# Tag Info

## New answers tagged lagrangian-formalism

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Whenever indices are repeated up and down and assumed to be summed over, they are always invariant under rotations because they are nothing but the scalar product between a vector and its dual (or more precisely, the action of a dual vector onto a vector). Such scalar product is invariant under orthogonal transformations, hence the eventual result will be, ...

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This is a statement of the principle of minimal action. Let $$\int\limits_{t_0}^{t_f}\text{d}t L (q(t),\dot{q}(t),t)$$ be the energy functional ($L = E_{kin} - E_{pot}$). First note that the energy is function of coordinate ($q$) and speed ($\dot{q}$). Then consider a infinitesimal variation of the trajectory from point $q(t_0)$ and $q(t_f)$: $$q(t) = ... 0 Suppose that the field B  can be decomposed into B=gB'+\phi \delta B with the coupling constant g  and \delta \phi = I where  \phi is indicator function on maximum set intersection such that \delta \phi  lies completely on the support of \delta B . Now the integration haar measure can be decomposed into d [B]=d [B'] d [\delta B \phi]. ... 0 Typically force should be included in your Lagrangian:$$L=M\dot\theta^2R^2+F \theta r$$But you could also have included it as F_{\theta}=F\,r They both equate to:$$2\,M\,R^2\,\frac{d\dot\theta}{dt}-FR=0$$2 Current trend today in MBD is towards writing code, doing simulations for some practical problems. That is not entirely true and it mostly depends on the actual areas and topics you are dealing with, as for all the other subjects. Of course, due to the industrial applications, the practical side always has more money and more academic positions, but ... 1 The charge associated to the U(1) symmetry you mention is called weak hypercharge. The relation with the electric charge is the following Q= T_3+Y/2 where T_3 is the third generator of the U(2) symmetry representing the weak isospin and Q the electric charge. This relation holds for all leptons. Neutrinos have weak isospin +1/2 and weak ... 0 It seems like you've got lost in the subject. To clarify some facts: The action for General Relativity (Einstein-Hilbert action) is, as usual, an integral of the Lagrangian density over spacetime:$$ S[g] = \frac{1}{16 \pi G} \int d^4 x \sqrt{-g} \cdot R, $$where \sqrt{-g} is the square root of the determinant of the metric tensor and R is the Ricci ... 1 The argument is basically that you can take an arbitrary plane \dot{\theta} = 0 so that on this plane you measure your \theta as \pi/2 (since rotations don't affect the physical phenomenon). And because of that the sin^2(\theta)  term becomes one and p_\phi  reduces to mr^2\dot{\phi} which is the usual angular momentum. This procedure ... 2 how do you find potential in a place where we have no intuition of force and are not allow to find it. Well I think this might be your problem; I've certainly never heard it said that you are not allowed to find forces. The Euler-Lagrange equations are simply another tool to finding the dynamics, but that doesn't mean you have to start from scratch and ... 1 In analogy to the ordinary calculus you need to look at the second or quadratic variation, Gelfand and Fomins Calculus of Variations do a good job of explaining it with the minimum of fuss. 1 The action is sometimes a saddle, but it is a minimum over small enough regions, but the reason it is a minimum instead of a maximum is due to a convention. The action depends on the endpoints, the path and the Lagrangian. So the stationary action(s) and the physical path(s) depend on the endpoints and the Lagrangian. How do you know which Lagrangian to ... 0 In a general Lagrangian formalism,  L  doesn't equal T - V . Rather it is a function of some field (be it scalar field, vector field, or whatever other field that is useful...) \phi, derivatives of \phi and spacetime (x,y,z,t). This function is chosen so that the equations of motion produce the correct physical phenomena. In general, canonical ... 3 First of all, my opinion is that the paper on your link is full of notational inconsistencies and therefore causes a great amount of confusion for someone who struggles to understand Noether’s theorem. So, allow me to formulate the field-theoretic version of Noether’s theorem in a more, according to my taste, charming way. Preliminaries 1: Lie Groups and ... 0 So you started with L' = L(|v|^2 + 2v\bullet \epsilon + \epsilon^2)  If you treat  2v\bullet \epsilon + \epsilon^2  as a variation of |v|^2, you may use the taylor's theorem treating |v|^2 as a variable, and you get the following expression:  L(|v'|^2) = L(|v|^2) + \frac{\partial L}{\partial |v|^2} (2v \bullet\epsilon + \epsilon^2) + higher order ... 1 Given$$ \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial q}\,\delta q + \frac{\partial L}{\partial v}\,\delta v \right)= \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial q}\,\delta q\right) + \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial v}\,\frac{d}{dt}\delta q \right) $$then the second contribution on the right ... 0 Partial integration is employed only for the second term in (1):$$\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\frac{\partial L}{\partial v}\delta v=\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\frac{d}{dt}\left(\frac{\partial L}{\partial v}\delta q\right)dt-\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\left(\frac{d}{dt}\frac{\partial L}{\partial ...

