# Tag Info

## Hot answers tagged lagrangian-formalism

5

In physics there is no general criterion on how to write down suitable Lagrangians, rather than a posteriori check on the equations of motions: all the Lagrangians generating the same dynamics are equally correct. For example, as an exercise, you may try to write down all the possible Lagrangians giving you back $F_j = m \ddot{x}_j$. This said, to directly ...

4

Yes, there exists a Legendre transformation from $g(p)$ to $f(x)$: $$f(x)=p(x)x-g(p(x))$$ with $x=dg/dp$. Here the notation $p(x)$ means $p$ written in terms of $x$. In your case, the Hamiltonian is a function of $p$ and you are transforming it to a function of $\dot{q}$, so you must use Hamilton's equation to get the velocity: $$\dot{q}_i=\frac{\partial ... 3 As demonstrated in this paper, the trajectory can never maximise the action but can in fact lie on a saddle point in cases where the potential has the appropriate spatial variation (at least partially repulsive) and where the final state is taken sufficiently far 'downstream' (beyond what these authors call the 'kinetic focus'). 2 The main idea. It makes sense that the "redundant" constraint modifies the result because when you introduce the Lagrange multiplier, you not only restrict your attention to the paths that satisfy the constraint, but you also restrict your attention to variations that preserve the constraint. In more detail. For clarity, let \mathscr P be the set of all ... 2 The correct way is to define the reparametrization-invariant action$$ S[X] = \int d\tau \sqrt{g_{\mu \nu} (X(\tau)) \cdot \frac{dX^{\mu}}{d\tau} \frac{dX^{\nu}}{d\tau} }. $$Note that the choice of \tau is arbitrary. The system has a large group of gauge symmetries - those are reparametrizations of the worldline (different choices of \tau). One way ... 2 First of all, the hamiltonian contains the coordinates q_i and their momenta p_i. You have to calculate the velocities \dot{q}_i. For that, you'll need the Hamilton-Jacobi equations$$\dot{q}_i = \frac{\partial H}{\partial p_i}$$The Legendre transform, as noted in the comments, is involutive, so the lagrangian is just the Legendre transform of the ... 2 You can either vary the action directly, or apply the classical field theory Euler-Lagrange equations. The latter for a Lagrangian \mathcal{L}(\phi^{\alpha}, \partial_{\mu}\phi^{\alpha}) read$$\frac{\partial \mathcal{L}}{\partial \phi^{\alpha}} - \partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi^{\alpha})}\Big) = 0.$$(Note that ... 2 Imagine this situation: at time t=0, we have a infinite long straight wire with current zero, and a charged particle q with zero velocity. at time t=T, we make the current to be I, thus we have a  \mathbf{B} field, and  \mathbf{A} field. during this process,  \mathbf{A} is build up from zero to some value, therefore we have induced electric field ... 2 A Lagrangian can easily be written down for a relativistic particle in a curved spacetime (i.e., under the influence of gravity.) Specifically, the "action" is the proper time between two events along a particle's world-line, and the particle's trajectory will extremize the proper time between these events:$$ S = \tau = \int \sqrt{ - g_{\mu \nu} dx^\mu ...

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If physics isn't an issue, you can add arbitrarily many terms. Once the physics comes in though, you will encounter a few restrictions : As said by Gennaro, it is assumed that the Poincaré symmetry applies. Higher derivative terms (second derivatives and above) are generally bad news. They can cause vacuum instability (energies can be arbitrarily ...

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We must distinguish between the gauge group $G$, stereotypically the Lie group $\mathrm{SU}(N)$, and the group of gauge transformations $\mathcal{G}$, which are all $G$-valued smooth functions of spacetime. There is no issue if you only write down quantities that transform in proper representations of the group of gauge transformations $\mathcal{G}$. The ...

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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 ...

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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 ...

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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, ...

1

$\require{cancel}I)$OP's is considering Dirac fermions in a curved spacetime. OP's action has various shortcomings. The correct action reads$^1$ $$S~=~\int\!d^nx~ {\cal L}, \qquad {\cal L} ~=~e L, \qquad L~=~T-V,\qquad e~:=~\det(e^a{}_{\mu})~=~\sqrt{|g|},$$ $$T~=~\frac{i}{2} \bar{\psi} \stackrel{\leftrightarrow}{\cancel{\nabla}} \psi, \qquad V~=~ ... 1 I got the answer reading a book from E. Poisson, what I was doing was indeed wrong, you have to start with the induced metric given by$$ h_{ab}= g_{\mu\nu}e^{\mu}_a e^{\nu}_b $$where$$e^{\mu}_a=\frac{\partial x^{\mu}}{\partial y^a}are the tangent vectors to curves of the hypersurface. Then, you just replace g by h in the usual relation ... 1 You're correct. To find the equations of motion, we have: \begin{align*}c_i&=\frac{\partial L(v^2)}{\partial v_i}\\ &=L'(v^2) 2 v_i \end{align*} so that L'(v^2) v_i is constant for all of time. Firstly, you could imagine a world in which all paths {\bf x}(t) are valid mechanical paths. Then the Galilean transform of a valid mechanical path ... 1 Comments to the question (v9): If we ignore the overall normalization, then OP correctly applies the Dirac-Bergmann^1 method, which leads to second-class constraints.^2 Normally the Majorana Lagrangian (1) is defined with a factor \frac{1}{2} in front. Then there will be no factor \frac{1}{2} in the anti-commutator relation (9), see e.g. Ref. 2. ... 1 I am doing my Bachelors on something like that and we even chose to include double boundary terms in the variation. In addition to this paper, Variational principle and 1-point functions in 3-dimensional flat space Einstein gravity by Stephane Detournay et al., we get one further term:-\frac{(1-\alpha)}{16\pi G}\int_{\mathcal{\partial ...

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There is a small misprint in the third line (the 3rd term). If you try to differentiate the second part, you will get exactly part of the force: $\frac{d}{dt}\frac{\partial}{\partial \dot{r}}\left(\frac{1}{r}+\frac{\dot{r}^2}{c^2 r}\right)=\frac{d}{dt}\left(\frac{2\dot{r}^2}{c^2 r}\right)=\frac{2 \ddot{r}}{c^2 r}-\frac{2 \dot{r}^2}{c^2r^2}$ And also your ...

1

Your sign is wrong when computing $$\frac{\partial{(B^2)}}{\partial{(\partial_{y} A_x)}}.$$ The only term in $B^2$ that contains $\partial_{y} A_x$ is $(B_z)^2 = (\partial_{x}A_y - \partial_{y} A_x)^2 ,$ and clearly by the chain rule, $$\frac{\partial{(B_z)^2}}{\partial(\partial_{y}A_x)} = -2B_z$$ which disagrees with what you have by a sign. Fixing ...

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