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I Hamilton's principle and minimum potential energy

  1. Aug 26, 2016 #1

    haushofer

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    Dear all,

    I was wondering what exactly the correspondence/relation is between Hamilton's principle (extremizing the action gives the allowed configurations) and the fact that a system wants to configurate such as to minimize its potential energy. Is there any? Somehow I can't find a decent treatment on this. Any suggestions or links are appreciated.
     
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  3. Aug 26, 2016 #2

    DrClaude

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    There is no such "fact"; energy is a conserved quantity. The minimization of energy comes about through interaction with an environment, because of the corresponding increase in entropy.
     
  4. Aug 26, 2016 #3

    haushofer

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    Yeah, perhaps I'm sloppy here.

    To consider a concrete example: a mass falling down on earth. To find the equations of motion, we can write down an action if we have formulated the kinetic and potential energy. The extremum of this action gives us the equations of motion. But the mass falls down because it wants to minimize its potential energy, right? Or do we really need the concept of entropy to explain this?
     
  5. Aug 26, 2016 #4

    DrClaude

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    I had something else in mind when answering your question. I'll have to think about this some more.
     
  6. Aug 26, 2016 #5

    vanhees71

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    I think what's meant by "minimizing the potential energy" is what's happening in the static case. Take some Lagrangian of the usual form in non-repativistic physics
    $$L=\frac{m}{2} g_{jk} \dot{q}^{j} \dot{q}^k-V(q).$$
    Then the equations of motion take the form
    $$\frac{\mathrm{d}}{\mathrm{d} t} [ g_{jk} \dot{q}^k]=-\frac{\partial V}{\partial q^k}.$$
    Now take the stationary case, i.e., look for solutions ##q^k=\text{const}##. Obviously you have to find the stationary points of the potential,
    $$\frac{\partial V}{\partial q^k}=0.$$
    Now you can ask what about solutions that are weak deviations from these stationary points. Around such a stationary point, say at ##q=0##, you have
    $$V(q)=V_0+\frac{1}{2} V_{jk} q^j q^k, \quad V_{jk}=\frac{1}{2} \left . \frac{\partial^2 V}{\partial q^j \partial q^k} \right|_{q=0}.$$
    Now it depends on the matrix ##V_{jk}##: if it is positive definite, the stationary point at ##q=0## is a minimum of the potential, and you can diagonalize ##V_{jk}## by a rotation, which shows you that in this case small deviations from the stationary point are stable, i.e., you get an approximately harmonic motion, and the amplitudes stay small for all times. If, however, ##V_{jk}## is not positive definite, i.e., if the stationary point is a maximum or a saddle point, the general small deviation from the stationary point is not stable, i.e., the particle moves away from the stationary point (in most of the cases rapidly since it's accelerated exponentially at a maximum of the potential).

    In short this analysis shows you that a stable stationary state in such a case exists if and only if the potential has a minimum at this stationary point.
     
  7. Aug 26, 2016 #6

    haushofer

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    Yes, this makes a lot of sense, thanks! But how about my example of a mass falling down? (or the well-known pencil standing on its point and falls down, used to illustrate spont.symm.breaking; that's how I came to this question in the first place :P) There the kinetic energy is non-zero, but the system still evolves such as to minimize V(r), right?
     
  8. Aug 26, 2016 #7
    vanhees71, I'm just starting a module on classical mechanics and spotted this thread, to digress for a second, if that's ok, I understand the Langrangian is L = T - V and the q (dot) values are generalised velocities, but what's the g matrix that you've written in your non-relativistic langrangian for?
     
  9. Aug 27, 2016 #8

    haushofer

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    It's the metric, which in this case is just the Kronecker delta ##g_{jk} = \delta_{jk}## because space is Euclidean (flat).
     
  10. Aug 27, 2016 #9

    ShayanJ

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    That's true, but in flat space you can also use spherical or cylindrical(or any other curvilinear) coordinates which have their own metrics.
     
  11. Aug 27, 2016 #10

    vanhees71

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    In Cartesian coordinates of a particle in unconstrained motion you have
    $$T=\frac{m}{2} \dot{\vec{x}}^2.$$
    Then you can choose for the generalized coordinates ##q^j## the Cartesian components of the position vectors ##x^j##.

