Tag Info

Hot answers tagged

21

Notice first that the phase space of any theory is nothing but the space of all its classical solutions. The traditional presentation of phase spaces by fields and their canonical momenta on a Cauchy surface is just a way of parameterizing all solutions by initial value data -- if possible. This is often possible, but comes with all the disadvantages that a ...


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, ...


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 ...


6

The phase-space represents the "number" of allowed final states (think of a discrete quantum system with degenerate final states, except that here we have a continuum). More final states makes the transition more likely to happen and thus gives it a shorter lifetime. Each of the reactions that you show has a three body final state and a single body ...


5

There are several different notions of microstates or distinguishability that might be relevant to your question. Coarse-graining of phase space into Planck cells. Consider two classical variables $x$ and $p$ with $x \sim x+x_0$ and $p \sim p+p_0$. You can think of this system as describing a particle that lives on a circle of radius $x_0$ and where ...


4

It is misleading to write $\rho_i$ for the components of $\psi$, and as they are complex numbers, you cannot use these in the formula for entropy. The space of wave functions is (not $[0,1]\times S^1$ but) the Poincare sphere (or Bloch sphere) S^2, parameterized by quaternions (corresponding to points on the complexified circle). ...


3

There are systems which are not integrable (in Poincaré sense) because interactions destroy the invariants. Consider a Hamiltonian $H = H_0 + \lambda V$, where $H_0$ is the unperturbed Hamiltonian and $\lambda$ the coupling constant. If you start with the interactions turned off you can find invariants of motion $\Phi^0$ by the usual Poisson bracket ...


3

Let there be given a $2n$-dimenional real symplectic manifold $(M,\omega)$ with a globally defined real function $H:M\times[t_i,t_f] \to \mathbb{R}$, which we will call the Hamiltonian. The time evolution is governed by Hamilton's (or equivalently Liouville's) equations of motion. Here $t\in[t_i,t_f]$ is time. On one hand, there is the notion of complete ...


3

Complete integrability is far stronger than solvability of the initial value problem. Complete integrability implies the absence of chaotic orbits. More precisely, all bounded orbits are quasiperiodic, lying on invariant tori. Perturbations of a completely integrable system preserve only some of these tori; this is the KAM theorem. ...


3

Gibbs' thought on this was (Elementary principles, page 204 footnote) "Strictly speaking, $\psi_{\rm gen}$ is not determined as function of $\nu_1,\ldots\nu_h$, except for integral values of these variables. Yet we may suppose it to be determined as a continuous function by any suitable process of interpolation." Here $\psi_{\rm gen}$ is the free energy of ...


3

The formula you write down is one of thermodynamics. In the statistical mechanics version it is valid in the grand canonical ensemble only if you interpret the extensive variables as expectation values. (See, e.g., Chapter 9 of my online book Classical and Quantum Mechanics via Lie algebras, arXiv:0810.1019.) But expectation values are continuous even when ...


3

Generally, coadjoint orbits of a Lie group provide important examples of global symplectic manifolds. In general such systems are obtained by symplectic reduction from a more fundasmental description. For example, the spinning top is modelled for constant $J^2$ on a symplectic manifold $S^2$ that is a coadjoint orbit of the rotation group $SO(3)$. It is ...


3

If you are looking for a detailed answer as to why in general Lagrangians depend only on first derivatives, then you should read the answer in this question, as Qmechanic rightfully said. However, I suspect that you are asking something else: Given that the equations of motion depend only on first derivatives (like Newton's law), why aren't second ...


3

The problem with the phase space flow in Hamiltonian mechanics is that the flow itself is non-dynamical, that is, the flow is immediately defined for a given Hamiltonian, so there is no independent equation governing its evolution. Thus, Liouville equation is simply a transport of a scalar variable in a given flow. So, dimensional analysis of the flow ...


2

''similarites/differences between phase space and Hilbert space?'' There are no similarities except that they are both infinite-dimensional vector spaces. How to use the two is completely different. Thus there is no book that can answer that. But if you mean the differences and similarities of classical and quantum physics, my book Classical and Quantum ...


2

Conservation of étendue is essentially the same thing as Liouville's theorem applied to the space of rays of light in geometric optics. This is central to non-imaging optics, for example in the design of car headlamps or in concentrating sunlight in photovoltaic cells.


