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Energy vs Stability 
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#1
Feb513, 12:22 PM

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I'm having trouble understanding the relationship between a system's energy level and its stability (in a general sense).
My understanding is that chemical and physical systems experience a driving force that pushes them toward the lowest possible energy state (ignoring quasisteady states and those things). Biochemistry calls this the Gibbs energy of reaction when the systems are chemical. The driving force represents the amount of nonPV (or useful) work that might be done by the system in moving toward that lowest energy level, and thus the system does the max amount of nonPV work by moving from an arbitrary energy level to the Gibbs energy minimum, or the energy level characterized by [tex] \begin{equation*} \frac{dG}{dt} = 0. \end{equation*} [/tex] But I do not understand why the driving force exists (i.e. why it is favorable thermodynamically for a system to minimize its free energy). In other words, I cannot tell from the the laws of thermodynamics why natural systems tend toward lowest energy states. And I also do not understand why a stable system corresponds to a lowenergy one. My engineering prof. calls the state of lowest energy the state of maximum stability. Stability is a term that is tossed around a lot it seems, but I don't really understand what it means. P.S. If the best answer would be something along the lines of "take a proper thermodynamics course," that would be fine. Thanks 


#2
Feb513, 06:13 PM

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Your picture is a little incomplete that's all.
A high energy state for a stationary ball would be on top of a hill. Give it a nudge and it rolls to the bottom of the hill  keeps going until it finds a depression, then it kinda rocks back and forth for a bit. So the depression is a lower energy state and stable. The driving force in this system would be gravity  mainly. The ball won't spontaneously roll to the top of a hill because there are other, dissipative, forces too  it makes a noise, heats up, shifts bits of the scenery it bumps into on it's way, that sort of thing. Taken together, all these things give the ball a "tendency to try to get to the lowest energy state available". The ball can sit in an identical geometry bowl and be stable at any height though... it may require more than a nudge of energy to get it out of the higher bowl  but once out the ball rolls to the lower bowl (there will be a highly unstable point where it could go either way). The "more than a nudge of energy" would be the reaction energy in chemistry  you have to give paper a bit of energy to start it burning but, once going, it will burn completely all by itself. That help? 


#3
Feb1813, 07:03 PM

P: 72

I agree with Simon.
As an example, in classical mechanics, when you have a particle in a central potential V(x), the force excerted to it will be: F=dV/dx Stability means that F=0, so dV/dx=0, V(x) is in a minimum or a maximum. However, when V(x) is in a maximum ( d(dV/dx)/dx<0), then a small pertrubation will lead the system in instability ( F#0). When, on the other hand, V(X) is in a minimum, every small pertrubation will lead it to oscillate back and forth from the minimum like an harmonic oscillator, and it will eventually return there. Thus, minimum potential energy equals greatest stability. 


#4
Feb1813, 09:22 PM

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Energy vs Stability
thanks e.chaniotakis  I'd just want to add that the oscillations die down only if there are losses in the system. The dissipation of kinetic energy is what provides the tendency to seek the lowest energy.
It would be nice to get some feedback from OP...(?) 


#5
Feb1813, 10:15 PM

P: 12

I'm sorry for my lack of feedback.



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