This dialogue is from the West Wing episode, "Evidence of Things Not Seen."

C.J. There's a spot on the earth where the temperature is exactly the same as it would be if you drilled through the earth to the other side.

WILL No, there isn't. How about six dollars if you do it with a face card?

C.J. Yes, there is, and it's called the antipode. And if that's true, then why can't it be that you could stand an egg on end at the equinox?

Now, the problem. Prove that at any given moment, what C.J. is saying is true: there are two points on the earth exactly opposite each other which have the exact same temperature. You can assume that temperature is continuous (i.e., there's no spot on the earth where the temperature just "jumps" from 0 degrees at one spot to 100 degrees at a spot infinitesimally close to it).

A related result is that there is at least one place on the Earth where there is no wind.

Freder Frederson -- no matter how precisely you measure, it will always still be true.

Anonymous Koward -- maybe it's true on the equator and maybe it's not. But unfortunately that doesn't a proof make...

Koward: on the equator, it's always dusk somewhere, and it's dawn on the other side. At those spots it's at least possible that they have the same temp; but the problem is actually mathematical.

Freder: actually, you have it backwards. A lack of precision would allow the statement to be false; with sufficient precision the statement is always true (assuming it's ever true).

However, I believe parts of Argentina and Chile are antipodal to China. Also, I believe parts of Norway are antipodal to Antarctica. Furthermore, if memory serves me, parts of Spain are antipodal to parts of New Zealand as well.

Just pick a diameter of globe. Pick some plane through the center of the earth to rotate it around. Let f(theta) be the difference in the tempreatures between the two antipodes. Then it's clear that f(theta + pi) = -f(theta). So, the intermediate value theorem tells us that there's a value of theta between zero and pi such that f(theta) = 0.

QED.

differencein temperature between that point and the point directly opposite it (more precisely, the temperature at that point minus the temperature directly opposite). If this difference-function is zero for every point on the equator, then we're done -- any two opposite points have the same temperature.If the difference-function is nonzero at some point (call it A), then the difference-function is the negative of that nonzero value at the exact antipode of A. Consider the segment of the equator from A to A's antipode. The difference-function goes from some positive value to the negative of that value as you go along the segment. At some point along the segment, the difference-function must be zero. That point and the point directly opposite it have the same temperature.

Consider two points A and B at opposite ends of the earth and any "equator" defined by them. Let t(theta) be the temperature at the point theta radians from A along that circle (so t(0) = t(2pi) = the temperature at A and t(pi) = the temperature at B). The claim in this one-dimensional case is that there is a point theta0 such that

t(theta0) = t(theta0 +- pi).

If t is constant, then the claim is true and we are done. If t is not constant, then it has at least one maximum or minimum. Let theta0 denote any such maximum or minimum; since t(0) = t(2pi), t(theta) must take on every value between t(0) and t(theta0) at least twice.

Here's the non-rigorous graphical intuition element of the proof. Graph f(theta) on a standard cartesian graph (with theta on the "x-axis" and f(theta) on the "y-axis") from 0 to 2pi. Imagine a line segment of length pi parallel to the theta-axis "moving" along the function's curve starting at theta = 0 and and ending at theta = pi (i.e. the line segment runs from (theta, f(theta)) to (theta + pi, f(theta))). By graphical intuition, you can see that at some point theta' in this process both ends of the line segment must intersect the curve of f. Since the line segment is pi long, this means that f(theta') = f(theta' + pi).

Since we have proved that the theorem holds along every equator of the earth, it is true on the earth itself. Q.E.D.

Here's the real question: demonstrate that on the earth there are two antipodal points with the same temperature _and_ pressure!

I think you are making this too difficult. As the above arguments show, this works on any circle, that is, in two dimensions.

Your argument about (essentially) not being able to comb a hairy coconut only holds in three dimensions, since we can comb a hairy circle.

Let T be a temperature distribution on the earth and let U(T) be the set of points whose temperature is the same as their antipodal points.

