Author |
Topic: Blending Water (Read 868 times) |
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Blending Water
« on: Dec 16th, 2006, 10:41am » |
Quote Modify
|
Let W be a blender of water containing n molecules of H_2 O. Turn on the blender and let it run as long as you like and then turn it off. Discover a way to show that there is at least one molecule m_k that is in the exactly same position it was before the water was blended.
|
|
IP Logged |
Regards, Michael Dagg
|
|
|
Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Blending Water
« Reply #1 on: Dec 16th, 2006, 5:38pm » |
Quote Modify
|
Umm... The direct physical interpretation is obviously false, and there are a number of possible mathematical models depending on how you define the word "position". I am particularly confused by the apparent blending of continuous and discrete. Could you be a little more explicit about the mathematical model you intend?
|
|
IP Logged |
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
|
|
|
Whiskey Tango Foxtrot
Uberpuzzler
Sorry Goose, it's time to buzz a tower.
Gender:
Posts: 1672
|
|
Re: Blending Water
« Reply #2 on: Dec 16th, 2006, 8:04pm » |
Quote Modify
|
Also, what does m_k mean here?
|
|
IP Logged |
"I do not feel obliged to believe that the same God who has endowed us with sense, reason, and intellect has intended us to forgo their use." - Galileo Galilei
|
|
|
physx
Newbie
Posts: 4
|
|
Re: Blending Water
« Reply #3 on: Dec 16th, 2006, 11:46pm » |
Quote Modify
|
This seems a direct application of the Banach Fixed Point theorem. You can state this problem in another way by supposing you have two identical pieces of paper superimposed on each other. Then, you take the top piece and crush it into a ball. You can prove there is a point on the crushed ball that is exactly on top of the point it was before it was crushed. This happens because crushing (or blending) is the same as mapping a function into another function. I've also seen this problem stated in yet another way: You observe a coconut, covered in its fibers. You can prove that there will always be at least one whorl of the fibers on the surface of the coconut. I'd really like to see other concrete applications to the fixed point theorem (mind you, there are many: http://mathworld.wolfram.com/FixedPointTheorem.)
|
|
IP Logged |
|
|
|
Michael Dagg
Senior Riddler
Gender:
Posts: 500
|
|
Re: Blending Water
« Reply #4 on: Dec 17th, 2006, 2:33pm » |
Quote Modify
|
m_k is just some kth modecule. I have a lot more to say about this problem I don't have much time at this moment to say more because I have to go somewhere but I will say that it is true for n >= 1, and discretely that is (but this is a trivial case) -- it is discoverable, however. The two-sheets of paper analogy is certainly related but the top sheet is a finitely multiply punctured domain as well and so it too in general is a discrete domain (I am not saying that a punctured domain is necessarily dscrete of course). There a couple of ways to get around this and really they are simple. A Banach space is not required here, i.e. so-to-speak, as there are domains that are Bananch space naturally (i.e. any Hilbert space works for this same problem, triivially nevertheless).
|
« Last Edit: Dec 17th, 2006, 4:14pm by Michael Dagg » |
IP Logged |
Regards, Michael Dagg
|
|
|
Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Blending Water
« Reply #5 on: Dec 17th, 2006, 7:54pm » |
Quote Modify
|
The problem with the fixed point theorem in this application is that he is discussing discrete molecules which, on that scale, are each surrounded by empty space. It is true that you can model the blending with a continuous function of which the molecules represent discrete points. But the fixed point thus guaranteed has 0 probability of happening to be at a point representing a molecule. If we represent the molecules as having larger than 0 volume, the probability is greater than 0, but is still not 1 unless we are unrealistic in our model in the opposite direction. And also in this case, can we say that a molecule is in the exact same position if it's orientation has changed? As for why I said that the direct physical interpretation is clearly false, all I need is one axis though the water on which no molecule falls (which - since matter is mostly empty space - is almost guaranteed to exist), and I can rotate about that axis, moving every molecule to a new position. Therefore, I must assume that either this problem is not meant to be interpreted discretely despite its reference to indexed molecules, and we really are meant to apply a fixed point theorem to obtain a point that does not move. Or else, we are intended to discretize space in some fashion, and the blending represents a rearrangement of the molecules in their various discrete spacial cubbyholes (which I doubt, as you can just move every molecule into the previous cubbyhole of the next higher molecule - with the highest numbered molecule moving into the 1st). Hence my confusion, and request for clarification.
|
|
IP Logged |
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
|
|
|
|