Tag Info

Hot answers tagged

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 ...


4

The mean free path can be meaningful quantity in quantum mechanics, although usually only in a semi-classical regime. It is particularly useful in the kinetic theory of quantum liquids at low temperature, where the excitations of the system can be described as quasiparticles that propagate approximately ballistically and interact only rarely. You can define ...


3

Where does that precise time come from? Is it just the energy / Hamiltonian $H$ that determines it? Yes. $\newcommand{\ket}[1]{|#1\rangle} \newcommand{\bra}[1]{\langle #1 |}$ Suppose I have an atom which can be in one of two states $\ket{0}$ or $\ket{1}$. Suppose that atom is coupled to another atom via an interaction Hamiltonian like this $$H_I = ...


3

You are missing the term $\frac{1}{n_s!}$ from the product, i.e. $$W = \prod_s \frac{a^{n_s}}{n_s!}$$ from which the wanted result follows. Obviously in a quantum sense classical particles are distinguishable, so the term does not immediately arise from indistinguishability per se. Rather, the reason the "extra" term appears is because we ought to be ...


3

The ground state of the toric code can be understand as a superposition of all loop configurations in the $z$ basis. The fact that these loops fluctuate at all length scales (and thus around the torus) leads to the topological order in the system. The $\sigma_z$ terms lead to a "tension" in the loops, penalizing long loops. Eventually, this tension will ...


2

The intuitive picture is that phase points move more quickly through regions of phase space where $|\nabla H|$ is higher. As a result if you have a constant-energy ensemble of phase points (a flat "packet"), their phase space area enlarges as they move through high $|\nabla H|$ regions. For the full Liouville's theorem this is not a problem. A ...


2

Where did the energy you spent go? You gained knowledge about the equation I gave. I'll try a different approach before this question is closed or migrated elsewhere. In the context of your question, the answer must be know (heh... no). Consider the case that you expend all the energy calculating incorrectly. Knowledge is, at least, justified ...


2

I have contributed this issue with this arXiv:1411.2425 and previous works. I stress that Gibbs entropy is not "defined" but rather "constructed". It is constructed on the expression of thermodynamic forces in the microcanonical ensemble, and is constructed in such a way that it reproduces them always and exactly. The construction is actually unique. ...


2

If we expand the argument of $\ln$ in $q_1, q_2$ around $\left(q_1,q_2\right)=\left(0,0\right)$ and set $q_1,q_2$ to zero, we find $$(1+x^2)^2-2x(1-x^2)2=(1+x^2)^2-4x(1-x^2)=(x^2+2x-1)^2$$ which is obviously minimized at the root of $x^2+2x-1$, giving $x_c = \sqrt{2}-1$. Because the model is taken in the $N\rightarrow\infty$ limit, the integral for $F/N$ ...


2

You can express energy in Joules. Then it is extensive (it will scale with the system). You can also call it specific energy and express it in Joules per kg. Then it is intensive. Or you could consider "molar" energy and have Joules per mole and it would also be intensive. Same goes for entropy, which is Joules per Kelvin (extensive), or enthalpy (Joules), ...


2

We're used to thinking of "most probable" and "mean value" as the same thing, but it need not be so. It's worth remembering that the "expectation value" of a six sided die is 3.5, but this is not a very probable result. You might object that this is due to discrete effects, but consider this example: you have two identical Gaussians, with width $\sigma$, but ...


1

In any probability distribution, there are many ways to find some kind of "average" value, that is, ways to define the "centrality" of the distribution. In discreet distributions you have almost certainly come across mean, median and mode, and perhaps also the different "flavours" of means - arithmetic, geometric, harmonic etc. For continuous distributions ...


1

Your requirement that $z\to 1$ (i.e. $\mu\to 0$) already requires the limit $N\to \infty$. If $N$ is finite, the chemical potential tends to a small but finite value at the transition temperature. This ensures that the occupation of the ground state is $\langle n_0\rangle \lesssim N$.


