Math Equations

Sunday, February 1, 2015

Central motion excludes the center

Central motion is the motion of a particle under the influence of a central force. I have known for some year that the particle cannot not pass through the origin. Mimicking a proof in my multivariable calculus book, I'll formalize the fact.

Theorem
If \(r:(a, b)\to\mathbb{R}^3\) is twice differentiable, \(r(t)\times r''(t) = \vec{0}\) for all \(t\) and \(r(t_0)\times r'(t_0) \not=\vec{0}\) for some \(t_0\), then \(r(t)\) is never \(\vec{0}\).

Proof
\[\frac{d}{dt}(r(t)\times r'(t)) = r'(t)\times r'(t) + r(t)\times r''(t) = \vec{0}\]Thus \(r(t)\times r'(t)=c\) for some \(c\). Because \(r(t_0)\times r'(t_0) \not=\vec{0}\), \(c\not=\vec{0}\). If \(r(t_1)\) were \(\vec{0}\) for some \(t_1\) then \(c\) would be \(\vec{0}\) so there is no such \(t_1\).

I wonder how to generalize this to higher dimensions.

1 comment:

  1. That's actually a pretty weak theorem because if r(t) is ever (0, 0, 0), then the acceleration won't exist in many cases such as r''(t) = 1/r(t)^2.

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