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Liouville Arnol'd theorem state that given an integrable Hamiltonian system and denoting with $$M_a'=\{(q,p)\in\Gamma:f_i(q,p,t)=a_i\}$$ the connected component of the level sets of all of the first integral, then the restriction of the foundamental form on $M_a'$ $$p\cdot dq\Big|_{M_a'}$$ equal $dS(q,a)$ where $S$ is the generating functional of the ...

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Note that various density kernels (like the gaussian kernel used in classical phase-space) have delta functions as a limit. Physicaly, clasicaly delta (dirac) distributions are the correct ones. Mathematicaly more smooth distributions (e.g gaussians) may be used to calculate integrals and take limits (the limits should be same as having delta ditributions)

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i will try this one. A Hamiltonian system is (fully) integrable, which means there are $n$ ($n=$ number of dimensions) independent integrals of motion (note that completely integrable hamiltonian systems are very rare, almost all hamiltonian systems are not completely integrable). What this states in essence (and intuitively) is that the hamiltonian system ...

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You are right that if you know exactly the initial conditions of your system that is the exact location of your system's state in phase space then its evolution is entirely determined. But that's where lies the issue; we don't know exactly the state of the system as described by a point in phase space. Instead we may know some values of macro- or ...

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This here is not exactly an answer, but a clarification of why Legendre transforms are not needed to get this result. Using exterior derivatives is much easier. If you have $pdq - PdQ = dF$, take the exterior derivative on both sides to get $dp \wedge dq - dP \wedge dQ = 0$ (the exterior derivative of a differential is zero). The wedge product is ...

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To start things off I'd say that noting the $L_z$ component is conserved seems to mean pretty much nothing, since you're considering the motion as restricted to the $\mathcal{X}\mathcal{Y}$ plane. If you had assumed the motion along the $\mathcal{Z}$ axis to be possible, then we'd be talking about the spherical double pendulum instead of the planar one ...