We know from the comments above that U(T) intersects each great circle on the earth. Of course U(T) is symmetrical. In general, what properties are sufficient for a set S on the earth's surface to equal U(T) for some T?

Or a spot on the equator during sunrise and sunset? I have absolutely no way of proving these.

"There's a spot on the earth where the temperature is exactly the same as it would be if you drilled through the earth to the other side."

You could obviously say that she (or the writers) mis-spoke, but what she is claiming has absolutely nothing to do with two spots on either side of the Earth. She is saying there is some point on the Earth's surface, let's call it 'X', at which the temperature would be completely uneffected if a hole were to be drilled through the Earth all the way to the other side.

It seems to me that such a hypothetical hole would have to have SOME effect, though I suppose that would depend on it's radius. I also don't see how such a thing could be proven or disproven.

Yes, it's definitely true on the Equator. (It's definitely true on any circle, in fact.) But I believe the earlier suggestion about the Equator was trying to rely on some sort of special temperature-related properties that are true on the Equator, which I was trying to discourage.

differencebetween the temperature of every point on earth and the point completely opposite it. This is, of course, still a continuous function. It can't be positive everywhere (because if it's positive in NY then it's negative in the point directly opposite NY), and similarly can't be negative everywhere. So it's positive in some places and negative in others. This means that it must be 0 somewhere. Wherever it's 0, the temperature is the same at two opposite points.For p a point on the sphere and -p its antipode, define the function

f(p) = (T(p) - T(-p), P(p) - P(-p))

This is a map from S^2 into R^2. Assume that it never hits the origin. Then we have a map into R^2 \ {0}. We also have that if (x,y) is in the image of f, then so is (-x,-y). Since the image of f is also connected, it contains a (non-contractible) circle that goes around the origin. Take the inverse image of this circle. This gives a region on S^2 that contains a circle which maps to the circle around the origin of R^2. However, any circle on the sphere can be "filled in" (ie, the sphere is simply connected). Apply f to this disc. It must map to a contractible region bounded by the unit circle in R^2 \ {0}. No such region exists, so we have a contradiction. Thus the image of f must include the origin.

But if you assume, as I think we are, that the notion of place is not necessarily continuous (i.e., temperature is a measure of a discrete place or field), then why would temperature be continuous?

You can assume that temperature is continuous (i.e., there's no spot on the earth where the temperature just "jumps" from 0 degrees at one spot to 100 degrees at a spot infinitesimally close to it).But this is not a safe assumption. Take, for instance, a lava flow or an iceberg. It is possible that the temperature may well jump or drop by 100 degrees from one point to the next. More mundanely, it depends on how you measure "on" the earth. A cliff may have a particular temperature at it's edge, while the next "spot on the earth", which is "infinitesimally close" for the purposes of determining an antipode could be the ocean below, with a significant rise or drop in temperature due solely to the change in altitude.

Maybe you live on a perfect sphere, but my planet is more interesting.

Let L=[0,360] be a parameter that functions as a generalized longitude on C.

Let T(L) be the temperature (or any continuous scalar field) on S.

Let d(L) = T(L)-T(180-L) be the difference between temperatures at L and its antipode L'.

Since d(L)=-d(L'), the intermediate value theorem says there must exist at least one L

_{o}for which d(L_{o})=0, i.e. for which T(L_{o})=T(L_{o}').The choice of equator was arbitrary; the proof holds for any great circle, or indeed

any circle or any closed loopon the surface of S. It's just that the tidy notion of "antipodes" becomes less tidy in that case.(For me, the fixed point theorem comes into play when I

visualizethe problem: a closed loop with a rising and falling line above it. Rotate the loop 180^{o}, i.e. map it onto itself. There must be at least one spot where the two lines cross.)Being a scientist, I repeated the experiment every day for 10 days.

It's a matter of eggs and steady hands, and the equinox is irrelevant.

I don't watch that TV show, and the comment by "Will" makes no sense to me. There must be some context missing...

