### decreasing entropy

I was thinking about the director of this previous post and how she
could use a mechanical device to generate the fire alarm.

A long and narrow cylinder has a slowly moving particle A on one side and a particle B at rest on the other (*).
She sets everything up on Sunday and simply waits until she hears the 'click' of the two particles colliding, which then triggers the fire alarm.

Here is my problem: It would seem that the entropy of this closed system decreases until we hear the 'click'; The entropy due to the unknown location of particle B is
proportional to lnV, but the volume V decreases with time as A moves
from left to right.

Notice that she can make the cylinder (and thus the initial V) as large as she wants and she could use more than one particle at B so that the initial entropy kNlnV can be arbitrarily large(**); And if the velocity of particle A is very small this decrease can take a long time ...

added 1 day later: I think I (finally!) figured out where the problem is with this puzzle. See the comments if you want to know my solution (instead of finding out for yourself.)

(*) While the initial position and momentum of A are well known (the
particles are heavy enough that we don't have to worry about quantum effects), the position
of particle B is unknown (but we do know that it is at rest).

(**) Of course the effort to set up the device will increase entropy by an even larger quantity, but all this occurs already on Sunday.

added later: I am no longer sure about that. She might have simply picked a cylinder
of unknown length, but shorter than 5m. The (right) end of that cylinder and the particle B would be identical. Now she sets up particle A on the left side to move with a (constant) speed of 1m/day and when A hits the other end (=particle B) it triggers the alarm (at which point she then knows the length of the cylinder).

I don't see how the act of picking up a cylinder of unknown length increased the entropy on Sunday.

#### 5 comments:

Cosma said...

What exactly are the macroscopic state variables here? One of them is "has the alarm gone off", and presumably the position and momentum of A are also supposed to be accessible degrees of freedom, but are there any others?

wolfgang said...

The only unknown (and thus constituting the micro-states) is the position of B.

wolfgang said...

It took a while, but I think I understand where the problem is.

If there is only 1 particle in the area B then on average the volume V will decrease to V/2 before the collision and the entropy decreases from log(V) to log(V/2), so the change dS is - ( log(V) - log(V/2) ) = -log(2)
The *average decrease* dS cannot be made arbitarily large by increasing V (as I initially assumed).

However, if there are n particles the volume V will decrease on average only to V*(n/n+1) and in the *thermodynamic limit* n-> infty we have dS -> -nlog(n/n+1) -> -1 , so that increasing n also does not increase -dS to arbitrarily large numbers.

The (constant) decrease of S seems to be of similar size as the "one bit" of information learned by hearing the 'click', which must increase the entropy of the person who learns about it (following Landauer etc.) by the same amount.

Anonymous said...

I cannot understand your solution, because the click would determine the position and momentum of both particles, so the V of your calculation would be zero?

wolfgang said...

The puzzle and the above calculation is about what happens *before* the 'click'.

In order to hear (or see) a 'click' the collision cannot be fully elastic, therefore the uncertainty about the momentum of both particles increases quite sharply (we assumed that we know the momentum of particles very precisely before the collision) and therefore it is natural to assume that the 'click' increases total entropy.

I hope this helps.

PS: This blog post was not one of my best and I guess quite confusing.
As I mentioned earlier, the quality of this blog continues to deteriorate (fully respecting the 2nd law).