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The Lagrangian can be constructed directly by performing a Dirac-Bergmann constraint analysis of OP's Hamiltonian (1). In eq. (3) OP has already correctly identified the primary constraint$^1$ $$\dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~\approx ~0, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\tau}, \tag{A}$$ where $\tau$ is the world-line (WL) ...


6

The first two definitions of the Lagrangian are not equivalent but very closely related in the sense that one is a generalization of the other. The second one may be a non-definition, but that can be easily helped with. I would argue that it is more logically sound to define a Lagrangian in the context of variation problems rather than in physics. The ...


6

First-class constraints generate gauge transformations (assuming the Dirac conjecture), i.e. map physically equivalent states onto each other. Even if you do not assume the Dirac conjecture, then first-class constraints still generate transformations that map allowed states, i.e. those on the constraint surface, to other allowed states. Second-class ...


6

In general the Legendre transformation$^1$ from the Lagrangian to the Hamiltonian formulation may be singular, which leads to primary constraints. This is e.g. the case for gauge theories like Yang-Mills (YM) theory with or without matter, which OP mentions. However, in case of a singular Legendre transformation, by performing a so-called Dirac-Bergmann ...


4

First, some clarifications. It does not make sense to talk about "primary first class constraints." You refer to "primary" and "secondary" constraints while going through the Dirac procedure. After the procedure terminates, there is no distinction between primary and secondary constraints; indeed the difference depends on how ...


4

OP's question is a quite broad topic, which is better studied in textbooks, such as Goldstein's Classical Mechanics, but let's give a few pointers: OP's definition (1) of a Lagrangian as $L=T-U$ is not the most general, cf. e.g. this & this Phys.SE posts. Although for huge classes of theories, it is actually true, cf. e.g. this Phys.SE post. OP's ...


4

On one hand, $$ 0~=~[0,0]~=~[f(x),\vec{p}\cdot\vec{\nabla}f]~=~i\hbar (\vec{\nabla}f)^2.$$ On the other hand, a constraint function $f$ typically satisfies a regularity condition $$ \left .\vec{\nabla}f \right|_{f=0}~\neq~\vec{0}.$$


4

Would an example suffice? If so, consider the case $f(\vec x) = x_1$. Then (1) says $x_1=0$, which is already inconsistent with the commutation relation, and (3) says $p_1=0$, which is again inconsistent with the commutation relation. If $x_1$ or $p_1$ is zero, then we can't have $[x_1,p_1]\neq 0$.


4

The fundamental problem here is that many people, and also Pitts in his paper, are not careful about what theory they are currently talking about. "Quantization of Gauge Systems" by Henneaux and Teitelboim is actually careful about this, and their chapter 3 shows the correct resolution of this problem, even though Pitts cites it as an example for ...


4

Most of the time homework questions are off-topic, but here you are asking a conceptual question where the homework question is just background. So I think it is a good/on-topic question. In this case, start by looking at pulley A. Since pulley A is massless then in order for the acceleration of A to be finite, the net force on A must be 0. Because of the ...


3

The canonical quantisation of general relativity stems from the replacement of the Poisson bracket by a commutation relation such that $$\begin{align} \left\lbrace A(p,q),B(p,q)\right\rbrace \rightarrow -\frac{i}{\hbar}\left[\hat{A}(\hat{p},\hat{q}),\hat{B}(\hat{p},\hat{q})\right] \;. \end{align}$$ When you choose a particular representation satisfying this,...


3

I will first discuss why in classical dynamics the form (kinetic energy minus potential energy) works out as being an effective Lagrangian. After that I will discuss how to apply the underlying pattern to other areas of physics. That is, what in general it takes to be an effective Lagrangian. For Classical mechanics I start with the Work-Energy theorem: $$ ...


3

If the rigid rod of the pendulum is replaced by a spring, then the constraint force is replaced with a spring force. The system has 2 position coordinates $(r,\theta)$. On one hand, the rigid pendulum has 1 constraint $r={\rm const}$, so virtual displacements have $\delta r= 0$ and $\delta t=0$. There remains 1 generalized coordinate $\theta$. On the other ...


2

I'll focus on the scalar $O(n)$ gauge theory that was mentioned in the question. The set of observables in the gauged version of the model is not a subset of the observables in the ungauged version. Gauging the $O(n)$ symmetry does eliminate some observables, but it also introduces others. Observables in the gauged version are required to be invariant under ...


2

Since the matrix $\omega_{\mu\nu}=-\omega_{\nu\mu}$ is antisymmetric, we must demand that $$\begin{align} \delta \omega_{\mu\nu}~=~&-\delta \omega_{\nu\mu}\cr ~=~&\frac{1}{2}\left(\delta \omega_{\mu\nu}-\delta \omega_{\nu\mu}\right)\cr ~=~&\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} - \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\...


2

since the bead is constrained to the wire, only $F(x)$ is acting on it (the normal force $N(x)$ is counteracted by the wire). The bead is constrained to move along the wire precisely because the normal force acts on it. The normal force is whatever it needs to be so that the motion follows the wire. Furthermore, the normal force is not just a function of $x$...


2

the constraint force is zero, because you don't have potential energy or external forces put for example external forces towards the x coordained you obtain $$L=\frac{1}{2}\dot{x}^2+\frac{1}{2}\dot{y}^2+\lambda(x-y-C)+F\,x$$ thus: $$\ddot x=\frac 12 F\\ \ddot y=\frac 12 F\\ \lambda=-\frac 12 F $$ so if $~\lambda~$ equal zero it mean, that you don't have ...


