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Question

In thermodynamics does hysteresis always cause an increase in entropy?

Additional information

This question came about when thinking about two different definitions of a reversible process:

  1. A reversible process is one that is quasistatic and has no hysteresis.
  2. A reversible process is one that has no change to the surroundings when you do it followed by its reverse.

I am trying to see how these two definitions are equivalent.

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    $\begingroup$ The answer depends strongly on what you mean by "hysteresis", so please, define it. I am not trying to avoid the answer but would you call a Lissajous curve hysteresis? $\endgroup$ – hyportnex Oct 15 '15 at 21:36
  • $\begingroup$ It is interesting and kind of tricky, if we reverse the cycle in hysteresis, should be entropy decrease? The work would. The only helpful thing that I can think of is that if we apply Jarzynski equality to a cycle, then $\langle\exp(-\beta W)\rangle = 1$, and since the energy does not change, $\langle\exp(\Delta S)\rangle = 1$. Then using $e^x \ge 1 +x$, we get $\langle\Delta S\rangle \ge 0$. $\endgroup$ – hbp Oct 26 '15 at 17:33
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As hyportnex points out, the contention is over the meaning of the words. Define quasistatic as a necessary condition for reversibility, i.e. all reversible processes are necessarily quasistatic, then the task boils down to the logical gap between the converse and the contrapositive of the claim. In other words, one must understand quasistatic processes that are irreversible.

It is easiest to think of a cyclic process. In a reversible cyclic process there can be a leg of the cycle where entropy is decreasing. One can go further. If any leg of a cyclic reversible process has positive entropy, then there must be some leg of the process with negative entropy, and of course the net sum is zero. In an irreversible quasistatic process, there is a dissipative piece of the process, which forces the net entropy to increase. The dissipation necessarily has an effect on its environment hence the wording in (2).

Enter the definition of hysteresis. Suppose as time strictly monotonically increases, the end state of a system is different for different starting conditions, and in particular different starting times. Suppose hysteresis means that entropy is decreasing. This is the second point of contention because one could define hysteresis as a history dependent process, and not specify whether entropy necessarily decreases or increases (locally). With hysteresis decreasing entropy, and a dissipative quasistatic process (not cyclic) the entropy change can be zero (in the quasistatic limit; otherwise this violates thermodynamics). But such a process would not be reversible, since the history determines the end state, and any two unequal start times result in different states. This implies hysteresis is incompatible with equilibrium. But equilibrium is necessary for reversibility. Consequently, since all reversible processes take place in equilibrium there can be no hysteresis, ergo the need for the added condition to quasistatic in (1).

In fact, many physicists, including myself, characterize hysteresis as possessing critical phenomena. In terms of time, this implies the existence of a critical time scale. So in this sense, hysteresis itself is incompatible with quasistatic. So the answer to the original question,

"In thermodynamics does hysteresis always cause an increase in entropy?"

The response, in terms of definitions given here, is that hysteresis always decreases entropy locally, i.e. it causes spontaneous ordering, but the full picture with the environment means there must be either no global change in entropy, or an increase. However such a process cannot be done while remaining at equilibrium, and so the entropy must increase (globally).

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