# Tag Info

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 ...

6

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 ...

6

A simpler example is Kepler's laws of planetary motion. In a spherically-symmetric gravitational field, the planets follow elliptical orbits. The orbits certainly do not have the full spherical symmetry of the potential. They may be extremely lopsided. :-D

5

I think the source of your confusion is mathematics, not physics. It is important here (and L&L did mention this) that the system of differential equations is autonomous. If this is the case, than along with solution $q_i=q_i(t,C_1,\dots,C_{2s}),$ it has a solution $q_i=q_i(t-t_0,C_1,\dots,C_{2s}).$ Because the former is the general solution, the later ...

5

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.

4

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 ...

3

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 ...

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

On the off chance there was any doubt about the numerics, I too wrote a code to follow these orbits. I use RK4 to take the first half timestep (a full timestep here is always $0.001$), and then perform the remaining integration via leapfrog. Below is the evolution of an asymmetric orbit. The potential is given by $v_0 = 0.9$, $q = 0.7$, $L = 0.05$. I set ...

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 ...

1

Well I was still confused after reading these answers so I found a book "Cosmology and astrophysics through problems" by T Padmanabhan that finally answered the question in a (fairly) clear way. Basically for a system with n degrees of freedom, you have $n$ Euler-Lagrange (E-L) equations. Since these equations are second order partial differential ...

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 ?

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 ...

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