states is always (anti)symmetric, but it is not necessarily decomposable. So, your $$\frac{1}{2}[(\chi(A)\psi(B)\pm\psi(A)\chi(B))+(\phi(A)\eta(B)\pm\eta(A)\phi(B))]$$ can be put in the form $$\frac{1}{\sqrt2}(f(A,B)\pm f(B,A)).$$ but in most cases there would exist no such functions $f_1$, $f_2$ that ...

3

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 ...

3

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 ...

3

Here is how Chandler does his counting: Take the (square) lattice to be of infinite extent or a finite lattice with periodic boundary conditions in both directions. The total number of edges is equal to $4N/2=2N$ if there are $N$ sites. The ground state corresponds to all spins being in the same state (all up or all down). The ground state energy is $-J$ ...

2

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 ...

2

[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 ...

2

First, you have to understand that Rudolf Clausius put together his ideas on entropy in order to account for the losses of energy that was apparent in the practical application of the steam engine. At the time he had no real ability to explain or calculate entropy other than to show how it changed. This is why we are stuck with a lot of theory where we ...

2

In classical thermodynamics only the change of entropy matters, $\Delta S = \int \frac{dQ}{T}$. At what temperature it is put zero is arbitrary. You have the similar situation with potential energy. One has to arbitrarily fix some point where the potential energy is put zero. This is because only differences of potential energy matters in mechanical ...

2

As a general rule, physics gets easier when the mathematics gets harder. For example, algebra-based physics comprises a bunch of seemingly unrelated formulae, each and every one of which needs to be memorized separately. Add calculus and wow! Many of those supposedly disparate topics collapse into one. Add mathematics beyond the introductory calculus level ...

2

As described on your link, we are trying to compute the probability that we will transition between two fully described states. So, by $i$ and $j$ they mean a complete specification of all of the attributes for each of the agents. Now, the probability that we will transition is determined by the rules of the simulation. In each time step we (1) choose a ...

2

Take a similar system of $N_A$ coins, where each can be either heads or tails with equal probability. The entropy of the system? $N_A k_B \ln 2$. This is to say that $k_B \ln 2$ is a natural "unit" of entropy (the information in one bit). Your system is slightly different (although both can be described as canonical ensembles with the fixed number of 1 mole ...

2

The formula for the pressure $$P=\frac{8\pi}{3h^3}\int^{p_0}_0 p^3\frac{dE}{dp}dp$$ is valid for a simple reason: $dE/dp$ is nothing else than the expression for the speed $v$. Check it for $E=p^2/2m$; the $p$-derivative is $p/m=v$. So the integrand only differs from the integrand for the electron number density $n$ in the first formula by the extra factor ...

1

Strictly speaking, you're using the relation $a\ln b = \ln(b^a)$ outside of its domain of validity. When $a$ is a quantity with units (and $b$ is dimensionless), it's perfectly valid to write $a\ln b$, but it is not equal to $\ln(b^a)$, because $b^a$ is undefined and so is its logarithm. If you want, for notational convenience you could specify that the ...

1

A higher entropy equilibrium state can be reached from the lower entropy state by an irreversible but purely adiabatic process. The reverse is not true, a lower entropy state can never be reached adiabatically from a higher entropy state. On a purely phenomenological level the entropy difference between two equilibrium states, therefore, tells you how "far" ...

1

I would strongly disagree with your statement that entropy has always been quite a mysterious quantity. Quite the contrary. What makes water (without something else interfering) always flow down the hill? Gravity. One does not have to know anything further about gravity to make good use of this statement, and people have used this fact for thousands of ...

1

There are indeed other assumptions in the derivation you quote. Namely, Chandler considers the classical limit of an ideal quantum mechanical gas with average particle number $<N> = \sum_j <n_j> = \sum [ e^{\beta (e_j - \mu)} ± 1 ]^{-1}$ (plus for Fermi-Dirac, minus for Bose-Einstein statistics). In the classical limit (low density) there ...

1

What it means is that the number of bits required to specify the exact physical state the system is in, increases by 100/log(2) bits after the gas is heated. I think measuring the temperature in energy units is a step in the right direction, but what is even better is to do without any units. I.e. while units may be introduced for convenience, the formalism ...

1

The answer given by Kyle refers of course only to the surface or photospheric temperature of the neutron star - the temperature of the layer from which photons can escape to reach an observer. In these outer layers the relationship between temperatures and particle motions is more-or-less consistent with the "everyday" Maxwell-Boltzmann picture referred to ...

1

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 ...

1

Maybe i found the source of my concern, the whole space phase have invariant measure so it's measure never change during the evolution and so i think i have a misleading idea of equilibrium approach, indeed it cannot be ''the system will spend it's future time in a set $A\subset\Omega$''.

1

Thermodynamically entropy is defined by $$\mathrm{d}S = \frac{\mathrm{d}Q_{rev}}{T} \, ,$$ 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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