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The action of a massless scalar field is given by: \begin{eqnarray} S(\phi)&=&\int{\cal{L}}dt\\ &=&\int d^{4}x \sqrt{-g}\left(g^{\mu\nu}\phi_{,\mu}\phi_{,\nu}\right) \end{eqnarray} Now choosing a tetrad, i.e., a basis of one form at each spacetime point $\{e^{a}=e^{a}_{\mu}dx^{\mu}\}$ we can rewrite the action as: \begin{eqnarray} ...

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Maybe this is an answer on the question you've asked. The main reason of applications of homotopy theory in QFT is the requirement of finiteness of an action. Suppose that we want to start from the field configuration for which an action is finite, and then to discuss perturbations near such configuration. The most simple way to satisfy the requirement of ...

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The Lagrangian is defined in the most simple case as a function of $q$, $\dot{q}$ and $t$: $L(q,\dot{q},t)$. This notation implies that $q$, $\dot{q}$ and $t$ are by definition independent variables. This is how you have to interpret the partial derivatives to $q$ and $\dot{q}$, it doesn't make sense to write: $q = f(\dot{q})$, because both are considered ...

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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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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 $(q_1, \ldots, q_E)$, ...

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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.

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OP is basically asking (v4) the following. How the equation $$\int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) \langle F^r\left[x,\Phi\right] \rangle_J\right) ~=~0, \tag{4.5}$$ or equivalently, $$\int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) F^r\left[x,\frac{\delta}{i \delta J}\right]\right)Z[J] ~=~0, \tag{4.5'}$$ ...

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David Morin - Introduction to Classical Mechanics

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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 $$x = ... 9 Comments to the question (v2): First of all, let us stress that OP is correct, that a given set of equations of motion does not necessarily have a variational/action principle, cf. e.g. this Phys.SE post and links therein. On one hand, if there exists a Lagrangian formulation, then one may in principle obtain a Hamiltonian formulation via a (possible ... 7 The field equations must be conservative in a fairly precise sense in order that this can be done in a physically appropriate sense. Then there are several Hamiltonian approaches to field theory: the De Donder-Weyl formalism and the multisymplectic formalism. Although both formalisms can accommodate Lagrangians, the can also be understood without any ... 4 The derivation in the given reference indeed seems confused and inconsistent. The crucial error seems to me that$$ 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 ... 0 There is a simple proof that the cancellation is impossible (at least unless you are willing to add to the classical a term proportional to \hbar), I am reformulating an answer by @Qmechanic in a simpler language: The anomaly, or the measure variation which is the typical source of the anomaly, contributes a term which is independent of \hbar, while any ... 4 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 ... 0 Thanks guys, i followed your seggestion. Using \dot{z} = const. =\alpha '  and \dot{\phi}² = (\frac{\beta - a\rho²}{m\rho²})². Now applying on H we have \frac{d\rho}{dt} = \sqrt{\frac{H}{m} - \alpha'² - \rho²(\frac{\beta -a\rho²}{m\rho²})²} then t = \int\frac{d\rho}{\sqrt{\alpha - \frac{(\beta - a\rho²)²}{m²\rho²}}} where \alpha = \frac{H}{m} - ... 0 "Is this causality condition equivalent to Lorentz invariance?" No. Lorentz invariance ensures that points in spacetime that are spacelike (timelike) separated in one frame stay spacelike (timelike) in all inertial frames. Causality, in this context, is the notion that an event can not have effect at any spacelike separated point in spacetime. "Now the ... 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. ... 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 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 ... 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 ... 0 Thanks for the hint. This is what I got so far:$$\delta W=\int d^4x' \mathcal{L}'(x')-\int d^4x \mathcal{L}(x)$$Now since d^4x'=d^4x(1+(\delta x^\alpha)_{,\alpha}) this equation reduces to:$$\delta W=\int d^4x \mathcal{L}'(x')-\mathcal{L}(x)+(\delta x^\alpha)_{,\alpha}\mathcal{L'}(x')$$Whic is just:$$\delta W=\int d^4x \delta\mathcal{L}+(\delta ...

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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 ...

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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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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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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, $\psi \equiv \left( \begin{array}{c} ... 3 Comments to the question (v2): Yes, OP is right. The classical path/stationary solution between$(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). ... 0 This normalization is convenient when you work in Fourier space, since it implies that $$\langle \phi(p) \phi(q) \rangle = \frac{1}{p^2 + m^2} \delta(p+q)$$ which is easy to remember. You normally derive this using path integrals/functional integration, and in that way the factor of$1/2$comes from a Gaussian integral -- where ultimately it's more ... 0 According to these Cambridge lecture notes it's "conventional". I'm guessing that choice of constant gives you the right T-V energy when you integrate the Lagrangian density over a certain volume. 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 $$j^{\mu} ~=~ p^{\mu} \cdot \bar{\delta}u + {\cal L} ~\delta x^{\mu}, \qquad ... 2 The lapse function is not defined by the metric alone, but instead depends on both the metric g_{ab} and its slicing into timelike hypersurfaces. One way to "slice" a spacetime \mathcal{M} into timelike hypersurfaces is to define a timelike coordinate f, which is just a function f: \mathcal{M} \to \mathbb{R} such that \nabla_a f is a timelike ... 0 Let me give a pedagogical answer. You're confused about the functional derivative$$ \frac{\delta}{\delta \phi(y)}\int dx\, \phi(x) \partial^2 \phi(x).$$We can compute such derivatives by perturbing a functional:$$ F[\phi + \epsilon \chi] = F[\phi] + \epsilon \int dx \, \frac{\delta F[\phi]}{\delta \phi(x)}\chi(x) + O(\epsilon^2).$$Now let$F[\phi] = ...

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