Q:

In physics, why does the potential energy of a system tend to decrease? I'm referring to the tendency of systems to decay to a lower energy state. For example if you drop an object why should it move to a position with a lower gravitational energy instead of staying where it is.
I expect that in macroscopic systems, it could be a consequence of the second law of thermodynamics but that doesn't explain microscopic systems such as why does an excited atom emit a photon and decay to a lower energy state.

- Connell (age 27)

Ireland

- Connell (age 27)

Ireland

A:

This is a deep and general question. Your expectation about macroscopic systems is correct. Let me explain that for other readers.

The basic principle (the second law of thermodynamics) is that in the long run all the possible quantum states become equally likely. That means that things tend more and more to be found in forms compatible with the largest possible number of states. We keep track of that number by a quantity called the entropy, the log of the number. Entropy is convenient because the number of states for two combined systems (think of dice) is the product of the numbers for the separate systems. Since the log of a product is the sum of the logs of the factors, the entropy of the parts is just the sum of the entropies. So the second law says that the sum of the entropies of all the interacting parts goes up as far as it can.*

Why does that make a low energy state of one part more likely? Energy is conserved, so if one part has low energy, that leaves more for everybody else. That increases their entropy.

Now that we've gone through the explanation for macro systems, we see that it applies also to small systems, like atoms, so long as they trade energy with macro systems.If an atom is in its lowest-energy state, that leaves more energy to make entropy in the surroundings.

If the surroundings happen to have a well-defined temperature, that leads to the Boltzmann factor for the probabilities of states of our system. P_{state} is proportional to e^{-Energystate}^{/kT}. where T is the absolute temperature and k is Boltzmann's constant. Together with the rule that all the P's have to add up to one, this gives the probability of each state when its surroundings are at T.

Mike W.

* See for a somewhat more sophisticated take on this process.

*(published on 10/06/2013)*