Fixed-Point Problem (2)

Originally posted by Palynka
A classical example of the Kakutani theorem is stirring a coffee cup. At least one molecule ends up in the same place it began.

(note: here molecule = point, so we suppose a continuity that doesn't really exist)

Assuming nobody moves the cup?

🙂

Originally posted by Palynka
A classical example of the Kakutani theorem is stirring a coffee cup. At least one molecule ends up in the same place it began.

(note: here molecule = point, so we suppose a continuity that doesn't really exist)

I think the function that describes the path of each molecule is completely continuous under classical physics and at least semi continuous (which fulfills kakutani requirements) under quantum mechanics

Originally posted by Palynka
A classical example of the Kakutani theorem is stirring a coffee cup. At least one molecule ends up in the same place it began.

(note: here molecule = point, so we suppose a continuity that doesn't really exist)

Take two pieces of 8*11 paper and lay them on top of one another so that every point on the top paper corresponds with a point on the bottom paper. Now crumple the top piece of paper in anyway that you wish and place it back on top. B's theorem tells us that there must be a point which has not moved, i.e. which lies exactly above the same point that it did initially.


I'm not sure the same applies to molecules in coffee. On the pieces of paper, each point remains in sort of the same place relative to it's neighbours.
In the coffee, molecules
ABCDEFGH could be re-arranged to
BCDEFGHA and all be in different places


with regard to the isotherms and isobars, first prove that there ARE 2 points with the same temperature (or pressure) when measured exactly.
There are an infinite number of temperatures between 14C and 15C

Originally posted by aging blitzer
I'm not sure the same applies to molecules in coffee. On the pieces of paper, each point remains in sort of the same place relative to it's neighbours.
In the coffee, molecules
ABCDEFGH could be re-arranged to
BCDEFGHA and all be in different places


with regard to the isotherms and isobars, first prove that there ARE 2 points with the same te ...[text shortened]... sure) when measured exactly.
There are an infinite number of temperatures between 14C and 15C

True. But in both cases, I think that we're assuming we're dealing with a continuum. This was stated in the coffee cup problem, but even in the temperature/pressure case is a pretty good model. After all, when you get down to small enough scales 'pressure' and 'temperature' don't really make a lot of sense - how can you have either in the gap between molecules?

And if we're dealing with a continuum, temperature and pressure are continuous functions. So there's no problem.

Originally posted by mtthw
OK, I think this is sound (though there may be some gaps to fill)...

Assume the pressure varies (if it doesn't the problem is trivial). Choose a pressure between the minimum and maximum values. Plot the points where the pressure has this value. There will be at least one closed loop (because pressure varies continuously).

Now plot a graph of ho ...[text shortened]... temperature.

If so, we've found our two points with the same temperature and pressure.

your answer only accounts for the two variables independently. for example, points A and B have the same temperature. and points C and D have the same air pressure. you can't really prove that the points are coinciding.

Originally posted by Stockton
your answer only accounts for the two variables independently. for example, points A and B have the same temperature. and points C and D have the same air pressure. you can't really prove that the points are coinciding.

No, it doesn't.

In the first step I'm restricting to points with the same pressure. So when I then find two of them with the same temperature, they have the same value of each.