7
$\begingroup$

Apart form the practical evidence that the systems that exist in nature try to attain lowest energy possible and hence, maximum stability, and atoms forms bonds to attain low potential energy but do we have any reason why is that ? Why low energy systems are stable?

What happens when a system loses its energy and why does losing energy (generally) signify stability?

This makes me wonder, what exactly is stability?

$\endgroup$
  • $\begingroup$ You have mixed atoms with thermodynamics. In thermodynamics, you should take account of entropy while estimating the stability of a system too. $\endgroup$ – Yashas Apr 23 '17 at 12:34
  • $\begingroup$ Yeah, i was aware of that. That's why i added the 'generally signify'. But u know what i mean by this question, $\endgroup$ – Mitchell Apr 23 '17 at 12:35
  • $\begingroup$ One can also argue in terms of $dU = -F.dr$ relation but that probably doesn't directly answer for the case of atoms. At potential wells, the system is stable because if it deviates from the local minima, the force acts in a direction to bring it back (unless you help it by giving energy to to overcome the potential barrier). If the system was at a potential hill at equilibrium, a small force would drive the system towards a potential well. Hence, systems have a natural tendency to end up in potential wells rather than hills. $\endgroup$ – Yashas Apr 23 '17 at 12:43
  • $\begingroup$ Another reason could be that as potential decreases, work is done by the system or energy is released in some form. In most cases, there isn't any constraint that prevents the release of energy. However, if something had to climb up the potential curve, you would have to provide energy and this is a constraint. A system can easily disperse energy but it is difficult for it to concentrate/obtain the right amount of energy to increase its potential energy. There are systems which are stable at very high temperatures as there is enough energy for a system to overcome potential barriers. $\endgroup$ – Yashas Apr 23 '17 at 12:45
  • $\begingroup$ @Yashas I think your 2nd comment is the perfect answer. We can also see it by expanding U at a point near the minima $\endgroup$ – Shashaank Apr 23 '17 at 17:44
1
$\begingroup$

Systems only settle into their lowest-energy state if they are in thermal equilibrium at zero temperature. If they are in thermal equilibrium at a nonzero temperature, then they settle into a statistical mixture of all possible energy states that minimizes the Helmholtz free energy. If they aren't in thermal equilibrium, then they can be in any energy state at all, determined by the initial conditions.

$\endgroup$
0
$\begingroup$

Think about what is required to change the state of a system.

A system in a low energy state requires energy be added from somewhere to bring it to a higher energy state. This therefore can't happen by itself.

However, a system in a high energy state doesn't need external energy to tranform to a low energy state. There may be a higher barrier between the high and low energy states so that a initial push is required. However, once that push occurs, the system emits overall energy. In some cases, this released energy can provide the push for other nearby systems to go from their high to low energy states. This releases more energy, which provides the push to more nearby systems, etc. This is called a chain reaction.

For example, a piece of paper sitting on your desk and the oxygen in the air around it are together in a high energy state. In this case, there is a high enough barrier between the two states that the paper doesn't spontaneously combust on its own. However, providing the initial push by, for example, lighting a match under one corner of the paper causes that corner to transform to a lower energy state. That release heat, which causes more of the paper to transform, etc. Eventually the whole sheet of paper burns up even though you only supplied a small push to one corner.

Now think of the reverse. What mechanism is there for the burnt paper to release its oxygen and return to the original paper state? Even if by some low probability of random states one molecule somewhere did release its oxygen. That absorbed energy overall, so there is now less energy available locally to kick nearby molecules into doing the same.

$\endgroup$
0
$\begingroup$

Atoms and molecules are coupled with the electromagnetic field. The electrons of the atom and molecule are accelerating and then it will emit electromagnetic radiation, making the system losing energy.

Classical mechanics will predict a total collapse of the atom. Turns out that Quantum Mechanics saves the day, introducing the concept of discrete spectrum and ground state, and the ground state will be a stable state for the system.

Quantum mechanics can describe the process of radiation of atoms and molecules, see this for more details of how this works.

Now, in general, systems are stable when they minimize the potential energy because there is always a force pushing the system towards the direction that minimizes the potential energy, i.e. the force $f_k$ applied on a generalized coordinate $q_k$ is equal $$ f_k=-\frac{\partial U}{\partial q_k} $$

if the system is conservative. If the system is perturbed at a local minimum of $U$, this force will always try to bring the system back to the minimum, and then will be stable if the perturbations are sufficient small.

Atoms and molecules are stable under sufficient small perturbation for other reasons. They are at the ground state, and any perturbation of the system will be reverted into electromagnetic radiation and sent to infinity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.