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The Lagrangian formalism treats $x$ and $\dot{x}$ as independent variables. In particular, you cannot write $\frac{\mathrm{d}}{\mathrm{d}t}x$ because $x$ is not dependent on time. What is dependent on time is a particular trajectory $x(t)$ that is the solution to the equations of motion (the Euler-Lagrange equations). Prior to solving the equations of ...

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Multiple classical solutions to Euler-Lagrange equations with pertinent/well-posed boundary conditions (such solutions are sometimes called instantons) are a common phenomenon in physics, cf. e.g. this related Phys.SE post and links therein. In optics, it is well-known that already e.g. two mirrors can create multiple classical paths.

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Actually, the extra path is not irrelevant. If you put a light bulb at A and a $4\pi$ detector (this means $4\pi$ steradian coverage, i.e. it detects incoming light in any direction) at B, the detector will see light along both paths: direct, and bounced off the mirror, which is exactly the result you got from Fermat's Principle. If you want to exclude the ...

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If you know the propagator, ie. $\langle x'|e^{itH}|x\rangle\,,$ then you could differentiate with respect to time at $t=0$ to get $\langle x'|H|x\rangle\,.$ From this we have, using the resolution of the identity, $H|x\rangle=\int_{-\infty}^\infty dx'\, |x'\rangle\langle x'|H|x\rangle\,,$ from which we have $V(x)|x\rangle=\int_{-\infty}^\infty \, ... 2 Here we assume that OP's question asks about$\phi^4$-theory in 1+1D, where the lagrangian density reads $$\tag{1} {\cal L}~=~\frac{1}{2}\dot\phi^2 -{\cal U}, \qquad {\cal U}~:=~ \frac{1}{2} \phi^{\prime 2} + {\cal V},\qquad \phi \in C^1(\mathbb{R}^2),$$ where the$\phi^4$-potential density $$\tag{2} {\cal V}(\phi)~\propto~(\phi^2-v^2)^2~ \geq~ 0$$ ... 2 The violation of gauge invariance by this term is the "only" reason why it's never written down – as long as we define the word "only" to include all other reasons that may be shown to be "physically equivalent" to gauge symmetry. Gauge symmetry is extremely important and its violation would make a similar theory inconsistent, especially at the quantum ... 2 You have a few different questions here, so let's try to go through them one by one. When we make the chiral symmetry local, have we introduced a gauge symmetry, or some analogue of a gauge symmetry? When you make the chiral symmetry local you introduce a gauge symmetry. The terms "gauge symmetry" and "local symmetry" are two different ways of saying the ... 2 You are confusing two definitions - closed system and conservation of energy. I'll clear them up for you. In classical dynamics a closed system is one where no force external to the system acts. In a closed system, the total energy, total momentum and total angular momentum must be conserved. This follows from Noether's theorem. If a has no interaction with ... 2 If one considers a dynamic system, which (from left to right) consists of a spring with constant k1, a mass m, a damper with constant c and the other spring with constant k2, all connected together, respectively If I understand your setup correctly, the damper is connected between the mass and the 2nd spring. Denote the extension of spring 1 with ... 2 The propagators themselves are not indicative for the form of the Lagrangian. They only provide information regarding the nature of the field - e.g. scalar / fermion / vector boson, etc (gravity metric?). Things that allude what the Lagrangian looks like are vertices / interactions. As a simple example: if you have a theory of field$\phi$with a 4-prong ... 2 The definition of the (integral of the) functional derivative (at least a definition that's good enough for physics level rigor) is the difference of the functional evaluated on a path$x(t)$plus an arbitrary variation$\epsilon(t)$and the functional evaluated on the path, to leading order in$\epsilon$. In other words ... 2 Draw an arrow to represent a vector, with its length representing the vector magnitude. Draw a coordinate system and get the components of the vector. Now draw another coordinate basis, rotated with respect to the first, and get the components with respect to the new basis. The length of the arrow is the same in both systems - i.e length is invariant - ... 1 Here is one way to imagine a variation. You have a path$\vec y(t)=(y_1(t), ..., y_n(t)),$where$y_i(t) = x_i(q_1(t),q_2(t),\dots,q_{n-k}(t),t).$So the idea is that for every time$t$you have some$q$s and they (together with the$t$) give you all your$y_i$hence give you your$\vec y.$Seems basic but the idea is that given a$t$you have an$\vec y.$... 1 Comments to the question (v3): Note that the gamma matrices are covariantly conserved$^1$$$\nabla_{\mu}\gamma^c~=~\omega_{\mu}{}^c{}_b\gamma^b+\frac{1}{4}\omega_{\mu}{}^{ab}[\gamma_{ab},\gamma^c] ~=~0,\qquad \nabla_{\mu}\gamma^{\nu}~=~0,\tag{1}$$ cf. e.g. Ref. 1. Consider the vector current $$J^{\mu}~:=~\bar{\psi}\gamma^{\mu}\psi,\tag{2}$$ where$\psi$... 1 The thing is that when you write the Lagrangian, you don't know the particle's trajectory yet. If you had a specified function$x(t)$, then of course$\dot{x}\$ is not independent. But if you only know the particle's position at a given time, its velocity can be anything, because you are free to set position and velocity as initial conditions how you ...

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