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[Puzzleblogger Kevan Choset, March 15, 2006 at 11:16am] Trackbacks
The West Wing & Math:

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).

Aaron Bergman (mail):
Ah, the joys of the intermediate value theorem.
3.15.2006 12:18pm
Very Anonymous Coward:
Suppose not; then construct the obvious retraction; Brouwer fixed point theorem; contradiction.

A related result is that there is at least one place on the Earth where there is no wind.
3.15.2006 12:44pm
Freder Frederson (mail):
With what precision are you measuring the temperature. If you are measuring the temperature with sufficient precision, I would imagine the statement is never true. If you are measuring it within a degree or two then I would imagine it is often true if you are looking at the tropics over the oceans and in temperate zones during spring and fall.
3.15.2006 12:47pm
Anoymous Koward:
There are probably serveral points on the equator where that's true. So what is her point?
3.15.2006 12:49pm
Anoymous Koward:
Except that when it's daytime on one side, it's night on the other. So maybe not.
3.15.2006 12:52pm
Kevan Choset (mail):
Very Anonymous Coward -- that all sounds fine, but would you care to explicate?

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...
3.15.2006 12:59pm
Kevan Choset (mail):
(And for those of you turned off by the math, I promise you don't need anything complicated. As for the Intermediate Value Theorem, you don't need anything more sophisticated than the fact that if it's 0 degrees in NY and 10 degrees in Boston then it must be 5 degrees somewhere between them. (And 4 degrees somewhere between them, and 9.99 degrees somewhere between them, and pi degrees somewhere between them, and 0.[TheNumbersFromLost] degrees somewhere between them, etc., etc.)
3.15.2006 1:01pm
HeScreams (mail):
I haven't wrapped my head around the proof yet, but I'm certian that this spot has never been in my kitchen.

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).
3.15.2006 1:03pm
Robert Cote (mail) (www):
Temperature is not a discrete number. All you need is two antipodes with daily highs and lows where one place has a higher high and lower low and sometime during that day their temeperatures passed each other. Duplicate except hold temperature is held constant and location is varied. Is this some kind of test? What did I win?
3.15.2006 1:12pm
A. Zarkov (mail):
Very Anonymous has got it. The key assumption is continuity of the temperature field. But we need to prove it and my topology is rusty.
3.15.2006 1:20pm
JLR (mail) (www):
This is technically not a "proof," but there aren't too many places where land-to-land antipodes are even possible. For example, I believe that anywhere in the lower 48 states, if you "drilled a hole through to the other side of the earth," you'd hit the Indian Ocean. About four-fifths or so of Earth's land is antipodal to water.

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.
3.15.2006 1:22pm
Aaron Bergman (mail):
I'm not sure what the Brouwer fixed point theorem has to do with this.

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.
3.15.2006 1:26pm
Dan Levine (mail):
Kevin, it has to be true around the Equator. Just as it has to be true that there is a place on the Arctic Circle where directly across the Arctic Circle (not going through the center of the Earth) the temperature is the same.
3.15.2006 1:27pm
James Grimmelmann (mail) (www):
Consider just the equator. For each point on the equator, consider the difference in 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.
3.15.2006 1:31pm
Dan Levine (mail):
I like James's proof. Here's a graphic way to think about it. Let the x-axis refer to degrees around a circle. Let the y-axis refer to temperature. Draw any continuous graph that crosses the x-axis at (0,0) and (360,0)--the idea being that when you make it all the way around the circle and wind up at the same point the temperature is the same. Now draw the same-shaped graph shifted over 180 degrees to the right---i.e., the same shape, but crossing the x-axis at (180,0) and (540,0). At any point where the two graphs cross, there will be a point on the first graph, 180 "degrees" to the left, that has the same y-value (i.e., the same temperature).
3.15.2006 1:48pm
SLS 1L:
This is slightly less obvious to me, because I don't remember the relevant bits of calculus very well, so I'm going to have a go at it by intuition. If it's true in the following one-dimensional case, then it's true on the earth as well:

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.
3.15.2006 1:53pm
FXKLM:
I think CJ is wrong about the egg thing though. And even if s/he is right, is it in any way related to the antipode?
3.15.2006 1:54pm
SLS 1L:
I see several people have posted the same intuitive idea in a less complicated way while I was writing mine. I like Dan's best.
3.15.2006 1:55pm
Gavin Peters (mail):
Ok, so any great circle has two antipodal points at the same temperature on it. That's boring first year analysis stuff.

