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## Hot answers tagged constrained-dynamics

15

The generalized coordinates of a system of $N$ particles apply to the system as a whole, not the individual particles, and accordingly they can (and often do) combine the coordinates of multiple particles. One common example is that of two-body orbital motion: one generalized coordinate is the position of the center of mass of the system, $$\mathbf{q}_1 = ... 10 Given a system of N point-particles with positions {\bf r}_1, \ldots , {\bf r}_N; with corresponding virtual displacements \delta{\bf r}_1, \ldots , \delta{\bf r}_N; with momenta {\bf p}_1, \ldots , {\bf p}_N; and with applied forces {\bf F}_1^{(a)}, \ldots , {\bf F}_N^{(a)}. Then D'Alembert's principle states that$$\tag{1} \sum_{j=1}^N ( ...

9

You've duplicated constraints because if any one particle is constrainined in all three dimensions with all the other particles this constrains all the particles. The number of constraints is 3(N - 1). To give an example, take three particles a, b and c. If a is fixed relative to b and is also fixed relative to c, then b and c are fixed relative to each ...

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You seem to be talking about the "old covariant quantization" in which $L_n$ for positive $n$ and $(L_0-a)$ annihilate physical ket states $|\psi\rangle$, right? It's analogous to the Gupta-Bleuler quantization http://en.wikipedia.org/wiki/Gupta-Bleuler_quantization which was a standard procedure used already in electromagnetism. The idea is that the ...

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1) According to usual terminology we wouldn't call a sliding friction force a constraint force as it doesn't enforce any constraint. (No pun intended.) In other words, a sliding friction does not by itself constrain the particles to some constraint subsurface, i.e., the particles can still move around everywhere. On the other hand, rolling friction and ...

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J.W. van Holten's "Aspects of BRST Quantization" arXiv:hep-th/0201124 might be what you're looking for...

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Let $Q$ denote the set of all possible configurations of the system (the configuration manifold). Consider a point $q_0\in Q$. For the sake of conceptual clarity, and to make contact with physics notation, let's work in some local coordinate patch around $q_0$. Suppose that $q_0$ represents the position of the system under consideration at time $t_0$. ...

6

Comments to the question (v2): To go from the Lagrangian to the Hamiltonian formalism, one should perform a (possible singular) Legendre transformation. Traditionally this is done via the Dirac-Bergmann recipe/cookbook, see e.g. Refs. 1-2. Note in particular, that the constraint $f$ may generate a secondary constraint $$g ~:=~ \{f,H^{\prime}\}_{PB} ... 6 Here we will for simplicity only consider the Schrödinger system. We will assume that$$\phi~=~(\phi^1+i\phi^2)/\sqrt{2}$$is a bosonic complex field, and that$$\phi^*~=~(\phi^1-i\phi^2)/\sqrt{2}$$is the complex conjugate, where \phi^a are the two real component fields, a=1,2. [Note the change in notation \psi\longrightarrow\phi as compared ... 6 Yes. There is a standard way to generalize to field theory. Let a theory of n\geq 1 fields \phi^i with a Lagrangian density \mathcal L = \mathcal L(\phi^i, \partial_\mu\phi^i) be given. Here we use that standard abuse of notation in which \phi^i denotes the vector whose components are the fields; \phi^i = (\phi^1, \dots, \phi^n). To obtain the ... 6 Hints to the question (v1): Let us parametrize the problem wrt. an arbitrary world-line parameter \tau (which does not have to be the proper time). The Lagrange multiplier \lambda=\lambda(\tau) depends on \tau, but it does not depend on the canonical variables x^{\mu} and p_{\mu}. Similarly, x^{\mu} and p_{\mu} depend only on \tau. The ... 5 Every rigid body has 3 translational dof. In addition, there are 0, 2, or 3 rotational dof, depending on the geometry, giving a total of 3, 5, or 6 dof. A spherically symmetric rigid body has no other dof. A rigid body with rotational symmetry around an axis has 2 rotational dof, namely two angles for orienting the symmetry axis along a direction. All ... 5 If you work with a smaller number of coordinates (usually "curved ones" in a sense) and no Lagrangian multipliers, you are simply considering a configuration space that is a submanifold of the full configuration space in the calculation that does include Lagrange multipliers. Extremizing the action S_{full} with Lagrange multipliers$$\delta S_{full} = ...

