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1

Hints: Then potential term $\frac{1}{2}(\nabla\phi)^2$ is semipositive definite and is only zero for a $x$-independent configuration $\phi$. If one completes the square of the potential $$V(\phi)~=~\frac{\lambda}{4}\phi^4-\frac{\mu^2}{2}\phi^2~=~ \frac{\lambda}{4} \left(\phi^2-\frac{\mu^2}{\lambda}\right)^2-\frac{\mu^4}{4\lambda},$$ then it becomes clear ...


0

With a Lagrangian like: $\mathcal{L} = \partial_\mu \phi^\dagger \, \partial^\mu \phi - V(\phi) = \mathring{\phi^\dagger} \mathring{\phi} + \partial_i \phi^\dagger \, \partial^i\phi - V(\phi) $, the Hamiltonian is: \begin{equation*} \mathcal{H} = \frac{\partial \mathcal{L}}{\partial \mathring{\phi}} \mathring{\phi} + \mathring{\phi^\dagger} ...


1

The velocity jacobian is $\vec{\omega}_B = J\, \dot{q}$ with $q=(\phi,\theta,\psi)$. This is used to transform between the generalized forces/torques $Q$ and the vector torques $\vec{M}_B=(\tau_x^B,\tau_y^B,\tau_z^B)$ $$ Q = J^\intercal \vec{M}_B $$ The power through the joint is $$ Q \cdot \dot{q} = Q^\intercal \dot{q} = \left(J^\intercal ...


0

Step back and ask how you know whether a Lagrangian, $L=L(Q_i,\dot Q_i),$ is correct. At the classical level the only answer is whether it gives the correct equations of motion as the Euler-Lagrange equations: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot Q_i}\right)= \frac{\partial L}{\partial Q_i}. $$ Which is unchanged if you replace ...


0

Conventions do not change physics. If they would, we would not call them conventions. When studying Lagrangian mechanics, you may have noticed that you can multiply a lagrangian by any constant, and receive the same dynamics. Thus, we often (Or always) choose the constant such that the term $(\partial_0\phi)^2$ appears with a positive sign. (And often with ...


2

No, the Lagrangian density is different: $$ \mathcal{L} = \pm \frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi. $$ The Hamiltonian density is actually the same in both conventions. However, this has no physical meaning. The choice of the signature is purely conventional.


1

While neither the Lagrangian $\mathcal{L}$ nor the action $S$ are invariant under boosts of the form $$\dot{q}(t) \to \dot{q}(t) + v, \quad v \in \mathbb{R},$$ the Euler-Lagrange equations are. The dynamics of the systems are unchanged for any transformation that preserves $\delta S = 0$, i.e. a transformation of the form $$ \mathcal{L}(q, \dot{q}, t) \to ...


1

It sounds like you are interested in symplectic reduction procedures. On of these methods is that of Routh's procedure to eliminate cyclic variables using a Legendre transform to a reduced-variable Hamiltonian called a Routhian. Forming a variational approach may be difficult for some reduction procedures, however we can view conserved quantities as ...


0

Yes, it is, but it is not easy to know the invariant quantities beforehand, so the usual way to use it is to write down the Lagrangian in terms of any (non-necessarily invariant) quantities, and then use the principle of least action to find the invariant quantities, Nother currents, etc.


0

Each of the components of $A^i$ depends on $(x,y,z,t)$ thus you must use the chain rule and evaluate $(dA^i/dt) + (dA^i/dj * dj/dt)$ where the j's are summed over x, y, and z.


0

I'd say the problem doesn't make much sense as it stands. Without terms that include a time derivative, the system would just immediately jump the the configuration with the smallest $L$, as the speed with which it does so is not "punished" (does not increase the action). In other words, the solution with the smallest action is the one with ...


2

Hints: By mean/average Feynman means temporal mean/average defined as $$\tag{1} \langle f \rangle ~:=~\frac{ \int_{t_i}^{t_f}\! dt~ f(t)}{t_f-t_i} . $$ Inequality: The mean square is always greater than the square of the mean $$\tag{2}\langle f^2 \rangle ~\geq ~ \langle f \rangle^2.$$ There are several proofs of ineq. (2), e.g. the variance is always ...


