# Tag Info

12

It is necessary to study Newtonian mechanics to truly understand Lagrangian mechanics since its underlying foundation is Newtonian mechanics. It is essentially a different formulation of the same thing. In a way when doing Lagrangian mechanics you are still doing Newtonian mechanics just in the way of energy. For example, under Lagrangian mechanics, say we ...

9

To apply Noether's theorem, which is what you are alluding to here, one needs to look at continuous symmetries of a Lagrangian description of a system's dynamics. The damped oscillation equation you have written, although it is invariant with respect to a time translation as you rightly say, is not a Lagrangian description. If you write the Lagrangian for ...

7

No, I would highly recommend studying Newtonian mechanics before Lagrangian mechanics. While, yes it is 'possible' to learn about Lagrangian mechanics before Newtonian, a lot of intuition would be lost beginning with one instead of the other which will, in the long run, do no more than harm you or, at best, possibly confuse you. But there are, indeed, many ...

6

Theorem: let $L$ be a homogeneous function of degree $k$; then the on-shell lagrangian is a total derivative. Proof: according to the Euler's homogeneous function theorem, $$k\ L(q,\dot q)=q\frac{\partial L}{\partial q}+\dot q\frac{\partial L}{\partial \dot q}\tag{1}$$ On the other hand, because of the Euler-Lagrange equations, $$(1)=q\frac{\partial ... 5 The generalised Lagrange equations are$$ \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial T}{\partial \dot{q}_j} - \frac{\partial T}{\partial q_j}=Q_j \tag{1} $$where T is the kinetic energy of the system and Q is the generalised force. This is the most general EoM, and is equivalent to Newton's F_j=m\ddot x_j. Now, if the generalised force can be ... 4 OK, as per your request…. My sense is you want to learn everything about integrability from here, and combine issues which confuses them, instead of separating them…. How about you supplement L&L with Arnold’s book? The seven additive integrals of L&L are the additive conservation laws of the isolated center of mass system, and standard center of ... 4 Actually, the metric variational definition for the stress-energy tensor (due to Hilbert, as remarked by Qmechanic) is an universal improvement procedure for the canonical stress-energy tensor (and hence not always concides with the latter), in a sense which will be made precise below. Such a procedure is necessary because the canonical stress-energy tensor, ... 4 Conservation of energy is related to time translation invariance for systems that can be described by a Lagrangian. Dissipative systems in general are not describable by Lagrangian mechanics (without altering the formalism that is) and so Noether's theorem cannot be applied to check whether or not energy is conserved. EDIT: the dissipative system the OP ... 3 It would be easier to answer your question clearly with a drawing. In the following, the angle coordinate of the pendulum is the angle it makes with the vertical line. When the pendulum swings right(left), the angle will be positive(negative). With this setting, I get the exact same answer as you by working out the equations of motion. However, there ... 3 The question that you have asked have some vague arguments as well as some partially true facts regarding Standard Model (SM). First, Yes SM describes physics up to some energy scale which is 14 TeV. On the other hand, if we accept Plank energy (~10^{18}GeV) as a fundamental energy scale, then we can possibly expect new beyond the SM energy scale. A ... 3 The conditions about (i) differentiability of the functions and (ii) the maximal rank of the corresponding rectangular Jacobian matrix are regularization conditions imposed to simplify the mathematical analysis of the physical problem, in particular to legitimate the possible future use of the inverse function theorem. In the affirmative case, the ... 3 You should study Newtonian mechanics before Lagrangian mechanics because Newtonian mechanics is more general than Lagrangian mechanics. In other words, while whenever a system allows a Lagrangian formulation it also allows a Newtonian formulation, the converse is not true; the quintessential case is dynamics in the presence of dissipative forces. Lagrangian ... 3 There are no contradiction at all and your questions can be answered at once. An isolated system with s degrees of freedom has 2s-1 integrals of motion since the solutions for the coordinates q_i involve 2s-1 constants (determined by the initial conditions). The 2sth can be solved in terms of t-t_0 where t_0 can be arbitrarily chosen. However ... 2 Yes, there is systematic way called the Noether procedure. Simply you write down all possible 2-derivative fermionic terms with arbitrary coefficients and vary the action using the SUSY transformation rules. Then, you fix the coefficients to obtain the invariance up to a total derivative. When you have 4-derivatives, there are two cases: a. Off-Shell ... 2$$ F^2 \to F^2 + \alpha_i f^{ijk} F_{j\mu\nu} F_k^{\mu\nu} $$f^{ijk} is completely antisymmetric in all its indices and is being contracted with something that is symmetric in jk. Thus, the action is invariant. We can see that f^{ijk} is totally antisymmetric as follows$$ [T^i , T^j] = f^{ij}{}_k T^k \implies \text{tr} \left( [ T^i , T^j ] T^k ...

2

The virial theorem is valid if few simple conditions are satisfied (system remains bound to finite region) and is thus quite general. Ergodic theorem requires evolution to be ergodic, i.e. evolution of state must be ergodic transformation: http://encyclopedia2.thefreedictionary.com/ergodic+transformation This is a special requirement that is not ...

