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Define the jolt/jerk to be

$$J(f)={{d^3 f} \over {dt^3}}$$

Why don't we use the concept of jolt more often considering that the change in acceleration is most definitely an important characteristic in the universe. Is there some mathematical proof that essentially kicks jolt out of the useful units club? However, it seems like initial conditions relying on jerk/jolt should be the norm rather than the exception. I mean, the universe doesn't care about what's easy to measure...

In addition, what about the other neglected members of the rate of change community. Why isn't

$${{d^4 f} \over {dt^4}}$$

or

$${{d^{55} f} \over {dt^{55}}}$$

Used in physics? In fact wouldn't having a function that just gives all order derivatives of a function useful? You could presumably study the rate of change of the spread rates of change!

$$S(f(x),n)={{d^nx} \over {dt^n}}$$

In fact, on a more math intensive note, using fractional calculus, you could treat this Spread function as a real continuous function, since fractional calculus defines fractional order derivatives.

Of course, lets not leave out the integral. Shouldn't an expression that integrates the position with respect to time have physical meaning?

$$\int x \ dt$$

what about $$\iint x \ dt$$ or even $$\iiint x \ dt$$

In fact, it seems you could derive a function very similar to the one mentioned above.

Questions: So why don't we use these other forms of rate of change? In addition, isn't the integral of position important. I mean, it has units of $L \cdot M$ which seems to me at least to be very important. In addition, what about this Spread function, what physical significance does it have. In addition aren't initial conditions subject to the effects of jerk/jolt etc...!!

Zach466920
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  • Well, clearly we've been able to do rather a lot without the need to refer to the derivative of acceleration! And I don't think the integral of position has any meaning (look at the units) – Carl Witthoft Apr 21 '15 at 20:27
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    Possible duplicates: http://physics.stackexchange.com/q/4102/2451 , http://physics.stackexchange.com/q/18588/2451 , http://physics.stackexchange.com/q/4471/2451 , http://physics.stackexchange.com/q/160711/2451 and links therein. – Qmechanic Apr 21 '15 at 20:28
  • I'm not quite sure what you're asking. For example, your statement "Shouldn't an expression that integrates the position with respect time have physical meaning?" Is a weird sort of opinion question. We can multiply that integral by the reciprocal of the total time so get average position, but I don't think that's what your asking. I would suggest cleaning up your question to seem like less of a rant. – Reid Erdwien Apr 21 '15 at 20:31
  • @Qmechanic that really doesn't address my other two questions. – Zach466920 Apr 21 '15 at 20:32
  • @ReidErdwien How does it look now? – Zach466920 Apr 21 '15 at 20:34
  • @CarlWitthoft I posted the units, you'll have to be more technical than that... – Zach466920 Apr 21 '15 at 20:35
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    It all boils down to the links Qmechanic posted: The equations of motion are second order, and hence no higher order derivatives and no integrals of position are needed to completely fix the dynamics. – ACuriousMind Apr 21 '15 at 20:37
  • @ACuriousMind initial conditions contradict that statement...and what about the spread function? – Zach466920 Apr 21 '15 at 20:38
  • How would initial conditions contradict that? A second order equation takes as initial values the values of the function and the first order derivatives to fix a solution. The trajectories in the $(q,p)$ phase space of Hamiltonian mechanics never intersect. – ACuriousMind Apr 21 '15 at 20:53
  • @ACuriousMind I'm saying its a feature of the universe. I mean if an object wants to decelerate, it has to decelerate at some rate correct? Hence, the need for jolt. In addition, initial conditions of a system may include jolt, and we just neglect the terms, since we don't have equations for that. – Zach466920 Apr 21 '15 at 20:59
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    No, we neglect the jolt because it is already fixed by giving the zeroth and first order conditions. The possible jolts are already constrained by the e.o.m being only second order - you are not free to choose any jolt as your initial condition, it has to be compatible with, and it is fact fully determined by, the zeroth and first order initial conditions. – ACuriousMind Apr 21 '15 at 21:05
  • Why not integrate with respect to y or z? – Jiminion Apr 21 '15 at 21:13
  • @ACuriousMind that's true for second order differential equations, but not for say fourth order equations. In that case, you have to specify jolt and acceleration conditions. – Zach466920 Apr 21 '15 at 21:58

2 Answers2

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In Physics what we do is try to understand reality based on the scientific method. What we do is build a mathematical model based on observations and then using it make predictions that are again verified. When we create a model and find out that it is compatible with nature we don't ask "why nature behaves like that", we simply found out that it does. And in truth, the model we came up with usually works under some circumstances.

Indeed, just for an example, Newtonian Mechanics for a long time could be considered as "right" since it was able to take care of most of the observed phenomena. But if you go to high speeds it breaks down and Special Relativity comes into play. In that sense a model is not right, nor wrong, nor is it the absolute truth about how nature works. It is just the way we found out to explain a certain situation and that was confirmed to not be wrong.

Knowing that, your question might be: "why in Newtonian Mechanics the derivative of acceleration is not important?" and if that is your question I'm afraid the best answer is "because we observed that nature works like that". Well, this is hardly a satisfactory answer, but it is the best one. The model we have, depending on two derivatives is satisfactory, since it makes right predictions and that's all we want. Notice I'm not saying that the higher order derivatives do not exist. They do and you can study them if you like. But to explain what we want, just the first two are enough.

One more interesting question might be "is there some situation where the derivative of acceleration becomes relevant to the model?" and the answer is yes. The Abraham-Lorentz force is one such case. It is the recoil force on an accelerating charged particle caused by the emission of electromagnetic radiation. The force is

$$\mathbf{F} = \dfrac{\mu_0 q^2}{6\pi c}\dfrac{d\mathbf{a}}{dt}.$$

Again, the derivative of acceleration appears there because it was found relevant for this particular phenomenon. So, you have to understand that the mathematical objects we use in Physics are those a particular phenomenon needs in order to be properly modeled and studied through math.

Gold
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It all boils down to the utility of a model (Physics) in explaining reality. In that case, higher derivatives of motion (jolt, jounce/snap, etc.) simply have less utility. It most (but not all) cases, assuming them to be 0 seems to work pretty well.

Jiminion
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  • As I tried to make clear in the question (I'll edit again), initials conditions don't care about what's accepted. What if they have jerk/jolt etc terms? – Zach466920 Apr 21 '15 at 20:48
  • @Zach466920 You said: "...considering that the change in acceleration is most definitely an important characteristic in the universe." What is your evidence of this? My evidence is that modeling works pretty well with just acceleration. – Jiminion Apr 21 '15 at 20:52
  • You literally ignored the question I just asked you, is it to hard to answer? – Zach466920 Apr 21 '15 at 20:53
  • Considering the acceleration changes all the time, yes, I expect jolt to be everywhere in the universe. So yes, I think that's proof enough of its importance. Perhaps not important enough to be in equations is what you meant? That's debatable at least. – Zach466920 Apr 21 '15 at 20:55
  • If they have jolt terms, then the model needs to use them. Some models DO require jolt and jounce terms. – Jiminion Apr 21 '15 at 20:56
  • Proof would be models failing if they don't account for jounce and jolt. – Jiminion Apr 21 '15 at 20:57