2

1) On a symplectic manifold $(M,\omega)$, Liouville's theorem is often stated as that every Hamiltonian vector field $X_f=\{f,\cdot\}$ is divergence-free $$ {\rm div}_{\rho} X_f~=~0 ,$$ where the volume density $\rho$ comes from the canonical volume form $$\Omega~=~\rho dx^1 \wedge \ldots \wedge dx^{2n}.$$ Here the canonical volume form ...


2

At a given point in space, there are several values of $f_i$, one for each discrete velocity $v_i$. So you have to apply the lattice-Boltzmann equation as many times. This is exactly the point of the lattice-Boltzmann method: each "element" is actually defined by its discrete value of the velocity. You apply the equation for each of these elements which have ...


2

The key inside to OP's question has already been provided by Ikiperu in above comments. Here we just want to show that the problem becomes very simple to study in the corresponding Lagrangian formalism. The Hamiltonian reads $$\tag{1} H(p,q) ~:=~ \frac{p^2}{2m} + \lambda pq + \frac{m\lambda^2}{2}\frac{q^6}{q^4+\alpha^4}. $$ Since there is no explicit time ...


2

Phase spaces which are not cotangent bundles can be realized in mechanical systems with phase space constraints . The phase space given by Arnold: the two sphere $S^2$ can be mechanically realized as the reduced dynamics of an energy hypersurface of a two dimensional isotropic harmonic oscillator: $ |p_1^2|+|p_2^2|+|q_1^2|+|q_2^2| = E$ We observe that the ...


2

We are not living in the $q$ part of phase space : we indeed live in the full phase space since we're definitely not fixed-momentum objects. However we give more importance to our position than to our velocity/momentum; then the question gets out of physics into psychology. In my opinion, part of it may be because of the way we gather knowledge : we fix it ...


2

The problem you are having is that there are two different uses of the word "phase". One is, as you point out, the argument of the $sin$ function. The other use of the word is the ordered pair $(x, p)$, that is, coordinate and momentum. $(x,p)$ specifies a point in a two-dimensional "phase space" where one axis is $x$ and the other $p$. The state of a ...


2

What is meant is that $$\frac{d\rho(q(t), p(t), t)}{dt} = 0$$ when $q,p$ are solutions to Hamilton's equations. While it is notationally convenient and space-saving to not write everything out in this detail, it is as you noted confusing. This particular confusion actually has a name -- it's the first fundamental confusion of calculus. (There's a second ...


1

Well you have detected a difference between a) and b) so you are doing better than me. But I will try to answer your question anyway. Let's review one way of doing a). Let's look at the trajectory of the oscillator in phase space. First let me explain phase space for completeness. Phase space is the space of states that the oscillator can be in. The state ...


1

Actually, the Liouville theorem is more general - it is valid even if the distribution function depends on time, and even if the Hamiltonian depends on time. http://en.wikipedia.org/wiki/Liouville%27s_theorem_(Hamiltonian) -> phase space volume preservation but no energy conservation: any Hamiltonian which depends on time, but you already know that. For ...


1

Liouville's theorem not only depends on the form of Hamilton's equations but also on the fact that $\partial\rho/\partial t = 0$, where $\rho$ is the statistical distribution function of the system. This is strictly true only for closed systems and is approximately true for quasi-closed systems when not observed for too long a time. Energy of a system is ...


1

Although I know little about this, other applications I have seen are in beam physics where one can make arguments about beam emittance. In particular one can argue that emittance remains constant in certain situations, using Liouville's theorem. See e.g. this article.


1

First, although it is common in some textbooks, I don't think it is a good thing to necessarily relate the equiprobability postulate to ergodicity. Second, what this postulate enables is to estimate the probability distribution for the macrovariable you want to look at. You can of course look at the most probable value for this macrostate and this will ...


1

I was looking for papers on the energy-time, momentum-length, and mass-tau trio of uncertainty relations when I ran across your question. The mass-tau uncertainty relation $\{mc, \tau\}$ certainly exists. In answer to the second part of your question, it is the source of the subpopulation of virtual particle pairs whose combined net momentum during their ...


1

If the dissipative system has a thermodynamic equilibrium state, then in general, the set of microscopic initial conditions is larger than the set of microscopic states in the thermodynamic equilibrium state. Imagine a melting ice with a final state of water at 10°C. The initial state (some microscopic configuration corresponding to ice, or in general a set ...



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