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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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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, ...

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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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I'm not aware of a deeper explanation of action from a classical mechanics point of view. In the classical view, the action is just a functional of the trajectory which is minimized (or maximized) at the physical trajectory. From a quantum mechanical point of view, and particularly the path integral formulation, there is a more physical interpretation: the ...

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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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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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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 ...

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$\require{cancel}I)$OP's is considering Dirac fermions in a curved spacetime. OP's action has various shortcomings. The correct action reads$^1$ $$S~=~\int\!d^nx~ {\cal L}, \qquad {\cal L} ~=~e L, \qquad L~=~T-V,\qquad e~:=~\det(e^a{}_{\mu})~=~\sqrt{|g|},$$ $$T~=~\frac{i}{2} \bar{\psi} \stackrel{\leftrightarrow}{\cancel{\nabla}} \psi, \qquad V~=~ ... 2 A Lagrangian can easily be written down for a relativistic particle in a curved spacetime (i.e., under the influence of gravity.) Specifically, the "action" is the proper time between two events along a particle's world-line, and the particle's trajectory will extremize the proper time between these events:$$ S = \tau = \int \sqrt{ - g_{\mu \nu} dx^\mu ...

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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 ...

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The second formula (where an overall minus sign is missing) is obtained using the product rule and discarding total derivatives. On the level of the action, it's just integration by parts assuming that surface/boundary terms vanish with fields decaying sufficiently fast at infinity. Thus: $$L = -\frac{1}{2} (A^\mu g_{\mu\nu} \partial^2 A^\nu - A^\mu ... 0 Not to mention, there are cases where the local extremum of the action isn't a physically realiziable path. Consider the plane with all the points satisfying y > |x| removed. now, consider a start point of (-2,1), and an end point of (2,1) on this manifold, with the Lagrangian \frac{1}{2}m\left({\dot x}^{2} + {\dot y}^{2}\right). Clearly, the ... 3 As demonstrated in this paper, the trajectory can never maximise the action but can in fact lie on a saddle point in cases where the potential has the appropriate spatial variation (at least partially repulsive) and where the final state is taken sufficiently far 'downstream' (beyond what these authors call the 'kinetic focus'). 0 Rather than a book, consider the Susskind Physics lectures made available on Youtube as well as Stanford on iTunes via iTunes-University. You want to watch the latest version of the course lectures on Classical Mechanics. After a small bit of introductory material, Susskind gets right into the Lagrangian and Hamiltonian approach to mechanics using a simple ... 1 Your sign is wrong when computing$$\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$$which disagrees with what you have by a sign. Fixing ... 2 The main idea. It makes sense that the "redundant" constraint modifies the result because when you introduce the Lagrange multiplier, you not only restrict your attention to the paths that satisfy the constraint, but you also restrict your attention to variations that preserve the constraint. In more detail. For clarity, let \mathscr P be the set of all ... 0 Let us suppress explicit time dependence t from the notation in the following. Hamilton's eqs. are the Euler-Lagrange (EL) eqs. for the so-called Hamiltonian Lagrangian$$\tag{1} L_H(q,\dot{q},p)~:=~ p_i\dot{q}^i-H(q,p).$$In other words, the solutions to Hamilton's eqs. are stationary points for the Hamiltonian action$$\tag{2} S_H[q,p]~:=~\int \! ...

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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: $$\dot{q}_i=\frac{\partial ... 2 First of all, the hamiltonian contains the coordinates q_i and their momenta p_i. You have to calculate the velocities \dot{q}_i. For that, you'll need the Hamilton-Jacobi equations$$\dot{q}_i = \frac{\partial H}{\partial p_i}$$The Legendre transform, as noted in the comments, is involutive, so the lagrangian is just the Legendre transform of the ... 2 The correct way is to define the reparametrization-invariant action$$ 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 ... 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 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 ... 0 It follows from L being a function \propto\dot{x}^2. With this at hand, you are left with two choices: \left(\nabla_{\dot{x}}L\right)'\sim\left(\dot{x}\right)'=0 implies \dot{x}=\rm const. L=0 implies \dot{x}=0=\rm const. Either way, you get that the velocity is constant in time (for this particular, free-particle case). 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 ... 1 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 ... 0 as far as i know these two expressions are used synonymically. and your coordinates h_{\mu \nu}=g_{\mu \nu}-\sigma n_{\mu}n_{\nu} arent gaussian coordinates in general. gaussian are for example flrw metrics like ds^2=-dt^2+h_{ij}(t)dx^idx^j with spacelike i,j where there is no mixed time/space basis term in the second part of the line element, only a ... 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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In a coordinate system rotating at constant angular rate $\omega$, neither energy nor angular momentum are conserved and one has coriolis and centrifugal forces. The bead is forced outwards by the centrifugal force: the energy increases by the work done on the bead by the rotating system. In fact, since the Hamiltonian $$H=\frac{p^2}{2m} - ... 2 You can either vary the action directly, or apply the classical field theory Euler-Lagrange equations. The latter for a Lagrangian \mathcal{L}(\phi^{\alpha}, \partial_{\mu}\phi^{\alpha}) read$$\frac{\partial \mathcal{L}}{\partial \phi^{\alpha}} - \partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi^{\alpha})}\Big) = 0. (Note that ...

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Lots of questions involved here, but I don't think you will get a complete derivation of all that you list without some digging through textbooks and on the net, on your part. in many fields of physics, we usually begin with an ordinary formulation (e.g Newton's Laws in classical mechanics), and then move on the Lagrangian, then Hamiltonion, and finally the ...

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