# Tag Info

## New answers tagged lagrangian-formalism

0

No, the principle of least action started well before quantum mechanics. It is a variational principle that, when applied to the action of a mechanical system, can be used to obtain the equations of motion for that system. The principle led to the development of the Lagrangian and Hamiltonian formulations of classical mechanics. this new formulation of ...

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Comments to the question (v4): The question formulation seems to talk about affine parametrizations before applying the principle of stationary action. In the context of Riemannian$^1$ geometry, an affine parametrization of a (not necessarily geodesic) curve means by definition that the arc-length $s$ and the curve parameter $\lambda$ are affinely related ...

1

What Qmechanic said in comments is pretty solid, "Lagrangian (2) is not bounded from below because the kinetic term of $A_0$ field has the wrong sign, and hence the theory is not physical in the first place", but I think your Question needs a change of emphasis. Your Lagrangian allows us to construct four equations of motion for four non-interacting fields. ...

0

After thinking about this a bit I realize the question was due to a rather silly misconception,the converse was not mentioned in the usual references because it's trivial from the proof of the usual direction. For future reference I will include a quick version here. $$F = L^2/2$$ EL[F] \equiv \frac{d}{ds}(\frac{\partial}{\partial \dot{x^\mu}}F)- ... 3 Orientation. Let a lagrangian L that is a local function of position and velocity be given. For a parameterized path x on M, define a corresponding action S as follows: \begin{align} S[x] = \int_a^b ds\,L(x(s), \dot x(s)). \end{align} Let \delta x be a fixed-endpoint variation, then a standard computation shows that the corresponding variation ... 0 Field \psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} with a given spin and mass (i.e. field which transforms under irrep of the Poincare group) must satisfy some determined conditions called irreducibility conditions: \tag 1 \hat{W}^{2}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} = -m^{2}\frac{n + m}{2}\left(\frac{n + m}{2} + ...

0

Hmm, cool problem. At each collision with the wall, $v_{r}=\dot{r}$ flips sign but $\dot{\theta}$ stays the same, and thus so does the angular momentum $L$. No idea if this works, but it seems worth a shot - I was inspired by quantum mechanics. You basically have a particle in a well. So \begin{align} L &= T-V\\ &= \frac{1}{2}mv^{2} - V\\ &= ...

3

If there is no external force with explicit time dependence, then the harmonic oscillator contains no explicit time dependence. Then the system has time translation symmetry, i.e. the result can only depend on the difference $T =t_b-t_a$, not on $t_a$ and $t_b$ individually.

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The Lagrangian provided is Maxwell's Lagrangian, supplemented by a gauge fixing term: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2$$ The equations of motion are, $$\partial_\mu F^{\mu\nu} + \partial^\nu (\partial_\mu A^\mu) = \partial_\mu \partial^\mu A^\nu = 0$$ Instead of making a gauge fixing procedure a ...

4

The extra term, in general $$\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}-\frac{1}{2 \xi}(\partial_{\rho}A^{\rho})^2$$ is called gauge fixing term. This term is needed in order to be able to quantize the field $A_\mu$. Without this extra term the photon propagator is ill defined $$D^{\mu\nu}={-i\over k^2+i0}\left(g^{\mu\nu}\,+\,(\xi-1){k^\mu ... 4 Problem: Given Newton's second law$$\tag{1} m\ddot{q}^j~=~-\beta\dot{q}^j-\frac{\partial V(q,t)}{\partial q^j}, \qquad j~\in~\{1,\ldots, n\}, $$for a non-relativistic point particle in n dimensions, subjected to a friction force, and also subjected to various forces that have a total potential V(q,t), which may depend explicitly on time. I) ... 0 Why these terms vanish? If$$L = \frac{mv^2}{2} - V(q) = \frac{m(\dot q)^2}{2} - V(q)$$Then$$\frac{\partial L}{\partial \dot q} = m \dot q$$Why? Because V(q) is a function of q thus the partial derivative of V with respect to \dot q is zero. Similarly$$\frac{\partial L}{\partial q} = -\frac{\partial V}{\partial q}$$Why? Because ... 2 The first term vanishes because you have assumed that the potential is independent of \dot{q}=v (as you have said, V(q)). The second term vanishes because speed and possition are independent. 3 It's Lie theory. When we have got a continuous symmetry, the means the symmetry group is a Lie group G. Saying that the Lagrangian \mathcal{L} is invariant under G means that \rho(g)(\mathcal{L}) = \mathcal{L}\;\forall\; g \in G, where \rho is the representation the Lagrangian transforms in. The "expansion to first order" physicists so often do ... 3 Look more closely at what you write. I'm not going to write it with all the partial derivative and Lagrangians, because the confusions are not dependent on that: The requirement for spatial infinity arises because \int_\mathcal{M} \mathrm{d}\omega \neq \omega, but \int_\mathcal{M} \mathrm{d}\omega = \int_{\partial\mathcal{M}}\omega  (Stokes' theorem). ... 2 The canonical momentum p is just a conjugate variable of position in classical mechanics, for we have the relation p=\frac{\partial L}{\partial \dot{r}}. When making the transition to quantum mechanics: we need substitute p by an operator -ih\nabla in the Hamiltonian; similarly, we need substitute r by i\hbar \nabla_p in momentum representation. ... 1 The version with the 'i' is correct. It is used in all references I checked, including printed books from pre-web days. The term without the 'i' may have come from some author using uncommon conventions or definitions, perhaps defining \gamma^5 = \gamma^0 \gamma^1 \gamma^2 \gamma^3 instead of the widely used standard definition \gamma^5 = i\gamma^0 ... 2 Recall that magnetic energies add to the Lagrangian while electric energies subtract from the Lagrangian--this is easily proven looking at the Lorentz force in the Lagrangian formalism. That is to say, the Lagrangian should be defined as$$ \mathcal L=T-V+W_b-W_e $$where W_n are the magnetic (b) and electric (e) energies. Thus, your Lagrangian is$$ ...

