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7

Yes, this is the opposite of Noether's theorem. So let's call our conserved quantity $A$ (we will consider just one conserved quantity for starters) and begin with $\left \{H, A \right \} = 0$ law for conservation. Because of the connection between Poisson bracket with flows on the phase space this tells you both that $\mathcal{L}_{V_H} A$ = 0 ($A$ is ...


4

There are already several good answers. However, the off-shell aspect related to Noether Theorem has not been addressed so far. (The words on-shell and off-shell refer to whether the equations of motion (e.o.m.) are satisfied or not.) Let me rephrase the problem as follows. Consider a (not necessarily isolated) Hamiltonian system with $N$ degrees of ...


3

I'll start with some terminological subtleties, that have to be accounted for, when it comes to "conserved quantities" or "integrals of motion". First of all it is important to state on which variables the quantities can depend. In the field of differential equations, Hamiltonian dynamics and dynamical systems, a conserved quantity by definition does not ...


2

I think you all your steps are correct, I would suggest adding units though, otherwise adding $\vec{x}$ and $\vec{v}$ can be misleading. For step 3. You can just calculate $a = F/m$, where $F$ can either be gravitation $F=m g$, so it does not depend on $x$, or for example for a spring dependent on $x$, so $F = -k x$, here you need the next position from ...


2

1) A constant of motion $f(z,t)$ is a (globally defined, smooth) function $f:M\times [t_i,t_f] \to \mathbb{R}$ of the dynamical variables $z\in M$ and time $t\in[t_i,t_f]$, such that the map $$[t_i,t_f]~\ni ~t~~\mapsto~~f(\gamma(t),t)~\in~ \mathbb{R}$$ doesn't depend on time for every solution curve $z=\gamma(t)$ to the equations of motion of the system. An ...


2

Symmetry of potential alone is not enough to guarantee a symmetric orbit. The initial and boundary conditions must also be symmetric. Also, for non-linear systems, numerical treatment has to be sufficiently accurate.


1

Hints: Write the 3D angular momentum vector $\vec{L}$ in spherical coordinates. Note that $\vec{L}$ depends on the angular coordinates $\theta$ and $\varphi$ (and their time derivatives), but not the radial coordinate $r$ (and its time derivative). Claim that $\vec{L}$ are three independent integrals of motion for a particle confined on the two-sphere ...


1

To integrate your equation, you need to know the force law: F(x)=-dU(x)/dx F(x)= (U(x)-U(x+Dx))/Dx What kind of potential to you deal with ?



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