# Tag Info

17

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 ...

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 ...

3

The instantaneous temperature of a system of $N$ particles of masses $m_i$ and velocities $v_i$ is $$T(t) = \sum_{i=1}^N \frac{m_iv_i^2(t)}{k_BN_f}$$ where $N_f$ equals the number of degrees of freedom, typically $N_f=3N-3$ for fixed total momentum. Note that the instantaneous temperature will fluctuate 5-10% about the true (thermodynamic) temperature ...

3

You can think of the whole thing as a "fluid of systems" and each one of them can be in any of the states $i$ available. $\pi_{ij}$ tells you what is the speed at which a system in state $i$ will go to state $j$ $p_i$ tells you how likely it is for a system (in this fluid or ensemble of systems) to be in state $i$ and is proportional to the number of ...

3

For gravitational systems one has to be careful making statements about entropy and the second law of thermodynamics. Your example is similar to the gravitational collapse of a gas cloud if you think carefully about it. In that case and in yours, the shrinking of the gas will raise it's heat. Now even though the increase of entropy due to the increased ...

3

TL;DR: The Gross-Pitaevskii equation is only applicable for very weakly-interacting bosons. At $a=\infty$ the gas displays universal physics. Strictly speaking, the Gross-Pitaevskii equation (GPE) is only valid for $$na^3 \ll 1,$$ where $n$ is the density of particles and $a$ is the $s$-wave scattering length. As it is a mean-field theory, one has to look ...

3

Stefano Bordoni's 2012 Taming Complexity (e-book from ResearchGate; review) is a good place to start.(Bordoni has a master's degree in physics and three PhDs, in the history of science, anthropology and epistemology of complexity, and philosophy.) Bordoni refers to Brush's 1986 The Kind of Motion We Call Heat: A History of the Kinetic Theory of Gases in ...

3

For the Laplacian $$\Delta ~:=~ -\frac{d^2}{dx^2} ~\geq~ 0,$$ the corresponding HS transformation reads $$\exp\left(-\frac{a}{2}\Delta\right) f(x)~=~\int_{\mathbb{R}} \!\frac{dy}{\sqrt{2\pi a}}\exp\left(-\frac{y^2}{2a}\right) f(x+y), \qquad a~>~0.$$ Proof: Use Fourier transformation.

3

As noted already, within classical physics, singularities such as $1/r^2$ signal a break down of the theory. If we are really interested in what is happening at the point of the singularity, we should use quantum physics. You can think of $1/r^2$ as the asymptotic scaling form of the quantum theory for large $r$. The actual singularity is not physical. On ...

3

I think one way to understand why this works is that the spectrum of energy levels $E_i$ has undergone a sort of transform (analogous to Laplace transform) which results in the partition function $Z(T)$. In principle if you know the function $Z(T)$ you can reverse the process, and reconstruct the original spectrum of energy levels. As such, all information ...

