By Hello lad

2014-12-09 15:30:35 8 Comments

My thinking:

  1. acceleration corresponds to a force which is instantaneous, so the acceleration of a rigid body can be rather spiky (non-smooth)

  2. velocity (angular velocity) describes the ratio of change of the distance(angle), so it is smooth in the real world.

Conclusion, it makes sense to smooth (e.g., simple averaging) a velocity signal (temporal velocity), but it does not make so much sense to do smoothing on acceleration signal. Am I right?


@Kyle Kanos 2014-12-09 16:02:19

As far as we know and can test, space is continuous, not discrete. Since space is continuous, then the labels we associate with it (i.e., positions) are also continuous. Calculus requires continuous functions to do the derivative and integral, so this implies that velocities and accelerations are also continuous because they are derivatives of positions: $$ v(t)=\frac{dx(t)}{dt}\qquad a(t)=\frac{dv(t)}{dt}=\frac{d^2x(t)}{dx^2} $$

If you have a discrete spectrum (e.g., measurements at different times/positions), then interpolation (whether linear or some higher-order method) is a necessary and useful tool to reconstruct the smooth distribution that we expect.

@Reinstate Monica --Brondahl-- 2016-11-11 18:13:58

Umm ... continuity of f(x) does not imply continuity of f'(x)? The Saw-Tooth function is continuous, but it's derivative is not. Now I can certainly believe that motion IS smooth, and thus that the Saw-Tooth function cannot truly represent movement, but your argument here (that space is continuous, which proves motion is continuous, which proves acceleration is continuous) is not sound.

@Kyle Kanos 2016-11-11 18:21:21

@Brondahl: This is physics, not mathematics. In physics, we assume changes in physical quantities occur smoothly (continuously), rather than discontinuously whereas this assumption holds no weight in mathematics.

@Reinstate Monica --Brondahl-- 2016-11-11 18:29:18

Ah, do you mean that your response to the question "Are these things smooth" is "In physics, we assume them to be smooth"? (No disrespect is intended ... this might be the best available answer to the question)

@Reinstate Monica --Brondahl-- 2016-11-11 18:30:44

Assuming that I've correctly interpretted that, then to what extent has that been examined/tested? I can see that we might test the continuity of space to the limits of our ability, but have the higher order derivatives been examined/thought about in detail? Is infinite jerk possible? If not, why not?

@Kyle Kanos 2016-11-11 18:48:11

@Brondahl: I probably should add an addendum to the last sentence before the equations mentioning the assumption, but I've got a few other things going for me (like work, family, etc)--feel free to recommend an edit though. I think my first link should answer your first question. The others, see, perhaps this post.

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