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I recently thought what if I wrote the $2$'nd law of thermodynamics as:

$$ dS_{\text{universe}} \geq 0 $$

Note when it is written in this form it is time asymmetric. Let, $dS = S(t+ d t) - S(t)=$ where $dt \geq 0$. Dividing both sides of the inequality with $dt$:

$$ \frac{S(t+ d t) - S(t)}{dt} \geq 0 $$

However, if time flows in the opposite direction , $ dS= S(t -dt) - S$, then:

$$ S(t-dt) - S(t) \geq 0$$

Dividing both sides with $-dt$ again:

$$ \frac{(S(t)- S(t-dt)}{dt} \leq 0$$

Hence, if time flows the other direction then one will see entropy decreasing.

However, there is an obvious flaw to this which is the limit will not be continuous unless $\frac{dS}{dt}=0$ (for example a system which has reached maximum entropy.) But all of known physics is essentially continuous and time symmetric with the exception of the collapse of the wave-function. Does this mean the collapse of the wave-function is responsible for the $2$'nd law of thermodynamics?

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  • $\begingroup$ What do you mean "the limit will not be continuous unless $\mathrm{d}S/\mathrm{d}t = 0$"? Which limit? That of $\mathrm{d}t\to 0$? In any case, you need to be more precise about what you mean by "if time flows in the opposite direction", and how the $S(t-\mathrm{d}t) - S(t)\geq 0$ comes about - since $S(t-\mathrm{d}t)$ is the entropy at an earlier time, so that there is not $\Delta S$ but $-\Delta S$, and so should be $\leq 0$, removing the apparent problem. $\endgroup$
    – ACuriousMind
    Apr 29, 2017 at 12:10
  • $\begingroup$ Yes, $dt \to 0$ ... It's easier to see $S(t-dt)$ if one reverses the flow of time, by drawing a diagram ... If time flows in the opposite direction then then instead of going forward in time one would expect to go backward or $t-dt$ in time $\endgroup$
    – drewdles
    Apr 29, 2017 at 12:15
  • $\begingroup$ Are you trying to pose Loschmidt's paradox here? If so, possible duplicates: physics.stackexchange.com/q/19970/50583, physics.stackexchange.com/q/314437/50583 and their linked questions. $\endgroup$
    – ACuriousMind
    Apr 29, 2017 at 12:18
  • $\begingroup$ I see I wasn't aware it was called that ... my bad .. $\endgroup$
    – drewdles
    Apr 29, 2017 at 12:23

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