# Tag Info

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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, $$\mathbf{q}_1 = ... 10 You've duplicated constraints because if any one particle is constrainined in all three dimensions with all the other particles this constrains all the particles. The number of constraints is 3(N - 1). To give an example, take three particles a, b and c. If a is fixed relative to b and is also fixed relative to c, then b and c are fixed relative to each ... 10 Given a system of N point-particles with positions {\bf r}_1, \ldots , {\bf r}_N; with corresponding virtual displacements \delta{\bf r}_1, \ldots , \delta{\bf r}_N; with momenta {\bf p}_1, \ldots , {\bf p}_N; and with applied forces {\bf F}_1^{(a)}, \ldots , {\bf F}_N^{(a)}. Then D'Alembert's principle states that$$\tag{1} \sum_{j=1}^N ( ...

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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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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 $t_0$. ...

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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 ...

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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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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 ...

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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 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 ... 6 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 rotational 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. ... 6 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, ... 6 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 ... 6 Hints to the question (v1): We cannot resist the temptation to generalize the background spacetime metric from the Minkowski metric \eta_{\mu\nu} to a general curved spacetime metric g_{\mu\nu}(x). We use the sign convention (-,+,+,+). Let us parametrize the point particle by an arbitrary world-line parameter \tau (which does not have to be the ... 6 So this depends very strongly on the shape of the slide. The easiest way to see this is to push it to its extreme: suppose one slide is purely vertical and has a length of 100 meters (i.e. H = L, then in the absence of friction getting to the bottom requires a free-fall time, which is gotten by solving H - \frac12 g t_1^2 = 0 to get a time t_1 = ... 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 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 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 $$It ... 5 If you work with a smaller number of coordinates (usually "curved ones" in a sense) and no Lagrangian multipliers, you are simply considering a configuration space that is a submanifold of the full configuration space in the calculation that does include Lagrange multipliers. Extremizing the action S_{full} with Lagrange multipliers$$\delta S_{full} = ...

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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 ...

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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 $\Omega$; ...

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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 $${\cal L}_{NG}~:=~-T_0\sqrt{{\cal L}_{(1)}},$$ $${\cal L}_{(1)}~:=~-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} ... 4 (1) You have a set of irreducible constraints, \lbrace \phi_j\rbrace, both primary and secondary This set of constraints defines a submanifold M within the "full" (unconstrained) phase space. (2) A function on the phase space is set to be weakly zero if it vanishes when restricted to the constrained submanifold M. A function is called strongly zero if ... 4 Adding to Lubos Motl's correct answer, it should be stressed that one may not always invert the relation p_i=f_i(q,\dot{q},t) to isolate \dot{q}^j, not even in principle, because of constraints. Such cases are known as singular Legendre transformations, and they are the starting point of the topic of constrained dynamics. Example. Consider e.g. the ... 4 The canonical momenta don't change if you add a total derivative to the Lagrangian. The particular total derivative you wanted to add to the Lagrangian as well as the Lagrangian itself has free i,j indices. You surely meant something else because the Lagrangian should have no free indices like that. Let me assume that you meant both expressions to be ... 4 I) Let us suppress position dependence q^i and explicit time dependence t in the following, and also assume that the Lagrangian L=L(v) is a smooth function of the velocities v^i, where i=1, \ldots, n. The Hessian matrix is defined as$$\tag{1} H_{ij}~:=~\frac{\partial^2 L}{\partial v^i \partial v^j}. Let us consider an open neighborhood$^1$ ...

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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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Basically, the multiplier method is a way to encode the constraint information of the system directly into the Lagrangian so that you don't have to worry about screwing up the physical requirements of the problem when you solve the equations of motion. In other words, instead of solving the equations of motion and constraining the results, you're ...

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The main point is that Goldstein is not saying we must exclude friction forces in our treatment, but we must place them in the tally of applied forces (that we keep track of in D'Alembert's principle) and not in the other bin of the remaining forces, see this and this Phys.SE posts. Of course, there does not exist a generalized potential $U$ for the ...

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