We (batch mates) were discussing about time reversibility and Karan pointed out an example of system that is not in equilibrium but seems stable: all the particles in a box are moving back and forth in one direction perpendicular to a wall. Now this system is not in equilibrium but in a meta stable state.

My question is that there are many-many such possible states and during the counting of the states of micro-canonical ensemble we add all such states. Even though such states can never be achieved by a system in equilibrium. Why even then we have correct results for observables ?

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Dont quite understand your question here. What do you mean the system is not in equilibrium but in a meta-stable state?

This may be be somewhat unrelated, but your use of the term microstate seems to be incorrect.

A microstate is characterized by the instantaneous position and momentaof all particles in the system and so by definition, a configuration in which particles move back and forth is not a single 'microstate'.

But, on the other hand, if you are referring to a microstate where all particles have momentum in the same direction, sure, we do count it when calculating macroscopic observables but I fail to see how this has anything to do with violating thermodynamic equilibrium.

Venkateshan

PS: I may have missed the whole point though.

Many reasons can be given for saying that system is not in equilibrium. One can be that we will find the different values of pressure if we measure it from different walls. You can also see that the probability of occuring other states is absolutely zero other then these microstates. Third if you once disturb the system slightly it will never come back to its original state.

(Assume you have changed the direction of some particles now from momentum conservation it can never be in the initial state.)

If i have used word microstate then it is just a mistake ,I am talking about all the microstateS occuring in this system.

Thanks Vanky here you have given me right hint for answer. Actually we don’t count these states. Because during counting we have Integration{[(dp1)*(dp2)*(dp3)]*position-terms}

for each particle in multiplication.(for microcanonical ensamble) In this case 1,2,3 stands for x,y & z axis. For the states i am talking about two of (dp1), (dp2),(dp3) will be zero hence giving zero probability for this state.

** (dp1) is small element of momentum in x-direction.

Still dont get it. Are you saying that each of those particular microstates where the momentum of all particles is in a particual direction (perpendicualr to the walls) is not in ‘equilibrium’?

That dosent sound right becuase equlibirum is defined in terms of the constancy of the macroscopic parameters which is obtained by averaging over all possible microstates available (with certain constraints like total energy in the case of microcanonical ensemble).

So, how would you define equilibrium for a microstate?

I still have a feeling we are talking about two different things here.

Venkateshan

Yes Vanky I am saying that if a system attends one of the microstates(in which momentum of all the particles is in the direction perpendicular to one wall) then it won’t be in equlibrium.

Ok here may be reason for our distinct consideration. I am considering that the particles are separated so much that a particle never interact with other particles during its journey from one wall to other wall and returning back to the same wall.

I was also considering that all particles travel together.( for argument about momentum conservation)

Sorry for my incomplete description of system. Well this is what my intution says about equilibrium ,correct me if you still feel wrong.

Of course, we do assume that there is sufficient interaction amongst the particles whenever we are treating the system using principles of equilibrium statistical mechanics.

Otherwise how then would the system visit all possible microstates that are taken into account when calculating the partition function.

Venkateshan