[it] is not a safe assumption [that temperature is continuous]. Take, for instance, a lava flow or an iceberg. It is possible that the temperature may well jump or drop by 100 degrees from one point to the next.Below the distance scale of the mean free path (~0.1 μm in room air), it is true that the continuity of the T field gets a little dicey, but (1) I think you can still define it using the KE of an individual molecule, and (2) outside of a physics lab you will not find a temperature measuring device capable of resolving at a distance scale and a time scale where this matters. Besides, even a lava/air system or ice/air system has a transition layer where temperatures match up.

Maybe you live on a perfect sphere, but my planet is more interesting.I realize you're joking, but (I think) the result holds for any closed loop on any surface equivalent to a sphere, including our beloved lumpy home. And far from reducing reality to a bland abstraction, math and science IMHO reveal aspects of its richness in a way that other ways of seeing do not.

It is only once the statement is relaxed to mean "any straight tunnel" that we can be assured of finding a solution.

CJ's quote doesn't require that it hold true for every (any?) point. Rather, she says that there is "a spot on the earth" for which this will be true, given that the line in question is through the center of the earth. If the spot is on the equator, runs through the center of the earth, and hits the equator on the other side, there is probably a solution (as other posters have pointed out).

Bergmann -

The first error in your proof is that it is not the case that a symmetric connected set in R^2 not containing the origin necessarily contains a circle.

Bergmann wrote:

Moving objects and the differences between materials complicate things, but to simplify, just replace "the earth" with "the atmosphere X meters above all buildings, ground, moving objects, and near-ground heat sources". The same argument applies to that layer of atmosphere.

In geography, the antipodes (from Greek anti- "opposed" and pous "foot") of any place on Earth is its antipodal point; that is, the region on the Earth's surface which is diametrically opposite to it. Two points which are antipodal to one another are connected by a straight line through the centre of the Earth.

At any given moment half of the Earth surface is absorbing energy (daytime) and the other half is releasing energy (nightime). The temperature will be the same at the point of equilibrium; where a spot has gained energy and it's antipode has released energy. Antipodal equilibrium will happen:

1) At noon when the rate of energy absorption and release are the same.

2) Towards sunset if the rate of energy absorption is faster than the rate of energy release.

3) Towards sunrise if the rate of energy absorption is slower than the rate of energy release.

In light of these three assertions, is temperature necessarily continuous?

(Not that this in any impugns the solutions offered to the actual problem posed - just an aside that seems interesting from here.)

But how would you account for cliffs and the like where an altitude difference might result in a significant temperature difference between two points that are adjacent for purposes of determining their antipode?

JNOV

Take the equator example. Picture the equator as a clockface--from 12 to 2 on the clockface (i.e. for 60 degrees of a circle)it is 100 degrees fahrenheit. At 2, it changes to 105. At 6 (180 degrees) it changes to 115 and at 7 is 110, and remains there until 11 when it goes back down to 100, returning to 100 at 12.

It makes more sense is you draw it on a piece of scrap paper. :)

There do not seem to be two antipodal points with the same temperature.

Now the condition "on the Earth" is ambiguous. But it clearly infers that any place "on the Earth" is fair game to satisfy the condition. Since the Earth is "sphere like" and further no condition AGAINST rotating the lines around the common midpoint exist... We define the possible range of probabilities as the area of a perfect sphere. Further though is "on the Earth" which contrast to "under the earth" or "above the earth" infers an area as infinitely as close to being "on the Earth" as possible. Let us decide as a tightest possible condition this area.

People think of the equator because the greatest chance of similar variables would occur to produce the same temperatures. Temperature is the final condition and so this is a valid source of points but not limited to as the asymmetrical nature of "the Earth" is a factor.

We have now defined the area that the matching temperature points could exist by refuting the any possible area where they could not exist. We do this at an infinite level because the statement to prove allows this. The places where the points can occur have extreme conditions BUT they are infinite because the movement of two points around a common point allows for infinite possibilities of success when MEASURING is the judge of success. Measuring can be as accurate as necessary, infinitely. So measuring temperature can in finely succeed at refinement(or grossness).

