# Tag Info

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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 ...

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Let us suppress time $t$ from the notation for simplicity. Let there be $n$ coordinates $q^j$ and $m$ (possibly non-holonomic) constraints $$\tag{A}f_{\alpha}(q)~=~0,$$ where $m\leq n$. Granted some regularity assumtions, we may in principle solve the $m$ constraints (A) locally so that the coordinates $$\tag{B}q^j~=~g^j(\xi, \varphi)$$ become ...

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This is a fact about the hamiltonian compared to the lagrangian which I find not trivial (and worth to keep in mind). Suppose that the lagrangian $L$ and hamiltonian $H$ are cyclic with respect to some coordinate $q_1$. Then we have a theorem (cfr. [1]): The evolution of the other coordinates $q_2,...,q_n$ is the one of a system with $n-2$ independent ...

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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 ...

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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 ...

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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 ...

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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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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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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 ...

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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 ...

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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 ... 0 If you do not yet master calculus, I recommend the book Zeldovich Ya.B., Yaglom I.M.: Higher Math for Beginners, Mir 1988 which is great. Then, you need to understand 1) ordinary differential equations of 1st and 2nd order and have some idea about 2) many-variable calculus and 3) linear operators and matrices. If you want to understand Hamilton's ... 0 In the UK, Lagrangian mechanics would normally be taught to first or second year undergraduate students who have a solid understanding of Newtonian dynamics and calculus with multiple variables. For an idea of the kind of texts you might need you could look at university syllabuses such as: University of Manchester: ... 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 ... 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, ... 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 ... 0 This approach to classical mechanics is named after Edward Routh (with ou rather than just u). In this approach the Legendre transform is only carried out for the cyclic coordinates. Goldstein covers it in section 8.3 (Routh's procedure). 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 ... 0 The geometric argument is clear: Consider a Lagrangian density {\cal L}=d_{\mu}F^{\mu} that is a total divergence. The action S[\phi] = \int \! d^dx~{\cal L} will then be a boundary integral, due to the divergence theorem. Therefore the corresponding variational/functional derivative,$$\tag{1} \frac{\delta S}{\delta\phi^{\alpha}(x)}$$which is an ... 0 A term \int\!d^4x\, \operatorname{Tr}\, F \wedge F can be added to a non-Abelian Yang-mills theory (it vanishes trivially for the Abelian case, because of the wedge), and it is a total derivative. This term doesn't influence the equations f motion. However, this is a topological charge that counts something akin to the "winding number" of the gauge field. ... 0 If you have a four divergence inside an integral over all of spacetime (which is what you get when you extremize the action), the result will be a term which will be some product of the field(s) and its/their derivatives, evaluated at the boundary of spacetime. Since we assume that all fields go to zero (sufficiently quickly, so that their derivatives also ... 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 ... 2 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) ... 0 I can think of several reasons for why using Hamiltonians is preferred, but the most important, I'ld say, is that you need to use path integral formalism in order to formulate (non relativistic) QM in terms of the Lagrangian, which, for an undergrad course, is a bit of an overkill. Also, many of the most renowned equations in QM like, say, the Schrödinger ... 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 ... 0 The cases with zero and nonzero angular momentum should be treated separately. If \theta ever crosses zero then the angular momentum at that time is zero. By the conservation law it means that the angular momentum vanishes at all times. This implies that \phi is constant and all its derivatives vanish. (That means that \ddot\phi\propto\dot ... 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 $$So we have ... 0 Yes, you can regard the expression$$-\frac12m^2\phi^2-\frac16R\phi^2$$as potential energy. Compare it to the harmonic oscillator: its potential energy is given by a term quadratic in position. The Ricci scalar can then be interpreted as simply contributing to the square of the mass of the scalar. To answer your question whether this is possible or not: ... 6 In a (classical) lagrangian field theory, the configuration space \mathcal C of the system is a space of field configurations. A field configuration (or just "field" for short) is usually taken to be a function \phi:\mathcal M\to T where M is a manifold and T is some set, often a manifold or vector space or both, called the target space of the ... 3 You can think of the Euler-Lagrange equations$$ \frac{\rm d}{\rm dt} \frac{\partial L}{\partial v} = \frac{\partial L}{\partial x} $$as a generalization of Newton's equations$$ \frac{\rm d}{\rm dt} p = F $$where momentum and force are functions on (velocity) phase space derived from a potential L via$$ {\rm d}L = p\,{\rm d}v + F\,{\rm d}x $$Now, if ... 0 Because it works. In a Newtonian system, if you take \mathcal L=T-V, and apply the Euler-Lagrange equations when your generalized coordinates are just the coordinates of the particle, you get$$\mathcal L=\sum \frac12 m\dot x_i^2- V(x_1,x_2,x_3,...))\therefore 0=\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial L}{\partial \dot ...

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We have to point out some simple remark of capital importance. First of all, the lagrangian is not defined as $L = T - V$. This turns out to be true only on a riemannian manifold; the fact this is almost always true in classical mechanics is an "accident" due to the postulates of classical (non relativistic) mechanics. For the most precise definition of the ...

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Well, it becomes a bit clearer when we see the final formulas of Ref. 1: $$\delta \langle a_f , t_f |a_i , t_i \rangle ~=~ \frac{i}{\hbar} \int_{t_i}^{t_f} \! dt \langle a_f , t_f | \delta L(t) |a_i , t_i \rangle \tag{7.126}$$  \delta^{\prime} \delta \langle a_f , t_f |a_i , t_i \rangle ~=~\frac{1}{2}\left(\frac{i}{\hbar}\right)^2 ...

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The Lagrangian for the system is whatever happens to give the right equations of motion when you apply the Euler-Lagrange equations, and for systems with conservative forces that is $T-V$. Though this isn't always the case, not even in classical physics. Even when dealing with electrodynamics the Lagrangian takes a different form.

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The Lagrangian is what it is. At root, it is a more fundamental concept than energy. After all, not all Lagrangian formulations predict paths that have conserved energy, but all systems with a conserved energy can be formulated with a Lagrangian.

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To answer your first question: Particles and fields are separate. Particles are the irreducible excitations of fields. You can only get particles after quantizing fields. However, you might often see people using particles without quantization of fields (in classical mechanics and GTR). You must understand that these are approximate models obtained by ...

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