# Is the second law of thermodynamics routinely violated?

**URL:** https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151
**Category:** Factual Questions
**Created:** [April 28, 2019, 12:15am UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151 "2019-04-28T00:15:23Z")
**Posts on this page:** 1
**Showing post:** 282

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### Author: ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)
#### Post date: [May 23, 2019, 4:15pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/282 "2019-05-23T16:15:34Z")

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> [@Max\_S](#):
>
> It is not obvious. If you observe the system for twenty seconds, no matter where the particle starts out it will only be at the far left position once while it will be observed at the 1 meter mark twice.

OK, let’s be precise: the probability of the particle being observed in any interval is proportional to the length of that interval.

> [@](#):
>
> And the “gas” is not at _internal_ equilibrium because we have moving particles within it. That’s why I don’t like thinking of thermal equilibrium as _internal_ thermal equilibrium. I could draw a line at 3 meters and considered the entire system to comprise of sub-system A to the left of that line [0,3] with sub-system B to the right of that line (3,10]. Therefore at every possible instant the temperature between the two subsystems is unequal, the two subsystems are not in thermal equilibrium, and the combined system as a whole is not in a state of _internal_ equilibrium. Not being in equilibrium, entropy can fluctuate (both ways if we aren’t assuming the second law).

But _all_ gases consist of moving particles. “Internal thermal equilibrium” simply means no heat flows between any two subsystems. If the temperature of two subsystems is unequal, then the entire system is still out of thermal equilibrium.

And, of course, a system at thermal equilibrium will experience microscopic fluctuations as usual. It is incorrect to say it is not at equilibrium; “equilibrium” does not mean “no motion”.

> [@](#):
>
> That is also what Mr. Eastman told me - I could not substitute _dQ[SUB]AuB[/SUB]_/_T[SUB]AuB[/SUB]_ for _dS[SUB]AuB[/SUB]_ because that formula only applies if the combined system is in a state of (internal) equilibrium. I hate bothering him but I still don’t understand _why_ the entropy formula requires internal equilibrium.

What is your reasoning that leads you to believe it does not? How exactly do you propose to define temperature of a system not at thermal equilibrium? Don’t say, “use a thermometer,” because one side might be hot and the other cold. And the microscopic definition assumes equilibrium as well.

Look, I honestly think you should carefully work through all the exercises on that web page (or any similar text). You can’t help but encounter all these definitions and concepts.

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