# Tag Info

64

When you say "why aren't things being destroyed", you presumably mean "why aren't the chemical bonds that hold objects together being broken". Now, we can determine the energy it takes to break a bond - that's called the "bond energy". Let's take, for example, a carbon-carbon bond, since it's a common one in our bodies. The bond energy of a carbon-carbon ...

33

Let's suppose you have a Rubik's cube that's made of a small number of atoms at a low temperature, so that you can make moves without any frictional dissipation at all, and let's suppose that the cube is initialised to a random one of its $\sim 2^{65}$ possible states. Now if you want to solve this cube you will have to measure its state. In principle you ...

24

The resolution to Maxwell's demon paradox is mostly understood to be through Landauer's principle, and it is one of the most compelling applications of information science to physics. Landauer's principle asserts that erasing information from a physical system will always require performing work, and particularly will require at least $$k_B T \ln(2)$$ of ...

23

The partition function is strongly related to a very useful tool in probability theory called the moment generating function(al) of the probability distribution. For any probability distribution $p$ of some random variable $X$, the generating function $\mathcal{M}(z)$ is defined as being: \mathcal{M}(z) \equiv \langle e^{zX}\rangle ...

22

The question isn't silly. The speed of each molecule in the liquid is much higher than the speed of either the piston or the water shooting out from the nozzle. At room temperature, for water molecules the average is on the order of 500m/s. And yet, the speed of sound in water is three times higher than that, which implies that pressure can propagate in ...

21

Adjacent molecules in a liquid all repel each other because of the electron clouds that surround the nuclei that they contain. In that sense these molecules never even 'touch' each other (at least not in the intuitive sense of the word). When you apply pressure to the liquid you're squeezing them into a (very slightly) smaller volume, thereby increasing the ...

17

$\hbar$ does not need to appear in classical statistical mechanics. You are free to replace it with any quantity with units of angular momentum, say $\hbar_{\mathrm{C}}$. As long as this is choosen smaller than the size you can experimentally probe (i.e., as long as you don't ask questions of the theory that contain structure on this length scale or below) ...

8

The partition function contains so much information because it is directly related to the free energy, $$F = - k_B T \ln(Z) \, .$$ The physical assumption behind considering $F$ as a thermodynamic potential is that the statistics of the system as described by the canonical ensemble. In turn, the applicability of the canonical ensemble is a direct ...

8

$h$ factor The factor of $1/h^{3N}$ is a total hack. The integral over phase space has dimensions, whereas $Z$ only makes sense if it's dimensionless. The $h$ factors are there to make $Z$ dimensionless. Suppose you have a system with only one particle in one dimension. Then the integral in phase space goes over one position variable, $dq$ and one momentum ...

6

In some sense yes. The temperature is defined as an imaginary time in Matsubara Green's functions or some path integrals. Thus, a negative inverse imaginary temperature can be considered as a time. Here is a quotation from Alexander Altland, Ben Simons "Condensed Matter Field Theory": "Thus, real time dynamics and quantum statistical mechanics can be ...

