Author |
Topic: Climate (opposite points on a sphere) (Read 740 times) |
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender: 
Posts: 7527
|
 |
Climate (opposite points on a sphere)
« on: Jul 5th, 2006, 2:01am » |
 Quote  Modify
|
I stumbled on the following problem on another forum. We couldn't come up with an elegant proof. I believe it is a well-known problem in mathematics. I'd appreciate comments, and I'm sure you'll enjoy discussing the problem.
Here it is:
Let's assume that the earth is a perfect sphere and that the temperature and air pressure changes continuously.
Show that there is at least one place on earth that has exactly the same temperature and air pressure as the place exactly opposite on the earth.
PS: funnily, just after posting the problem here, I found a nice proof. I didn't realize how efficient this forum is. Feel free to discuss it anyway.
|
| « Last Edit: Jul 5th, 2006, 2:53am by Grimbal » |
 IP Logged |
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Climate (opposite points on a sphere)
« Reply #1 on: Jul 5th, 2006, 9:34am » |
Quote Modify
|
This claim is an interpretation of Borsuk-Ulam Theorem, which in turn implies Brouwer Fixed Point Theorem.
The 1-dimensional case (e.g. equator) is easily analyzed.
|
|
IP Logged |
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Climate (opposite points on a sphere)
« Reply #2 on: Jul 5th, 2006, 9:58am » |
Quote Modify
|
Indeed. Looks like it.
I could prove it for the sphere, but it is probably much harder to prove it for higher dimensions.
|
|
IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Climate (opposite points on a sphere)
« Reply #3 on: Jul 5th, 2006, 7:05pm » |
Quote Modify
|
Temperature or air pressure alone is easy to prove for any number of dimensions: let T(x) = temperature at x - temperature at antipode of x. Connect a pair of antipodes by any curve. T is continuous along the curve (an assumption about the nature of temperature), and moves from positive to negative. Ergo, there is some point along the curve where T = 0.
One approach to including air pressure would be to note that the points for which T=0 must include a connected submanifold of dimension n - 1 preserved under taking the antipode. Defining A(x) to be the difference in air pressures between points on the submanifold and their antipodes, we can procede in the same fashion as with T to obtain A(x) = 0. Since this x is restrained to the submanifold T(x) = 0 as well.
Alas, actually proving that {T=0} contains the needed submanifold, while not overly hard, is rather messy. And you are bringing in some big guns with the manifold idea, anyway.
There are of course easier methods, and I assume Grimbal's "nice proof" is one of them.
|
|
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
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: Climate (opposite points on a sphere)
« Reply #4 on: Jul 6th, 2006, 12:37am » |
Quote Modify
|
Anyway, here is the proof I sent on the other forum.
First, to simplify the problem, consider the function
p(x) = (the pressure at x) - (the pressure at the opposite point)
t(x) = (the temperature at x) - (the temperature at the opposite point)
By definition, p and t have the property that for opposite points, they have the same value but with opposite sign.
The problem to find a point such that the antipodal point has the same pressure and temperature translates to the problem of finding a point x where p(x) = t(x) = 0.
OK, consider the equator. Go once around and map p and t on a XY chart. You will get a closed circuit. Due to the property on p and t, you will get a symetrical chart. Half of the loop will be the symetric image of the other half.
If the loop passes through (0,0), you've got a point where p=t=0. If not, due to the central symetry, you have to go around the center (0,0) at least once.
Now, move the equator to the north up a bit. It becomes a parallel. The loop transforms continuously into another loop. (It is not necessarily symetric any more). As long as the loop does not touch the (0,0) point, it has to loop around it. But as you move the parallel higher the loop will eventually become smaller, and when the parallel reaches the north pole, by continuity, the loop collapses to a point. Either it collapses on the (0,0) point or, at some point, the loop crossed over the (0,0) point. In any case, when the loop touches (0,0), that means there is a point x on the parallel such that (p(x),t(x)) = (0,0), which is exactly saying that the point has p(x) = t(x) = 0. It proves that such a point must exist.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print