Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anybody please explain what is the difference between equilibrium state and steady state, as quoted by book by Degroot and Mazur. Also, does violation of Principle of Detailed Balance means the system will not reach equilibrium?

share|cite|improve this question

The meaning of the word equilibrium depends on its context; it occurs in, inter alia, biology, chemistry, economics, and physics.

In physics, a good definition is that interaction rates are time symmetric - a process occurs just as fast forwards as backwards. As a consequence, the system's macroscopic properties do not change with time.

A steady state is a state with properties that do not change with time, but it isn't necessarily time symmetric.

share|cite|improve this answer

When looking at gas dynamics in particular, it is also possible that the steady state is not in equilibrium because the flow has become frozen. In other words, the energy is locked in each molecule and unable to transfer to others for various reasons, usually because the time between collisions is too large.

This is in obvious contrast to an equilibrium state.

An example of the difference between equilibrium and frozen flow for a rocket nozzle shows that the two are bounding cases for the real situation. The flow is in equilibrium in the subsonic region, non-equilibrium through the throat and eventually becomes frozen as the flow accelerates considerably through the nozzle.

To answer the second part, a violation of detailed balance does not mean that it will not reach equilibrium, just that it is not yet in equilibrium. Detailed balance is valid only at equilibrium -- if the system is not yet in equilibrium, it is not possible for the processes to be reversible.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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