1

If we consider temperature to be due to translational motion of the molecules and we assume the system has reached equilibrium, then the velocity distribution of the molecules is given by the Maxwell distribution: $$ f(v) = \sqrt{\left(\frac{m}{2\pi k T}\right)^3} 4 \pi v^2 \exp\left(\frac{m v^2}{2 k T}\right)$$ which will give you the velocity ...


1

Correlation between two variables (or objects) is, very simply put, how much a change in one variable affects or determines a change in the other. Replacing variables with spins, highly correlated spins would mean that, due to some interactions between them, a change in the direction of one spin will cause a change in the direction of the spin it is ...


1

One way to understand it is to write it using the Dirac measure to express the phase space in the microcanonical ensemble (because that's what it is about). In this ensemble the idea is to say that the energy $H(\textbf{r})$ (where $r$ is a point in phase space) is fixed at some value $E$ (actually it belongs to a very small interval $[E,E+\delta E]$). One ...


1

Yes, any molar quantity it is considered intensive. You can find a list at this wikipedia page: http://en.wikipedia.org/wiki/Intensive_and_extensive_properties


1

I dont see why we dont just write $\Delta E=\Delta Q +\Delta W$ and $\Delta Q=0$ gives $\Delta E=\Delta W$. That's exactly what you do. $\delta Q = 0$ is the definition of an adiabatic process. Pressure and volume are well defined in the case of a quasi-static adiabatic process, in which change the work done on the system (and hence the change in ...


1

You have forgotten to include the chemical potential $\mu$, which enforces the constraint of a fixed average number of particles in the system. The correct result for the thermal occupation of an ideal Fermi gas at temperature $T = 1/k_B \beta$ is $$ n(E) = \frac{1}{e^{\beta(E-\mu)} + 1}.$$ At zero temperature, you have $\beta\to\infty$, so that for $E=0$ $$ ...


1

What is most intuitive to me is to look at what we're asking. What is the probability that we find the system in a state with total energy $\hat{E}$? It is just the fraction of all possible states that have total energy $\hat{E}$, i.e. $$p(E) \equiv \frac{\int \delta\left(E(\Omega)- \hat{E}\right)d\Omega}{\int d\Omega}\sim N_\hat{E}$$ But $N_\hat{E}$ is ...


1

You've answered your own question: The pitfall is obviously due to exponentiating a Taylor series with higher order terms dropped. To bring this into sharper focus, witness that the method, as you say, does indeed have some validity for any function analytic in the neighbourhood of the point in question. Unless we are at a stationary point, there is a ...


1

The analytic continuation of $S^d$ is de Sitter space, often denoted as $dS_d$. Euclidean QFT on $S^d$ then corresponds to Lorentzian QFT on $dS_d$. This can be seen a number of ways, but the quickest is to simply note that the sphere is the maximally symmetric Euclidean signature space with positive curvature, and de Sitter is the maximally symmetric ...


1

Ok. Let's put in this way: We know the average energy: $\langle E\rangle=\sum P_k E_k $ $P_K$ is a probability distribution: $\sum P_k=1$ We don't know nothing more about the system. The question is, what probability distribution $P_k$ capture the previous assertions in an unbiased way? The answer is, the distribution that maximize the Shannon's entropy ...


1

I think the wrong step was the assumption that entropy increases, where in fact maintaining the temperature would require a an outflow of heat, which means the entropy of the gas is decreasing. To see how this relates to your formula, notice that this decrease in entropy would also increase the enthalpy by the same amount, however enthalpy will also ...


1

The wave function for a pair of independent, indistinguishable bosons/fermions with no internal degrees of freedom can be written in the position representation as $$ \Psi(r_1,r_2) = \frac{1}{\sqrt{2}} \left[\psi_1(r_1)\psi_2(r_2) \pm \psi_2(r_1)\psi_1(r_2) \right ].$$ You can calculate your desired probability distribution from $P_{12}(r_1,r_2) = ...


1

I don't know if this is standard, but consider a pendulum that can swing a full circle in a plane. Vibrate the point of suspension up and down at the appropriate frequency. The pendulum will gain energy and spin either clockwise or counterclockwise.



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