# Tag Info

## Hot answers tagged lagrangian-formalism

9

Comments to the question (v2): First of all, let us stress that OP is correct, that a given set of equations of motion does not necessarily have a variational/action principle, cf. e.g. this Phys.SE post and links therein. On one hand, if there exists a Lagrangian formulation, then one may in principle obtain a Hamiltonian formulation via a (possible ...

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The field equations must be conservative in a fairly precise sense in order that this can be done in a physically appropriate sense. Then there are several Hamiltonian approaches to field theory: the De Donder-Weyl formalism and the multisymplectic formalism. Although both formalisms can accommodate Lagrangians, the can also be understood without any ...

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This has nothing to do with "bras" or "kets" and more with the elementary observation that a complex number has two real degrees of freedom, and that derivatives are with respect to one real degree of freedom. The $\frac{\partial}{\partial\phi}$ and $\frac{\partial}{\partial\phi^\ast}$ are the Wirtinger derivatives, which in particular fulfill ...

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Comments to the question (v2): Traditionally, the classical action $S$ sits in the Boltzmann factor $\exp\left[\frac{i}{\hbar} S\right]$ behind an inverse power of $\hbar$ in the path integral, while the path integral measure is independent of $\hbar$. In the conventional way of counting, we say that the Jacobian $J$ from the path integral measure is a ...

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The derivation in the given reference indeed seems confused and inconsistent. The crucial error seems to me that $$S[\phi + \epsilon\delta\phi] = S[\phi]$$ is just not true for an infinitesimal symmetry. The definition of a symmetry is that $S[\phi']=S[\phi]$ (modulo boundary terms) for the finite transformation $\phi\mapsto \phi'$. Writing this ...

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Extremum of potential vs. extremum of action Yes, the action principle is in a special case equivalent to the principle of extremum of potential energy (the maximum of a potential also presents an equilibrium, even though it is an unstable one!). Consider the action principle of a point particle in a potential $V(\vec{x})$, then the Hamilton's action ...

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Comments to the question (v2): Yes, OP is right. The classical path/stationary solution between $(q_i,t_i)$ and $(q_f,t_f)$ does not necessarily exist nor is it necessarily unique. See e.g. this and this Phys.SE posts. However, existence and uniqueness is often true in sufficiently small neighborhoods (if the path is not allowed to leave the neighborhood). ...

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All the alternatives to the Dirac Lagrangian are actually forbidden by the requirement of requiring the hamiltonian to be well behaved (bounded from below and unbounded from above) and hermiticity of the action. To see this most simply we write the Lagrangian in terms of the fundamental left and right handed fields, $\psi \equiv \left( \begin{array}{c} ... 2 There are conserved quantities which don't come from Noether's Theorem. For instance, the topological numbers that characterize the so called topological solutions such as vortices, monopoles, instantons, etc. In general these topological solutions arise in non-linear, vacuum degenerate and spontaneously broken theories. For gauge theories these ... 2 The lapse function is not defined by the metric alone, but instead depends on both the metric$g_{ab}$and its slicing into timelike hypersurfaces. One way to "slice" a spacetime$\mathcal{M}$into timelike hypersurfaces is to define a timelike coordinate$f$, which is just a function$f: \mathcal{M} \to \mathbb{R}$such that$\nabla_a f$is a timelike ... 2 Free theories can be built out of non-interacting scalars, fermions and vectors, and therefore have a Lagrangian description. There may be exceptions for higher-spin fields or exotic SUSY multiplets etc. but those are not so interesting for your question. Next, a weakly coupled fixed point normally means starting with a free theory (which always has a ... 2 I think what they mean is FI-term is not gauge invariant under the full gauge symmetry of the theory, but under this remaining gauge freedom after WZ gauge, which is$U(1)$. 1 The case that$E$does not exist cannot happen on topologically trivial configuration spaces, and even then, it exists locally by the Poincaré lemma. Your "complete conditions" say nothing but that the form$\mathcal{E}=\mathcal{E}_i\mathrm{d}q^i$is closed, i.e.$\mathrm{d}\mathcal{E} = 0$. By the Poincaré lemma, every closed form is locally exact, i.e. ... 1 Writing$x=iy-z, y=(z-x)/i$doesn't help you very much because your goal is to introduce new coordinates and then write$(x,y)$in terms of these. It's nice to start by writing the map in the other direction, i.e. $$z=x+iy$$ The complex conjugate is then$\bar{z}=x-iy$. These can be inverted to write$(x,y)$in terms of$(z,\bar{z})\$ as $$x = ... 1 David Morin - Introduction to Classical Mechanics 1 You can have a look at Jean Zinn-Justin's book "Quantum Field Theory and Critical Phenomena", which is far from being fun to read, but has all the technical details you may want. 1 Disclaimer: Renormalization is a huge subject with many facets, such as, e.g. overlapping divergences of subgraphs, regularization, renormalization group, etc. Here we will only elaborate on OP's quote from Ref. 1. Ref. 1 is considering a Feynman diagram {\cal F}(q_1, \ldots, q_E) in momentum Fourier space, with external 4-momenta (q_1, \ldots, q_E), ... 1 The first one is never valid. Even when considering the relative motion of C to B with the point of contact not moving, then center of C should move with v_C -v_B instead of v_C. The correct picture is the second one. The disk C is specified to have velocity v_C at its center, and v_B at the contact point, and the only way to achieve this is through ... 1 The answers to both your questions basically boil down to "because the kinetic energy is only zero when the system is at rest, and is positive otherwise." However, the answer to the second question in particular gets pretty deep into the linear-algebraic weeds, so fasten your seatbelt. 1. Why is M invertible? This is basically an assumption about the ... 1 Comments to the question (v3): OP's last formula is the standard expression$$ \delta W~=~ \int \! d^4x~ \left[ {\rm EL} \cdot\bar{\delta}u + d_{\mu} j^{\mu}\right], \tag{A} $$for the variation of the action W=\int\! d^4x~{\cal L} (= Wirkung in German). Here$$ j^{\mu} ~=~ p^{\mu} \cdot \bar{\delta}u + {\cal L} ~\delta x^{\mu}, \qquad ...

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Killing vector fields correspond to infinitesimal isometry generators of the spacetime manifold and any physical action including the Polyakov action should be preserved under it. In fact, any physical action should be invariant under the (infinitely) larger group of diffeomorphisms of a manifold. Isomotry transformations are just a finite subset of these ...

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