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Hysteretic phenomena are often linked to dissipation. When there is a hysteresis loop, the dissipated energy can usually be computed as the area of the cycle.

For example, in ferromagnetic materials, the relationship between the magnetization and the magnetic field can exhibit a hysteresis loop, corresponding to the microscopic dissipation by Joule effect; in elastic materials, there is a hysteresis in the relation between the constraint and the extension, corresponding to the internal friction.

There are lots of other examples where I do not know exactly the dissipation processes: in all first-order phase transitions (e.g. liquid-gas), in the contact angle, and so on. I feel like hysteretic phenomena cannot appear without dissipation, because hysteresis needs memory as well as the possibility to lose this memory (which is an irreversible process). However, perhaps I miss some other possibility.

So, is hysteresis always linked to dissipation? Is it due to irreversibly? Is there a means to prove that formally?

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  • $\begingroup$ Is there a corresponding Joule effect in ferroelectric systems? $\endgroup$
    – John M
    Jul 22, 2015 at 19:50
  • $\begingroup$ I am not sure that you can interpret it as a Joule effect, but there are indeed dissipative processes in ferroelectrics, which are essentially the same as in ferromagnetic systems, but with ferroelectric domains of constant polarization instead of ferromagnetic (Weiss) domains of constant magnetization. In both cases, the reorganization of those domains is a dissipative process which causes the hysteresis of the system. $\endgroup$
    – Georg Sievelson
    Jul 23, 2015 at 12:20
  • $\begingroup$ To answer the second question you asked: to be reversible, a process must be quasi-static and undergo no hysteresis. Hence hysteresis is required for a process to be irreversible, given it occurs quasi-statically. $\endgroup$
    – inya
    Dec 25, 2016 at 19:26
  • $\begingroup$ @inya That's very interesting. Do you have a reference or explanation on that? I am not sure to see the link. $\endgroup$
    – Georg Sievelson
    Jan 2, 2017 at 17:10
  • $\begingroup$ @GeorgSievelson Blundell and Blundell, Concepts in Thermal Physics. think about a piston in a cylinder, original state variables $P_1, V_1, T_1$. If we add small pebbles to the top of the piston to compresses the gas, we can do this quasi statically. friction in the cylinder, will lead to some energy being lost. We reach $P_2, T_2, V_2$. We cannot go back to our original state along the same path, as taking pebbles off the top of the piston, quasistatically, does not give us our original energy lost due to friction back. As you can see here, Hysteresis has occured. $\endgroup$
    – inya
    Jan 2, 2017 at 19:18

2 Answers 2

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Bridgman in "The Thermodynamics of Plastic Deformation and Generalized Entropy", REVIEWS OF MODERN PHYSICS VOLUME 22. NUMBER 1 JANUARY, 1950, is discussing specifically stress-strain hysteretic cycles:

During the part of the cycle during which heat would be flowing in from the outside if there were no hysteresis, less heat flows in than otherwise would because the irreversible internal generation of heat takes the place of the heat of external origin, so that during this part of the process the entropy of the external universe decreases less than it otherwise would (that is, there is an equivalent algebraic increase). On the other hand, during the part of the process during which without hysteresis heat would flow out of the body to the surroundings, more heat flows out than otherwise would, the excess being generated by the irreversible transformation within the body. Again the result is a greater than normal increase of entropy of the external universe. On balance, therefore, the total entropy increases as it should after every cycle.

Later he generalizes the irreversible entropic description beyond stress-strain cycles, as well, but it seems that Bridgman associates dissipation with hysteresis if not explicitly then at least implicitly everywhere.

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  • $\begingroup$ As an aside: what does the phrase "algebraic increase" mean? I quick google search does not reveal its meaning. $\endgroup$
    – Kvothe
    Aug 5, 2021 at 8:32
  • $\begingroup$ @Kvothe "algebraic increase" ~ addition. Bridgman talks about how "heat" is going in/out during the various legs of the hysteresis cycle and if the cycle were reversible the amount of "heat" would equal going in or out but this is not the case when irreversible. Bridgman illustrates how at the end of the cycle the entropy returns to its original value in the magnet while increases in the outside world. $\endgroup$
    – hyportnex
    Aug 5, 2021 at 13:18
  • $\begingroup$ @hypornex Thanks, so "algebraic increase" just means "increase"? Is this common? Have you seen it before elsewhere? $\endgroup$
    – Kvothe
    Aug 5, 2021 at 15:00
  • $\begingroup$ @Kvothe old style, en.wikipedia.org/wiki/Percy_Williams_Bridgman $\endgroup$
    – hyportnex
    Aug 5, 2021 at 16:51
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In the control of certain systems, hysteresis is built into the open loop control logic to prevent the system from oscillating. For example in convertible cars, sometimes there is a fence extending above the windshield to change the airflow. The fence is extended at 35 mph and retracted at 25 mph, for example. If it was extended above 35 and retracted below 35, then it could be continuously going out and back in again when one cruises around 35 mph. If there is dissipation of a certain quantity, then it's at least not immediately obvious.

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  • $\begingroup$ What is a "loose running mechanical fit"? $\endgroup$
    – N. Virgo
    Aug 23, 2015 at 11:07
  • $\begingroup$ Noticeable spaces between parts. The interior part is smaller in dimension than the piece holding or containing it. $\endgroup$
    – jjack
    Aug 23, 2015 at 11:19
  • $\begingroup$ Ok, but the hysteresis in that case most certainly does involve dissipation. Without it all collisions would be elastic and the inner piece would continue bouncing around inside the outer one for ever. $\endgroup$
    – N. Virgo
    Aug 23, 2015 at 12:39
  • $\begingroup$ That's true. I'll think about it some more. $\endgroup$
    – jjack
    Aug 23, 2015 at 12:41
  • $\begingroup$ In control systems terminology, what you are describing is also known as gap control. $\endgroup$
    – David White
    Jan 26, 2017 at 3:44

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