Here's the real question: demonstrate that on the earth there are two antipodal points with the same temperature _and_ pressure!
3.15.2006 2:10pm
LTEC (mail) (www):
Coward --

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.
3.15.2006 2:14pm
Philip S:
Consider just the equator. Let the temperature fcn. at time t be defined as f ranging from 0 to 2pi. If f is constant the result is trivial. If f is not constant define f over the whole real line periodically in intervals of 2pi. Now define g(x) = f(x) - f(x+pi). f is continuous by assumption, and since g is the difference of two continuous fcns. it too is continuous. Since f is continuous but not constant, therefore g must have both pos. and neg. values. Since g is also continuous, therefore there exists x s.t. g(x) = 0. QED
3.15.2006 2:16pm
Philip S:
whoops. Instead of "Since f is continuous but not constant" I should have written "Since f is continuous, periodic, and not constant." I think that should do it.
3.15.2006 2:21pm
Siona Sthrunch (mail):
In fact, there are antipodal points on the earth that have the same temperature and barometric pressure.

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?
3.15.2006 2:23pm
rbj:
Perhaps the Arctic after 6 months of winter and the Antarctic after 6 months of summer?
Or a spot on the equator during sunrise and sunset? I have absolutely no way of proving these.
3.15.2006 2:25pm
Jim T:
I don't think that you are reading the quote correctly. Her first sentence is:

"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.
3.15.2006 2:45pm
Kevan Choset (mail):
Dan,
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.
3.15.2006 2:49pm
Kevan Choset (mail):
James's solution was the one I had in mind. You can actually view the whole earth instead of just the equator, but use his trick of looking at the difference between 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.
3.15.2006 2:54pm
Aaron Bergman (mail):
There might be an easier way to do the pressure/temperature thing, but I came up with this (which is essentially an obstruction theory argument).

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.
3.15.2006 3:03pm
Tennessean (mail):
Is there any sense to the assumption that temperature is continuous? Maybe there is; certainly, it makes sense to assume that there is no place where the temperature is 0 immediately next to a place where the temperature is 100.

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?
3.15.2006 3:07pm
Juan Notwithstanding the Volokh:
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.
3.15.2006 3:20pm
John (mail):
There are an infinite number of such points. Connect any two points that are at the two ends of a diameter of the earth. Assume they are of different temperatures. Now rotate the diameter line until the ends are reversed. As each point was traversing the temperature range from what it started at to what the other end started at, the two temperatures must, at some point be the same.
3.15.2006 3:27pm
exfizz (mail):
Let C be a great circle (e.g. the equator) on the sphere S.
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 Lo for which d(Lo)=0, i.e. for which T(Lo)=T(Lo').

The choice of equator was arbitrary; the proof holds for any great circle, or indeed any circle or any closed loop on 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 visualize the problem: a closed loop with a rising and falling line above it. Rotate the loop 180o, i.e. map it onto itself. There must be at least one spot where the two lines cross.)
3.15.2006 3:51pm
lpdbw:
I have personally balanced an egg on end during an equinox.

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...
3.15.2006 3:52pm
exfizz:
Juan Notwithstanding the Volokh:
[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.
3.15.2006 4:18pm
eng:
The trouble is the phrase "through the Earth" which can quite reasonably be taken as through the center of the Earth. If you make such a restriction that there must be a line passing through the center of the earth, then it is false that there must be a pair of points on the perimeter intersecting such a line that are of equal temperature.

