Hot answers tagged

9

Translational symmetry in the sense of the standard formulation of Noether theorems means that the Lagrangian is invariant under the action of the group of spatial translations. This is not the case in your example because $U$ does not admit such invariance. However there is another, more physical, version of the idea of translational invariance for a ...


9

Indeed the problem with boundary conditions, generally speaking, is not well-posed. There are boundary conditions admitting no curves or admitting many curves, satisfying both these conditions and Euler-Lagrange equations. Examples. (1) Think of a particle constrained to stay on a smooth sphere where it can freely move. If you assign the North and the ...


4

You are asking four questions, whose answers are routinely provided in textbooks. We consider it in QCD since there is no fundamental reason to exclude it, and topological configurations such as instantons, etc.. might well generate it in an effective low energy theory: the rule of thumb is that anything that is not prohibited has to emerge out of the ...


4

All self-adjoint differential equations are consistent with the Principle of Least Action. See p. 226 of Lanczos' Linear Differential Operators. Lanczos explains that those physical systems which exhibit no loss of energy automatically provide a scalar quantity which can be minimized/maximized. For more advanced cases see How do I show that there exists ...


4

First of all I guess that what you wrote is the Hamiltonian and not the Lagrangian of the system and $\dot{x}$ stays for $p_x$ and $\dot{y}$ stays for $p_y$. You can decouple the problem redefining $$(X,Y)^t = R(x,y)^t$$ for a suitable $R\in O(2)$ diagonalizing the symmetric matrix in the potential part of your Hamiltonian. This way you see the final ...


3

I) Let the Lagrangian be $$\tag{1} L~=~\frac{m}{2}v^2-U(x), \qquad v~:=~\dot{x}.$$ Let the force $$\tag{2} F~=~-U'(x) $$ be a constant. II) Infinitesimal translations $$\tag{3} \delta x~=~\varepsilon $$ is a quasi-symmetry $$\tag{4} \delta L ~=~\varepsilon \frac{df}{dt}, \qquad f~:=~Ft $$ of the Lagrangian (1). Here $\varepsilon$ is an ...


3

Going the way stated in the question's title is easy: The Euler-Lagrange condition is, inherently, a condition on the action -- the statement is that the classical path is the path for which the action takes a minimum value for the path. Since this is a statement about the value of the action, and the action is Lorentz-invariant, then this minimum value is ...


3

Going from action to EOMs is simple: it is just (functional) differentiation. Going the other way from EOMs to the action is hard: It is (functional) integration, and sometimes impossible! OP is now essentially asking: Can we integrate one more time? Well, not the action itself. But if we replace the EOMs and the Lagrangian $L$ with their dynamical ...


3

The issue is that the underlying classical physics is determined by equations of motion (EOMs) (i.e. Newton's 2nd law), which are common for initial value problems (IVPs) and boundary value problems (BVPs). For BVPs , the EOMs can often alternatively be formulated as Euler-Lagrange (EL) equations of a stationary action principle. The latter approach does ...


3

123hoedjevan gives you a wrong answer. The principle of least action states that the physical configuration of the system of fields realizes a minimum of the action with respect to compactly supported variations of the fields which, by the very definition of compactly supportedness, must then vanish on the boundary of the support itself. This in turn means ...


2

I assume that by "potential $V$ of the scalar field" is meant "everything that stands on the right of $\partial \overline \phi \partial \phi$ in the scalar field's lagrangian", e.g. $V=-m^2 \overline \phi \phi$ for the complex KG field. Assuming that $V$ is invariant under $U(1)$ gauge transformations, you can introduce a covariant derivative in the usual ...


2

Comments to the question (v2): OP is considering the higher-derivative Lagrangian density $$ {\cal L}_1~=~ \frac{1}{2}(\partial\phi)^2 +\frac{g\phi}{2} (\partial^2\phi)^2,\tag{1} $$ where $g$ is a coupling constant. We use Minkowski sign convention $(+,-,-,-)$. Quantum mechanically, the model is not unitary and therefore ill-defined, cf. the Ostrogradsky ...


2

Comments to the question (v2): On one hand, the Kuramoto-Sivashinsky (KS) equation is a dissipative differential equation (DE). Each term has an even number of spatial derivatives. It's a non-linear version of the heat equation. Dissipative systems rarely have variational action formulations nor Hamiltonian formulations. On the other hand, in the Korteweg ...