Basically because of "on the Earth", "two points exactly opposite", and a measurement(temperature) the condition to prove contains many infinites and so an infinite number of successful conditions can occur if only one condition of success can occur.

perfectlyverical, there will always be some slope and thus a smooth gradient. I guess you have to assume that the earth's surface is relatively smooth (sorry, I'm sure there's a technical term for this and I'm sure I'll know it by the end of the semester when I finish my topology course).Philistine: we assumed that the temperature was distributed continuously. You have four points of discontinuity there, which is why it doesn't work. But physically, you can't have a discontinuous temperature field for a non-infinitesimal duration.

Nope, it's not pretty trivial to find a counterexample, since there are no counterexamples!

In the case you describe, the antipodes will be between 2 and 6 and between 7 and 11. Between 2 and 6 the temperature changes from 100 to 115 and between 7 and 11 the temperature changes from 100 to 110. Somewhere in there, the points opposite will have the same temperature as each other.

I still don't see it. In my hypothetical, there

couldbe equivalent antipodal temperatures--but only coincidentally. The vast majority of situations would not have them.For instance--keep my hypothetical, but refine it. All temperature changes are not "instantaneous" but occur in the 1 minute (6 degrees of the circle) before they are given. (Thus from 100 to 105 occurs just before 2, 105 to 115 just before 6, 115 to 110 just before 7 and 110 to 105 just before 11.

If this is the case, I don't see antipodal equivalents. Am I missing something?

--Philistine

In your refined example, it is 105 degrees at both 11:00 and 5:00, which are antipodal. You can change your setup to avoid this problem, but it will create another problem. The beauty of a mathematical proof (like James's, above) is that we don't have to worry about trying lots of different examples, since his logic holds for

all possibleexamples.As for your claim about the A-B difference, t(theta) (the temperature at angle theta along a given great circle as we rotate around the circle) is a continuous function. So is t(antipode(theta)). Since both are continuous, t(theta) - t(antipode(theta)) is also continuous. If t is nonconstant, then the difference is positive at some point and negative at some other point, and by the intermediate value theorem it must take on the value zero at some point. No assumptions about linearity are required.

(Again, this is not to detract from the elegance of the proof adduced above by many.)

Here, however, the proofs all depend on the fact of continuity. So, it is a reasonable to question to ask. Just saying that it is a good assumption does not make it so (especially where we've already discussed that temperature is a statistic, not an attribute, as far as the temperature of a particular point at a particular time).

threeor more points.-dk

Let U be a path connected subset of R^2 \ {0} s.t. if p \in U, then -p \in U.

Pick a point p. Then, by assumption, there exists a path q(t) in U s.t. q(0) = p and q(1) = -p.

Define the following path

r(x) = q(2x) for x \in [0,1/2]

r(x) = -q(2x - 1) for x in [1/2,1]

By assumption again, r(x) is in U. It's easy to see that r is continuous and that r(0) = r(1), so it is a circle. It can be seen to be noncontractible in R^2 \ {0} by looking at polar coordinates, for example.

It is not true that r must be a circle. For example, r could be an ellipse, correct? Or any closed antipodally symmetric loop.

So let’s leave out of it the "real" world for the sake of the conditions in the problem to solve. I don't have a problem with the mathematical proof given so far for opposite points from an equal midpoint. Dr.T adds some excellent points I think. The thing is (and I wish someone could put this in mathematical terminology for me) that proving that is not enough. The problem calls for equal temperatures as well as antipodes.

The problem calls for a cross correlation of "same area" between the math proof given for the equidistant points AND the ranges where the temperatures are the same at any given instant. It is worth noting that it doesn’t say the antipodes(as far as space-time are concerned)must remain in any sort of continuity. They could exist in any of possible area that the conditions are able to be satisfied in at any given instant.

I think the key to the problem is given in the final statement about the temperature being continuous. It then even defines continuous for us by saying you cannot find a place to measure temperature, "at no spot on the earth", where a "spot infinitesimally close to it" has a non-continuous jump of value.

