New answers tagged

0

Lagrangian mechanics can be derived directly from Newton's second law using only algebraic manipulation and a some calculus. This includes both the general form of the Euler-Lagrange equation and the specific form Langangian $L = T - V$. No assumptions of stationarity, use of the calculus of variations, or even any reference to the concept of action are ...


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 ...


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 ...


1

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.


0

Firstly, given a differentiable Lagrangian $L(q,\dot{q},t)$, we can always form the Lagrangian energy function $$\tag{1} h ~:=~\sum_ip_i \dot{q}^i-L ,\qquad p_i ~:=~\frac{\partial L }{\partial \dot{q}^i }. $$ Secondly, make the assumption that $$\tag{2} \text{The Lagrangian } L=L(q,\dot{q}) \text{ has no }{\it explicit} \text{ time dependence.} $$ ...


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

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

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 ...


0

You can "derive" the Lagrangian formulation from Shannon entropy arguing Liouville's theorem in reverse.


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 ...


0

Given a Lagrangian the states of the theory are given by the solutions to the linearised equations of motion, i.e. the quadratic part of the Lagrangian. Assuming that you have a standard kinetic term the solutions are plane waves $e^{i k x}$. Then an efficient way to find the Feynman rule for a vertex is to replace each field by its plane wave solutions. In ...


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 ...


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

If a constraint can be expressed as f(r1,r2,...t) =0 . position vectors are connected with time then the constraints are called holonomic ,but if they can not be expressed as above then the constraints are non-holonomic


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 ...


0

If you have a Lagrangian which is Lorentz and SUSY invariant, it means that when you perform a SUSY transformation the Lagrangian does not change (apart from surface terms). If it was Lorentz invariant before the transformation it would be so after. Otherwise your theory would not be SUSY invariant. You do not simply exchange fermions with bosons. You ...


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 ...


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

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 ...


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.


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 ...


0

Making use of Einstein's convention so that we sum implicitly over the repeated index in a single term, we have $\frac{\partial{L}}{\partial{A_{\nu}}}=\frac{\partial{L}}{\partial{A_{\mu}}}\frac{\partial{A_{\mu}}}{\partial{A_{\nu}}}$ directly by the chain rule, where the greek index goes from $0$ to $n$ ($n$ being the total number of variables). Now, 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 ...


0

The partial derivatives appear as a result of the chain rule for partial derivatives; this takes place implicitly when the variation is applied to the Lagrangian functional. The big O notation says that there are higher order terms which can be ignored in the variational limit. The web page shows a couple of ways to obtain the result. Euler used purely ...


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 ...


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 ...


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 ...


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 ...


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

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)} = ...


0

@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.


0

First of all, the Lagrangian is not unique, Multiplying by a constant will give the right Equations of motion (EOM's) when there are no constraint forces. In case of when there are constraints (holonomic) the variation of action would be $$\delta S=\delta \int_{t_1}^{t_2}L+\lambda f \space \mathrm dt = 0$$ and the new Lagrangian $L'$ would be ...



Top 50 recent answers are included