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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?

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    $\begingroup$ What is your definition of smooth? $\endgroup$
    – Hritik Narayan
    Commented Dec 9, 2014 at 15:42
  • $\begingroup$ @HritikNarayan thanks for the response and sorry for the ambiguity, I am a physics newbie. I mean "doing smoothing" here: the possibility to use imputation (if at some point the measurement is missing) and smoothing (using techniques for example, simple averaging) $\endgroup$
    – Hello lad
    Commented Dec 9, 2014 at 15:47
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    $\begingroup$ What do you mean by "a force which is instantaneous"? It's true that acceleration at any given instant is determined by the force at the same instant, but in realistic scenarios we do expect the forces on an object to vary in a continuous way, not change instantaneously, so the acceleration should vary continuously too. $\endgroup$
    – Hypnosifl
    Commented Dec 9, 2014 at 16:04
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/35674/2451 , physics.stackexchange.com/q/9720/2451 , physics.stackexchange.com/q/1324/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Dec 9, 2014 at 20:45
  • $\begingroup$ I think this could be a duplicate of the last of those but it doesn't seem like a duplicate of the others. $\endgroup$
    – David Z
    Commented Dec 10, 2014 at 5:22

1 Answer 1

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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.

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    $\begingroup$ 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. $\endgroup$
    – Brondahl
    Commented Nov 11, 2016 at 18:13
  • $\begingroup$ @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. $\endgroup$
    – Kyle Kanos
    Commented Nov 11, 2016 at 18:21
  • $\begingroup$ 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) $\endgroup$
    – Brondahl
    Commented Nov 11, 2016 at 18:29
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    $\begingroup$ 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? $\endgroup$
    – Brondahl
    Commented Nov 11, 2016 at 18:30
  • $\begingroup$ @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. $\endgroup$
    – Kyle Kanos
    Commented Nov 11, 2016 at 18:48

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