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If the contact force applied to a physical object (ex. empty bucket) is given by the Heaviside function:

$$ F(t) = F_0~H(t)=\begin{cases} 0, t<0 \\ F_0, t \geq 0\\ \end{cases} $$

Then, assuming constant mass, the rate of change of acceleration is given by $$\lim_{t \to 0} \frac{da}{dt} \rightarrow \infty$$ since $$\frac{dF}{dt}=\delta(t),$$ the Dirac delta function.

How am I supposed to interpret this result physically? My intuition suggests that a mechanical system would have a minimum reaction time and that we can't have an 'instantaneous' force.

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closed as unclear what you're asking by ACuriousMind, Cosmas Zachos, CuriousOne, knzhou, Gert Jul 15 '16 at 1:46

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ You're explicitly setting a discontinuous second derivative of the position. Why are you worried that $d^3 x/dt^3$ has some funky behaviour there? (Also, why are you worrying about $d^3 x/dt^3$ in the first place?) $\endgroup$ – Emilio Pisanty Jul 13 '16 at 10:02
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    $\begingroup$ The Heaviside function simply means that you are turning a force on. There is nothing wrong with that in classical mechanics, since everything happens instantaneously. Unlike e.g. in electrodynamics we don't have any objections against such functions. In practice all this means is that "it happens so fast that we don't have to care how long it really takes to turn that force on". The acceleration has a step, the velocity is continuous and the position is (twice) differentiable. $\endgroup$ – CuriousOne Jul 13 '16 at 10:07
  • $\begingroup$ @EmilioPisanty I'm not worried. Just curious. $\endgroup$ – user29305 Jul 13 '16 at 10:07
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    $\begingroup$ Sure, but the point stands - why would an undefined (or distribution-based) third derivative be a problem at all in this situation? For sure, what you say is correct, but I'd say the appropriate response is more along the lines of "oh, ok", and moving on to something more interesting. $\endgroup$ – Emilio Pisanty Jul 13 '16 at 10:09
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As a general rule, whenever you are setting sudden discontinuities (as in your example with the Heaviside function) it is not a surprise that these may reflect in discontinuous distributions, or derivatives, or infinities here or there. Do keep in mind that it is just a result of the mathematical simplifications that we are introducing (although distributions may look scary, they are actually the easiest we can do to model).

In physics and in nature there are no sudden discontinuities, everything changes (more or less) smoothly; as a consequence $0$ and $\infty$ are to be interpreted as very small and very big, instead. A correct and precise smooth approximation would yield finite results, instead.

Example: a standard example of the above is the Coulomb law between two point particles that sit very far away from each other: $$ \textbf{F}_{1,2} = k \frac{q_1q_1}{r^2}\textbf{u}_{1\to 2} $$ what happens when $r\to 0$, is the force diverging? In such a case the above approximation does not hold anymore and the two charges cannot be considered point like and very far away from each other: charge distribution comes into play and the effect of the density modifies the Coulomb law to a form that is totally regular in $r=0$ (there are plenty of exercises on that in the literature).

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  • $\begingroup$ One of the things I'm wondering about is whether there's a 'fastest' reaction time for a system in Classical Mechanics. If so, then the use of the Heaviside function is actually a mistake. $\endgroup$ – user29305 Jul 14 '16 at 2:16
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    $\begingroup$ @AidanRocke : Classical Mechanical systems are abstract ideas. They can in theory have zero reaction time. If relevant, the reaction time of the model should be close to that of the system being modelled. The Heaviside and Dirac Delta Functions are only mathematical models of (approximations to) reality. If the results they give aren't realistic enough, we don't use them. $\endgroup$ – sammy gerbil Jul 16 '16 at 15:14