What does this mean if not that two points(or the range of area defined by the points in the surface of a sphere intersecting with the surface of the earth) are connected by a continuum of temperature? That you can not MEASURE any two spots "infinitesimally close", or more to the point, any two points NOT that close(range of min/max variable in earth topology) without finding a gradient of values between the two points. It doesn't just jump from 0 to 100, 25 to 75, or anything to any other thing without being able to MEASURE a gradient between the two points in space.

I am sure there is a mathematical way to formulate this. To me it means since there are an infinite number of gradients between two points equidistant from each other then measuring for value in those gradients(temperature) can give infinite results. Somewhere in infinity will be a matching condition at any given instant?

I also think you would never be able to prove this type of "problem" by observation of nature. It's mostly mental masturbation.

Prove to me that Greenwich and The Antipodes never have the same temperature.

"Prove to me that Greenwich and The Antipodes never have the same temperature."That's a fun example, Lev, and contra SenatorX, easy to investigate.

Considering that air temperature is a roughly periodic function (strong components at 1d

^{-1}and 1y^{-1}) varying widely about the mean, and that antipodal points share the same solar warming (just 180^{o}out of phase), I'd wager that T_{G}(t)=T_{A}(t) many times each year. (Of course, applied to the proxy data below the IVT guarantees at least twice/year.)It won't happen tomorrow. Using proxies for the Antipodes, tomorrow Auckland will range from 52-71 and Wellington from 54-68, vs. London only from 37-42; thus the ranges are disjoint. But April-June and again September-October the monthly average temperature ranges overlap significantly and we might expect to see T

_{G}(t)=T_{A}(t) more often. A simple "bot" could scrape the weather pages repeatedly and compare temps during those months. Good project for a smart HS kid.Appendix: Monthly avg temp for Wellington, NZ / London, UK

Jan 58-67 / 33-44 F

Feb 57-67 / 33-45

Mar 56-65 / 35-50

Apr 53-61 / 38-55

May 49-56 / 43-61

Jun 46-53 / 49-67

Jul 45-51 / 52-71

Aug 45-52 / 52-71

Sep 47-55 / 48-66

Oct 49-57 / 44-59

Nov 51-60 / 37-50

Dec 55-64 / 34-46

"If the difference-function is nonzero at some point (call it A), then the difference-function is the negative of that nonzero value at the exact antipode of A."

I missed that insight, which makes his solution much more elegant than mine. I proved it by showing that:

1 if your difference function ever returns 0, then the points match

2 if it's always negative, that means temp always decreases as you go around the circle, but that's no good because if you go around and around multiple times it keeps getting colder--but at 0* it must be the same temperature as at x* when x approaches 360.

3 if it's always positive, that means that means the temp always increases as you go around, with the same problem.

4 if it's both positive and negative at times, then it must pass through the origin (be equal to zero at some place), because it is continuous. (this is where the Mean Value Theorum actually comes into play)

But my proof is better, because it can be expanded to show to show that for ANY continuous cyclical path with continuous temperatures, for ANY distance between the two points (not necessarily "halfway" but 10%, or 10 miles, etc.) you will be able to find some two points that satisfy the distance requirement AND are at equal temperature. So there :-)

On the equator, early morning and early evening could have equal temperatures, but only if heating and cooling were exactly symmetrical and local weather didn't interfere.

In the tropical and temperate zones it would be possible that a winter daytime temperature would equal a summer nighttime temperature, but given the scarcity of antipodal land areas, variations of elevation, and local weather conditions, a rare coincidence.

Let's try the polar regions: thick icecaps provide good insulation, probably even from extreme seasonal variations, though no guarantee against geothermal variations.

Jim T's interpretation of CJ's original assertion is the cleanest answer: the temperature at a spot on the Earth would be unaffected by drilling (figuratively) through to the other side.

None of these proofs demonstrates that these two points are antipodes except in very special cases.I'm too tired to check if all of the proofs demonstrate this, but the majority of them definitely show that the antipodal temperatures are equal.

picking a pair of points A and A*, seeing that they do not satisfy T(A) == T(A*), then rotating the diameter until A

goes to A* and A* goes to A.

