# Tag Info

## Hot answers tagged phase-space

26

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 ...

10

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 ...

9

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

Introduction Let us define the density of particles of species $s$ in a volume element, $d\mathbf{x} \ d\mathbf{v}$, at a fixed time, $t$, centered at $(\mathbf{x}, \mathbf{v})$ as the quantity $f_{s}(\mathbf{x},\mathbf{v},t)$. I assume this function is non-negative, contains a finite amount of matter, and it exists in the space of positive times and $\... 7 Your question has, indeed, been beaten to a pulp in the 70 years of the formulation, and, as you suggested, the necessary conditions are not all independent, so parts are redundant. For a pure state real$f(x,p)$the sufficient condition is straightforward, eqn (6) of Ref. 1: Given its Fourier transform (the cross-spectral density) must left-right" ... 6 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 ... 6 I) Let us for simplicity work in 1D with$\hbar=1$. (The generalization to higher dimensions is straightforward.) Moreover, let us for simplicity take an operator$\hat{f}(\hat{X},\hat{P})$without any ordering ambiguities, i.e., each monomial term in the symbol$f(x,p)$depends only on either$x$or$p$, but not on both. Then one possible motivation of ... 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 ... 6 See this article on the history of phase space. Assuming the article is to be trusted, Boltzmann noted that in a 2-D system the trajectories looked like Lissajous figures, and the shape of the Lissajous figure is determined by the relative phase of the two input signals. He then used the work phase to refer to that part of the configuration that was ... 6 Consider a non-relativistic massless particle with charge$q$on a 2D torus $$\tag{1} x ~\sim~ x + L_x , \qquad y ~\sim~ y + L_y,$$ in a constant non-zero magnetic field$B$along the$z$-axis. Locally, we can choose a magnetic vector potential $$\tag{2} A_x ~=~ \partial_x\Lambda, \qquad A_y ~=~ Bx +\partial_y\Lambda,$$ where$\Lambda(x,y)$is ... 6 The essential idea of a PoincarĂ© map is to boil down the way you represent a dynamical system. For this, the system has to have certain properties, namely to return to some region in itâ€™s state space from time to time. This is fulfilled if the dynamics is periodic, but it also works with chaotic dynamics. To give a simple example, instead of analysing the ... 6 Look carefully at it, especially at the arrows. You should think of the "intersections" as limits of the two different situations (the loops and the waves). Trajectories that actually reach the unstable equilibrium will remain there indefinitely, so there is no crossing over the potential peak. If you miss that peak by a tiny amount it will look like you ... 6 Notice that the "speed" in the phase space is given by $$v_{\mathrm{ps}} = \sqrt{\dot{q}^2 + \dot{p}^2} = \sqrt{H_{p}^2 + H_q^2}.$$ Here,$H=H(p,q)$is the Hamiltonian, and$H_p$and$H_q$are shorthand for$\partial H/\partial p$and$\partial H/\partial q$. Provided that there exists a closed constant-energy curve in the phase ... 6 In brief: phase space is not made into a vector space because that additional structure provides no benefit; quantum mechanics uses a Hilbert space because that additional structure does provide benefits. Any time you relate a mathematical structure to a physical concept, you need to ask how useful that relation is. The mathematical structure will have ... 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 ... 5 First, let's take a look at one-dimensional systems with phase space dimension$2$. The volume form is just the symplectic one, ie any volume-preserving flow is symplectic and thus at least locally Hamiltonian (but not necessarily globally so). Now, consider an arbitrary phase space of dimension$2n\geq4$with canonical coordinates$q^i,p^i$. Up to a ... 5 I) More generally, OP is essentially pondering: Let$X\in \Gamma(TM)$be a given vector field on a$2n$-dimensional manifold$M$. Under what conditions is the evolution equation $$\tag{1} \frac{df}{dt} ~=~ X[f]+\frac{\partial f}{\partial t}$$ a Hamiltonian system? In other words, under what conditions is$X$a Hamiltonian vector field? Globally,... 5 A metric structure$g$and a symplectic structure$\omega$are two very different structures, although sometimes they can co-exist in a compatible way. Unlike a symplectic structure, there are no Jacobi-like identity and no Darboux-like theorem for a metric structure. There exists a unique torsionfree metric connection$\nabla$on a pseudo-Riemannian ... 5 Non-equilibrium systems are most often considered in the approximation where local equilibrium is valid, yielding a hydrodynamic or elasticity description. Local equilibrium means that equilibrium is assumed to hold on a scale large compared to the microscopic scale but small compared with the scale where observations are made. In this case, one considers a ... 5 $$\frac{\partial \rho }{\partial t}= -\sum_{i}\left(\frac{\partial \rho}{\partial q_i}\,\dot{q_i}+\frac{\partial\rho}{\partial p_i}\,\dot p_i\right)$$ This means that if we have a function of$t, p, q$namely$\rho(t,\vec p,\vec q)$and we have a trajectory that is a curve in$(p,q)$space, namely$q_i(t), p_i(t), i=1\ldots N,$then: $$\frac{\mathrm d}{\... 4 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 ... 4 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 ... 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). http://en.wikipedia.org/... 4 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 ... 4 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. http://en.wikipedia.org/... 4 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 ... 4 U(1) Chern-Simons theory with (physical) space a 2-torus is such an example. Its phase space is the gauge equivalence classes of flat connections on the 2-torus. These are specified by the holonomies around two 1-cycles forming a basis of H_1(T^2). This is of course a 2-torus U(1) \times U(1). Because of the form of the Chern-Simons action, these ... 4 i will try this one. A Hamiltonian system is (fully) integrable, which means there are n (n= number of dimensions) independent integrals of motion (note that completely integrable hamiltonian systems are very rare, almost all hamiltonian systems are not completely integrable). What this states in essence (and intuitively) is that the hamiltonian system ... 4 With$$ \tag{1} \det(J) = 1 + \mathcal{O}(dt^2)$$he means that when dt \to 0 the quantity$$ \tag{2} \frac{ \det J -1}{(dt)^2}$$is bounded. Now what is the derivative of \det J? Using the definition we have$$ \tag{3} \frac{d \det J}{dt} = \lim_{dt \to 0} \frac{ \det J - 1}{dt} = \lim_{dt \to 0} \frac{ \det J - 1}{(dt)^2} dt = 0,$\$ because is the ...

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