2

First a couple general results: given a function $f(t)$ and an interval $T = [t_1,t_2]$, the square of the mean of $f$ on $T$ is $$\langle f\rangle_T^2 = \biggl(\frac{1}{t_2 - t_1}\int_{t_1}^{t_2}f(t)\,\mathrm{d}t\biggr)^2$$ and the mean of the square of $f$ on $T$ is $$\langle f^2\rangle_T = \frac{1}{t_2 - ...


0

Intuitively, I can offer two hopefully helpful ways to think about a purely time-dependent term. First, since such a term depends on neither positions nor velocities, it has no spatial gradient, so there is no corresponding force. Neither the potential nor the kinetic energy are changed if we displace the system or change its velocities. So if I may be ...


0

The volume form is $dt f(\theta,\phi)d\theta d\phi=(dt)(Rd\theta )(R\sin\theta d\phi).$ However, you already have that factor from $\sqrt{|h|}$ so your original equation might have meant to have the coordinate differential, not the 3-volume differential, in which case you should just use $d^3x=dtd\theta d\phi$ instead of the volume element $R^2\sin\theta ...


1

In the end this is physics, so maybe you should try it less formal? (and not confuse eulerian and langrangian methods) Initially the mass element at $a$ had mass ( I use $\delta$ to make clear that we really should consider finite differences and then perform a limit at the end) $$m = \rho(a,0) \delta a$$ now, we follow its motion and after some time t we ...


2

Classical Lagrangian field theory deals with fields $\phi: M \to N$, where $M$ is spacetime and $N$ is the target-space of the fields. We shall for convenience call $M$ and $N$ the horizontal and the vertical space, respectively. OP is in this terminology essentially asking Q: What is the meaning of horizontal transformations? A: It is a (horizontal) flow ...


1

Try to think of this problem using a polar coordinate system. $x$ is essentially the radius $r$ or $\rho$, measured from pivotal. $w$ is simply the angular velocity. So the position vector of the object is $x\hat{\vec r}+\theta\hat{\vec \theta}$ So the velocity vector is $\dot x\hat{\vec r}+w\hat{\vec \theta}$ The hatted vectors are unit. So the ...


3

It is well known that adding a total time derivative to the Lagrangian does not change equations of motion. The Lagrangian above adds a term $$-q\dot q=-\frac{1}{2}\frac{\mathrm{d}q^2}{\mathrm{d}t}$$ (a total time derivative) to the free particle Lagrangian $\dot q^2$. It is thus fully equivalent to to the standard free particle Lagrangian (up to an ...


1

\begin{equation} \mathcal{L}\left( q,\dot{q},t\right)= \dot{q}^{2} - q\dot{q} \tag{01} \end{equation} \begin{equation} \dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{q}}\right)-\dfrac{\partial \mathcal{L}}{\partial q}=0 \tag{02} \end{equation} \begin{equation} \dfrac{d}{dt}\left[\dfrac{\partial \left(\dot{q}^{2} - q\dot{q}\right)}{\partial ...


1

When you integrate the Lagrangian density over a certain region $\Omega$, this is in principle allowed to change and this gives you a "boundary" term in the variation. This is well discussed in, e.g., the book of Goldstein (3rd edition), where the correct proof of the Noether theorem is given.


0

You may be interpreting the explanation of the coordinates wrong. "The angle between the top of the bigger cylinder and the position of the smaller one" does not tell us where the center of the angle is measured from, but if it is measured from the center of the larger cylinder, which perhaps is itself at a height $y$, then the proper formula would be $m g ...


1

The Einstein-Hilbert lagrangian coupled to a matter action $$ S_m[\varphi,g] = \int d^Dx\, \sqrt{-g}\mathcal L_m(\varphi,\partial_\mu\varphi), $$ i.e. $$ S[g,\varphi] = \frac{1}{16\pi G}\int d^Dx\,\sqrt{-g} R + \int d^Dx\,\sqrt{-g} \mathcal L_m(\varphi,\partial_\mu\varphi), $$ satisfies $$ \delta S=-\frac{1}{16\pi G} \int d^Dx\, \sqrt{-g} \mathcal ...


1

May I ask what text you are reading? My understanding of the stress energy tensor is as follows. The Noether condition is written as,\begin{equation} \partial _\mu \bigg[\frac{\partial \mathcal L}{\partial (\partial _\mu \phi )}\delta \phi +\mathcal L \delta x^\mu\bigg]=0 \end{equation} In the discrete case we can imagine separate infinitesimal time and ...


