Professor Ogurtani has already given the answer due to Gibbs. At constant (internal) energy, the entropy of an isolated nonequilibrium system can only increase. This is a rule that applies to a statistical system but not a purely mechanical system;. One can show that if one tries to define the entropy of a purely mechanical system, it must be a constant during its temporal evolution. Thus, if we can keep the entropy of a statistical system constant such as in an isentropic process, the statistical system appears to be formally identical to a purely mechanical system. Therefore, it is not surprising that the energy of such a nonequilibrium system will only decrease. The result can be proven for a system out of equilibrium, provided the entropy can be defined as a state variable in the spirit of an equilibrium entropy.

Hope this answer will provide some intuitive understanding of the issue.

Mechanics and thermodynamics are different . In the case when come together can be applied one or another approach, which depend on phenomena (regular, Gaussian , Levy's flight ).

When the entropy of a system is reach to the maximum, it means every subsystem in the whole system have the same influence on the system. And if the entropy is reach to the minimum, it means that only one of the subsystem have the most important influence on the whole system and other subsystem have no influence on the system. 

That was clearly stated by Gibbs under the title of ‘‘Criteria of Equilibrium and Stability’’ in p.56 The Scientific Papers of J. Willard. 1875. Dover Publ.

One has to be very careful in reading the constrained on the system such as:

1) In isolated system it is necessary and sufficient condition that for any possible variations in the state of the system, which keeps the energy invariant, the variations of entropy shall either vanish or negative. Maximum Entropy Hypothesis.

2) In isolated system it is necessary and sufficient condition that for any possible variations in the state of the system, which do not alter its entropy, the variations of its energy shall either vanish or be positive.  Minimum Energy Hypothesis. 

Don't forget this is for the reversible processes. What happens for the Irreversible Processes Gibbs doesn't say anything.

I have shown by computer simulation  in the conclusion section of  following  paper that for the dynamical dissipative systems,  for  the isobaric isothermal process excluding the final catastrophic stage, the global Gibbs free energy change,  which is negative with respect to time shows monotonic increase in the rate of variations that means the second variance in Gibbs free energy is negative. This means the second’s variation in the internal entropy variation is positive, due to well known Planck relationship extended by Ogurtani for the connection between the dissipation function and the Gibbs free energy in isobaric systems.    This behavior neither verifies nor opposes the universal validity of the Minimum Internal Entropy Production Hypothesis of Prigogine, MIEP.  Prigogine presets a minimum condition in the generalized force space keeping the conjugate fluxes constant, such as the second variance in entropy negative definite for those processes that may exhibit stable non-equilibrium states. The situation is exactly same for the maximum entropy production hypothesis of Ziegler, which does not furnish any rule for the time variations of entropy production rate such the second variance in entropy for the isobaric isothermal systems. '' 

Conclusion: the initial transient stage is control by Ziegler and the approach to the stationary non-equilibrium stage by Prigogine and the final stage most cases end up decimation or by catastrophe!!! 

                                                                                                                                                                                  Gibbs. , Volume One Thermodynamic

Professor Ogurtani has already given the answer due to Gibbs. At constant (internal) energy, the entropy of an isolated nonequilibrium system can only increase. This is a rule that applies to a statistical system but not a purely mechanical system;. One can show that if one tries to define the entropy of a purely mechanical system, it must be a constant during its temporal evolution. Thus, if we can keep the entropy of a statistical system constant such as in an isentropic process, the statistical system appears to be formally identical to a purely mechanical system. Therefore, it is not surprising that the energy of such a nonequilibrium system will only decrease. The result can be proven for a system out of equilibrium, provided the entropy can be defined as a state variable in the spirit of an equilibrium entropy.

Hope this answer will provide some intuitive understanding of the issue.

For isothermal non-equilibrium processes the Dissipation is represented either by Gibbs free energy or Helmholtz free energy depends whether one has isobaric  or isochoric systems. Then the  internal entropy production is connected directly to the DelSin=-1/T Del G>0  or DelSiin-1/T Del F >0, as I have shown in many places in my various publications. Where Del indicated Global variations including the surfaces.  In the case of isentropic system (that is not exactly the same as adiabatic system: SEE: R. Haase for his refined arguments on this topic) really one has the direct connection between the energy dissipation and internal entropy production that is the reason people like R. Haase uses dissipation function.  

 del S (internal) =- 1/T  Del E >0  Isentropic isothermal nonequilibrium Processes

I also use the rate of energy loss per unit time as power dissipation which is connected to  -1/T Del E/dt=  dSin/dt>0.   REGARDS

'Dear Dr. Gujrati;  All we know from the written literature are followings connections:  MODERN FORMULATION OF THERMODYNAMIC adopted from  MAX BORN,  Physik Z. 1921,v.22, 218 and M. PLanck , Ann.Phys. Lpz. 1887 v.30, p.583. 

.q= dU- w=dF + d(TS) -w= dF +TdS + SdT -w  for isothermal dT=0

 q- TdS =dF-w    where   q -TdS= 'q - TdSex - TdSin=-TdSin 0   unnatural , which never take place

For isochoric isothermal process  dV=0   w =0

dF< 0  Natural

dF=0 Reversible etc.

İsobaric case:

w= -SUM (P dV)  = - dSUM PV   = d(F-G) = dF-dG

w-dF=-dG< 0  Natural   Planck  etc......