It is amusing that the same proof applies to this puzzle:

Given a true chair and a warped floor (continuous and

with some requirements on max error) show that you can

positon the chair so it won't wobble.

Bob Ayers

The current version of your proof (of Borsuk-Ulam) is still either malformed, substantively incorrect, or both.

It's malformed because you wrote in an earlier draft "unit circle" and there is no "unit loop." Frankly, the logic of your proof is still not written clearly, although it does use many standard mathematical terms.

It's substantively incorrect, to the extent that I can analyze its nonstandard terminology, because there is a continuous map from S^2 onto the unit circle in R^2, which sends an equator onto the unit cirle, and the unit circle is of course antipodally symmetric. I believe the point of your proof may have been that there is no such map, since your construction of the map f:S^2->R^2 from the vector field on S^2 has only the named properties, and yet is in fact existent, contradicting your analysis.

I would be curious to see this since the fixed point theorem depends on the topology of R^n. Brouwer does usually come from the homology or homotopy chain functors (assigning sequences of groups to embeddings of topological spaces); I don't see how this can be replicated at a high-school level using category theory.

/feels a little better now

I am assuming we are limited to land areas because “antipodes” means “feet growing out of heads” (as necessary for walking down under), and feet are for walking on land, aren’t they? So you’d have to look pretty hard for antipodal equal temperatures during much of the year, except maybe in the polar regions. Is the proposed problem intended to apply to any pairs of opposite points on the globe, whether land, water, mountains, or icecaps?

It would take a lot more investigation to look for daily temperature range overlaps between water/water opposite points and water/land opposites points. Opposite-point equal-temperature moments might be quite common in some seasons, But do they occur somewhere “at any given moment”? I can’t say for sure, and there is little in the comments that is even remotely relevant. It’s amazing to me how little the math discussion in these comments has to do with understanding of seasons and weather conditions on a real Earth.

Perhaps it's just for the 2-dimensional case? There are a number of cute ways of proving it that don't require very much machinery at all: Sperner's lemma, the game of Hex, some easy covering space arguments to get the fundamental group of the circle, etc.

Yes, that was my first impression as well. But I agree now; the IVT is all you really need.

I believe it is correct that there are not only at least two such points, but there are an infintite number of points (in fact, I think that these points must form a continuous path, that crosses the equator at least twice--and so it also crosses any circumference--but I could be wrong about that. In fact, there may be more than one such path, but there is at least one.)

The trick to thinking about it is, the points on this path need not all be the same temperature. We're only talking about equivalent temperatures--but any temperature can have an equivalent. We're also assuming that a temperature scale is smooth, so you can slice it up into infinitely fine degrees. This also means that it's not necessarily true for every single temperature on the scale--in other words, it might only be true at any given time for the range between 50F and 51F, for example.

It's mathematically true, but in this case the mathematics can reflect the real world.

Continuity of temperature requires a careful definition. It's true that at the sub-molecular level temperature loses it meaning. So we can reasonbly say that the temperature of a point in space is defined as the average for a sphere centered on that point. (Pick any size sphere, as long as it contains enough vibrating molecules to make the concept of temperature meaningful.)

But ...

In the real world there are truly discontinuities in the surface. The obvious example is an OVERHANG, noted by Dick above. If your great circle passes over an overhang, you truly get a discontinuity in elevation. And therefore you truly get a discontinuity in temperature.

Say the overhang is 10 feet high and in the sun. Say you step over the edge to the bottom, which is in the shade. The temperature change may be continuous over this 10-foot drop, but that 10-foot path is not part of the great circle.

So does this allow the conjecture to possibly fail??

Alright, here's another good one for you guys to flame over.

Prove (minimally!) that given the existence of some wind somewhere, there must be a cyclone somewhere on earth. You may assume that the earth is a perfect sphere with a 2D (flat) atmosphere (ie: wind has no radial component).