Tag Info

Hot answers tagged statistical-mechanics

21

Actually, temperature is defined as $$\frac{1}{T} = \frac{\partial S}{\partial E} = \frac{k_B}{\Omega}\frac{\partial\Omega}{\partial E}$$ So in order to have zero temperature, you would need a system with either zero multiplicity, which you can't have by definition, or an infinite derivative $\partial\Omega/\partial E$ even though the multiplicity itself ...

14

This is a very interesting question which is usually overlooked. First of all, saying that "large scale physics is decoupled from the small-scale" is somewhat misleading, as indeed the renormalization group (RG) [in the Wilsonian sense, the only one I will use] tells us how to relate the small scale to the large scale ! But usually what people mean by that ...

13

A convenient operational definition of temperature is that it is a measure of the average translational kinetic energy associated with the disordered microscopic motion of atoms and molecules. The underlying framework of all matter is quantum mechanical. This means that the Heisenberg Uncertainty principle holds. Even for a single particle the HUP ...

11

Here's a self-contained (hopefully clear) derivation. Step 1. Setup and definition of differential scattering cross-section Let $\mathcal L$ denote the incident luminosity (number of incident particles per unit area, per unit time) of a beam to be scattered. We assume that we have a spherical detector at infinity with which to measure scattered particles, ...

10

I think gatsu's right: it's because of an entropic attraction resulting from the fact that two spheres whose centers are less than $2a$ apart leave more room for other spheres. To see why this happens, it may help to draw a picture: Here, the blue spheres all have radius $\frac12a$. The gray dashed circle around each sphere has radius $a$, and shows the ...

8

No, it's not a problem. The reason is that, in order for expressions like $$\mu=-T\left(\tfrac{\partial S}{\partial N}\right)_{E,V}.$$ to be meaningful, you have to be using the grand canonical ensemble (or a generalisation thereof), in which particles are able to enter and leave the system. Consequently, $N$ stands not for an integer number of particles, ...

8

Bogoliubov proved long, long ago that the condensate is stable against weak interactions. The interactions scatter some fraction of bosons out of the lowest-energy single-particle state ("depleting" the condensate), but off-diagonal long range order remains. For a nice introduction to Bogoliubov's theory see Ben Simon's lectures ...

8

I prefer to see it in the following way: $$Z_{quantum} \equiv \sum_{m} e^{-\beta E_m}$$ Where $m$ is a quantum microstate eigenstate of the Hamiltonian. Now, you can split the sum into two parts; a sum over quantum microstates that yields the same energy eigenvalue $E_n$ and a sum over all possible values of $E_n$: ...

8

Right, there is a small probability that the entropy will decrease. But for the decrease by $-|\Delta S|$, the probability is of the order $\exp(-|\Delta S| / k)$, exponentially small, where $k$ is (in the SI units) the tiny Boltzmann constant. So whenever $|\Delta S|$ is macroscopically large, something like one joule per kelvin, the probability of the ...

8

I'll state one version of the theorem, valid for classical systems. I'll not give the most general framework, as things become messy, but this should still give you an idea of how general the result is. We need the following ingredients: Spins: to each vertex of the lattice $\mathbb{Z}^2$, we attach a spin $\phi_x$ taking values in some compact ...

8

If two particles are close to each other, there is more space for the rest of the particles to move. This gives rise to an effective entropic attraction between the particles because when looking at two particles for different separations while "tracing out" over the degrees of freedom of the rest of the system, the entropy of the rest is higher when the two ...

7

Your interpretation is not quite right. One sharp interpretation one can give to this "cutting" of phase space into cubes of size $h^{2N}$ (here $N$ is the dimension of the system's configuration space), is that it allows one to use classical phase space to count the number of energy eigenstates of the corresponding quantum hamiltonian. Instead of trying ...

7

Sir James Jeans has an excellent answer, without using the word "order" or "disorder". Consider a card game...¿is there anyone else on this forum besides me who still plays whist? His example is whist. I had better use poker. You have the same probability of being dealt four aces and the king of spades as of being dealt any other precisely specified hand, ...

7

If we take a system, and let it evolve for some indefinite amount of time, it will be in an incoherent mixture of energy eigenstates. Many systems we encounter in nature have been sitting for some time, and not interacting with the environment (much). These can be considered to be in energy eigenstates. For example, suppose we consider an atom in a gas. ...

7

I could be wrong, but in my understanding, you're describing and justifying steady state, not detailed balance. In thermal equilibrium, steady state is true always, and detailed balance is true sometimes. Detailed balance means that the rate $X \rightarrow Y$ is always the same as the rate $Y \rightarrow X$. If a system is both in thermal equilibrium and ...