2

The fact that, $$ \phi_{1}=p_{x}+\frac{qB}{2c}y\approx0 $$ Does not imply that, $$ p_{x}=-\frac{qB}{2c}y $$ But only, $$ p_{x}\approx-\frac{qB}{2c}y $$ and conversely for $y$, $x$. That is, $p_x$ is a function of $y$ only on the constraint surface. Because the Dirac bracket is a derivative operator built by design to keep you on the constraint surface, while ...


2

OP is demoting a gauge theory (cf. the Dirac conjecture) with a first class constraint (=Gauss law) and a gauge-fixing condition (=Coulomb gauge) to a non-gauge theory with a pair of second class constraints. (This demotion is always possible; the opposite lifting is not always possible.) The Dirac-Bergmann prescription states that preservation of gauge-...


2

Calling a phase space that is not $T^\ast M$ for some configuration manifold $M$ "curved" is intuitive but likely to lead to confusion - a phase space as a symplectic (or Poisson) manifold does not carry a metric and hence has no notion of curvature in the usual sense of (pseudo-)Riemannian manifolds. Elimination of constraints in the Hamiltonian ...


2

A path integral version of the delta function identity $$\int\limits_{-\infty}^\infty \frac{\mathrm{d}p}{2\pi}\; \mathrm{e}^{\mathrm{i}\, p\cdot x}=\delta(x)$$ reads $$ \int \mathrm{D}{\phi}\ \mathrm{e}^{\mathrm{i} \left<\phi,\psi\right>}=\delta(\psi),$$ where $\left<\bullet,\tilde{\bullet}\right>$ is an inner product on the space where $\phi$ ...


2

I) More generally Ref. 1 observes that if we start with a Lagrangian $$L_0~=~\sum_{I=1}^{2n}\vartheta_I\dot{z}^I-H\tag{A}$$ on the Faddeev-Jackiw form with a non-degenerate symplectic 1-form potential $$\begin{align}\vartheta~=~&\sum_{I=1}^{2n}\vartheta_I\mathrm{d}z^I,\cr \mathrm{d}\vartheta~=~&\omega~=~\frac{1}{2} \sum_{I,J=1}^{2n}\omega_{IJ}\...


2

But can we really know the 6th coordinate? You are right -- given knowledge of five coordinates, there is a discrete choice (corresponding to a reflection) for the remaining coordinate. However, by convention we typically (but not always) use the word "degree of freedom" to mean a continuously varying quantity. This expresses the idea that you ...


2

If you really move $M$ very slowly then $T=\frac12Mg$ so $\mathrm dx_1=2\mathrm dx_2$ in all other cases you have acceleration up or down and so some kinetic energy, so for arbitrary $T$ your energy balance is wrong.


2

$\mathbf{Q1}$ There are 2 kinematic constraints for this system, using your notation, $$dx=R\sin\phi\, d\theta,\tag{1}\label{1}$$ $$dy=-R\cos\phi\,d\theta,\tag{2}\label{2}$$ which become, after divided by $dt$ on both sides, $$\dot x=R\sin\phi\;\dot\theta,$$ $$\dot y=R\cos\phi\;\dot\theta,$$ and you can use $v=R\,\dot\theta$ to write the constraint equations ...


2

Generically OP's action principle (1) gets destroyed if we apply EOMs in the action, cf. the linked Phys.SE post. One the other hand, in the massive case $m\neq 0$ of the Pauli-Fierz action (2) we are allowed to use the EOM for $h_{0i}$. The reason is similar to the following simplified toy model: $$\widetilde{L}(q,B)~=~ L(q,\dot{q},t) + \frac{\xi}{2}B^2+B\...


2

I'm working on this regularization too. The triangulated Hamiltonian $H_{E}^{\Delta}$ you wrote it's such that in the limit it tends to (here I take $k = 1$) \begin{equation} 2 \int_{\Delta} \mathrm{d}^{3}x \, N(x) \epsilon^{abc} \mathrm{Tr}(F_{ab} \, \{ A_{c}, V \}), \end{equation} namely in this infinitesimal limit ($\epsilon \rightarrow 0$) you should ...


1

If you take a particle $m$ with coordinates $x,y,z$ in a field $U(x,y,z)$, then the Lagrangian $$ \mathcal L = \frac m2(\dot x^2+\dot y^2+\dot z^2) - U(x,y,z) $$ will lead you to the well-known equations: $$ m\ddot x=-\frac{\partial U}{\partial x},\qquad m\ddot y=-\frac{\partial U}{\partial y},\qquad m\ddot z=-\frac{\partial U}{\partial z}. $$ However, if ...


1

The Hamilton-Jacobi-Einstein equation corresponds to the HJ equation $$ H[g_{ij},\frac{\delta W}{\delta g_{kl}}]~=~0$$ for Hamilton's characteristic functional$^1$ $W[g_{ij}]$ with vanishing energy $E=0$. The Hamilton-Jacobi-Einstein equation is closely related to the Wheeler-DeWitt equation $$ \hat{H}[\hat{g}_{ij},\hat{\pi}^{kl}]|\Psi\rangle~=~0.$$ Since ...


1

Definition. A coordinate $q^j$ is cyclic if the Lagrangian $L$ doesn't depend on it: $\frac{\partial L}{\partial q^j}=0$. The corresponding momentum $p_j=\frac{\partial L}{\partial \dot{q}^j}$ is not necessarily conserved unless the Lagrange equations is of the form of the Euler-Lagrange equations, cf. e.g. my Phys.SE answer here.


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