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The phrase "the function is spherically simmetrical" means that, if $G$ is an orthogonal transformation (that sends spheres into themselves), then $$f(G\mathbf r, G\mathbf p,t)=f(\mathbf r , \mathbf p, t).$$ If you know $\mathbf r^2$, $\mathbf p^2$, $\mathbf r \cdot \mathbf p$ you can calculate $f$ by taking an orthogonal transformation which maps $\mathbf ... 2 You can start by reading the wikipedia article on the method of characteristics. You will see that in our case the tangent of the characteristic is$(1,-\partial H/\partial q,\partial H/ \partial p)$where the components are in order$t,p,q$. When you formulate the equation of the characteristic, you will actually find out you get equations of motion of a ... 0 We interpret OP's question as essentially asking the following. (If this is not what OP is asking we expect at least the proof method to be very similar.) Given $$\tag{1} \det \{\varphi_{i}, \kappa_{j}\}_{PB}\neq 0, \quad \{\kappa_{i},\kappa_{j}\}_{PB} ~=~ 0, \quad \{\varphi_{i}, \varphi_{j}\}_{PB} ~=~ \sum_{k}d^{k}_{ij}\varphi_{k},$$ in strong ... 7 I'll write here a list of my personal favorites plus some commonly used books. I wouldn't be surprised if your teacher chose either one of the books below as a textbook: i) Mechanics, the first volume of the Landau course on Theoretical Physics; ii) Goldstein's book "Classical Mechanics"; iii) Taylor's book "Classical Mechanics"; iv) Marion's book ... 1 Isn't it just a convention you have to choose once and for all. It's all about how to pass from Maxwell to lumped element circuits. Especially, I can choose the two different conventions (all quantities are vectorial in the following) $$E=\pm\nabla V$$ since I didn't choose my field-to-potential rules yet. Usually one chooses to conform to the classical ... 0 Partial answer here about what the author means by flux: We now introduce the flux$\phi_n$of a node$n$which is defined by the time integral of the voltage measured along the path connecting the node to the ground on the spanning tree. (Defined a few pages prior to 363.) It then seems$\dot\phi$is voltage as we would normally define it. 1 Intuitively, Legendre transform is just "integrate by parts". So, from$pdq - PdQ =F_1$, we have$pdq -d(PQ)+QdP =dF_1$, i.e.,$pdq +QdP =d(F_1+PQ) \equiv dF_2$. The "Transform" means after "integration by parts", we need to change the independent variables. For example, in$F_1$,$p=p(q,Q)$but in$F_2$,$p=p(q,P)$. So we need to solve$Q = Q(q,P)$and ... 0 The differential equation is given by, $$\ddot{x} + \gamma \dot{x} + \omega^2 x = 0$$ If we propose the ansatz$x(t) = e^{rt}$for some$r \in \mathbb{C}$, then we obtain a quadratic equation for$r$: $$r^2 + \gamma r + \omega^2 = 0$$ The solutions are given by, $$r = \frac{1}{2} \left( -\gamma \pm \sqrt{\gamma^2 -4\omega^2}\right)$$ Let us choose one ... 0 You're simply meant to plot$x$against$\dot{x}$. You can rewrite the DE as$p\,\mathrm{d}_q p + \gamma\, p + \omega^2 q=0$where$p = \dot{x}$and$q=x$, but I can't see a way of solving this one directly: without the dissipative term you'd get$p\,\mathrm{d}_q p + \omega^2 q = \mathrm{d}_q (p^2 + \omega^2 q^2/2) = 0$, which simply defines an ellipse. A ... 2 Hamiltonian formalism won't help so much because the problem is dissipative. You can solve the homogeneous linear diff. equation with constant coefficients by supposing$x=C e^{\lambda t}$. You will get complex solutions, but you should be able to add them to get a real solution. Now you should have$x(t)$with two arbitrary constants determined by initial ... 0 Pages 39 and 40 of this paper should provide a partial answer. You may need to read pages 35-39 first to understand the procedure. 1 In this answer we will consider a Lie algebra$L$(rather than a Lie group). Then: If$M$is a manifold, let there be a Lie algebra homomorphism$\rho:L\to \Gamma(TM)$into the Lie algebra of vector fields on$M$. The map$\rho$is called an anchor. If the manifold$(M,\{\cdot,\cdot\}_{PB})$is a Poisson manifold, it is natural to require that the vector ... 1 I) In general, for a given choice of boundary conditions, it is important to adjust the action with compatible boundary terms/total divergence terms in order to ensure the existence of the variational/functional derivative. As OP observes, the problem is (when deriving the Euler-Lagrange expression) that the usual integration by part argument fails if the ... 0 Consider the kinetic energy operator $$\hat{K}=\frac{1}{2m}\hat{p}^{2}.$$ Then$[r^{-1}, \hat{K}]$is ambiguous depending on the coordinate system. Moreover, in Cartesian coordinates, it is quite peculiar compared to its Poisson bracket... (At least, if I did my math correctly, which is possible considering how sloppy/quick it was done, as I am ... 2 I don't know how elementary you consider a simple position dependent mass, but due to ordering ambiguity in the kinetic term$\hat{p}^2/2m(\hat{r})$such a system will have a quantum Hamiltonian different from the classical one. For example: Analytic results in the position-dependent mass Schrodinger problem Position-dependent effective masses in ... 4 The basic idea is the following. For the shake of simplicity, I henceforth assume that every function does not depend explicitly on time (with a little effort, everything could be generalized dealing with a suitable fiber bundle over the axis of time whose fibers are spaces of phases at time$t$). On a symplectic 2n dimensional manifold (a space of ... 0 Great question! One way to look at what is going on is to use the Hamiltonian version of Noether's theorem. The Noether procedure generates a conserved charge$Q$associated with the symmetry with parameter$\theta$. It turns out that$Q$is the generator of that symmetry, in the sense that for some function$A$of phase space variables ... 2 The problem here is how to quantize systems whose classical hamiltonian involves factors of the form (for example)$p^nx^m$, because these cannot be unambigously represented in a formalism where$p$and$x$do not commute. As such there are many alternatives (all of which are classically equivalent) but only one is quantum-mechanicaly relevant. In most ... 1 I think expecting "fluid" behaviour in terms of a material that does not support shear, is not useful in the context of the various systems you have listed in the question. Instead, I believe you are intuitively connecting ideas and concepts pertaining to conservation laws. So the idea that in specific systems, conserved charges (in the sense of Noether) ... 2 Consider an element$g$of the symmetry group. Say$g$is represented by a unitary operator on the Hilbertspace $$T_g = \exp(tX)$$ with generator$X$and some parameter$t$. It acts on an operator$\phi(y)\$ by conjugation $$(g\cdot\phi)(y) = T_g^{-1}\phi(y) T_g = e^{-tX}\phi(y) e^{tX} = \big[ 1 + t[X,\cdot]+\mathcal{O}(t^2)\big]\phi(y)$$ On the other ...

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Here is one line of motivation: On one hand, in the Lagrangian formalism, the Lagrangian energy function $$\tag{1} h(q,v,t)~:=~v^i \frac{\partial L(q,v,t)}{\partial v^i}- L(q,v,t)$$ is defined as the Noether charge for time translations. Noether's theorem states that if the Lagrangian is invariant under time translations, which implies that ...

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