Equatorial temperatures

This one strikes me, and apparently others, as highly counter-intuitive, but it’s true because of mathematics!

Take any two places in the world; call these points A and B. Take any two paths between A and B that are the same distance; call these paths C and D. Let C(x) be as far down path C from A and D(x) is from B, and let T be any continuous function, such as temperature, altitude, or humidity. For all T, there is at least one value x such that C(x) = D(x).

For instance, if C and D are both five miles long, there is some point, perhaps one mile from A along C and one mile from B along D, where the temperatures are the same. There has to be.

The specific version of this states that, at any given time, there are at least two opposite points on the equator (or any great circle) that are the same temperature. But it holds regardless of the lengths or shapes of the paths.

Indeed, it can be generalized even further: C and D don’t have to be the same length; we could instead say that the ratio between C(x) and the entire length of C is the same as the radio between D(x) and the entire length of D. As long as that holds, there will be at least one value of x where it’s true. However, this is a little trickier to actually visualize, so let’s stick with the case where C and D are the same length.

If C and D are the same length, then each point on C can be cleanly mapped onto each point on D. Let T(C(x)) be the temperature (altitude, humidity, whatever) at point C(x) and T(D(x)) be the temperature (etc.) at point D(x). Then let ΔT(x) = T(C(x)) – T(D(x)): This is the key to understanding the proof. We are creating a continuous function of values which is the difference between the corresponding points on C and D.

At x = 0, that is, point A (for C) and point B (for D), ΔT(0) is either 0 or not. If it’s 0, then we have our pair of points: A and B have the same temperature (etc.), and we’re done.

If ΔT(0) > 0, then let ΔT(0) = L > 0. At x = length(C), that is, point B (for C) and point A (for D), ΔT(length(C)) = –L. As x moves from 0 to length(C), it moves from L to –L. It could go there smoothly, or it could meander stochastically, but it must go from a positive to a negative. And, by the intermediate value theorem (and just plain common sense), that means it must equal 0 at some point. The same argument holds if ΔT(0) < 0 (and therefore ΔT(length(C)) > 0).

At that point, the continuous value of interest is the same. Perhaps it’s three feet from A along C and three feet from B along D; perhaps it’s only inches. But it must exist.

Surely it’s possible, you might think, for C to be entirely below sea level and for D to be entirely above it! Therefore, you might think, the theory is incorrect, since every point on C is lower than every point on D. But in that case, the altitudes and A and B are the same. So even in the case where C is a valley and D is a hill, there must be some point the same distance from A (along C) and from B (along D) at the same latitude.

How does this apply to the specific case of great circles? Well, path C is one of the two semicircular paths from A to B and D is the other one; together, they form the great circle. Granted, this analysis superficially implies that the earth is a sphere (which it isn’t), but recall that we could be using ratios instead of raw distances: It doesn’t matter if C is exactly the same length as D, because we’re referring to “opposite points” on the planet, not to points that are exactly the same distance along the great circle.

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