Why do systems tend to favor the state with the lowest potential energy

1. Apr 12, 2010

marol

When looking at a macroscopic system (a pendulum for instance) we usually describe it in terms of terms of their general coordinates and assume Energy is conserved. We know this is an idealisation as real macroscopic systems are dissipative, they lose Energy in the form of heat due to friction. In fact this heat generation is nothing more than a transfer of the mechanical energy associated with the macroscopic degrees of freedom (E_kin of the Centre of Mass) to the mechanical energy associated with the microscopic degrees of freedom (movement of individual molecules). Now due to this unavoidable "energy loss" macroscopic objects tend to favor the state with the lowest potential energy, something known as a stable state.

Now it is very clear to me based on this logic why macroscopic systems tend to move towards equilibrium but it is unclear to me why this is also true for microscopic system like say an atom.

Let me elaborate, if we have a hydrogen atom in an excited state (thus not the lowest potential) it will fall back to it's ground state after a certain time by emitting a photon. Thus in analogy with the macroscopic system it loses energy, this time in the form of a photon instead of in the form of heat.

Now what principle is responsible for the fact that systems favor the lowest potential and where does it derive from? Or am I plainly wrong and do microscopic systems not favor the lowest potential at all.

2. Apr 12, 2010

Eynstone

This won't happen unless there is an external disturbance.
Also, consider a block with unequal sides.It can stand on any of its face & the equilibrium is stable. The lowest energy configuration is not inevitable , only a stable equilibrium is.

3. Apr 12, 2010

cesiumfrog

Another way of thinking about this is that by the equipartition theorem, the system will tend toward states where the energy is distributed evenly between all degrees of freedom. (So even the macroscopic modes should contain only microscopic quantities.)

Another way of thinking about this is to note that in an isolated enclosed box, you would expect the excited atom to periodically emit a photon, and then to reabsorb the photon back again, in a continuous cycle. But in the unenclosed situation you're imagining, after the atom eventually cycles to the unexcited state, the photon will rapidly depart, stranding the atom in the ground state. The asymmetry arises when (your attention is restricted to the case where) the background radiation energy density so happens to be very low compared to the atomic excitations.

Last edited: Apr 12, 2010
4. Apr 12, 2010

mikeph

It can happen in the form of spontaneous emission.

5. Apr 12, 2010

Andy Resnick

There's a difference between equilibrium and minimum energy. For example, any process where there is an energy barrier to overcome (for example, chemical reactions), the system will remain in a metastable state at a local minimum, not the global minimum.

But I like to think of the *observed* tendency of *all* systems to minimize their energy simply as "everything is lazy".

6. Apr 12, 2010

SpectraCat

This is just another way of stating the second law of thermodynamics, which basically says that the entropy of the universe must increase or stay constant over time. When a system has high potential energy, that means that it has stored some energy internally in an "organized" form. Releasing that stored energy into the surroundings allows that stored energy to become more "disorganized", which is the same as saying the entropy increases.

For a microscopic example like your H-atom, classical definitions of entropy don't really apply, since they are based on the statistics of large numbers. In this case, it is easier to understand the system in terms of the first law of thermodynamics (conservation of energy). If the atom is in an excited state, then it has some well-defined energy ... any other configuration of the system which has the same energy is at least theoretically possible. There are no states that are "more excited" that fit that description, since they are all higher in energy. However, lower energy states are accessible, provided that the atom spits out a photon via an allowed transition to balance the energy. This happens via the process of spontaneous emission, which according to quantum field theory (QFT) is what happens when a vacuum fluctuation perturbs the nominally stable excited state. Thus, during any time interval, there is a finite probability that the H-atom will emit a photon and go to a state with lower internal energy, but there is zero probability that it will spontaneously go to a state with higher internal energy.

7. Apr 12, 2010

marol

Thanks for all the answers. If I understand correctly now both macro- and microscopic systems move to states of increased Entropy due to the laws of thermodynamics.

Especially the post of SpectraCat helped me understand why this also happens for microscopic objects like the H-atom. Now the only thing left for me to do is learn QFT ;)