2

Yes. You are correct. A non-relativistic theory would be invariant under the Galilean group. Lorentz invariance (specifically, invariance under Lorentz boosts) is what defines a relativistic theory.


2

The Noether charge is the generator of the symmetry it belongs to, see e.g. this answer by Qmechanic. This relationship is also preserved in the quantum theory, see this question, in the sense that the quantum Noether charge $Q$ must commute with the Hamiltonian $H$, at least in the absence of anomalies and if we do not run into "quantization issues" when ...


2

Supersymmetry generators are not always Hermitian. If you impose SUSY, and then compute de corresponding Noether's currents, and then you calculate the conserved charge, i.e., the fermionic Lorentz generators, you will get two non-Hermitian conserved currents. (By the way, the relation $Q^\dagger=\bar Q$ is only valid in Lorentzian signature, in Euclidean ...


2

I) Assuming that the variational problem for the action $S=\int \! d^nx~{\cal L}$ is well-posed (with appropriate boundary conditions), the field-theoretic Euler-Lagrange (EL) equations read in general $$\tag{1} 0~\approx~\frac{\delta S}{\delta \phi^{\alpha}} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} -\sum_{\mu} \frac{d}{dx^{\mu}} \frac{\partial ...


2

The Lagrangian is \begin{equation} L = \frac{1}{2} \dot{x}^2 - ax-b. \end{equation} Introducing spatial translation $x \rightarrow x+\Delta$ for constant $\Delta$ we see that \begin{equation} L \rightarrow L' = \frac{1}{2} \dot{x}^2 - ax - a\Delta -b. \end{equation} Therefore the action changes as \begin{equation} \delta S = \int{dxdt \; (L'-L)} = ...


2

Noether's theorem tells us that a conserved quantity is related to a symmetry of the action, where the action $S$ is given by: $$ S = \int L dt $$ where $L$ is the Lagrangian given by: $$ L = T - V $$ Since the potential $V$ is a function of position the Lagrangian and hence the action is not symmetric under displacements in space.


2

$1$. Instead of using the standard Dirac Lagrangian, I'd use the hermitian version $$ L=\frac{i}{2}\bar\psi \not\partial\psi-\frac{i}{2}\partial_\mu\bar\psi\gamma^\mu\psi-\bar\psi m\psi $$ which differs from the standard Lagrangian by a total derivative. It is just a matter of taste, but from this Lagrangian you get a hermitian $T_{\mu\nu}$ (and probably ...


2

The metric for $AdS_3$ is $$ds^2=\frac{1}{cos^2\rho}(dt^2-d\rho^2-sin^2\rho d\theta^2)$$, because $d=2$ is $AdS_3$. So $$g=\frac{1}{cos^2\rho}\times\frac{-1}{cos^2\rho}\times\frac{-sin^2\rho}{cos^2\rho}=\frac{sin^2\rho}{cos^6\rho}.$$ That's why in the mass term there is an extra $\frac{1}{cos^2\rho}$. And I think there is a typo in the expression of the ...


1

For example, if the first particle is moving on a spring and its position sets the electric potential that controls the second, electrically charged, particle. This way you'd have the potential energy of the coupling in the form of the product of coordinates.


1

@Andrew answered the question in a comment. Here a summary: $\sigma$ is a real field while the initial field was complex. Therefore to obtain the canonical propagator a factor of 1/2 is in fact necessary in the kintetic term.


1

The Lagrangian is in fact an equation of $\vec \Omega$, however, in general it will be a quadratic function of $\vec \Omega$, as the rotational kinetic energy would be given by $$\frac{1}{2}\vec \Omega ^T \mathbf{I}\ \vec \Omega$$ This would give you the desired generalized momenta as a function of the general velocity vectors, as the diagonal entries of the ...


1

You cannot just substitute fermions for bosons to determine the SUSY equivalent of a given interaction. For example, the kinetic terms for the bosons and fermions look quite different--first order in derivatives for the fermions, versus second order for bosons. As another example, the SUSY equivalent of a four-boson interaction is a Yukawa interaction ( a ...


1

And regarding why it's called a "free" theory, it's not specific to a momentum-space formulation. It's "free" because the Lagrangian is quadratic in the fields, and therefore the equations of motion (what you get from plugging the Lagrangian into the Euler-Lagrange equation) are linear in the fields. Therefore you can superpose different classical ...



Only top voted, non community-wiki answers of a minimum length are eligible