How Do Jumps and Discontinuities in Functions Affect Real-World Motion?

[Nikarasu M]

Hello,

Something on my mind today...

As you keep differentiating functions that are sometimes used to represent the displacement of objects you eventually end up with a function that has discontinuities and jumps in its path.

Simple example for the sake of illustration - an object at rest starts accelerating at 1 m/s^2 at t=0.

f(x) = 0, for x \leq 0
f(x) = x^2, for x>0

f'(x) = 0, for x \leq 0
f'(x) = x, for x>0

f''(x) = 0, for x \leq 0
f''(x) = 1, for x>0

How are these jumps and discontinuities dealt with a real-world physical sense ? A jump in position would infer instantaneous/greater than light speed travel, but why is there no issue for acceleration ?

I read about jerk and jounce and so on (the 4th and 5th derivatives), but what are the rules and mechanisms for when and how each derivative is 'allowed' to make these instantaneous changes in value ?

Maybe there was an infinite regression of them - an infinite derivative maybe and there was some kind of mathy limit involved ?

Maybe I'm looking at simplified math textbook examples/models (polynomials) and reading too much into them and the real world physical equations account for my conundrum ?

Is t=0 the big bang ? and everything is in deterministic sense always moving already and these jumps don't exist ?

 

[JHamm]

How do you get from one point to another without jumping? Also your position and velocity graphs won't have discontinuities and your acceleration will just "jump" when you start applying your force but in reality it too will gradually grow as the force is applied up to its maximum.

 

[Nikarasu M]

ok,

so then jerk and jounce have the kink then discontinuity ?

Or some other nth derivatives ?

Or you're saying the kinks and discontinuities end at the infintiy-th derivative ? (what is it called ?)

 

[chrisbaird]

In the real physical world, there is no discontinuity in the position versus time graph, nor in the velocity, acceleration, jerk, etc. graphs. The discontinuities only arise mathematically when we use an idealized functional expression to represent these parameters which is only an approximation (although often a very good one). For example, if I have a shopping cart sitting at rest, then at some time t, I start pushing it with a constant force so that it accelerates, you could model the acceleration as jumping from 0 to a at time t. But if you actually measured the acceleration with a high-resolution accelerometer, you would find when you zoom way into the data that the acceleration smoothly changes from zero to a very quickly. Not only that, but also there is a finite time required for the affect of your push on the handles to propagate through the shopping cart to the front, so you will actually generate vibrations in the shopping cart (not enough to feel though - the vibrations you feel when pushing a shopping cart are from a bumpy floor/non-round wheels).

This actually leads to significant errors in numerical modeling of physical systems. For instance, in a similar way, three-dimensional objects are using represented in a computational system as having perfectly sharp edges, meaning that there is a discontinuity in the mass as a function of space at the surface going from object to no-object. If you look close enough at real objects, their surfaces are fuzzy. Depending on the frequency and resolution, this idealization can lead to significant effects. Some computational models have added algorithms that smooth out such effects in the end.

 

[Nikarasu M]

ok,

I fear an infinite regression of questioning now ;)

so, you're saying if you zoom in on the step function you'll see it's actually curved - differentiate this and eventually you'll end up with what looks like another step function - zoom in again - it's also curved - and so on ... (infinitely?)

My question maybe is getting more philosophical (?) - but how does anything begin ?

It's like the step function in its digital on-off sense sense is the 'decision' at time t to make a move.

Say you have something moving and you reverse it's direction - which level of differentiation down all the functions first crosses the abscissa (time axis) ? The infinite one ? This is hard to for me to get my head around.

 

[Nikarasu M]

Just realized my calculus is a bit off in my original post - but it shouldn't affect the logical flow of my query ...

 

[chrisbaird]

I believe the problem is that you are still thinking objects have hard edges. Consider a glass marble traveling at constant speed, then it strikes a wall and reverses direction. The collision of the marble and the wall is actually an inter-atomic electromagnetic field interaction. The electromagnetic fields don't just extend to the radius of an atom and then drop to zero, they extend out to infinity. The fields of an atom will be very weak far away, but not zero. They die down smoothly out to infinity. That means that even when the marble is two feet away from the wall, it is already feeling some force from the wall (amazingly weak, but non-zero). So there is no "begin" or switch-on point for a collision (which would lead to a discontinuity in some derivative), because in reality, the interaction is always takes place. A marble two miles away heading for a brick wall is already interacting with it, and in a sense, experiencing part of the collision process (although in reality such a force at that distance will be buried in the noise of other stronger forces, e.g. thermal, friction, wind, seismic). For practical purposes, we must make some approximating thresholds. For instance, approximate that when two atoms are farther apart than 100 atomic radii, their interaction strength is so week that it can be treated as zero.

 

[Nikarasu M]

Thanks Chris,

I see what you're saying here - I'll ponder it for a while and see how it sits (first read, I'd say it's sitting well)

Nick

 

[torquil]

It is not impossible for a function that starts at zero to become nonzero and at the same time have all continuous derivatives. Consider e.g:

x(0) = 0
x(t) = exp(-1/t^2) for t>0

All its derivatives at t=0 are zero, but the function still deviates from zero at any t>0. So this is not a mathematical impossibility.

 

[Nikarasu M]

My question was mostly re. the simple example you find in math textbooks - maybe I need to read up more on physics. Does this function model a physical process ?