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One of the most important insights you take away from a physics undergrad is that you can model much of physical phenomena as a harmonic oscillator. The reason

One of the most important insights you take away from a physics undergrad is that you can model much of physical phenomena as a harmonic oscillator. The reason for this is quite simple 1. Every closed system has a fixed total energy, so many systems just settle into an oscillating state, where kinetic energy converts into potential and back. 2. Most real world systems are approximately closed, so they leak energy till they have low total energy (this also follows from the second law). 3. An oscillating system with low total energy can have its potential energy accurately approximated with a quadratic function. Or in other words a harmonic oscillator. So, while I can't say if there are many interesting/useful coupled classical oscillator systems that need an exponential speedup for us to study, it is nevertheless exciting to hear that such systems do admit a quantum speedup.

noduerme 3y ago
Would a 3-or-more body gravitational problem be one of these that could use a speed up?
- dchftcs 3y ago
  For that, accuracy seems to be a stronger limiting factor than speed.
  - noduerme 3y ago
    My lay understanding of the problem with classical algorithms is basically that a lack of resolution means you need to monte carlo the thing millions of times... which is why it's slow. If you could model it as a set of quantum states of similar inaccuracy, wouldn't that by definition be just as (in)accurate but faster? [edit] this reminds me of something I read about how NASA doesn't predict solar eclipses by trying to keep an exact model of the solar system, but rather uses pattern matching algorithms.
    - deleted 3y ago
      [deleted]
    - valine 3y ago
      I don’t think there’s many three body problems in the real world where compute is the limiting factor. If you’re projecting so far out into the future that simulation speed is a problem, you’re initial measurement error will have compounded to the point that your solution is meaningless. We struggle to predict the exact path of asteroids because of measurement errors, not because computing is slow. Minuscule changes to the initial condition manifest as massive differences in the outcome.
    - fgoesbrrr 3y ago
      People were predicting solar eclipses thousands of years ago with no calculators or even a modern understanding of math. Can't be that hard.
      - lippel82 3y ago
        The reason this worked is, that the back-action of the moon on the sun is very insignificant. For a "real" chaotic three-body system, you need three bodies that interact with each other on comparable scales.
      - noduerme 3y ago
        I found the article I was referencing[0] >> In summary, it is clear ancient people could predict timings for lunar eclipses and partial solar eclipses, but there is no convincing evidence of people predicting the times and locations of total solar eclipses. >> Today, we don’t rely on calculating the orbits of the whole Solar System to predict eclipses. For example, NASA uses a highly advanced form of an ancient technique – pattern recognition. Using some 38,000 repeating mathematical terms, NASA can predict both solar and lunar eclipses for 1,000 years into the future. Beyond that, the Moon’s wobble and Earth’s changing rotation make eclipse prediction less accurate. [0] https://www.astronomy.com/observing/humans-have-been-predicting-eclipses-for-thousands-of-years-but-its-harder-than-you-might-think/ https://www.astronomy.com/observing/humans-have-been-predict...
- AnimalMuppet 3y ago
  Probably not (at least by this breakthrough). You can't model a 3-or-more body gravitational problem as coupled harmonic oscillators.
mjburgess 3y ago
Well that's an illusion of physics education -- you can model almost nothing in the world this way. Rather we engineer circumstances (devices, experiments, etc.) where nature is forced to imitate this otherwise useless approximation. And mostly this fails. You cant really analyse the world using simple closed-form formula -- all the stuff that this worked for is studied and reported in books.
- ndriscoll 3y ago
  You can model any conservative system with stable equilibria this way (i.e. anything with local minima in the potential). As another poster pointed out, just do a Taylor expansion around the equilibrium state, and you have a locally quadratic potential, i.e. a harmonic oscillator. Explicitly, WLOG the potential is `V(x) = V_0 + V_1 (x-x_0) + V_2 (x-x_0)^2 + ...`. But the first derivative is 0 near a minimum, so V_1 = 0. The constant V_0 term doesn't affect dynamics, so choose it to be 0. Then the higher order terms go to 0 as x->x_0, so you have `V(x) = V_2 x^2` near any stable point. And it's pretty obvious that lots of real-world phenomena 1. do have stable states and 2. do oscillate around them before settling, so the model is pretty good for lots of real-world systems (with the error being that the system is generally not conservative, so eventually it settles into the equilibrium because friction is stealing the energy).
  - mjburgess 3y ago
    How much of reality is a stable system under equilibrium?
    - ndriscoll 3y ago
      Look around you. Is the building you're in falling down? Or is it standing still? Are the things on your desk bouncing about, or are they just sitting there? Are the atoms in your body fissioning, or are you still here? Stable systems are extremely common. The world would be a much more violent place to live in if they weren't. And it makes sense that stable points exist. That means there's a well somewhere in the potential (and the stable point is at the bottom of the well). It seems reasonable that if you didn't know anything about the potential of a system, you might suppose it has at least one squiggle somewhere, and in that squiggle you'll find your equilibrium. Edit: I guess actually it's not so obvious in the higher dimensional case since you can have saddles. I'd have to go back and look at my nonlinear diffeq book to see, but there might be some topological result (similar to the hairy ball theorem) that explains it, at least for compact configuration spaces.
pa7x1 3y ago
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 3y ago
  > 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.
  - thfuran 3y ago
    Most things jiggle for a while if you poke them a bit?
    - godelski 3y ago
      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 3y ago
      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.
  - arbitrandomuser 3y ago
    > > 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.
stared 3y ago
There is a saying that "physics is a subset of reality that can be approximated by coupled harmonic oscillators". If they are purely linear - it is simple; just write a matrix of their spring coefficients and diagonalize it. If there are non-linear terms, well, then it isn't. This is the case in one of the Quantum Field Theory (and its instance, the Standard Model), the most fundamental theory describing particles. One of the takes on Feynman diagrams is that it is a series approximation of non-linear interactions in the QFT.

