# Tag Info

## New answers tagged lagrangian-formalism

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Peter Kravchuk has already given a good answer. Here we will follow the programming hint given in the Exercise 1.6. How would one program this minimization problem? By discretization. So the positions ${\bf r}_n$ live on discrete times $$t_n~=~n\Delta t,\qquad\Delta t ~:=~\frac{T}{N},\qquad n\in\{0,\ldots,N\};$$ and velocities are discretized as e.g. ...

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Hints: The Majorana spinor is real. For instance $\bar{\psi}=\psi^T\rho^0$ without complex conjugation. The SUSY transformation $\delta{\cal L}$ of the Lagrangian density ${\cal L}$ does not have to vanish. It is enough if it is a total divergence. See the notion of quasi-symmetry, cf. e.g. this and this Phys.SE posts.

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If your action only depends on first time derivatives, it is then not required for the trajectory to have second time derivative -- i.e. an abrupt change in velocity does not by itself give a contribution to the action. In other words, there is no penalty for changing your speed instantaneously. It then means that you can ignore the boundary values for ...

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1) Configuration space is, in a sense, the possible "positions" of the mechanical system. The states of motion, eg. velocities/momenta are not part of the configuration space. The configuration space (especially when constraits are in the picture) is modelled as some real, $n$-dimensional differential manifold, which I'll denote as $\mathcal{C}$. The ...

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Gauss's principle of least constraint Principles of Least Action and of Least Constraint (a review paper by E.Ramm) If I remember correctly, this principle has been used to derive equations of motion for Gaussian isokinetic thermostat (i.e., a computational algorithm for maintaining a fixed temperature of the system). Please see, for example, Statistical ...

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Point in configuration space represents configuration of the system, i.e. positions of the constituent particles. Point in phase space represents state of the system, i.e. positions and velocities of the constituent particles together. No. Liouville's theorem has no simple analogue in the configuration space. Depends on what is the task at hand and what are ...

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You may know about it already, but you can find an excellent account of Lagrangian Mechanics on manifolds in the book Mathematical Methods of Classical Mechanics by V. Arnold. Also to specifically address your question: $L:TM \rightarrow \mathbb{R}$ so that $L$ is a 1-form; ie $L \in \Omega^1 M$ which is neccessary to integrate over a ...

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Your professor is telling you something that is absolutely fundamental to a proper understanding of relativity. Suppose we draw out the trajectory of some object on a space time graph, we may get something like this: The path traced out by the object(the blue curve) is called the world line. The length of the world line, $s$, is equal to $c\tau$, where ...

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The shortest world line (geodesic path) is given by the GR coordinate system. GR has $x_0$ (ct), $x_1$ (x), $x_2$ (y), $x_3$ (z). Often in GR, c is set to 1 (c = 1). Distance squared ($ds^2$) is given by $$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$. If one second passes, the shortest distance is to stay in the same spatial location, that is $$ds^2 = -dt^2$$ ...

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With a strong grasp of Lie Algebra and Calculus of variations, "Invariante Variationsprobleme" should provide all the foundation one needs to build Newtonian Mechanics (and so much more). The deeper reason that we use either of these formalism is that they agree with experiment; that either formalism predicts the other is far less valuable than that they ...

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Here is an outline of the reduction from the Nambu-Goto (NG) action to the light-cone (LC) formulation from a Hamiltonian perspective: The starting point is the Hamiltonian formulation of the NG string, cf. e.g. this Phys.SE post. The Hamiltonian density is of the form "Lagrange multipliers times constraints"$^1$ $${\cal H}~=~\lambda^{\alpha} ... 4 "The number of degrees of freedom can be defined as the MINIMUM number of independent coordinates that can specify the position of the system completely" (wikipedia) In your case the number is ONE, because you only need to know the position of the particle along the curve. It doesn't matter if the curve is not a line, or even contained on a plane, because ... 3 I'm not altogether sure what you are asking, but I suspect the following may help. To represent rotations, spins and vectors in SU(2) we work as follows. Rotations live in SU(2). Vectors (in the physicist's sense) live in the algebra \mathfrak{su}(2). The position vector (x,\,y,z) is:$$X =x\,\hat{s}_x+y\,\hat{s}_y+z\,\hat{s}_z = ...

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I) In this alternative answer we resolve the singular Hessian $H_{\mu\nu}$ of the Nambu-Goto string action by introducing two auxiliary variables from the onset, thereby indirectly showing that the Hessian $H_{\mu\nu}$ must have co-rank 2. The target space metric has $(-,+,\ldots,+)$ sign convention, and $c=1=\hbar$. Consider the extended Nambu-Goto ...

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As the counter example given by Herr_Mitesh shows it is not true and this is because the lagrangian is not uniquely determined. In physics sometimes you don't have to think like in mathematics and in this case you must content yourself thinking that if the lagrangian does not contain x as a variable that is enough for the condition of homogeneity to be ...

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The following might help: $H = \frac{1}{2}(mv^2 + kx^2) + \gamma mkvx$ decays exponentially with time along the solution of the damped system. Check by integrating $H$ with respect to $t$ and using the equations of the system. So the "energy" $H$ decays exponentially instead of remaining constant.

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Perhaps the most enlightening is just to show how it goes in OP's example. If the Lagrangian reads $${\cal L}_1(A,\phi)~:= ~{\cal F}(A)- Ay(\phi),\qquad F~=~{\cal F}^{\prime}(A),\tag{1}$$ then the eom for the "auxiliary" variable $A$ reads $$F(A)~\approx~ y(\phi) \qquad \Leftrightarrow\qquad A~\approx~ F^{-1}(y(\phi)),\tag{2}$$ where we have assume that ...

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If by 'equivalent' you mean equal, then no. They can clearly differ by a constant, but they moreover can differ by a total time derivative. So if two lagrangians $L_{1}$ and $L_{2}$ are such that $L_{1} - L_{2} = \frac{\mathrm{d} \Phi}{\mathrm{d}t}$ for some function $\Phi$, then they lead to the same equations of motion. You can find a proof of this in Jose ...

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