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1

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 ...


1

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.


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 - ...


0

Here's how I think about it. An action is a functional: It eats a function and returns a number. The functional derivative asks: "For very small changes in the function fed to the functional, how how the functionals value change?" First let's think of a trajectory, $x(t)$. This is what we will feed to the functional. Now let's consider a smooth family of ...


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 \begin{equation} ...


0

Let express the equations of motion and the Euler-Lagrange equations with zero right hand sides \begin{equation} \bbox[#FFFF88,12px]{\ddot{Q}_{k}+\omega^{2}_{k}Q_{k}-2\dfrac{\dot{q}}{q}\sum_{j}g_{kj}\dot{Q}_{j}-\dfrac{\ddot{q}q-\dot{q}^{2}}{q^{2}}\sum_{j}g_{kj}Q_{j}-\dfrac{\dot{q}^{2}}{q^{2}}\sum_{j\ell}g_{jk}g_{j\ell}Q_{\ell}=0} \tag{01a} \end{equation} ...


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

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 $$\tag{2} {\cal V}(\phi)~\propto~(\phi^2-v^2)^2~ \geq~ 0$$ ...


0

The energy density of the state $\pm v$ is going to be something like $\propto μ^4$, if you are using the basic $\varphi^4$ theory. While the energy of the domain wall is finite, the energy of the vacuum state is not, and so the transition to the vacuum state iver all space will be infinite.


3

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

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 ...


0

I have a hunch that it might not be possible in the general case. Since you integrated over the fields already, it would be similar to trying to find the original integral from a real number. Also the basic path integral $Z[0] =1$ no matter the field, for instance.


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 ...


0

I'm not sure where the term with the third derivative came from; taking the equilibrium point of our system to be $x=0$ we find the force due to the leftmost spring is given by: \begin{equation} F_{\textrm{spr,1}}\left(x\right)=-k_1x \end{equation} the force due to the rightmost spring is given by: \begin{equation} F_{\textrm{spr,2}}\left(x\right)=k_2x ...


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

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 ...


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 $\psi$ ...


2

Let the equations of motion be expressed in a frame with coordinates $q$. We now want to switch over to another (arbitrarily moving) frame, whose corresponding coordinates are $Q$, given by: $$Q = f(q, t)$$ For example, if the frame itself is moving with position $x(t)$, we will have: $$Q = q - x(t)$$ (where $x$ is not dynamic, but is completely specified in ...


8

Particles do not minimize their action. Instead, they minimize their action given certain boundary conditions. We can only apply the action principle when we know the start and end points ahead of time. If we know that a particle will be at location $x_i$ at time $t_i$ and that it will be at location $x_f$ at time $t_f$, then the particle takes a path of ...


4

Since it's a quiet Sunday morning let me dust off my brain cells and see if I can remember how to do this. We start by noting that if the initial and final positions for the particle as $\mathbf{r}(t_1)$ and $\mathbf{r}(t_2)$, then the action is: $$ S[\mathbf{r}(t)] = \int_{t_1}^{t_2} \tfrac{1}{2}m\dot{\mathbf{r}}^2 $$ We'll make the change $\mathbf{r} ...



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