# Tag Info

## Hot answers tagged statistical-mechanics

5

A quick google search leads one immediately to the wikipedia page on this particular theorem. The first paragraph of this page states: The Bohr–van Leeuwen theorem is a theorem in the field of statistical mechanics. The theorem states that when statistical mechanics and classical mechanics are applied consistently, the thermal average of the ...

5

No. You wouldn't say that pair of beams has a temperature. Temperature is defined by the zeroth law of thermodynamics, which states that if $A$ is in thermal equilibrium with $B$ and $B$ is in thermal equilibrium with $C$ then $A$ is in thermal equilibrium with $C$ and $A$, $B$ and $C$ are said to have the same temperature. Temperature is fundamentally a ...

4

It's the differential relationship between internal energy and entropy: \begin{align} dU &= T\,dS + \cdots \\ \frac{\partial S}{\partial U} &= \frac 1T \end{align} As energy is added to a system, its internal entropy changes. Remember that the (total) entropy is $$S = k \ln\Omega,$$ where $\Omega$ is the number of available microscopic states that ...

3

I will give you the argument that is presented in the book I learned from, Physical Gas Dynamics. For reference, it's page 104, Chapter 4 section 6. I present it instead because it does not rely on the Boltzmann distribution for the derivation so perhaps there is some additional insight for you. They established in section 5 that: $$\frac{N_j^*}{C_j} = ... 3 What is entropy, more than disorder. Mathematically, entropy is just a measure of spread of a probability distribution: The lower the entropy, the more spiked the distribution. In statistical mechanics, a state is generally only partially defined via some macroscopic constraints, and entropy is a measure of microscopic indeterminacy. Maximizing entropy ... 3 I believe it was Boltzmann who first made the connection between entropy and micro states. chapter 12 of "Classical and Statistical Thermodynamics" by Ashley H. Carter discusses Boltzmann's arguments. To summarize from that book: Entropy (S) corresponds to a particular configuration of an ensemble of particles called a macro state. A macro state can be ... 3 Hints: Define difference \delta:=\Delta-\Delta_0. Deduce from |\delta|\ll |\Delta_0| that the lhs. of eq. (1) is$$\tag{A}\text{lhs}~\approx~ -\frac{\delta}{\Delta_0}.$$Substitute \xi=x\Delta  in the integral on the rhs. of eq. (1). Deduce using \hbar \omega_D \gg \Delta that the rhs. is$$\tag{B} \text{rhs}~\approx~ \int_{\mathbb{R}} \! ...

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Let us decompose the formula to better understand what it means: $\hbar \mathbf k/m$ is the velocity of the electrons $\epsilon_k - \mu$ is the energy relative to the temperature, so if particles with higher energy than the temperature move in a direction, they produce a positive thermal energy flux, particles with lower energy an negative flux. ...

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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 ...

2

For BE or FD statistics you have the additional constrained that you are dealing with a fixed number of particles (or fixed mean number). That gives you a normalization condition $$\int d\epsilon \ n(\epsilon) = N$$ Now, for large T $\beta \epsilon$ is small, i.e. a large fraction of the possible energy states is actually accessible, i.e. could possibly ...

2

inelastic collisions are necessary in order to properly distribute energy among the internal molecular degrees of freedom (vibration, rotation). They are, but the ideal gas (nominally) does not have such internal degrees of freedom.1 Allowing them is a simple extension that is often made without announcing it2 and with that extension you get some ...

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Yes, these equations exist and can be derived from the partition function in JGab's answer. The internal energy per spin is: $$u(\beta) = - \frac{\partial}{\partial \beta} \left( \ln(2) + \frac{1}{8 \pi^2} \int_{0}^{2\pi} \mathrm{d} q_1 \int_{0}^{2\pi} \mathrm{d} q_2 \\\ln \left[ \big( 1 - \sinh(2 \beta J) \big)^2 + \sinh(2 \beta J) \left( 2 - \cos q_1 - ... 2 What is entropy, more than disorder? Don't look at entropy as disorder. Thinking of it as disorder has long been a source of confusion. Many texts are moving away from using the disorder description. Macroscopically, it's better to think of entropy as a measure of energy dispersion rather than as a measure of disorder. Microscopically, it's better to think ... 2 1. Hawking-Bekenstein entropy of a black hole is given by S_{\text{BH}} = \frac{kAc^3}{4\hbar G} where A is the area of the event horizon. Assuming a non-rotating black hole, there holds r_s=\frac{2GM}{c^2} for the Schwartzschild radius, and therefore A=4\pi r_s^2=\frac{16\pi G^2M^2}{c^4}, which results in$$ S_{BH}=\frac{4kGM^2}{\hbar c} $$For ... 1 Thermal velocity is the velocity that a particle in a system would have if its kinetic energy were equal to the average energy of all the particles of the system. Take an ideal gas in three dimensions as an example. In equilibrium there is a distribution of velocities among all of the particles. Some move "fast" others move "slowly". That means that ... 1 For a system of atoms to have a meaningful temperature, I would say there has to be uncertainty in their state, and there must be a way (at least hypothetically) for the system to exchange energy with some other system. The first example you give of the two beams is an idealized state with no uncertainty, so assigning a temperature would be impossible. You ... 1 This may be the relevant paper that would answer your question: http://journals.aps.org/prb/abstract/10.1103/PhysRevB.57.8472 1$$-\left(\frac{\partial^2\Omega}{\partial H^2}\right)_\mu=\left(\frac{\partial M}{\partial H}\right)_\mu=\left(\frac{\partial M}{\partial H}\right)_N+\left(\frac{\partial M}{\partial N}\right)_H\left(\frac{\partial N}{\partial H}\right)_\mu In principle, there is a difference between these quantities, at least in ferromagnets (otherwise the magnetization ...

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 ...

1

I'm not sure if you're talking about probabilities here but if you are then I found some problems online: 1) In how many ways can 8 people line up for concert tickets? 2) There are 5 women running a race. How many different ways could 1st, 2nd, 3rd place finishers occur? 3) There are 13 members on a board of directors. If they must form a subcommittee ...

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When studying dynamical systems you consider a low-dimensional phase space that only includes macroscopic variables. In this phase space dissipation does indeed cause volume contraction - but only because the energy that has been lost has been transferred to the far bigger phase space including all the microscopic variables. In this space energy must be ...

1

I made a very simple mistake of plotting Specific heat and not Specific heat per spin. It is specific heat per spin that scales as $L^{\alpha/\nu}$. And hence, the actual value from the data of my previous simulations is $2.44-2 = 0.44$. Using system sizes $10$ and $12$, one in fact gets a value $0.3$.

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According to my limited understanding of density functional theory. Coulomb interaction is one of the correlation effects. Besides Coulomb interaction, there are interaction due to Pauli exclusion principle and change of kinetic energy compared with that of non-interacting electron gas.

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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 $dS=δQ/T$ can only depend on imaginary reversible ...

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