2
$\begingroup$

A system has maximum entropy when it has reached thermal equilibrium. But as statistical mechanics say, there is always an otherwise infinitesimal probability of particles to confine at a corner of the system. Definitely , this condition is not a state of thermal equilibrium. And thus , it has less entropy. So, the energy fluctuation lets the system to have a state which has less entropy. So, it is contradicting the Second law of thermodynamics which tells that a system must adopt a state where the entropy is maximum. So,what is the solution to this apparent contradiction? Where am I mistaking?? Please help.

$\endgroup$
0

2 Answers 2

6
$\begingroup$

The second law is statistic, and can be shown mathematically to be exact in the limit when the number of particles is infinite. In a real gas, or in a gas composed of a finite number of particles there will be fluctuations that will seem to violate the second law. for instance, the Poincaré recurrence theorem states that certain systems will, after a sufficiently long but finite time, return to a state very close to the initial state. The time needed to reach a state with smaller entropy increases exponentially to the difference in entropies. This means that the time you need to wait before a fluctuation increases with the size of the fluctuation, so the likelihood of a large fluctuation is very small. So small that we regularly do not see them in our daily experience. These times are really long: I do not have the exact numbers, but I do remember that the chance of a glass full of gas to spontaneously reduce its volume to fit to a corner of the glass would take more than the lifetime of the universe. So the second law is well and healthy for any practical purposes.

$\endgroup$
8
  • $\begingroup$ Ok,sir,at the time of fluctuations, though come very lately, does the Second law get violated? $\endgroup$
    – user36790
    Jan 2, 2015 at 18:14
  • $\begingroup$ The wiki article, it is written that second law which speaks for macroscopic bodies,doesn't get violated at all. And this is my question,sir,how??? $\endgroup$
    – user36790
    Jan 2, 2015 at 18:17
  • $\begingroup$ @user36790 It is violated in the classical sense (before statistical mechanics was developed). The "new" understanding, as the wiki article says, is that the second is only statistical in nature, which means that the prediction is statistical not deterministic, so there are circumstances where it "fails" (it is not an actual fail, as the "new" statistical formulation allows for these fluctuations). $\endgroup$
    – user65081
    Jan 2, 2015 at 18:25
  • $\begingroup$ I didnt find the quote you mention, but if it says that the second law is never violated for macroscopic objects: it is wrong, unless by macroscopic it means the limit when the number of particles is infinity. $\endgroup$
    – user65081
    Jan 2, 2015 at 18:26
  • $\begingroup$ Sir, can you please define me the word statistic that you and the wiki used frequently? And sir what is deterministic? $\endgroup$
    – user36790
    Jan 2, 2015 at 18:39
1
$\begingroup$

The solution to that question is to read the second law of thermodynamics, again. It says nothing about entropy. It says nothing about fluctuations. It does say something about heat flowing from hot to cold only, at least in the absence of other effects. Since this is the first time in the laws of thermodynamics that temperature was even mentioned, that qualifies the second law as the definition of temperature. Since it's not completely sufficient for that, we also need the third law to ground the temperature scale at zero. It's not that hard... but one has to actually read the laws.

Now lets talk about fluctuations in statistical mechanics and why they don't contradict the second law. For one thing the second law doesn't contain any language whatsoever about microscopic systems. More importantly, though, the term equilibrium characterizes the ability of a system to perform work. A system in equilibrium can not perform work. So why can we measure fluctuations (which imply that microscopically the system can perform work, even though it can't do that on average, either), if we go microscopic? Very simple: the measurement tools that we use to measure these fluctuations are assumed to be at a temperature of 0K! By going microscopic we changed the system from all warm to all warm plus a classical observer with an infinity of position and momentum probes that operate at absolute zero. That's not the same system. Strictly speaking, it's not even a thermodynamically allowed system.

$\endgroup$

Your Answer

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