2

As already mentioned, the action corresponds to massive electrondynamics, including external sources, in Minkowski spacetime. This is also known under the term Proca action. As you mention, the corresponding equations of motion can be found using Euler-Lagrange equation, that is $$0 = \frac{\partial \mathcal{L}}{\partial A_\mu} - \partial_\nu \frac{\partial ... 2 OP is observing that in Minkowski space g_{\mu\nu}=\eta_{\mu\nu}, it doesn't matter whether we write$${\cal L} ~=~\sqrt{|g|}\partial\phi\partial\bar{\phi} \tag{1} $$or$${\cal L}~=~-\sqrt{|g|}\phi\Box\bar{\phi}\tag{2} $$for the Lagrangian density, if we don't care about total divergence terms. OP is pondering what happens in curved spacetime (M,g)? ... 2 The covariant divergence of a vector is$$\nabla_\mu V^\mu = \frac{\partial_\mu (V^\mu \sqrt{-g})}{\sqrt{-g}}$$Meaning that adding a covariant divergence to the Lagrangian will result in the following change :$$\Delta S = \int d^4x \sqrt{-g} \nabla_\mu V^\mu = \int d^4x \partial_\mu (V^\mu \sqrt{-g})$$which is once again easy to see that it vanishes ... 2 Well, you cannot take any ol' matter theory in flat Minkowski space and stick in a curved metric tensor g_{\mu\nu} in the matter action as you like, if that's what you're implying. The caveat is that the resulting matter action S_{\rm m}[\Phi, g] should be a general relativistic diffeomorphism-invariant functional. Then the Hilbert stress-energy-momentum ... 2 It depends on what you mean by vacuum. If you mean a field configuration that has F=0, then the gauge potential is locally pure gauge (cf. this answer by Qmechanic), so there can only be global obstructions to the gauge equivalence class of the A corresponding to F=0 being A=0 everywhere. On \mathbb{R}^{1,3}, there is no such obstruction. If ... 2 There are two kinds of derivatives we should differentiate:$$ \frac{\mathrm d\mathcal L}{\mathrm dx}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x+h),\phi'(x+h),x+h)-\mathcal L(\phi(x),\phi'(x),x)\big]\tag{1} $$and$$ \frac{\partial\mathcal L}{\partial x}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x),\phi'(x),x+h)-\mathcal ...

2

Consider a smooth 2D sphere and a point of positive mass constrained to move on that without friction. The only force is the reactive force normal to the sphere. The trajectories of the motions of the point are geodesics of the spherical surface. Therefore if you fix the north and south poles as boundary conditions you find infinitely many solutions ...

2

I) When Ref. 1 writes let there be $\ell$ equations (17'') of constraints, it is implicitly assumed that they are independent, as also noted in Peter Diehr's answer. Obviously this implies that $\ell\leq 3N$. Moreover, the rectangular $\ell \times 3N$ matrix $$\tag{A} \left( \frac{\partial f_k}{\partial x_i}\right)_{1\leq k\leq \ell, ~1\leq i \leq ... 2 There is a well-known isomorphism between the linear space {\mathcal M}_{m, n} of m\times n matrices and typical (vectorial) linear spaces {\mathcal L}_{m\times n} of dimension \text{dim}({\mathcal L}_{m\times n}) = m\times n. Everything that is valid in {\mathcal L}_{m\times n} has an equivalent in {\mathcal M}_{m, n} and conversely. For this ... 1 Everything is a function of the angle \theta and its derivatives \dot{\theta} and \ddot{\theta}. From there use the chain rule of differentiation.$$\begin{align} x & = \ell \sin \theta & y & = \ell (1-\cos \theta) \\ \dot{x} & = \ell \dot{\theta} \cos \theta & y & = \ell \dot{\theta} \sin \theta \\ \ddot{x} & = ...

1

As you noticed, if we use Euler-Lagrange equation on $L= \frac 1 2 (\dot x^2 + \dot y^2) -mgy$ we get $$\ddot x=0$$ $$\ddot y = -g$$ Something is clearly missing: gravity is not the only force acting on our mass: we have to take into account the tension of the rod/string. But why doesn't it come out from the equations? The point is that system only has ...

1

The equations of constraint must be independent of each other - otherwise they would represent the same constraint; this independence is what guarantees that Lagrange's theorem is satisfied. Joos' method is correct if the conditions are met. An Introduction to Lagrange Multipliers provides a geometric analysis which is often helpful in understanding ...

1

We do not start from the gauge fixed path integral in the BRST construction. What you describe (once one adds the missing Faddeev-Popov determinant) is the original Faddeev-Popov trick to get the ghosts, not the systematic BRST construction. The (Hamiltonian) BRST construction crucially first introduces the ghosts as parts of the extended phase space, and ...

1

The BRST symmetry cannot be seen without introducing auxiliary variables. The fastest way to realize the BRST symmetry is to "exponentiate" the delta function $$\delta(G)~=~\int \!{\cal D} B ~\exp\left[iB_{\alpha}G^{\alpha}\right]$$ and the Faddeev-Popov (FP) determinant \det\Delta ~=~\int \!{\cal D} c ~{\cal D} \bar{c} ~\exp\left[\bar{c}_{\alpha} ...

Only top voted, non community-wiki answers of a minimum length are eligible