Adiabatic process:  According to Guggenheim classification:

if   du =w  then  dS> 0   Natural Process ,  if du =w then dS= 0 Reversible 

if   dV= 0  , du = 0   then   dS > 0  Natural .  İsochoric İsentropic

Similarly

if  dH=0  , d P =0    then   dS> 0  Natural İsobaric Isentropic

Dear Siddhartha Pany

For dissipative systems, the principle of least action is unacceptable, since this principle is derived under the condition of holonomic constraints. A more detailed explanation can be found in: The dynamical entropy. International Journal of Science. – 2014. – P.30-36; Somsikov V. and Mokhnatkin  A. "Non-Linear Forces and Irreversibility Problem in Classical Mechanics," Journal of Modern Physics, Vol. 5 No. 1, 2014, pp. 17-22. doi: 10.4236/jmp.2014.51003.

Dear Vyacheslav, what do you think about local linearization ?

Dear Danilo Rastovic

Dissipative processes, entropy change are determined the second-order terms.

Therefore linearization eliminates the possibility of describing them, eliminates the evolutionary processes.

 Dear Professor Ogurtani:

I am not sure if I disagree with what you have said in the response you addressed to me directly, except what you said at the end when you say dS>0 for processes you identify as isentropic. To me, an isentropic process is the one in which S does not change so dS=0. It appears that you think of an isentropic process in which the exchanged entropy deS with a heat bath is zero . This means that you take  dS=diS>0. Our definition of an isentropic process is the same for a reversible process but differ for an irreversible process.

Puru

Thermodynamics is more general than dynamics (mechanics), in the sense, the former concerns not only with processes involving purely mechanical interactions, but also heat interactions. In principle, all ideal mechanical processes are reversible. A spontaneous mechanical process (no heat interactions) occurs so as to minimize the energy of the system.

However, when we consider adiabatic free expansion of an ideal gas, a process where no heat interactions are involved, we find the process is irreversible/spontaneous, and that the entropy of the universe increases. Internal energy of the system remains constant (does not decrease). This gives an example where mechanics does not predict the spontaneity of the process because the energy of the system does not change. We thus find that there can be no exceptions to thermodynamic results, and therefore, thermodynamic considerations always prevail over those from mechanics.

Radhakrishnamurty Padyala

We take energy of SP as a sum of internal energy and motion energy. Differentiating the SP’s energy and assuming that Newton’s laws describe the dynamics of each MP; we obtain the motion equation of dynamics of structural particles (SP) in a non-homogenous field of forces. This motion equation is irreversible

(see 1.                  Somsikov V.M. Why It Is Necessary to Construct the Mechanics of Structured Particles and How to do it. Open Access Library Journal, 2014, 1PP. 1-8, DOI: 10.4236/oalib.1100586).

Dear Gujrati,  You are perfectly right  since the god fathers of this term they claimed that no friction and no dissipation of energy should be considered  in isentropic process: dS=dSex+ dSin=0   not only dSin=0  but   also dSex=0.   That means it an  adiabatic reversible process.  For me it is an unnecessary   terminology. Haase in one place in his book tells that adiabatic process looks the same if the variations or departure from the equilibrium   state is very small? Even he is not  convenient very much what does it means!  

Actually he defines two different  reactions or processes: 1: adiabatic -isochoric process where U  and V stay constant. 2: adiabatic -isobaric process where H  and P stay constant.  He defines  entropy  S(u,v, Phi)  or S(H,P, Phi)  where  Phi is the decrease of advancement of the reaction or process.  He writes the Maclaurin expansion of entropy with respect to Phi in vicinity of the equilibrium state denoted by  Phi-bar, and shows that first derivative of entropy is  equal A/T, where  A  is the affinity of the reaction, which is zero at the equilibrium.  Then as a first order approximation the  entropy change becomes zero. That is the story of an isentropic or adiabatic process where the  reaction is really taking place in the proximity of the equilibrium but still entropy is invariant !!!!!!!  He calls this adiabatic isochoric or isobaric depends on the constraints on the system  Since both cases the first derivative of  entropy is A/T.   Best Regards.

Professor Ogurtani: We both agree with the conventional nomenclature. And I will not dare challenge you regarding classical thermodynamics of which you are a master. I think that with the current emphasis in the literature on nonequilibrium thermodynamics and our microscopic understanding, we need to define terms with the care that they deserve so as not to create more confusion. Isentropy is one such term that I believe needs to be properly defined so that the term means what the language assigns to it. I believe we should try to do so. If you are in the vicinity of equilibrium, one can treat approximately diS to be zero. This is not valid for a state far from equilibrium. If we insist on introducing the entropy S for such a state, I find it convenient to treat an isentropic process to be simply the one for which dS=0. However, there is no harm in considering deS=0 to imply an isentropic process. I prefer to call such a process to refer to a thermally insulated process. As long as we understand the qualification(s), we will all come to the same conclusion(s). 

First of all, I must thank you all for the in-depth replies. I never imagined it to evolve into an interesting discussion. 

The system I had in my mind while thinking of this question is about a bunch of free-electrons moving at non-relativistic velocities and experiencing forces from time-invariant electromagnetic sources. While external EM sources do influence the electron beam, the beam, with high current and energy can have a significant contribution to fields in the surrounding space as well - a typical particle-in-a-cell problem.

Frankly, I am clueless about the notion of "entropy" for this system. So, if any of you would enlighten me on the interpretation of entropy for this system. 

Thank you again. 

Entropy: A concept that is not a physical quantity

https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity

Dear Shufeng Zhang

I do not think so. Please see in attachments.

Calculus is not "take for granted", ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.

It is well known that calculus has a definition.

Any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.

Based on the definition of calculus, we know:

to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.

As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,

∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.

1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.

2) On the other hand, In fact, Q=f(P, V, T), then

∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).

We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.

that is, the so-called "entropy " doesn't exist at all.