6

I think that the most prominent example of "prediction before observation" in statistical physics is the Bose-Einstein condensate. It was predicted in ~1925 by, well, Bose and Einstein, obviously. Then after more than ten years it was proposed as an explanation for superfluidity and superconductivity. And the actual BEC of atoms (as a new state of matter) ...

6

Firstly, the logarithm needn't necessarily be to base 2. Changing the base just introduces a (scale) factor, so log10, log2 and ln are all equally useful. Log2 is convenient for people working with binary systems. Let's deconstruct the formula. I will define entropy to be $H = E[-\log(p)]$. You can see that this will reduce to a weighted average which ...

6

The more natural relationship between the two distributions is the opposite one. The Boltzmann distribution $$\exp(-E/kT)$$ is the more general one (connected with the microscopic definition of the temperature $T$ in any system in physics) and one may simply substitute the kinetic energy $mv^2/2$ for $E$ to get the Maxwell part of the distribution. The ...

6

The third law of thermodynamics states that a quantum system has absolute zero temperature if and only if its entropy is zero: meaning it is not reacting with anything including its environment, which is impossible to achieve. In the QM sense, 0K would be achieved when all the motion of all the particles comprising matter stops and everything comes to a ...

6

Mathematical/formal answer: In classical statistical mechanics the answer is "no return to the exact same state". Although the Poincare recurrence theorem indicates that a given isolated system will return to a state that is arbitrarily close to a selected initial state, phase space is continuous and the probability of reaching any specific state is zero.

6

I think it will depend the kind of statistical mechanics. For classical statistical mechanics, there is no time, so it is really hard to imagine a nice physical picture of the propagation of something. But nevertheless we still talk of loops as propagating "particles" (we give the "momenta", for instance, which is conserved, etc.). Interestingly, ...

5

Off the top of my head, the example I can think of is the whole work that Boltzmann did. He based his entire theory of statistical mechanics on the concept of indivisible particles (i.e. that all matter is made up out of atoms). Doing this, his theory (using theoretical mathematical methods as you said) was able to predict how the atoms determine the visible ...

5

The mathematician John Baez recently wrote a long series of blog posts about using quantum techniques for non-quantum stochastic systems, in which chemical reaction networks played a central role as an important special case. This culminated in a paper entitled Quantum Techniques for Reaction Networks, which might be something close to what you're looking ...

5

None. The grand canonical partition function is $$Z_\mathrm{GC} = \mathrm{tr}(e^{-\beta (H - \mu N)})$$ where $H$ is the hamiltonian and $N$ is the number operator. Interacting particles simply means that the Hamiltonian needs to involve terms that take into account the interaction energy between particles. The same is true for the hamiltonian in, ...

5

The most physical and understandable definition of Nekrasov's partition function to me uses five-dimensional gauge theories. Namely, any 4d N=2 susy gauge theory has a 5d version with the same matter content, so that compactifying it on a small $S^1$ brings it back to the original 4d theory. Then we put the theory on the so-called Omega background: it is ...

5

Good question. I think that you're incorrect to assume the chemical potential is independent of the zero point. I think it's easiest to see that the chemical potential must change by using $\mu = (\partial U/\partial N)_{S,V}$. Suppose I first define the zero point as 0 so that the energy is $U(N)$. Then I redefine the zero-point energy (for a ...

5

There simply doesn't exist any container with $\mu\gt \epsilon_0$; that's what the quoted sentence says. What you could try is to try to increase the chemical potential. But the Bose-Einstein distribution says $$\langle n_i\rangle \sim\frac{g_i}{e^{(\epsilon_i-\mu)/kT}-1}$$ and if you chose values $\mu\gt \epsilon_i$, then the exponent in the denominator ...

5

Within mean-field, the critical point can be calculated as an analytical function of $z$, the number of nearest neighbours for each lattice point. See for example the Physical Review B paper of Fisher, Grinstein, Weichman and Fisher (1989), or the book 'Quantum Phase Transitions' by Sachdev. Within mean-field, the way to do it is to calculate the particle ...

5

Is there a more sophisticated answer than "This is done because isolated low-temperature systems tend to give up energy more often than they receive it, and thus, gravitate toward their ground states. Thus, even if they start in a mixed energy state, the system will radiate away energy until it reaches its ground state, which is obviously an energy ...

5

I do not think Mainwood makes any argument against what he calls the "theoreticians case", much less a compelling one. The "theoretician's case" is that phase transitions do not exist in finite size systems but only as features which become infinitely sharp in the infinite size limit (also user10001's comment). In fact Mainwood briefly dismisses the case and ...

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