# Tag Info

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There are several reasons for using the Hamiltonian formalism: 1) Statistical physics. The standard thermal states weight pure states according to $Prob(state) \propto e^{-H(state)/k_BT}$. So you need to understand Hamiltonians to do stat mech in real generality. 2) Geometrical prettiness. Hamilton's equations say that flowing in time is equivalent to ...

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Some more comments to add to user1504's response: For a system with configuration space of dimension $n$, Hamilton's equations are a set of $2n$, coupled, first-order ODEs while the Euler-Lagrange equations are a set of $n$ uncoupled, second-order ODEs. In a given problem it might be easier to solve the first order Hamilton's equations (although sadly, I ...

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In a (classical) lagrangian field theory, the configuration space $\mathcal C$ of the system is a space of field configurations. A field configuration (or just "field" for short) is usually taken to be a function $\phi:\mathcal M\to T$ where $M$ is a manifold and $T$ is some set, often a manifold or vector space or both, called the target space of the ...

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I) A first integral to a system of second-order ODEs $$\tag{1} f^i(\ddot{q},\dot{q},q,t)~=~0$$ is usually an equation of the form $$\tag{2} g(\dot{q},q,t)~=~0,$$ that (when differentiated wrt. the parameter $t$) produces an equation $$\tag{3} \left(\frac{\partial }{\partial t} +\dot{q}^i\frac{\partial }{\partial q^i} +\ddot{q}^i\frac{\partial ... 5 First of all, Lagrangian is a mathematical quantity which has no physical meaning but Hamiltonian is physical (for example, it is total energy of the system, in some case) and all quantities in Hamiltonian mechanics has physical meanings which makes easier to have physical intuition. In Hamiltonian mechanics you have canonical transformations which allows ... 5 Just a pair of remarks. The second is the most interesting, in my view. (1) The Lagrangian of a charged particle in an assigned electromagnetic field still has a Lagrangian {\cal L}= T-U, but here U is not a standard position-depending function, since it generally depends also on \dot{q} and t as is well known (see Jackson's textbook, for ... 4 It's because they're based on the historical approach: Schroedinger's equation. Schroedinger's equation was discovered on its own before we knew about canonical quantization. Dirac came up with the canonical quantization rules which re-wrote (and generalized) Schroedinger's equation into the familiar one we have today, \hat{H} \psi = i \dot{\psi}. That ... 3 It seems that OP is pondering the following. What happens in a field theory [in OP's case: GR] if spacetime M has a non-empty boundary \partial M\neq \emptyset, and we don't impose pertinent (e.g. Dirichlet) boundary conditions (BC) on the fields \phi^{\alpha}(x) [in OP's case: the metric tensor g_{\mu\nu}(x)]? I) Firstly, it should stressed ... 3 Basically, the multiplier method is a way to encode the constraint information of the system directly into the Lagrangian so that you don't have to worry about screwing up the physical requirements of the problem when you solve the equations of motion. In other words, instead of solving the equations of motion and constraining the results, you're ... 3 Your question actually is one of the most important questions in analytic mechanics. This is because, when you explicitly write the Eulero-Lagrange equations for any constrained system with n degrees of freedom and Lagrangian of the form:$$L(t, {\bf q},\dot{\bf q}) = T(t, {\bf q},\dot{\bf q}) - U(t, {\bf q},\dot{\bf q})$$where T is quadratic in ... 2 I assume the case that you can write L=T-U has the structure$$ L=T(\dot{q})-U(q) $$with T(\dot{q}) as kinetic energy depending on momentum/velocity \dot{q}, and U({q}) as potential energy depending on coordinates {q}. 2+1D Chern-Simons theory is an example which cannot be written in this form. For non-Abelian Chern-Simons has the action$$ ...

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Also note that there are many examples in fluid mechanics where $L\neq T-V$. Particularly when the Eulerian reference frame is used. For instance, for irrotational deep water surface gravity waves, the Lagrangian is written as $L = \int \left(\int_{-\infty}^{\eta} \phi_t +\frac{1}{2}(\nabla \phi)^2 \ dz \right)+ \frac{g}{2}\eta^2 \ dx$ where $\phi$ is the ...

