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Energy vs Stability

by Cole A.
Tags: energy, stability
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Cole A.
#1
Feb5-13, 12:22 PM
P: 12
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 quasi-steady states and those things). Biochemistry calls this the Gibbs energy of reaction when the systems are chemical. The driving force represents the amount of non-PV (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 non-PV 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 low-energy 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
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Simon Bridge
#2
Feb5-13, 06:13 PM
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Your picture is a little incomplete that's all.
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).
It is statistical - it is easier to go at random to a lower energy than a higher one for a similar reason that it is easier to randomly get a pile of sand than a sand-castle. The "driving force" is kind-of a blanket term to cover "whatever it is about the system that tends to shift it from stability" which sounds like a circular definition doesn't it? See below.
I also do not understand why a stable system corresponds to a low-energy one.
The instability of a system is it's propensity to move away from it's previous state if it is given a little nudge.

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?
If the best answer would be something along the lines of "take a proper thermodynamics course," that would be fine.
That would help - but even then a decent picture does not emerge until you start to contemplate postgraduate work. You are still at the stage of using the results rather than working the underlying math.
e.chaniotakis
#3
Feb18-13, 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.

Simon Bridge
#4
Feb18-13, 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...(?)
Cole A.
#5
Feb18-13, 10:15 PM
P: 12
I'm sorry for my lack of feedback.

The instability of a system is it's propensity to move away from it's previous state if it is given a little nudge.
This one line provided a great deal of clarification. I appreciate your input, both of you.


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