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The fact that $p = \large \frac{\partial L}{\partial \dot{q}} = 0$ introduces a problem in the equivalence between Lagrangian and Hamiltonian representations. The idea is that the Hamiltonian representation plus the constraint $p = 0$ is equivalent to the Lagrangian representation The Lagrangian $L$ is a function of $q$ and $\dot q$, that is $L(q, \dot ... 5 Yes, of course that the$p$-$v$relationship may be transcendental so that it cannot be inverted in terms of elementary functions. That doesn't mean that the inverse function doesn't exist, however. Even functions that can't be written down in terms of elementary functions may exist. For example, consider the Lagrangian $$L =\exp(bv^2)\cdot mv^2$$ It ... 5 Constraints are handled in Lagranian mechanics through either of two approaches: 1) The constraint equation is used to reduce the degrees of freedom of the system. For example, if a particle is constrained to the surface of a sphere, then the Lagrangian can be written entirely in terms of two generalized coordinates and their associated momenta (typically, ... 5 An equation of motion is a (system of) equation for the basic observables of a system involving a time derivative, for which some initial-value problem is well-posed. Thus a continuity equation is normally not an equation of motion, though it can be part of one, if currents are basic fields. 5 Well, when canonically quantizing a system with constraints, you have two methods: Dirac's approach "Quantize, then Constrain"; Reduced Phase Space approach "Constrain, then Quantize". Although these two approaches have analogs with path integral quantization, the Path integral approach sweeps a lot of problems under the rug when you pick a particular ... 5 A modern treatment of this subject can be found in Segreev's book on the Kahler geometry of loop spaces also available online. This line of research started with the seminal work of Bowick and Rajeev: The holomorphic geometry of the closed bosonic string theory and$Diff S^1/S^1$(Spires) (and independently Kirillov and Yuriev (Please see the reference in ... 5 Let there be given a (configuration) manifold$M$. Often in physics one assumes that a constraint function$\chi$obeys the following regularity conditions:$\chi: \Omega\subseteq M \to \mathbb{R}$is defined in an open neighborhood$\Omega$of the constrained submanifold$C\subset M$;$\chi$is (sufficiently$^1$many times) differentiable in$\Omega$; ... 5 I) For a general Lagrangian$L(q,v,t)$, the Legendre transformation may be singular, i.e. the velocities$v^i$in the momentum relations $$\tag{1} p_i~:=~\frac{\partial L(q,v,t)}{\partial v^i}$$ cannot be isolated. How to perform a singular Legendre transformation to achieve the corresponding Hamiltonian formulation goes under the name Dirac-Bergmann ... 4 The principle of Least (Stationary) Action (aka Hamilton's Principle) is derived from Newton's axioms plus D'Alembert's principle of virtual displacements. Because D'Alembert's principle allows to account for the (reactions of the) bonds between the components of a system in a transparent way, the Lagrangian and Hamiltonian formulations are possible. ... 4 Comment to the question (v4): Classically, the Lagrangian for a fermion system reads $$L ~=~ \int\! d^3x~ i\psi^{\dagger}\dot{\psi}-H.\tag{A}$$ The Legendre transformation from the Lagrangian to the Hamiltonian formalism is tricky for at least three reasons: The traditional Dirac-Bergmann analysis leads to constraints. See e.g. my Phys.SE answers here ... 4 I) In this answer we will consider the standard Nambu-Goto (NG) string and show that the Hessian has co-rank 2. The target space metric has$(-,+,\ldots,+)$sign convention, and$c=1=\hbar$. The NG Lagrangian density is $${\cal L}_{NG}~:=~-T_0\sqrt{{\cal L}_{(1)}},$$ $${\cal L}_{(1)}~:=~-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} ... 4 (1) You have a set of irreducible constraints, \lbrace \phi_j\rbrace, both primary and secondary This set of constraints defines a submanifold M within the "full" (unconstrained) phase space. (2) A function on the phase space is set to be weakly zero if it vanishes when restricted to the constrained submanifold M. A function is called strongly zero if ... 4 The canonical momenta don't change if you add a total derivative to the Lagrangian. The particular total derivative you wanted to add to the Lagrangian as well as the Lagrangian itself has free i,j indices. You surely meant something else because the Lagrangian should have no free indices like that. Let me assume that you meant both expressions to be ... 4 I) Let us suppress position dependence q^i and explicit time dependence t in the following, and also assume that the Lagrangian L=L(v) is a smooth function of the velocities v^i, where i=1, \ldots, n. The Hessian matrix is defined as$$\tag{1} H_{ij}~:=~\frac{\partial^2 L}{\partial v^i \partial v^j}.$$Let us consider an open neighborhood$^1$... 4 Adding to Lubos Motl's correct answer, it should be stressed that one may not always invert the relation$p_i=f_i(q,\dot{q},t)$to isolate$\dot{q}^j$, not even in principle, because of constraints. Such cases are known as singular Legendre transformations, and they are the starting point of the topic of constrained dynamics. Example. Consider e.g. the ... 4 Off-shell, meaning without assuming the Lagrange equations and the constraints, the Lagrange multipliers$\lambda^a(t)$does by definition not depend on the dynamical variables$q^j(t)$. On-shell, meaning using the Lagrange equations and the constraints, the Lagrange multipliers$\lambda^a(t)\$ may, as a consequence, depend on the dynamical variables ...

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