# Tag Info

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, $$\mathbf{q}_1 = ... 9 You've duplicated constraints because if any one particle is constrainined in all three dimensions with all the other particles this constrains all the particles. The number of constraints is 3(N - 1). To give an example, take three particles a, b and c. If a is fixed relative to b and is also fixed relative to c, then b and c are fixed relative to each ... 8 You seem to be talking about the "old covariant quantization" in which L_n for positive n and (L_0-a) annihilate physical ket states |\psi\rangle, right? It's analogous to the Gupta-Bleuler quantization http://en.wikipedia.org/wiki/Gupta-Bleuler_quantization which was a standard procedure used already in electromagnetism. The idea is that the ... 7 J.W. van Holten's "Aspects of BRST Quantization" arXiv:hep-th/0201124 might be what you're looking for... 7 Given a system of N point-particles with positions {\bf r}_1, \ldots , {\bf r}_N; with corresponding virtual displacements \delta{\bf r}_1, \ldots , \delta{\bf r}_N; with momenta {\bf p}_1, \ldots , {\bf p}_N; and with applied forces {\bf F}_1^{(a)}, \ldots , {\bf F}_N^{(a)}. Then D'Alembert's principle states that$$\tag{1} \sum_{j=1}^N ( ...

6

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 ...

6

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 $t_0$. ...

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 ...

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

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

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

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} = ... 5 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 ... 5 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 ... 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 ... 4 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 ... 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 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, ... 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 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 ... 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 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; ... 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 It seems OP's question (v4) is related to the proper handling of derivatives of Dirac delta distributions. Reductions are performed with the help of (the appropriate 3D generalizations of) the following formulas:$$\tag{A} \{\partial_x+\partial_y\}\delta (x-y)~=~ 0,\tag{B} \{f(x)-f(y)\}~\delta (x-y)~=~ 0,\tag{C} \{f(y)-f(x)\}~\partial_x \delta ...

3

A general remark. Goldstein is not saying that the applied forces vanish when one "transforms to generalized coordinates," he is simply saying that the equation \begin{align} \sum_i \mathbf F_i^{(a)} \cdot \delta\mathbf r_i = 0 \end{align} does not necessarily imply that the applied forces are zero. The virtual infinitesimal displacements must respect ...

3

Yes, it should be $x/z = tan \theta$, this is probably a typo. The constraint should be $a - \sqrt{x^2 + z^2}=0$ for the argument to make sense. $r$ is a coordinate which is variable but due to the constraint it will always be equal to $a$, so we can use $a$ in the equations instead. ($\dot{a}=0$). You know that a gradient of $f$ is always perpendicular to ...

3

I think that your description that the points of the configuration manifold are possible states of the system is as close to a precise definition as one will find. So for $n$ particles in three dimensions, the configuration manifold is just $(\mathbb{R}^3)^n$. As for how this relates to constraints, consider the simplest example: two particles attached with ...

3

Strength of materials is affected by defects. A perfect crystal of iron would be extremely strong. Once a crack starts, it is not so hard to make it advance one atom deeper. Think of tearing open a plastic bag. Much easier once the tear starts. Brittle materials can be easier to break because they stretch less. It is easier to tear a sheet of paper than a ...

3

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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