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I am currently studying non-linear dynamics on my own time. One of the theorems in the material is that systems that can be written as gradient problems cannot have closed orbits i.e. systems like $$\dot{x}=-\nabla V.\tag{1}$$

Isn't this the general form of a gravitational system with $V$ being the gravitational potential (or other conservative systems) and $x$ being the momentum? What am I missing here, knowing that such problems (gravity and like) often have closed orbits?

See this for reference http://www.cds.caltech.edu/archive/help/uploads/wiki/files/224/cds140b-perorb.pdf

Qmechanic
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curiousny
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2 Answers2

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  1. OP's eq. (1) is Aristotelian mechanics $$ m\dot{q}^i~=~-\frac{\partial V(q)}{\partial q^i} \qquad\Rightarrow\qquad V_i-V_f ~=~2 \int_{t_i}^{t_f} \! \mathrm{d}t ~E_{{\rm kin}} \tag{A}$$ This is dissipative. There are no$^1$ closed orbits.

  2. In contrast Newtonian mechanics $$ m\ddot{q}^i~=~-\frac{\partial V(q)}{\partial q^i}\qquad\Rightarrow\qquad V_i+E_{{\rm kin},i} ~=~ V_f+E_{{\rm kin},f} \tag{N}$$ preserves mechanical energy for conservative forces.

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$^1$ Indirect one-line proof: A closed orbit would mean that the LHS of the second equality in eq. (A) is zero, but the RHS is clearly positive. Contradiction. $\Box$

Qmechanic
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What's confusing you is the fact that you have implicitly omitted the argument of $V$. The definition of gradient system (http://www.cds.caltech.edu/archive/help/uploads/wiki/files/224/cds140b-perorb.pdf) is such that

$$\dot{x}=-\nabla V(x)$$

so if you define $x$ to be momentum, then you now have a potential which is dependent on momentum, which is not equivalent to a gravitational system (it has a potential that depends only on position).

probably_someone
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