Question

Prove that for a given ring, there exists at least two diametrically opposite points, which will have the same temperature (temperature is continuous along the ring)

Answer 1

Explanation:

Okay, so lets just consider points along the equator, and let t:[0,2π]→R be the temperature at a point with angle θ from some predetermined point on the equator.

Now, we are given that t is continuous in θ on [0,2π], and we see that t is 2π periodic.

Define T:[0,2π]→R to be the antipodal difference in temperature, that is T:=t(θ+π)−t(θ)

Then T is also continuous on [0,2π], and we have that:

T(0)=t(π)−t(0) and T(π)=t(2π)−t(π)

So as t is 2π periodic, we get that T(0)=−T(π)

If T(0)=0 then we are done and we have our antipodal points with equal temperature, otherwise if T(0)≠0, then as T is continuous on [0,π]⊂ [0,2π] and without loss of generality T(0)<0<T(π), then ∃α∈[0,π] such that T(α)=0.

And then t(α)=t(α+π) so we have found our antipodal points with the same temperature.

This is really just a specific case of a really interesting theorem called the Borsuk–Ulam Theorem, which makes similar sorts of statements for n-dimensional spheres mapping to n-dimensional planes. Here we have a 2 dimensional sphere mapping to a 1 dimensional plane, but we considered a 1 dimensional subsphere (our equator), and the Borsuk–Ulam Theorem says on any continuous mapping of an n-dimensional sphere to an n-dimensional plane, there will be two antipodal points who get mapped to the same point.