6

I can see different subtleties in Landau's argument. First of all, it isn't entirely clear what is meant by "there are only seven additive constants of motion". To give an example, consider a single particle hamiltonian: $$H=\frac{\mathbf p^2}{2m}+m\omega ^2\frac{\mathbf q ^2}{2}.$$ For this hamiltonian there are several conserved quantities: $$e(\mathbf ... 6 This phenomenon has been studied by Vella and Mahadevan and written up in the American Journal of Physics (http://scitation.aip.org/content/aapt/journal/ajp/73/9/10.1119/1.1898523). It's called the Cheerios effect. If the cereal pieces clump together away from the edges of the bowl, they gravitate toward a slight concavity in the surface caused by milk ... 6 The notion of temperature is all about how the equilibrium an otherwise isolated system shifts when the system's internal energy changes. So you do not need to worry about whether this internal energy is kinetic, potential, whatever. Actually the temperature is not quite the ensemble average kinetic energy. Your statement is true for an ideal gas and also ... 6 It depends on the parameter you consider. The Maxwell-Boltzmann distribution (as the name is applied in the Wikipedia article you link) is the distribution of the particle's speed (absolute value of the velocity) in a gas and this is a chi distribution. You probably were thinking of the distribution in terms of the vectorial velocity and then indeed it ... 6 There's really two questions here, one about the definition of distribution functions and one about the derivation of the BBGKY hierarchy. I will address them in turn. Definitions Let's define, for convenience, \mathbf{r}^n = \mathbf{r}_1,\dots,\mathbf{r}_n and \mathbf{r}^{(N-n)} = \mathbf{r}_{n+1},\dots,\mathbf{r}_N. Next, let's denote the ... 6 Let's simplify things down to the barest minimum: one dimension, one particle, and a wall. O | The particle moves to the right, hits the wall, and rebounds, perfectly elastically. If the wall is fixed in place, the particle will leave the collision with exactly the same kinetic energy as it came in with. But what if the wall is moving to ... 6 OP's question (v1) is essentially asking Does the operator identity$$ e^{\frac{it}{\hbar}[\hat{H},~\cdot~]}\hat{A}~ =~ e^{i\hat{H}t/\hbar}\hat{A}e^{-i\hat{H}t/\hbar} \tag{1} $$have an analog using functions/symbols H and A rather than operators \hat{H} and \hat{A}, respectively? The answer is: Yes, in terms of the Groenewold-Moyal star ... 5 It is not possible to have a state with four indistinguishable particles such that P_{12} \psi = -\psi and P_{34} \psi =\psi, for an algebraic reason. Namely, the exchange operators have to form a representation of the permutation group S_4. It is rather well known that there are exactly two representations of S_n: the trivial representation where ... 5 Yes, neutrinos should obey Fermi-Dirac statistics and yes, the Pauli Exclusion Principle should operate for neutrinos. But let's examine how dense the neutrino population has to be for this to be important. The Fermi momentum is given by$$ p_F = \left( \frac{3}{8\pi}\right) h n_{\nu}^{1/3} $$where n_{\nu} is the neutrino number density. In order to be ... 5 Giving the value simply of k_B T is generally more useful, because I can plug that into anything. Sure, I might need to know the ideal gas energy, and multiply by 3/2. But maybe I need to put it into a partition function, and I just need k_B T. Maybe I'm worried about a harmonic oscillator and I just have the two degrees of freedom. The 3/2 is ... 5 There are of course many books out there, quantum statistics is a really well-established field, so regardless of suggestions here you should really look further on your own as well and find one that suits you best. But here are a few that I've used in the past that you may find useful: Statistical mechanics: A survival guide by Mike Glazer and Justin ... 5 First let me repeat what Yvan Velenik says in the comment: The terminology is somewhat unfortunate, because you don't need that much statistics, rather you'll need some probability theory. To elaborate, quoting Wikipedia, Statistics is the study of the collection, analysis, interpretation, presentation, and organization of data. [...] Statistics deals ... 4 It's true. Special equipment and a long time is required to mix helium and nitrogen. According to one study, a mixture of 2.7% He, 93.3% N at 800 p.s.i.g. required a special cradle to repeatedly upend the cylinder, and 20.5 hours to reach equilibrated gas, which then remained mixed: http://pubs.acs.org/doi/abs/10.1021/je60005a002. The helium repeatedly ... 4 Characteristically, a critical point occurs somewhere anytime you have a continuous phase transistion. That is, if you have two phases of a substance that themselves share their intrinsic symmetries. The classic example is the critical point associated with the liquid gas transition, as you note. Liquids are isotropic and homogenous, gases are isotropic ... 4 In statistical mechanics, at least when you can ignore the spin of the particles you're dealing with, the occupation number of a quantum state (that is, the number of particles in the state, or probability of a particle being found in the state) is proportional to e^{-E/kT}, where E is the energy of the state. If you have a harmonic oscillator at ... 4 The ratio between avaible microstates is given by the following ratio:$$ \frac{n_{B}}{n_{A}}=\frac{\int_{V/2}d^{3N}q d^{3N}p}{\int_{V}d^{3N}q d^{3N}p}=\frac{\int_{V/2}d^{3N}q}{\int_{V}d^{3N}q}=\frac{\int_{V/2}d^{3}q_1 \int_{V/2}d^3 q_{2}\dots }{\int_{V}d^{3}q_1 \int_{V}d^3 q_{2}\dots}=\left( V/2 \right)^N/V^N=\frac{1}{2^N}$$where N is the number of ... 4 The equipartition theorem is a mathematical consequence of very specific kind of Hamiltonians. It states that any 'squared' term of deegree of freedom in the Hamiltonian gets \frac{1}{2}k_bT of energy (it is a statement about the energy distribution for this kind of Hamiltonians). For example - classical ideal gas Hamiltonian - ... 4 You're missing a minus in the entropy definition - S=-Tr(\rho\ln\rho) Entropy of a unitarily evolving system (doesn't matter in which picture) is conserved (The entropy is a trace of a function of the density matrix "operator" thus it depend solely on the eigenvalues of it's input operator, but the eigenvalues of the density matrix don't change under ... 4 In macroscopic units it should be$$S=-R\alpha \log(\alpha e^{-S_1/R})-R(1-\alpha)\log\Big(1-\alpha)e^{-S_2/R}\Big) \\=\alpha \Big(S_1-R\log\alpha\Big)+(1-\alpha)\Big(S_2-R\log(1-\alpha)\Big), where $R$ is the universal gas constant. In the pure case, this reduces to the textbook formula. But such a formula cannot be true in general. The general formula ...

4

wsc's answer (i.e., Onsager's computation of the free energy) provides one alternative road to a proof of a phase transition in the Ising model. It implies the existence of a phase transition in dimension 2 (for the nearest-neighbor model). Combined with correlation inequalities, this implies existence of a phase transition in any dimension d≥2, and ...

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