noduerme 3y ago

Would a 3-or-more body gravitational problem be one of these that could use a speed up?

dchftcs 3y ago
For that, accuracy seems to be a stronger limiting factor than speed.
- noduerme 3y ago
  My lay understanding of the problem with classical algorithms is basically that a lack of resolution means you need to monte carlo the thing millions of times... which is why it's slow. If you could model it as a set of quantum states of similar inaccuracy, wouldn't that by definition be just as (in)accurate but faster? [edit] this reminds me of something I read about how NASA doesn't predict solar eclipses by trying to keep an exact model of the solar system, but rather uses pattern matching algorithms.
  - deleted 3y ago
    [deleted]
  - valine 3y ago
    I don’t think there’s many three body problems in the real world where compute is the limiting factor. If you’re projecting so far out into the future that simulation speed is a problem, you’re initial measurement error will have compounded to the point that your solution is meaningless. We struggle to predict the exact path of asteroids because of measurement errors, not because computing is slow. Minuscule changes to the initial condition manifest as massive differences in the outcome.
  - fgoesbrrr 3y ago
    People were predicting solar eclipses thousands of years ago with no calculators or even a modern understanding of math. Can't be that hard.
    - lippel82 3y ago
      The reason this worked is, that the back-action of the moon on the sun is very insignificant. For a "real" chaotic three-body system, you need three bodies that interact with each other on comparable scales.
    - noduerme 3y ago
      I found the article I was referencing[0] >> In summary, it is clear ancient people could predict timings for lunar eclipses and partial solar eclipses, but there is no convincing evidence of people predicting the times and locations of total solar eclipses. >> Today, we don’t rely on calculating the orbits of the whole Solar System to predict eclipses. For example, NASA uses a highly advanced form of an ancient technique – pattern recognition. Using some 38,000 repeating mathematical terms, NASA can predict both solar and lunar eclipses for 1,000 years into the future. Beyond that, the Moon’s wobble and Earth’s changing rotation make eclipse prediction less accurate. [0] https://www.astronomy.com/observing/humans-have-been-predicting-eclipses-for-thousands-of-years-but-its-harder-than-you-might-think/ https://www.astronomy.com/observing/humans-have-been-predict...
AnimalMuppet 3y ago
Probably not (at least by this breakthrough). You can't model a 3-or-more body gravitational problem as coupled harmonic oscillators.

mjburgess 3y ago

Well that's an illusion of physics education -- you can model almost nothing in the world this way. Rather we engineer circumstances (devices, experiments, etc.) where nature is forced to imitate this otherwise useless approximation. And mostly this fails. You cant really analyse the world using simple closed-form formula -- all the stuff that this worked for is studied and reported in books.

ndriscoll 3y ago
You can model any conservative system with stable equilibria this way (i.e. anything with local minima in the potential). As another poster pointed out, just do a Taylor expansion around the equilibrium state, and you have a locally quadratic potential, i.e. a harmonic oscillator. Explicitly, WLOG the potential is `V(x) = V_0 + V_1 (x-x_0) + V_2 (x-x_0)^2 + ...`. But the first derivative is 0 near a minimum, so V_1 = 0. The constant V_0 term doesn't affect dynamics, so choose it to be 0. Then the higher order terms go to 0 as x->x_0, so you have `V(x) = V_2 x^2` near any stable point. And it's pretty obvious that lots of real-world phenomena 1. do have stable states and 2. do oscillate around them before settling, so the model is pretty good for lots of real-world systems (with the error being that the system is generally not conservative, so eventually it settles into the equilibrium because friction is stealing the energy).
- mjburgess 3y ago
  How much of reality is a stable system under equilibrium?
  - ndriscoll 3y ago
    Look around you. Is the building you're in falling down? Or is it standing still? Are the things on your desk bouncing about, or are they just sitting there? Are the atoms in your body fissioning, or are you still here? Stable systems are extremely common. The world would be a much more violent place to live in if they weren't. And it makes sense that stable points exist. That means there's a well somewhere in the potential (and the stable point is at the bottom of the well). It seems reasonable that if you didn't know anything about the potential of a system, you might suppose it has at least one squiggle somewhere, and in that squiggle you'll find your equilibrium. Edit: I guess actually it's not so obvious in the higher dimensional case since you can have saddles. I'd have to go back and look at my nonlinear diffeq book to see, but there might be some topological result (similar to the hairy ball theorem) that explains it, at least for compact configuration spaces.

pa7x1 3y ago

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 3y ago
> 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.
- thfuran 3y ago
  Most things jiggle for a while if you poke them a bit?
  - godelski 3y ago
    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 3y ago
    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.
- arbitrandomuser 3y ago
  > > 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.

stared 3y ago

There is a saying that "physics is a subset of reality that can be approximated by coupled harmonic oscillators". If they are purely linear - it is simple; just write a matrix of their spring coefficients and diagonalize it. If there are non-linear terms, well, then it isn't. This is the case in one of the Quantum Field Theory (and its instance, the Standard Model), the most fundamental theory describing particles. One of the takes on Feynman diagrams is that it is a series approximation of non-linear interactions in the QFT.