1

Comment to the question (v2): The association (2) is not correct. To find the Hilbert SEM tensor, one varies the action wrt. the metric $g_{\mu\nu}$; not wrt. the gauge potential $A_{\mu}$ (or the field strength $F_{\mu\nu}$).


5

The energy momentum tensor is found by varying the metric and holding all other fields constant. Since clearly $$\frac{\partial F}{\partial g}=0\longleftrightarrow \delta_gF=0$$ we end up with $$\delta_g S=\frac{1}{2}\int\mathrm{d}v\,\left(F^2g_{\mu\nu}/4-F^\tau{}_\mu F_{\tau\nu}\right)\delta g^{\mu\nu}$$ and comparison with ...


0

In general you can write the kinetic energy of a free particle as: \begin{equation} T = \frac{1}{2} m \,\vec{v}\cdot\vec{v} \end{equation} which holds whatever coordinate system you choose (could a physical quantity such the trajectory of a particle depend on the coordinate system that you choose?). We can rewrite this: \begin{equation} T = \frac{1}{2} m ...


0

I'm not sure why you're talking about an $x$ and $y$ component of velocity when you're working in a polar coordinate system. Maybe you're confusing $x(t)$ (the extension of the spring as a function of time) with the Cartesian coordinate $x$. These are very different things. To understand what the radial component of the velocity is, assume the pendulum isn't ...


2

Velocities in the kinetic part of Lagrangian The variable $\;x\;$, that represents the displacement of the string from its position at rest, has been replaced by the variable $\;s\;$ in order not to be confused with the coordinate $\;x\;$ of a Cartesian system. The velocity of the particle $\:\mathbf{v}\:$ is analysed as follows \begin{equation} ...


1

I) The action principle of a theory is the usually taken as the first principle of a theory, and therefore it can strictly speaking not be derived. Nevertheless, the Nambu-Goto action is a natural a generalization of the following line of thought: In a Riemannian space $(M,g)$ [with Euclidean signature], a geodesics are (locally) the shortest path between ...


0

We are assuming that this is the case. In general, one can think that every Lagrangian corresponds to some equations of motion, so the NG action corresponds to some sort of motion of a string. The powerful justification for this action is that it is manifestly Lorentz covariant, and one can show that the equations of motion that it leads to are those one ...


1

The action principle holds by assumption. It is assumed that all equations of motion follow from this principle with the appropriate action. By introducing an auxiliary tensor field $h_{\alpha\beta}$, one may write down the so-called Polyakov action $$S_\mathrm{Poly}=-\frac{T}{2}\int_\Sigma ...


0

I'd like to point out that knowing $2n$ quantities and the equations of motion are not enough to determine the solution. Even at the level of $L=T-U$ for just one particle ($n=1$). Consider $T=\frac{1}{2}m\left(\frac{dx}{dt}\right)^2,$ and $U=-C\frac {9m}{2}x^{4/3}.$ Then, for your equations of motion, you get ...


3

Observe that, knowing $\ddot{q}$, to get $\dot{q}$ and then $q$ you have to integrate twice. This introduces $2n$ integration constants you have to know to fully describe the system, which is the same amount of freedom you get when solving the Euler-Lagrange equations, where you need initial conditions for $q$ and $\dot{q}$.


0

$\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$You can reduce the problem to a problem in classical physics without having to use the lagragian formalism. Going back to classical physics, you had $$-\nabla V = \begin{pmatrix} V_x \\ V_y \\ V_z \end{pmatrix} = \vec F = m \ddot {\vec x} \tag 1 \, ,$$ where $V_{x_i} = \partial V / \partial x_i$. If ...


0

@DanielSank has given the answer in the comments already! Your question was, where can you find the motivation to conclude that equilibrium points follow that particular equation. To start with, define equilibrium point as the coordinate where the generalized force is zero. The Euler-Lagrange equation is $$\frac{d}{dt}\left(\frac{\partial L}{\partial ...


0

I similarly find $$\frac {\partial \mathcal L}{\partial(\partial_\eta A_\phi)} = \lambda ~ g^{\eta\phi} (\partial_\mu A^\mu) - 2 ~\partial^\phi A^\eta .$$But now when you bring $\partial_\eta$ over this expression you have to not skip steps when you write out$$\partial_\eta~ \frac {\partial \mathcal L}{\partial(\partial_\eta A_\phi)} = \partial_\eta~(\lambda ...