    Now if you have some motion, maybe subject to (holonomic) constraints, you start the calculation by parametrizing the position vector in terms of the independent generalized coordinates ##q^k##,
    $$\vec{x}=\vec{x}(q),$$
    and then evaluate the kinetic energy as
    $$T=\frac{m}{2} \dot{\vec{x}}^2=\frac{m}{2} \frac{\partial x^j}{\partial q^k} \frac{\partial x^j}{\partial q^l} \dot{q}^k \dot{q}^l,$$
    where the Einstein summation convention is applied (i.e., over twice appearing indices you sum). Now you can write
    $$g_{kl} = \frac{\partial x^j}{\partial q^k} \frac{\partial x^j}{\partial q^l} $$
    and get the exrpression for ##T## I wrote down in my previous posting.
     
  12. Aug 27, 2016 #11
    Thanks guys :)
     
  13. Aug 28, 2016 #12

    haushofer

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    Yes, I assumed Cartesian coordinates, but that´s of course not mandatory.

    My question still stands:
     
  14. Aug 28, 2016 #13

    ShayanJ

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    There are two kinds of equilibrium points, stable and unstable. You can see how they appear in post #5. The case you're asking about is an unstable equilibrium point. In the absence of external forces, there is no difference between them and the system just sits there. But if even a small external force is applied to the system, a system in a stable equilibrium point will start to oscillate around that point but a system in an unstable equilibrium point accelerates exponentially and gets away from the equilibrium point. This doesn't happen with zero kinetic energy although it may appear so. If you balance a pencil on its point in a world where no force acts on the pencil, it will remain in that situation forever. But in the real world there are a lot of noises.
     
  15. Aug 28, 2016 #14
    I know this will be a very unsatisfactory answer, but I think the fact that objects act to minimise their potential energy is just a experimentally proven fact of nature, I don't think it's actually known why it happens. Something worth investigating though if you're interested in it. Maybe you could research it and plan some experiments to test the generality of the assumption? See if there's any situation where an object in a force field doesn't tend to minimise it's potential energy etc.
     
  16. Aug 28, 2016 #15

    haushofer

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    Ok. So let´s take a marble in a halfpipe like construction, swinging from one point to another. Due to friction the kinetic energy will be converted in heat, the kinetic energy will become less and less, and because of VanHees71's comments the system will strive to the stable equilibrium of the potential.

    I have to think about my falling mass further, but I only realise now that the equilibrium point is unstable as you say.
     
  17. Aug 28, 2016 #16

    haushofer

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    Well, I just want to understand from which principles it follows. I understand that we probably won't find the 'why' for such principles. I'm writing a popular science book and was writing about the Higgs mechanism, which made me realize I never thought this question through.
     
  18. Aug 28, 2016 #17

    ShayanJ

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    I think I should correct the above sentence. This is better:
    If you balance a pencil on its point in a world where no force acts on the pencil,except for gravity and the normal force of the surface, it will remain in that situation forever.
     
  19. Aug 28, 2016 #18

    vanhees71

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    I've given the answer in #5. You define a class of forces occuring quite often in non-relativistic physics, namely conservative forces, which by definition are given as the gradient of a scalar field,
    $$\vec{F}(\vec{x})=-\vec{\nabla} V.$$
    Then from Newton's Laws (or even more fundamental Hamilton's principle) you get the equation of motion
    $$m \ddot{\vec{x}}=-\vec{\nabla} V.$$
    For a stationary point, i.e., a solution ##\vec{x}=\vec{x}_0=\text{const}## this implies that
    $$\vec{\nabla} V(\vec{x}_0)=0,$$
    i.e., a stationary point of the system is a stationary point of the potential.

    A stationary point can be a maximum, minimum, or a saddle point. A stability analysis, i.e., the treatment of the deviation of the solution from a stationary point in linear approximation, (see #5) then shows that only for a minimum of the potential the stationary solution is stable, i.e., the (locally) stable stationary points of the system are precisely the minima of the potential.
     
  20. Aug 29, 2016 #19

    haushofer

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    Yes, I see it now. Many thanks!
     
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