1

When you write this: $$D_{t_2} S=-x_2^2\csc^2t_2.\tag6$$ You obviously mean total time derivative by $D_{t2}$. Which means: $D_{t2}S=dS/dt_2=\frac{\partial S}{\partial q}\frac{\partial q}{\partial t_2}+\frac{\partial S}{\partial t_2}\frac{\partial t_2}{\partial t_2}$ In our case: $q=x_2$ and you're missing this part: $\frac{\partial S}{\partial ... 2 An intuitive way to see why a term proportional to$F_{\mu\nu}F^{\mu\nu}$is natural is recognizing the fact that this term introduces dynamics for the gauge field. If you make the dependence on the gauge field explicit, you will see that it contains a properly normalized kinetic term, allowing us to treat it as a field propagating in spacetime. This, as you ... 0 In gravitation and in electromagnetism, the introduction of potentials greatly enhances our problem solving power because it is usually much easier to sum the potentials due to sources and apply relationships between the relevant fields and potentials, than it is to try and find the vector sums of the fields they produce. In gravitation this is quite obvious ... 6 In this answer we formally show that a (quasi)symmetry of an action implies a corresponding symmetry of its EOM$^{\dagger}$. The answer does not discuss form covariance of EOM. For further relations between symmetries of action, EOM, and solutions of EOM, see e.g. this Phys.SE post. Let us first recall the definition of a quasi-symmetry of the action ... 3 (This is a largely a response prompted by your comment.) You can get the answer by just remembering the commutation/anti-commutation properties of the$\gamma$matrices, and the fact that${\bar \psi} = \psi^{\dagger} \gamma^0$. To see the following, you would have to expand the exponential factor, up to linear order$e^{M} = I + M + \ldots$. (I'm not ... 3 Example: Consider an action functional $$\tag{1} S[q]~=~\int_{t_i}^{t_f} \! dt ~L, \qquad L~=~\frac{1}{2}m\dot{q}^2-V(q),$$ with Dirichlet boundary conditions (BC) $$\tag{2} q(t_i)~=~q_i \qquad \text{and}\qquad q(t_f)~=~q_f,$$ where the potential$V$has a repulsive pole $$\tag{3} V(q_0)~=~+\infty$$ at$q=q_0$. Then it is possible to choose a ... 3 In optics, you can take the example of a concave mirror : the optical path chosen by the light to join two fixed points A and B is a maximum. 2 If we solve equations of motion for a particle with mass$m=1$in some potential, e.g.$U=x^4-4x^3+4.5x^2$, fixing two points like$x(0)=0$and$x(1)=2.651$, we'll get infinite number of solutions, here're some of them: They differ by initial velocity. Now each of them satisfies equations of motion, but only one makes the action minimal, here's how$S$... 0 Thanks to Qmechanic's answer, I've understood that$\partial S/\partial x=p(t_2)$, while I was under the illusion that it would somehow equal$p(t)$. Now follows the finalization of my attempts. First, consider the second part of the expression inside the integral for$\partial S/\partial x$in the OP, $$A=\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot ... 1 Comments to the question (v1): The main point is that Ref. 1 is considering the (Dirichlet) on-shell action function$$ \tag{1} S(q_f,t_f;q_i,t_i)~:=~I[q_{\rm cl}], $$not the (off-shell) action functional$$\tag{2} I[q]~:=~ \left. \int_{t_i}^{t_f}\! dt \ L(q(t),v(t),t)\right|_{v(t)=\dot{q}(t)}. $$See e.g. this Phys.SE post. The Lagrangian ... 0 In the case of no potential, the speed of your particle is constant. Over any interval S = \tfrac{1}{2}mv^2 T where T is a length of time. Hamilton's Least Action principle should state:$$T_{AC} + T_{CB} > T_{AB}$$This will definitely be true since d = vt so we can multiple both sides by v:$$ \overline{AC} + \overline{CB} > ... 2 As said in this answer, velocity and position are not varied independently. Indeed, when deriving Euler-Lagrange equations, we explicitly use the fact that$\delta v=\frac d{dt}\delta x$. So, when I add the constraint$v_i=\frac{x_{i+1}-x_i}{\Delta t}$, specifying$x_1$and$x_n$remains the only additional thing to converge to the solution. For example, ... 1 Hint to the question (v2): For a velocity-dependent force${\bf F}$(such as e.g. the Lorentz force), the relationship between force${\bf F}$and potential$U$is $${\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}.$$ See e.g. Goldstein, Classical Mechanics, Chapter 1. See also e.g. this and this Phys.SE ... 5 The Lagrangian provided is given by, $$L=\sqrt{t+\dot{y}^2t}$$ Clearly,$\partial L/\partial y = 0\$. We compute the second term in the Euler-Lagrange equations: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} = \frac{d}{dt} \left( \frac{t\dot{y}}{\sqrt{t+\dot{y}^2 t}} \right) = \frac{t(\dot{y} + \dot{y}^3 + 2t\ddot{y})}{2[t(1+\dot{y}^2)]^{3/2}} = 0$$ ...

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Usually there's a great deal of overlap between the definitions of the momenta you've listed, so your confusion is understandable, but nonetheless there are cases (that I know of at least) where the distinction is more clearly enunciated: Momentum as known in Newtonian mechanics: The momentum is a vector quantity (its vectorial superposition for many ...

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