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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.

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.
    - smallnamespace 3y ago
      I think your explanation only seems clear because you’re eliding some quite relevant detail > Look around you. Is the building you're in falling down? You can say it’s stable, or that the building is in the act of slowly falling down without human input (maintenance). It’s a bit misleading to sneak in time scale. > Are the things on your desk bouncing about, or are they just sitting there? At a small enough scale bits of the desk are bouncing around due to thermal energy. Also same as the house, your desk will eventually decay into dust due to that bouncing. > Are the atoms in your body fissioning, or are you still there? This one is actually stable. The other two systems are better characterized as metastable - they appear stable at certain time and length scales. In fact metastability can be pretty tricky to explain, and sometimes requires a detour into thermodynamics or other fields (e.g. biology if you’re asking why a person’s body seems rather stable).
      - ndriscoll 3y ago
        Sure but the point is in a closed system, thermodynamic fluctuations alone are going to take eons to make your house fall down. What will actually get it is that the system is not fully closed, and eventually something from outside the system will give it the activation energy needed to escape the well it's in (e.g. a windstorm causing a tree to fall on it). But it's useful and practical to think of that larger system as decomposing into a product of pretty much independent systems (until it isn't). The point is wells exist and are common in day-to-day life, including ones that are deep enough that things don't randomly thermodynamically pop out of them. If something is sitting on your desk when you go to bed, you can be pretty certain it'll be there in the morning too. The scale of the well is orders of magnitude larger than the thermal fluctuations, even if eventually a fluctuation could excite your computer monitor to suddenly vaporize. The atoms in your body and all matter will eventually decay to nothing too, but it's not useful to think about that. And even if it's metastable, as long as the system stays inside the well, the potential is locally quadratic. So you see oscillations around the equilibrium in the meantime.
        mjburgess 3y ago
  - godelski 3y ago
    The vast majority of it. By definition. Unstable equilibria are unstable to perturbations. By definition.

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.
  - smallnamespace 3y ago
    I think your explanation only seems clear because you’re eliding some quite relevant detail > Look around you. Is the building you're in falling down? You can say it’s stable, or that the building is in the act of slowly falling down without human input (maintenance). It’s a bit misleading to sneak in time scale. > Are the things on your desk bouncing about, or are they just sitting there? At a small enough scale bits of the desk are bouncing around due to thermal energy. Also same as the house, your desk will eventually decay into dust due to that bouncing. > Are the atoms in your body fissioning, or are you still there? This one is actually stable. The other two systems are better characterized as metastable - they appear stable at certain time and length scales. In fact metastability can be pretty tricky to explain, and sometimes requires a detour into thermodynamics or other fields (e.g. biology if you’re asking why a person’s body seems rather stable).
    - ndriscoll 3y ago
      Sure but the point is in a closed system, thermodynamic fluctuations alone are going to take eons to make your house fall down. What will actually get it is that the system is not fully closed, and eventually something from outside the system will give it the activation energy needed to escape the well it's in (e.g. a windstorm causing a tree to fall on it). But it's useful and practical to think of that larger system as decomposing into a product of pretty much independent systems (until it isn't). The point is wells exist and are common in day-to-day life, including ones that are deep enough that things don't randomly thermodynamically pop out of them. If something is sitting on your desk when you go to bed, you can be pretty certain it'll be there in the morning too. The scale of the well is orders of magnitude larger than the thermal fluctuations, even if eventually a fluctuation could excite your computer monitor to suddenly vaporize. The atoms in your body and all matter will eventually decay to nothing too, but it's not useful to think about that. And even if it's metastable, as long as the system stays inside the well, the potential is locally quadratic. So you see oscillations around the equilibrium in the meantime.
      - mjburgess 3y ago
- godelski 3y ago
  The vast majority of it. By definition. Unstable equilibria are unstable to perturbations. By definition.