The Meteorologists' Theorem

Date: 6 Jan 1995 15:48:34 -0500
From: Mizuki Nishisaka
Subject: Extreme Math Problems

Please help me on math problem.  This problem is extremely hard, 
I don't  even know where to start. Well, here is the problem:

Prove the "Meteorologists' Theorem":

At any given moment, there are two diametrically opposite points on the 
(spherical?) Earth's surface where the temperatures are equal and the 
barometric pressure are equal.  

I hope you will come up some kind of answer for me. Thank you.......

                                                   Mizuki Nishisaka
                                                   mnishisa@walrus.mvhs.edu

Date: 6 Jan 1995 19:56:53 -0500
From: Dr. Ken
Subject: Re: Extreme Math Problems

Hello there!

I really like this question.  It's a really amazing consequence of some neat
mathematics.  I'll see what I can do to explain it to you.  You're right, it
is hard.

Basically, what you've got here is an application of the Intermediate Value
Theorem.  It says that if you've got a continuous function f, and two
different input values (call them a & b) for the function that give you two
different output values (call them v & w) for the function, then for every
output value you pick between v & w, there's some input value between 
a & b that you can plug into f to get that output value.  In other words,
everything between v & w is hit by f.

So what does that mean?  Well, we're going to define a neat function, and
then show that it has some neat properties.  Take two diametrically opposed,
or "antipodal", points, and define a function f that gives you the difference
between their temperatures.  For instance, if we plug in the north pole and
the south pole, we'll get a difference of -30 degrees or something, and if
we plug in the south pole and the north pole (the order is important in
subtraction) we'll get 30 degrees, or whatever the opposite of the first
value was.  Note that I haven't done anything with the barometric pressure
yet.

Okay, are you still with me?  I hope so, because this is where it gets
interesting.  For the sake of argument, let's say that the temperatures at
the north pole and the south pole are different.  Now pick a continuous 
path that goes from the north pole to the south pole.  It doesn't matter 
which one you pick, because we'll eventually be picking a bunch.

Now travel along that path.  When you start out at the north pole, your
value for f is on one side of zero, and when you end up at the south pole,
it's on the other side of zero.  So somewhere in the middle there, it had to
be zero.  VOILA!  THE INTERMEDIATE VALUE THEOREM!!  So 
what you've found is a pair of antipodal points with the same temperature, 
because the difference between them is zero.  

Now pick another path, and do the same thing.  You'll get another pair of
antipodal points like that, that have the same temperature.  If you keep
doing that, with lots of different paths, you'll eventually get a whole
band's worth of such points.  When I say band, I mean that there is a whole
closed curve (a loop on the surface of the Earth) that contains only this
kind of points, points where the temperature on the other side of the earth
is exactly the same.  Also, keep in mind that if a point is on this curve,
its antipodal point is, too.  (Something to think about: why do these points
form a closed curve?  Think about the continuous nature of temperature.)

Okay, on to barometric pressure.  We'll essentially do the same thing here
as we did with temperature, and come up with a closed loop in which pairs 
of antipodal points have the same barometric pressure.  

Now we want to show that these two curves share a point, i.e. that they
intersect somewhere (actually, they'll intersect in two antipodal points;
think about why that's true).  Then the place they intersect will be a pair
of antipodal points whose temperature AND pressure are identical, since
these points lie on both curves.

Well, we can do that.  Let T be the temperature curve, and let P be the
pressure curve.  Then take any point x which is on P but not on T, and 
look at the pair of points x & the antipodal point to x, which I'll call x'.
Then x is on one side of the curve T, and x' is on the other side.  So
somewhere along the curve T, you have to cross P.  So that's where P 
and T intersect.  So that's the deal.

This is a pretty involved result.  I hope you followed me the whole way
through.  Write back if you have any more great questions like that!

-Ken "Dr." Math

Date: 7 Jan 1995 23:27:18 -0500
From: Dr. Ken
Newsgroups: local.dr-math
Subject: Re: meteorologists thm.

As t.foregger wrote,
> 
> Ken,
>        I read your proof. I think it has some parts that
> are not correct. In the intermediate value theorem
> there could be several points for which the function
> f is 0. Which one do you choose?
> I don't think you can rigorously show this result
> Borsuk-Ulam theorem) without use of algebraic topology
> results (see Massey, Algebraic Topology an introduction, p 170-172)
> But I think your "proof" as least makes the result seem plausible.
> It would be nice to have a proof just based on calculus.
> 
> tom foregger

Hello there.

Yes, I'm aware that the proof isn't completely rigorous.  That's because 
I was replying to the question of a high school student who I couldn't 
even be sure had taken calculus.

But about the intermediate value theorem, I don't really see the problem
with there being several nice points along the path from the north to the
south pole.  The only thing you have to do is show that the points that you
do get are arranged in a continuous fashion, i.e. that they form a closed
loop on the surface of the sphere.  Isn't that right?  That's the part that
I felt like I swept under the rug in answering the student, because it
looked pretty hairy.  Perhaps this would require some algebraic topology.
And I'll be honest here, I haven't studied very much topology yet, since I
haven't yet completed my undergraduate schooling.  But it looks like the
kind of thing you could do with a standard two-dimensional continuity
argument from calculus.

-Ken "Dr." Math