1

First of all, you have some indices that randomly get lowered or raised. I'd advise you check for consistency. But mostly, it seems that you have the wrong definition for $\Box$. In your work, it seems that you think this is correct: $$\partial^\phi\partial_\eta A^\eta=\Box A^\eta$$ And that cannot be true, since your indices don't match. On your left side ...


3

It isn't necessary to introduce the effective potential in orbital mechanics but it is really useful. Let's say we have a particle moving in a central gravitational potential. Newton's laws give you a vector equation of motion \begin{equation} m \ddot{\vec{x}} = - \nabla U \end{equation} where $U = - G M m /r$. In a general coordinate system this is a ...


0

i only know the effective potential when talking about central forces situations in classical mechanics, there it is defined as: $V'(r) = V(r) + \frac{L^2}{2mr^2}$ with V(r) being the radial potential, the second term can be considered a centrifugal potential which results from considering the azimuthal part of the kinetic energy. $E = V + E_{kinetic} = ...


2

As to your question, yes, the QED Lagrangian is indeed invariant under charge conjugation. You may have found differently because your transformations under charge conjugation are faulty. The prefactors are correct, however, under charge conjugation $\psi$ goes to $\bar{\psi}$ and vice versa, i.e. $$ \hat{C} \, \psi \, \hat{C} = -i(\bar\psi \gamma^0 ...


1

The use of the Lagrangian density is a convenience, and it is not directly related to causality or relativity, and neither strictly to quantum theories. What I mean is that it is possible to formulate non-relativistic (quantum or classical) field theories using exactly the same language. The difference between "mechanics" and "field theory" is that, instead ...


2

Locality is a physical requirement we impose (for good reasons). Locality is implemented in the theory by using fields, with local interactions in a Lagrangian density (ie, the Lagrangian only depends on products of fields and derivatives at a single point). I would definitely not say that locality occurs because Lagrangian densities show up in field theory, ...


4

The difference is that in classical mechanics positions are exactly the fields you are looking at, whereas in general field theories they are the variables the actual fields depend on. In classical mechanics the solution of the dynamics is given by the knowledge of the position and the velocity $(q(t),\dot{q}(t))$ at any time $t$. Time plays the role of the ...


1

Tips: 1) Remember that $\mu$ and $\nu$ are dummy indices. It will be easier to see if you lower all indexes, but with pratice this won't be necessary anymore. 2) For terms like $(\partial_ \mu A^\mu)^2$, write them as $g^{\mu \nu} g^{\sigma \rho} (\partial_\mu A_\nu) (\partial_\sigma A_\rho$) and use Leibniz's rule . 3) For terms like $A^\mu A_\mu$ just ...


0

Building on the responses from ACuriousMind and Gennaro Tedesco, I will make an attempt to provide a satisfactory, though not mathematically rigorous, answer. Question: Does there exist a nontrival non-Legendre transformation T such that the function defined by F(q,p,t)=T[L(q,q˙,t)] contains the full dynamics of the system? Yes, any invertible ...


3

This is a particular example of a general theorem in effective field theory: if you have an operator that is proportional to the lowest order equations of motion, you can push that operator to higher order in perturbation theory by a field redefinition. This is especially useful if you are working to a fixed order in perturbation theory, in which case you ...


3

Question: Does there exist a nontrival non-Legendre transformation T such that the function defined by F(q,p,t)=T[L(q,q˙,t)] contains the full dynamics of the system? Answer: any function that produces the equations of motion under some sort of rules that you state is an allowed function to describe the dynamics. In particular any function that you can ...


1

Comment to the post (v2): It is true that if $t\mapsto q(t)$ is a solution to the Euler-Lagrange (EL) eqs. for the action $$S[q]~=~\int_{t_i}^{t_f} \! dt L(q^1,q^2,q^3, \ldots; (\dot{q}^1)^2,(\dot{q}^2)^2,(\dot{q}^3)^2, \ldots)$$ of a time-symmetric Lagrangian, then the time-reversed path $t\mapsto q(-t)$ would also be a solution to the EL eqs. In fact, ...


0

The only reason to define things as they are in classical mechanics is that they give rise to the correct equations of motions that can be directly measured and observed. Given a field theory described by $\phi(x)$, its action is defined as $$ S(\phi,\dot{\phi})=\int_{\mathcal{D}}d^4x\,\mathcal{L}(\phi(x),\dot{\phi}(x),x) $$ and the equations of motions for ...



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