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If initially the mass is at $x=0$ and the initial velocity is $V$ then the (underdamped) position response is: $$x(t) = X \exp(-\beta t)\sin(\omega t) = \frac{V}{\omega} \mathrm{e}^{-\zeta \omega_n t} \sin(\omega t)$$ where \begin{aligned} \omega_n & = \sqrt{\frac{k}{m}} \\ \zeta & = \frac{d}{2 m \omega_n} = \frac{d}{2 \sqrt{k m}} \\ \omega ... 0 It seems to me you are making this more complicated than it needs to be. When the cable first becomes taut, the spring force is not yet in play and the only force will be v\cdot k - by the definition of the drag in the dash pot. You can compute the subsequent motion by solving the damped harmonic oscillator. Let me know if this is enough? 1 First, you should be careful with your choice of indices. What you have written in Eq. 1 implies a summation over \mu that I don't think you actually want. It is true that $$\frac{\partial F_{\mu\nu}}{\partial A_{\sigma}}=0,$$ but that is just because F_{\mu\nu} depends on the derivatives of A_{\mu} and not A_{\mu} ... 0 Now, someone tried to mark this question as a duplicate of this other one. From my point of view, having derived the Euler-Lagrange equations before even mentioning the least action principle or the action, it didn't seem too related. The point was, I wanted to have a physical interpretation of the Lagrangian, and leave the action and the principle as ... 3 is there any reason for this quantity to be introduced? It is a quantity for which the actual dynamics makes the integral of the thing be stationary with respect to changes of paths when you consider alternate, but nearby changed paths. Does it have any physical meaning? One problem is that many Lagrangians give the same equations of motion, so ... 1 As you have already mentioned, L is NOT in general T-V. T-V only holds in classical mechanics. And I will try to motivate the construction of T-V in classical mechanics following "The Variational Principles of Mechanics" by Cornelius Lanczos. To start out, let's talk about statics. It it well known that the condition for a physical system to be at ... 0 In order to describe newton's law in a 2d universe, you need a lagrangian with two generalized co-ordinates, x and y, and their repspective velocities. You'll get two differential equations when you solve it, one for x, and one for y. In your question, you only mentioned the one for x. So your statement that the two equations are equivalent is incorrect. ... 4 The Lagrangian is defined over a bunch of different position variables and their derivatives, which is basically why you don't lose any information. That is, for a single particle in 3D, you have \mathcal L = \frac 12 m \dot x^2 + \frac 12 m \dot y^2 + \frac 12 m \dot z^2 - U(x, y, z), which contains all of the information you need to calculate p_x = ... 4 There are already several good answers showing the algebra. Here we will make some comments to the question (v4) concerning terminology and notation, which may clarify a thing or two. (In the following we refer to the q position space as the vertical space and the t time axis as the horizontal space.) Usually, the principle of least action refers to ... 4 You can break \int_{t_1+\Delta t_1}^{t_2+\Delta t_2} L(\alpha) dt into \left( \int_{t_1 + \Delta t_1}^{t_1} +\int_{t_1}^{t_2} + \int_{t_2}^{t_2+\Delta t_2}\right)L(\alpha) dt. Then of these three pieces, the \int_{t_1}^{t_2} piece combines with the -\int_{t_1}^{t_2} L(0) dt piece to give you the \int_{t_1}^{t_2} \delta L dt. This means that ... 0 For a classical field theory or a classical field and particle theory you want the dynamics of that stationary path. (Or the dynamics of one of the stationary paths.) But you consider all kinds of dynamics, and just reject the ones that don't have a stationary action. If you know the Euler-Lagrange equations you are aiming for are going to just have/be ... 1 The question seems okay to me. It's more or less giving you the time-independent Klein-Gordon equation for a 1D lattice, right? I think a clarification on how to discretise the derivative would go miles on this question. If you write \frac{\partial \phi}{\partial x} = \frac{\phi_{j+\frac{1}{2}} - \phi_{j-\frac{1}{2}}}{a} $$then you can easily just ... 0 Just as a starting-point, it might help clarity if we're a little more careful about what it would be for a solution to be "gauge-invariant". I'll use the example of electromagnetism to illustrate. So suppose that we have written Maxwell's equations in terms of potentials, and having done everything in terms of coordinates:$$\partial_\mu\partial^\mu A^\nu - ...

