Newton's law with Cantor potential!

  • Thread starter jostpuur
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  • #1
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Main Question or Discussion Point

I want that [itex][0,\infty[\to\mathbb{R}[/itex], [itex]t\mapsto x(t)[/itex] satisfies

[tex]
\ddot{x}(t) = -\partial_x U(x)
[/tex]

where [itex]U:\mathbb{R}\to\mathbb{R}[/itex] is some potential function. Then I set the initial conditions [itex]x(0) < 0[/itex], [itex]\dot{x}(0)>0[/itex], and define

[tex]
U(x) = \left\{\begin{array}{ll}
0,&\quad x < 0\\
\textrm{Cantor steps},&\quad 0\leq x\leq 1\\
1,&\quad x > 1\\
\end{array}\right.
[/tex]

What's going to happen? How should the Newton's law be interpreted?

I've got a feeling that this has something to do with weak solutions, but it doesn't seem clear to me. For example

[tex]
0 = \int\Big(\ddot{\rho}(t) x(t) + \rho(t) \partial_xU(x(t))\Big) dt
[/tex]

with some smooth test function [itex]\rho[/itex] doesn't solve the problem, because you cannot get rid of [itex]\partial_x U[/itex] with integration by parts.
 

Answers and Replies

  • #2
14
0
By energy conservation you see that a step in U makes an instantaneous change in velocity of the particle. You represent the Cantor steps as a limit of step functions of increasing detail. For each such approximating 'pre Cantor' potential the motion is thus well defined. My further proceeding would be to produce the (x,t)-diagram for a few pre Cantor potentials (e.g. with Mathematica) and let me inspire in guessing a limit for the final U. The behavior of x(t) will, of course, depend decisevely on the initial velocity: is the particles kinetic energy large enough to hopp over the first step of U.
Good luck for your research, looks like an interesting problem.
 
  • #3
14
0
What I proposed in my previous post as an computer experiment can be done by mere thinking:
I assume that 'Cantor steps' means indicator function of Cantor's ternary set. In particular
[itex] U(x) \in \{0,1\}[/itex].
Two cases:
[itex](a) \quad \dot{x}(0) < \sqrt{2}[/itex] (i.e. [itex] E_\text{kin} < U_\text{max}[/itex])
[itex](b) \quad \dot{x}(0) > \sqrt{2}[/itex]
The case of equality is unphysical i.e. corresponds not to a
real world situation, even after accepting reasonable idealizations.
Solution:
(a) The particle will become reflected at x=0 and will travel along the negative x-axis with constant velocity.
(b) The particle will traverse the Cantor steps region without any change of the original velocity and will at x=1 instantaneously become slowed down to velocity
[itex]\sqrt{{\dot{x}(0)}^2 - 2}[/itex]
and will continue to travel with this constant value of velocity.
The reason that the Cantor steps region is traversed with unmodified velocity is as follows:
Consider any member of the sequence, the limit of which defines Cantor's set. Then the potential is a sum of finitely many plateaus (hight 1) with valleys (hight 0) in between. We see the particle slowed down to the previously mentioned value during traversing a Cantor plateau. At the end of the plateau, the original velocity will suddenly restituted. When following the sequence towards the limit, the cummulative length's of plateaus tends to zero (this is an elementary propery of Cantor's ternary set, explained e.g. in Wikipedia) so that the effect of the occasionally reduced velocity tends to zero.
 
Last edited:

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