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Why do objects always move to a state of lower potential energy?

  1. Jul 8, 2013 #1
    Is there no answer given by QM till now?....
    Also, on a related note, when objects move through space we consider it as a point mass for the purpose of interactions. Why this simplification? Is it due to the distribution of potential energy throughout the object when a force(contact/non-contact) is applied to one end of it?
     
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  3. Jul 8, 2013 #2

    Bill_K

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    They don't. Not always. Balls don't always roll downhill!

    It's true, a ball may roll downhill, converting its potential energy into kinetic energy. But just as often a ball may roll uphill, converting kinetic energy into potential energy.
     
  4. Jul 8, 2013 #3

    kith

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    The reason why systems seem to evolve towards lower energy is that usually, you have an environment with many degrees of freedom which are weakly excited. For a given energy of your system, there are many ways to excite these environmental degrees of freedom. For a given energy distribution in the environment, there are few ways to excite the system to a state with the initial energy. So if all ways of distributing energy have equal probabilities -which is a central assumption from which we can derive thermodynamics-, systems tend to go to states with lower energy.

    There is no difference between QM and classical mechanics wrt this. The argument applies both to the ball rolling down a hill and to the excited state in an atom which decays by emitting a photon. In the first case, the environment is given by the atoms in the ground. In the second case, the environment is the elctromagnetic field and the photon is an excitation of this field which can take many forms given an initial excited state of the atom (different momenta, etc.).

    /edit: note that this isn't a contradiction to what Bill said. He was talking about the easier situation without environment/friction.
     
    Last edited: Jul 8, 2013
  5. Jul 8, 2013 #4
    This can be answered statistically, consider all possible microstates for the system. We know by the postulate that, the system goes to that macrostate which has maximum number of micro states, it turns out that a macro state with lower potential energy has more microstates!!!. I may be wrong!! Please correct me if I am. I just realized that I am not able to remember the intuition that I had previously in this regard!!
     
  6. Jul 8, 2013 #5

    Bill_K

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    Translation: It's not that the potential energy decreases. rather what drives things is that the entropy increases. While this applies to situations encompassing many degrees of freedom, such as a decaying atom, it does not address simple situations like a ball rolling down a hill.
     
    Last edited: Jul 8, 2013
  7. Jul 8, 2013 #6

    SteamKing

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    It would be bloody awful if things kept jumping off the floor at you for no particular reason.
     
  8. Jul 8, 2013 #7

    kith

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    For a layperson, a simple situation is an everyday situation. If you roll a ball down a hill, it won't be able to roll up another hill to the same height, hence it loses what we call energy. I strongly believe that good explanations have to relate to the experience of the person who wants to learn.

    Sure, it's conceptually simpler to ignore friction. But talking to laypersons, this should be highlighted and not be taken as self-evident. After all it's one of the the key insights underlying Newton's axioms.

    But maybe we just have a different impression of the original poster.
     
  9. Jul 8, 2013 #8

    naima

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    Do excited atoms return to ground level due to entropy ?
     
  10. Jul 8, 2013 #9

    mfb

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    You can use entropy as an argument, indeed.

    In terms of classical motion, it is useful to define energy via the forces acting on objects - and forces are related to accelerations. In classical mechanics, the answer is "that's how energy is defined". In quantum mechanics, we first have to define what "move" means.
     
  11. Jul 8, 2013 #10

    Bill_K

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    Let state A = excited atom, and let state B = decayed atom plus photon. A and B have the same energy, and there's a small interaction between them so that each of them will gradually evolve into the other.

    First suppose you model this situation as a double well with the particle initially in well A. The probability of finding it in well B will gradually increase. But B can just as easily evolve into A, so the particle will also gradually come back to A. Depending on the initial conditions, the particle will either oscillate back and forth between A and B, or reach an equilibrium half in A and half in B.

    This, of course, is not what a decaying atom does. Once decayed, it stays decayed! The difference is entropy - in the case of the atom, "well B" is actually many states, e.g. all possible decays where the photon heads in a different direction. Consequently in decaying from A to B the entropy increases, and while it's still theoretically possible for the photon to return and make the atom "undecay", it's highly unlikely.
     
  12. Jul 9, 2013 #11

    hilbert2

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    This is like asking "why do objects always accelerate to the direction of applied force?". The concept of potential energy is defined in such a way that particles accelerate to the direction of decreasing potential energy. It's just a concept, invented by man, that makes mechanics easier than the older newtonian method where force instead of energy was the basic physical quantity.
     
