[pa7x1]

This is overly complicated. The reason harmonic oscillators pop up everywhere is even simpler and more general than that.

It models first order perturbations over a stable equilibrium. For sufficiently small perturbations around a stable equilibrium everything is an harmonic oscillator.

It's basically taking the first order perturbation of a Taylor expansion around a local minima.

[robertlagrant]

> It's basically taking the first order perturbation of a Taylor expansion around a local minima.

I'm not sure that's less complicated, at least for my level understanding.

[arbitrandomuser]

> > It's basically taking the first order perturbation of a Taylor expansion around a local minima. > > I'm not sure that's less complicated, at least for my level understanding.

You know Taylor expansions ? So every complicated function of x can be written as soe expansion of infinite powers of x, a0 +a1 ( x )+a2 ( x^2 ) ... And so on Every complicated force can then be written as a function of displacement x like this , now if x is small ignore the sqaure and higher order terms what you get is the linear part , which looks something like F = a1x + a0 , the a0 part only shifts the default position (you can rewrite x as x-c so that a0 goes to zero,) lo and behold you have F=Kx everything is like your familiar hooks law spring as long as x is small enough.

[ridiculous_fish]

This is on the right track but not entirely right. In your F = a1x + a0, here a0 is a force, not a position, so you cannot neglect it on the basis of "default position." Instead you set it to zero because the system is at a stable equilibrium.

Here's the presentation I've seen. Usually we like to work in potentials, not forces, because potentials are nice scalar functions, while force is an ugly vector function. So say you have a potential V which is at a local minimum at position x.

Expand the potential at x around a small displacement dx: V(x + dx). This gives us the Taylor series V(x+dx) = V(x) + a1 V'(x) dx + a2 V''(x) dx^2 + a3 V'''(x) dx^3...

We can neglect V(x) since it's just a constant, and adding a constant to the potential does not affect the physics. And (the crux) we can neglect V'(x) because the potential is at a minimum, so the derivative is zero.

That leaves the quadratic and higher-order terms. Neglecting the higher order terms on the basis that dx is small, we get the harmonic potential, or Hooke's Law in the language of forces.

[thfuran]

Most things jiggle for a while if you poke them a bit?

[godelski]

Most things jiggle even if you don't poke them. In fact, for them to not jiggle you'd have to be at absolute zero (not technically correct, but good enough. Damn you quantum and your jigglyness)

[taneq]

Yep pretty much. Everything's jiggling a bit, if you poke them they jiggle, if you poke them twice has hard they jiggle twice as much, if you don't then they gradually slow down.

[abdullahkhalids]

That's what I said, without using the word Taylor expansion, as this audience might not necessarily know it. I also explained in points 1 and 2 why many systems have can be modeled via small perturbations (i.e. Taylor approximation) around a stable equilibrium - they need to have low energy.

Also, just a reminder that it is the second-order term in the Taylor expansion that is relevant for harmonic oscillators. Zeroth-order term (constant) does not determine the dynamics. The first-order term (linear) is zero only for low energies, as any potential well at sufficiently low energies will be symmetric. The second-order term (quadratic) is what provides the restoring force towards the equilibrium.

[pa7x1]

The basic difference is pointing that there is something more fundamental and it doesn't have to do with physics. It has to do with math, with calculus, and with approximating complex functions via perturbative expansions. This is very general and applies to other kinds of systems, not necessarily physical. That's the key insight.

The first relevant term is quadratic in the potential around a stable equilibrium but linear in the force. That's why they are called linear harmonic oscillators, I thought that was obvious in my description.

[godelski]

Honestly I kinda laughed when I saw the response because it reminded me of the classic nerd flexing (that happens especially when conveying something to laymen) where two nerds go at it saying the same thing but using heavier and heavier jargon as a means to flex (increasingly losing the laymen) rather than clarify or admit that language requires inference and the whole show is ironically just a demonstration at accurate communication or else one would not have been able to "um acktually" the other. (I think a lot of us have been guilty of this, often unintentionally, but we can still laugh at ourselves)

[lagrange77]

> It models first order perturbations over a stable equilibrium. For sufficiently small perturbations around a stable equilibrium everything is an harmonic oscillator.

That's the same reason, why we linearize nonlinear systems around the equilibria to apply linear control theory, right?

While in control, this makes sense to me, since the goal is often to stabilize the system, how does this help with modeling the whole system in general (far away from any equilibrium point)?

[pa7x1]

Yes, near a stable equilibrium you can always linearize a system to model its behavior quite accurately for small perturbations. That's the crux of what it was pointed above.

Outside of equilibrium things get more complicated. In principle, you can still do a first order expansion to understand the dynamics in a vicinity of that regime, the problem is that outside equilibrium you are not going to stay near the point of the solution space you started at. You will keep drifting, at least until you reach another stable equilibrium if there is one.

Systems outside equilibrium are much harder to study because we cannot linearize. Basically.

[dustingetz]

seems circular - if you limit the world to small perturbations over a stable equilibrium of course you get simple harmonic motion, push a spring down a bit it pushes back and starts bouncing, the interesting insight is why can you limit the world to that?

i suppose it’s because most interesting forces are conservative, i.e. if motion doesn’t dissipate energy (like friction, unlike electromagnetism) then energy is conserved so perturbations must bounce. OP contains the key insight- dissipating systems dissipate proportional to total energy, so low energies are approximately stable, IIUC.

[godelski]

I don't think fewer words means less complicated. I know we're on HN but I don't think most of the crowd here is going to be familiar with dynamic systems to parse your words and certainly not DEs or PDEs. I mean you won't get to this till what, Classical Mech in 3rd year of a physics undergrad? Maybe I'm too pessimistic but I feel like having sufficient background to parse your statement implies sufficient background to already have this knowledge. I definitely think OP's version is far clearer to the layman.