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I don't have an answer for why there is no simple Lagrangian formulation, but I can explain some of why a Hamiltonian one is easy. Part of the way to go from Classical Mechanics to Quantum is by replacing Poisson brackets with commutators, and observables with operators on Hilbert space and their expectation values. So the equation $\frac{d}{dt} f(q, p, t) ... 2 The expression for$\delta S_m$that you're expecting holds provided the variation you are performing is the variation with respect to the inverse metric only; there should be no$\delta\varphi$terms. In other words; set$\delta\varphi = 0$, and you obtain the desired expression. See, for example, Carroll Spacetime and Geometry p.164, he does the same ... 2 I) At least three different quantities in physics are customary called an action and denoted with the letter$S$: The off-shell action$S[q;t_i,t_f]$, The (Dirichlet) on-shell action$S(q_f,t_f;q_i,t_i)$, and Hamilton's principal function$S(q,\alpha, t).$For their definitions and how they are related, see e.g. this Phys.SE answer. II) OP's ... 2 I) In case of a point particle with mass$m$(and no moment of inertia), the best one can do seems to be to model the friction/drag via a Rayleigh dissipation function${\cal F}(v^2)$with a friction/drag force $$\tag{1} {\bf F}_f~:=~-\frac{\partial {\cal F}(v^2)}{\partial {\bf v}} ~=~-2{\cal F}^{\prime}(v^2){\bf v},$$ i.e. the Lagrange equations read ... 2 Hints to the question (v1): Let us parametrize the problem wrt. an arbitrary world-line parameter$\tau$(which does not have to be the proper time). The Lagrange multiplier$\lambda=\lambda(\tau)$depends on$\tau$, but it does not depend on the canonical variables$x^{\mu}$and$p_{\mu}$. Similarly,$x^{\mu}$and$p_{\mu}$depend only on$\tau$. The ... 2 Let there be$n$coordinates$q^j$. Ref. 1 is discussing in Section 2.4 a type of non-holonomic constraints that is known as semi-holonomic constraints. However we interpret OP's question (v2) as mostly being about counting independent degrees of freedom in constrained systems, and not so much about semi-holonomic constraints per se. Therefore, to gain ... 1 There are two ways to deal with a linear term in$\phi$: Complete the square, as was suggested in the comments. This is very often possible, but sometimes you do not want to do that. Interpret it as an interaction term with a$\phi$particle popping out of the vacuum or vanishing. This will lead to non-zero tadpoles in your Feynman diagrams, so additional ... 1 One way to see the relationship of Hamilotian classical mechanics and Quantum mechanics is not to look for a direct translation of Hamiltionian -> Quantum Hamiltionian (which exists: Geometric Quantization), but consider the reverse relationship. Given a Hamiltion operator and evaluating it on wave functions of the form$e^{\frac{i}{\hbar} \phi}$(which can ... 1 The canonical (Hamiltonian) formalism offers one of the main paths for quantizing gravity. General Relativity can be expressed in terms of the ADM 3+1 decomposition of spacetime: http://en.wikipedia.org/wiki/ADM_formalism And Hamiltonian's underlie quantum mechanics: http://en.wikipedia.org/wiki/Hamiltonian_(quantum_mechanics) Not only does this provide ... 1 I think it's misleading for the text to refer to the four-velocity identity as just another first integral of the equations of motion. Instead, I'll argue that, in this context, this identity is better thought of as an initial condition. The same issue arises in special relativity, so I'll first stay in flat-land. There the Euler-Lagrange equations of ... 1 Variations of the action must be performed with respect the field of which you want to get the equations of motion. You have varied with respect the metric field and the$\phi$field at the same time. What have you done is not strictly incorrect, since the variations are independent, but led you in confusion. In fact, you are not sure how to behave. However, ... 1 For instance, a Lagrangian$L = \partial_i \phi \partial^i \phi + m^2\phi^2$has the same equation of movement that the Lagrangian$L' = \partial_i \phi \partial^i \phi + m^2(F\phi - \frac{F^2}{2})$. The Euler-Lagrange equation for$L'$simply give$\Box \phi +m^2F=0$and$ F = \phi$, so we have$\Box \phi +m^2\phi=0$, which are the Euler-Lagrange ... 1 The Euler Lagrange equations just give the differential equations that determine the motion of the object. The end points are boundary conditions for the differential equation. The differential equation which the brachistochrone curve satisfies will have its constants fixed so that it reduces to a line when you give it the boundary conditions of the object ... 1 First of all, the Pauli matrices are not space-time dependent, so of course you can pass the derivative right through them. Second of all,$\operatorname{Tr} [\partial(\vec{\pi}\cdot\vec{\tau})]^2 = \operatorname{Tr}\partial_\mu \pi^i \partial^\mu \pi^j \tau^i \tau^j $Now remember$\tau^i \tau^j = i \epsilon_{ijk} \tau^k + \delta^{ij} I_{2x2}$So ... 1 I can answer (1) and (2). The answer is: NO. Passing form classical mechanics to quantum one requires, in general, to add more information. There is no rigorous machinery allowing one to write the quantum corresponding of a classical object. Physically speaking, this is because quantum structures are more fundamental in Nature than classical ones. ... 1 The problem is that you assume the system is in equilibrium in your first line. Apparently the pendulum is not in equilibrium if the dot product of gravity and motion is not zero. But actually d'alembert principle states the following for general cases, $$\sideset{}{}\sum_{i}(F_i-\dot{p})\cdot \delta x_i=0$$ So we have ... 1 Another example, expanding on Qmechanic answer, can be the 2D harmonic oscillator, with the Lagrangian:$L = m\dot{q}_1\dot{q}_2 - m\omega^2q_1q_2$this lagrangian has the same EoM than the usual standard harmonic oscillator, but it si quite diferent, the Noether theorem makes a big mess of the usual symmetries and the conserved quantities, for example the ... 1 I) If you just want a simple example, here is an example of a free point particle in two dimensions [1] $$\tag{A} L~=~m\dot{x}\dot{y}.$$ This Lagrangian (A) is different from the kinetic energy & standard Lagrangian $$\tag{B} L_0~=~T~=~\frac{m}{2}(\dot{x}^2+\dot{y}^2).$$ Yet the Euler-Lagrange equations are the same:$\$\tag{C} ...

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