  13. Jul 9, 2013 #12
    I've got a bunch of answers which seem quite different. Degrees of freedom, definition, and some vague answers. Could someone please sum up?
     
    Last edited: Jul 9, 2013
  14. Jul 9, 2013 #13

    Bill_K

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    Why don't YOU sum it up? Part of the reason you got a variety of answers is that your question could be interpreted in various ways. Which ones do you think were closer to what you actually meant?
     
  15. Jul 9, 2013 #14

    naima

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    I like the explanation with the photon.
    The common point betwen the excited atom and the ball on the top of the hill is that
    they will warm the environment in multiple undistinguished different ways
     
  16. Jul 9, 2013 #15

    hilbert2

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    This is how I would put it:

    Entropy is a thermodynamical concept and can only be applied to a system that has a huge number of coupled degrees of freedom. In the idealized situation of a ball rolling down a hill one can't say that the entropy increase is what drives the ball downhill towards smaller potential energy. It's the Newton's or Lagrange's equations that drive the ball. The only mechanical degrees of freedom in that situation are the coordinates of the ball. In the real world, though, the potential energy of the ball rolling downhill is lost to frictional heating due to the motion, which excites the huge number of vibrational modes of the molecules on the surface of the hill, and entropy really increases.

    In the case of a decaying excited atom, the combined system of the atom plus the surrounding electromagnetic field has very many degrees of freedom because EM field has an infinite number of normal modes (Fourier components). In that case one can argue that the atom decays because of the entropy increase that results.

    Let's say we have a system of two coupled oscillators, initially at rest. Then at some moment we give oscillator A a 'kick' of kinetic energy. Because of the coupling, energy is tranferred to the oscillator B too, but the system still periodically returns to a state where oscillator A has all the kinetic plus potential energy. This is called 'Poincare recurrence'.

    On the other hand, if we have a system of, like, a billion coupled oscillators, and we kick one of the oscillators, we will notice that energy will eventually be distributed pretty evenly to all of the oscillators, and the system never seems to move back towards a state where a minority of the oscillators have most of the total energy. The tendence of entropy to increase is what causes this. The Poincare recurrence time of the system of billion oscillators is so large that it's essentially infinite.
     
  17. Jul 10, 2013 #16

    kith

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    As an interesting side note, a similar phenomenon may occur for an excited atom. A state which decays via spontaneous emission can get re-excited under certain circumstances (unfortunately, the corresponding article in wikipedia doesn't explain much).
     
  18. Jul 10, 2013 #17
    Okay. I don't understand kith's first answer. I got lost reading this part:
    "For a given energy distribution in the environment, there are few ways to excite the system to a state with the initial energy. So if all ways of distributing energy have equal probabilities -which is a central assumption from which we can derive thermodynamics-, systems tend to go to states with lower energy."

    Then there is the entropy argument, but we're just replacing one question with another, right? Why does entropy always increase?

    I also got a microstate argument, I understood that.

    Don't we define things because we see them being the way they are? How can the reverse be possible?

    That's my summation(at least that's all I understood).
     
  19. Jul 10, 2013 #18
    What about the second part of my question? when objects move through space we consider it as a point mass for the purpose of interactions. Why this simplification? Is it due to the distribution of potential energy throughout the object when a force(contact/non-contact) is applied to one end of it?
    Thanks for all the replies. This thread is pretty interesting! :)
     
  20. Jul 10, 2013 #19

    kith

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    Bill elaborated on this in post 10. My first answer was about the entropy argument. I should have mentioned this.

    Let's consider a simple situation. We have one atom A which we track in our experiment and a whole bunch of atoms Bn in the environment. Let the energy of the whole system be E, which is the difference between the excited state of the atom and its ground state. There is only one configuration, where our atom A is excited but millions of configurations where an atom of the environment is excited. If the energy quantum E gets emitted and re-absorbed randomly throughout the system, the chance to find it in A at an arbitrary time t is incredibly small.

    Since we don't distinguish between the states of the environment in the experiment (we just ask, "is the atom excited or is the environment excited") it is meaningful to define a measure for the probability that A is excited at time t. This is done by defining Boltzmann entropy which essentially counts the possible (micro) states which belong to the macro state "the environment is excited".

    Your question is a bit ambiguous. Most answer's assumed that you intended to ask why the atom stays in the ground state and doesn't get re-excited (which corresponds to the question why the ball stays downhill / in a basin). Bill's and hilbert2's first posts took your question literal.
     
  21. Jul 12, 2013 #20
    Great. I understand why elementary particles always move to a state of lower potential energy, but I still don't understand why macro objects do this. Is it the same principle on a macro scale?
     
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