3

I learned I can calculate the entropy $$S=-k_B\sum_jp_jln(p_j)$$ where $$p_j$$=probability at j state but I saw that the entropy is also can be calculated by $$S=-k_Bln(Z)$$ I think this equation applicable for both of isolated system and non isolated system these two equations are same ? Given the number of states $\Omega(E,\Delta ... 3 I can only offer a partial answer to the first portion of your question. one comes across functions that diverge at a given point, typical examples would be the Coulomb or the gravitational forces being ∝ 1/r², clearly they diverge at r=0... Gravitational force isn't actually proportional to 1/r². Take a look at the plot of gravitational potential on ... 2 I can't speak to singularities in the sense of general relativity, but your example of$1/r^2$laws in classical physics is actually solved most of the time by internal structure. One of my physics professors used to always say that nature solves infinities with internal structure. For example, for a charged sphere of uniform charge density, the electric ... 2 Yes, you will definitely increase the temperature of the water. Lets say that we increase the speed of the container, with an amount of energy of 1kJ. If we stop the container (does't matter if we do it suddenly or progressively), the 1kJ speed energy will be converted to heat, both to the water inside and to the device used to stop it (e.g. a wall or our ... 2 Actually, that's exactly what James P. Joule demonstrated in 1845: the Mechanical Equivalent of Heat. He used brass paddles to mix a barrel of water. He found a direct relationship between his calculated amount of work and measured temperature. Boiling point is highly improbable. For a simply detectable change, Oocities.org reckons: 'The amount of heat ... 2 I'm guessing you're actually asking how you prove that the motion is chaotic rather than periodic. If so, the answer is with great difficulty! The three body problem was solved by Karl Sundman in 1909, but the n-body problem was only solved in 1991 by Qiudong Wang$^1$. I say solved but atually both proofs really only showed the motion could be described by ... 2 The grand canonical ensemble allows the number of particles of your system to fluctuate but makes the assumption that it is constant amongst the reservoir and system combined i.e. $$n_{res} + n_{sys} = const$$ For the case of photons this is not true. 2 Singularity in Force Laws If force laws were fundamental to nature, this would be a serious problem. Imagine, for example, the gravitational energy between photons. They are Bosons and can hence occupy the same quantum state; crucially, more than one of them can be and stay in the same position where the gravitational force (they have energy and hence, ... 2 I think your concern is why to use the fundamental "symmetrization postulate" and not only the hamiltonian symmetry. The thing is that doesn't matter, suppose i'm trying to describe two particles (fermions for example) with Hilbert space$\mathfrak{H}_1$and$\mathfrak{H}_2\$, the Hilbert space of the two particles is $$\mathfrak{H} = ... 2 The answer lies in probability theory. Roughly, the probability of an event or macro state A to happen is the number of instances \Omega(A) in which it is fulfilled divided by the total number of possible instances or micro states \Omega i.e. p(A) = \frac{\Omega(A)}{\Omega}. So the reason why you want to maximize \Omega(A) is because you seek ... 2 response function = susceptibility = (pure or mixed) second derivative of a (Helmholtz, Gibbs, etc.) free energy. Magnetization is not a response function as the free energy is not observable, so one cannot observe the response to a change of some variable. 2 "The double summation (first over the numbers n_\epsilon constrained by a fixed value of the total number N, and then over all possible values of N) is equivalent to a summation over all possible values of the numbers n_\epsilon independent of one another" Why is this true? I have been stuck on this argument for hours... The ... 1 While lemon's answer is of course correct, it is not the only way to calculate the temperature from a molecular dynamics simulation: it can also be obtained from the configurations, that is, the particle coordinates, of the system. This is called "configurational temperature" (see, e.g., this article; pay-walled). The key identity is (for a canonical ... 1 Consider the general diffusion equation of the form,$$ \frac{\partial \psi}{\partial t}=D\frac{\partial\Phi}{\partial x}=D\frac{\partial^2\psi}{\partial x^2}\tag{1} $$where \Phi=\partial_x\psi is the probability current (flux). In order to replicate a reflecting boundary, we establish an infinite potential at the boundaries, ... 1 My original answer was incorrect and Mark gave a good answer above. Here is a slightly different approach by considering the total energy of the system. Say we have \alpha total harmonic oscillators. We can write the total energy of the whole system as$$ E = \sum_\alpha \frac{1}{2}\hbar\omega_\alpha + \sum_\alpha n_\alpha \hbar \omega_\alpha .$$For ... 1 Quantum mechanically the general expression you want for the partition function is$$ Z = \mathrm{Tr} \left( \mathrm{e}^{-\beta H} \right),$$where \mathrm{Tr} means the trace (i.e. sum over micro-states). Now you can use the fact that the modes are independent, so that quantum Boltzmann operator \mathrm{e}^{-\beta H} factorises into a product. This ... 1 You should start by the definition of temperature$$\frac{1}{T} = \frac{\partial s}{\partial E }= k \frac{\partial \ln \Omega}{\partial E} $$Also you have the partition function$$Z_1=\sum \exp(-bE_r) \text{ where } b=1/kT $$Let's now assume a single magnetic dipole in a magnetic field inside a heat tank( we can assume the heat tank as the rest of the ... 1 The partition function \mathcal Z is the sum over all those these exponential indexed by the state index. In German it's called "\mathcal Zustandssumme", literally "state sum". You could device a scheme of indexing the possible overall energies so that you could write it as \sum_n, i.e. finding an injection {\mathbb N}\to{\mathbb N}^2, but if the ... 1 A short answer to questions 2 and 3: In Mermin-Wagner's paper the short-range condition is stated as \sum_{\bf R} {\bf R}^2 |J_{\bf R}|<+\infty. For interactions with (or more precisely majorized by a) power law decay |J_{\bf R}| \sim R^{-\alpha}, this requires \alpha > D+2, where D is the space dimensionality (i.e., \alpha >4 for D=2 ... 1 Basically, you need a higher dimensional version of the Leibniz rule to differentiate under the integral sign. Suppose you have an integral of this kind$$ \int_{\Omega(s)} f(s,\vec x) \, d\vec x $$That's it, your region of integration and your function depends of a parameter s. (in your case s = x_i). Then if you want to calculate$$ ...

Only top voted, non community-wiki answers of a minimum length are eligible