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This question already has an answer here:

It's obvious that for every particle velocity is smooth i.e it cannot undergo sudden finite change in its position in infinitisiminal time.

Similarly any particle's velocity cannot undergo a change instantaneously (Infinite acceleration can't happen, intuitively).

Does this pattern apply to higher time derivatives of position like jerk? If yes then till how much higher derivative? 10th? 100? Infinite?

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marked as duplicate by Kyle Kanos, Qmechanic May 22 at 1:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Possible duplicate of Are velocity and acceleration smooth quantities? $\endgroup$ – Kyle Kanos May 21 at 23:00
  • $\begingroup$ This question cannot be answered at the classical level. You have to go quantum, but then these concepts loose their meaning and all you have left is conservation laws that don't care for higher derivatives. $\endgroup$ – safesphere May 21 at 23:50
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    $\begingroup$ We’re not even sure that position is smooth! There is no evidence that it isn’t, but there are theoretical reasons to think that spacetime itself is quantized. Particles may move by jumping from one quantum grain of spacetime to another. $\endgroup$ – G. Smith May 22 at 0:06
  • $\begingroup$ @Kyle Kanos I agree my question is similar to the one you posted above but it is not the same. My question is bit more. I'm not simply asking whether velocity and acceleration is smooth but I'm trying to ask till what n-th derivative this smoothness is there? You see it's a bit more broad than the question you've marked. But now I've got the real answer, as many good people pointed out quantum mechanics tells that even position is not smooth. Moreover we can't determine exact position and velocity simultaneously at a moment (Uncertainty principle). I hope I've understood correctly. $\endgroup$ – Shivansh J May 22 at 5:02
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Typically these higher derivatives are assumed to be smooth.

The key question will be what causes a discontinuity in the n-th derivative. If you focus on classical mechanics, the forces on an object boil down to the positions of particles in the system, which are continuous. This means there would need to be a discontinuity (such as a divide by zero) in the equations of motion in order to have a non-smooth higher derivative.

When you push towards quantum mechanics, the terms get really murky quickly, because position ceases to be a single observable number. But if you stick to classical mechanics, we find things stay nice and smooth.

Now that being said, when modeling real systems, we very often assume instantaneous changes in velocity or higher derivatives. This is because, in many cases, we can get away with ignoring the precise acceleration or jerk function and treat it as-if it were a simple discontinuous system. A straight forward example of this is a billiard ball collision. For most intents and purposes, this collision is "instantaneous" and the velocity of the balls changes in a discontinuous manner. However, if we look closer, with a slow motion camera, we find that the collision is not actually instantaneous -- position and its derivatives smoothly change over time. In fact, if you look hard enough, you can even see the ripple as the effects of the impact race across the surface of the ball. But, for the purposes of determining the result of a trick shot, these fine details are immaterial, and calculations assuming an instantaneous change in velocity are used.

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  • $\begingroup$ Okay so there is no meaning of exact position and velocity in quantum mechanics? (Are you talking about uncertainty principle?) Don't you feel sad that we can't measure exactly position of a particle in reality, despite having best measurement tools? I do $\endgroup$ – Shivansh J May 22 at 5:05
  • $\begingroup$ @ShivanshJ It bothered me in the past. Then I started exploring other ragged edges of the philosophy of science. The rabbit hole goes even deeper than that. For example, you use the phrase "position of a particle in reality." Did you know that, at the deeper philosophical levels, science doesn't even claim that particles exist in reality? It merely claims that if you model reality as-if it were made of particles, you can make very good predictive statements about reality. $\endgroup$ – Cort Ammon May 22 at 15:35
  • $\begingroup$ We could be living in The Matrix, where all is bits and bytes in reality, not particles, and science would still arrive at the same conclusions about how the world works. When you think in those terms, troublesome things like wave/particle duality becomes slightly less freaky (still pretty weird though) $\endgroup$ – Cort Ammon May 22 at 15:36
  • $\begingroup$ I never thought we could be living in artificial world. Wow! I still wish that one day will come when we would have complete knowledge of everything. $\endgroup$ – Shivansh J May 22 at 17:27
  • $\begingroup$ Although I know it's very difficult to have a perfect equation that describes every phenomenon. I'm just a high school student but till now I've experienced that Mathematical formulas quickly turn very complicated in simplest of cases. For example: Finding an electric field at a general point due to uniformly charged ring is very very complex, but think that entire universe comprises soooo many entities and phenomenon. It gives me chills to even imagine how much effort we will have to do in order to understand everything. One thing I know is that we're a long way from discovering that formula. $\endgroup$ – Shivansh J May 22 at 17:43
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In order for jerk to be non-zero, the acceleration $a=\frac{\mathbf{d}v}{\mathbf{dt}}$ must be time-dependent:

$$a=\frac{\mathbf{d}v}{\mathbf{dt}}=f(t)\tag{1}$$

That's because the derivative of any number, no matter how large, is always zero.

But if $(1)$ applies the jerk becomes:

$$j=\frac{\mathbf{d}a}{\mathbf{dt}}=f'(t) \neq 0$$

If we take a very simple case, where:

$$a=a_0+a_1t$$

then: $$j=(a_0+a_1t)'=a_1\tag{2}$$

And the even higher derivatives all become zero.

$(2)$ also shows that $a$ could be very large (large $a_0$) and yet $j$ might be very small (small $a_1$)

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    $\begingroup$ How does this answer relate to the question? The OP is not asking what jerk or snap is. The question is if they can be zero in real world, but not in a simplified mathematical model. $\endgroup$ – safesphere May 21 at 23:46

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