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I am not too sure of the homework policy on this forum so I won't answer your question directly, but I hope this helps you :) Starting from the Lagrange density $\mathcal L$; $$\mathcal L=\frac 12 \rho _0\dot \eta^2 +P_0\nabla \cdot \eta -\frac 12 \gamma P_0(\nabla \eta )^2$$ The equation of motion for the $\eta$ field is given ...

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Fermat's principle says the path with minimum optical path or minimum time is chosen by light. It can be direct or indirect (containing reflections or refractions). But as others said, it's not necessarily unique, because there might be paths all with minimum optical paths, that means, all with equal minimum optical paths. That's exactly what happens in an ...

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If I understood well, you want to find the second law of Newton using the Lagrangian and the Euler-Lagrange equation? To make it easy, I'll only take one direction "$x$". Even if you can work with more degrees of freedom, it's the same application of Euler-Lagrange equation, so like that the writing will be lighter. So you have the Lagrangian ...

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Given the Lagrangian $L(q, \dot{q})=\sum^{N}_{j=1}{\frac{1}{2}m_{j}\dot{q_{j}}^{2}}-V(q)$, $$\frac{\partial L}{\partial \dot q_{i}} =\sum^{N}_{j=1}{\frac{1}{2}m_{j}\frac{\partial (\dot{q_{j}}^{2})}{\partial \dot q_{i}}} = \sum^{N}_{j=1}{\frac{1}{2}m_{j} 2 \dot{q}_{j}\delta_{ij}} = m_i\dot{q}_i.$$ I guess you are stuck because of the dot product, but the ...

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You are correct that the potential energy of this system is: \begin{align*} U=mgl(1-\cos{\theta}) \end{align*} when you take the potential energy to be zero at the bottom of the pendulums swing. However, because ($mgl$) is a constant and will play no part in the dynamics of the pendulum, you can throw it out. You can see this by finding the equations of ...

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On the other hand, considered non-mathematically (i.e. physically) the PLA seems to imply that Nature is thrifty in all its actions. Here, thrifty is taken in the sense that Nature avoids waste, avoids doing anything unnecessary or needless. Hey, now, let's not get ahead of ourselves. The principle is actually one of stationary action, or action which ...

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None English speaking person here, sorry by advance. Principle of least action is also call the principle of stationary action. And should be called like that all the time... The action is none intuitive to understand... And honestly i'm not sure to understand it myself, so take what i'm going to say carefully, i might be wrong. The action's unit is J.s, ...

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As noted in the comments, the transformation from the $(q_{i},\dot{q}_{i})$ coordinates to the $(q_{i},p_{i})$ coordinates is an example of a Legendre transform. Informally speaking, this allows you to use different coordinates to describe the system, while maintaining all information about the system. In accordance with the general formulation of a ...

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Well, if you have a term like $\partial_\mu \mathcal{J}^\mu$, the divergence theorem lets you convert it into a surface term upon integrating to find the action, and since variations are assumed to vanish at the boundary, this term goes away. The Euler-Lagrange equations don't change because they come from setting the variation of the action to zero. ...

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