According to the second of thermodynamics

$$dS\geq 0\ \ \ \ \text{Isolated Process}$$ Is it write to say,

If the given system $S$ goes from state $A$ to state $B$, then the path taken by the system is the one in which entropy is maximized.

To me, it's No, Since the law says that entropy can't decrease but it doesn't say that it has to change by maximum.

  • 2
    $\begingroup$ maximized with respect to what? What is the range of possibilities from which the entropy maximizing one might be selected? $\endgroup$ Jan 14, 2022 at 13:26
  • $\begingroup$ Is System S and isolated system? $\endgroup$
    – Bob D
    Jan 14, 2022 at 13:39
  • $\begingroup$ @BySymmetry Maximized with respect to other path that can be taken from A to B $\endgroup$ Jan 14, 2022 at 15:12
  • $\begingroup$ @BobD Yeahhhhhhhhhh! $\endgroup$ Jan 14, 2022 at 15:13

3 Answers 3


According to the usual thermodynamics, the cited sentence, without further qualifications, is meaningless.

Entropy is a function of the state. It is meaningful to speak about the initial and final entropy for any transformation starting and ending in equilibrium states. There is no reason to believe, in general, that during the transformation, the system is continuously close to equilibrium states. Therefore, there is no control over the path. Even worse, in a non-equilibrium transformation, there is no path at all in the space of thermodynamic variables.

  • $\begingroup$ Is that mean, In non-equilibrium, the state of the system is not defined? $\endgroup$ Jan 15, 2022 at 10:46
  • $\begingroup$ @GiorigioP What about quasi-static process, Can we talk of path when the process in quasi-static? $\endgroup$ Jan 15, 2022 at 10:47
  • $\begingroup$ @YoungKindaichi Yes, in that case, there is a path in thermodynamic space. However, the path is pre-defined by the change of external conditions. There is no alternative to that path. $\endgroup$ Jan 15, 2022 at 14:28

A statement such as "If the given system S goes from state A to state B, then the path taken by the system is the one in which entropy is maximized," cannot be true in general without specifying that in a sufficiently small neighborhood of the path taken there is only one stable equilibrium point. In that case the statement of maximum is really just that the entropy as function of the state variables is a convex one, that is its second derivative matrix $\frac{\partial ^2 S}{\partial x_k \partial x_m}$is negative definite where $x_k$ are the extensive parameters of the system.

On the contrary, if there are multiple locally stable equilibrium points accessible then there is no guarantee that the natural path compatible with the constraints will seek out the one with the largest entropy. One can say only then that the entropy is maximized within a sufficiently small neighborhood of the equilibrium point.


If a given closed system goes from equilibrium state 𝐴 to equilibrium state 𝐵, then the change in entropy between these states is the maximum of the integral of $dq/T_{boundary}$ over all the possible paths taken by the system between these two end states. Here, $T_{boundary}$ is the temperature at the boundary of the system (with its surroundings) through which the heat dq flows.

  • $\begingroup$ But you are talking of change in entropy. I'm asking if we can sort out the correct path that system is going to take by maximizing something. $\endgroup$ Jan 14, 2022 at 15:15
  • $\begingroup$ There is not just one path that maximizes the integral; there are many paths, and they all give the same value for the integral. All other paths give a lower value. All the paths that give the maximum value are reversible paths. $\endgroup$ Jan 14, 2022 at 15:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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