It is only once the statement is relaxed to mean "any straight tunnel" that we can be assured of finding a solution.
3.15.2006 4:19pm
SLS 1L:
eng:
It is only once the statement is relaxed to mean "any straight tunnel" that we can be assured of finding a solution.
I don't quite follow. If you consider e.g. my "proof" above, I think it's clear that any great circle has two points of equal temperatuer separated by pi radians - i.e. on opposite sides of the earth. I am hardly 100% confident of my reasoning or anyone else's here, but can you explain?
3.15.2006 4:28pm
james tierney (mail):
Eng--

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).
3.15.2006 4:31pm
Siona Sthrunch (mail):
For p a point on the sphere and -p its antipode, define the function

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:

Since the image of f is also connected, it contains a (non-contractible) circle that goes around the origin.
3.15.2006 4:32pm
DK:
The temperature _is_ continuous, although this gets into physics, not math, and you have to measure very precisely. In any medium allowing heat diffusion, diffusion instantly smooths any transient heat discontinuity into a continuous but initially steep shape. Thus, if you have a magic wand that can create a temporary temperature discontinuity, the discontinuity will vanish right way.

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.
3.15.2006 4:39pm
Jam (mail):
http://en.wikipedia.org/wiki/Antipodes
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.
3.15.2006 4:49pm
Tennessean (mail):
As an uneducated remembrance, it was my impression that temperature and pressure were closely related, and that pressure was a sampled statistic of a point _oer time_ used to "represent" an area. Also, although there is diffusion, it is not clear that this is an instant process.

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.)
3.15.2006 4:59pm
Juan Notwithstanding the Volokh:
exfizz:

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
3.15.2006 5:34pm
Philistine (mail):
Maybe I'm just confused, but isn't it pretty trivial to find counterexamples to the contention that there would mathematically always be an antipode with equal temperatures?

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.
3.15.2006 6:09pm
SenatorX (mail):
I think John is right and there are an infinite number of points. Basically that if the condition to refute is " there are two points on the earth exactly opposite each other" it infers that there are two points that share a common midpoint. This midpoint defines two equal lines.

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.
3.15.2006 6:21pm
Jadagul (mail) (www):
Juan notwithstanding the Volokh: since no cliff will be perfectly verical, 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.
3.15.2006 6:29pm
Kevan Choset (mail):
Philistine-
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.
3.15.2006 6:34pm
Philistine (mail):
Kevan Choset


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 could be 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
3.15.2006 7:05pm
Kevan Choset (mail):
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 possible examples.
3.15.2006 7:28pm
Dr. T (mail):
I do not believe that any of the given math proofs work. They make too many assumptions about the physics of temperature distributions that are not true. They assume that because one spot is temperature A and its antipode is temperature B, the difference A-B can be brought to zero by rotating around a plane that passes through the center of the earth. As others have noted, this assumes no temperature discontinuities. The only way that we can guarantee no temperature discontinuities is if we assume that the spots at which temperatures are measured are infinitely small, a physical impossibility (because temperature requires moving molecules, and molecules have size). Another false assumption is that the A-B difference as we rotate around the bisecting plane is linear. Why should it be? If the gradient is non-linear, then there may never be a place where the antipodal temperature difference is zero.
3.15.2006 7:58pm
SLS 1L:
Dr. T - if I understand the physics correctly, temperature becomes undefined, not discontinuous, when you try to consider it on a small enough scale. But continuity is a reasonable physical approximation.

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.
3.15.2006 8:13pm
Tennessean (mail):
SLS - Concedering arguendo that temperature becomes undefined when you try to consider it on a small enough scale (i.e., for a point as opposed to for a volume), why would you assume that "continuity is a reasonable physical approximation"?

(Again, this is not to detract from the elegance of the proof adduced above by many.)
3.15.2006 8:21pm
The Real Bill (mail):
I think uniformity is required as well, not just continuity.
3.15.2006 8:25pm
The Real Bill (mail):
Nope, uniformity is not required.
3.15.2006 8:49pm
SLS 1L:
Tennessean - few, if any, physical quantities that we treat as continuous without running into problems are actually continuous. Charge is quantized, but that doesn't mean "charge is continuous" is a bad physical approximation. Basically the same point holds for temperature.
3.15.2006 9:02pm
Tennessean (mail):
SLS: The normative judgment "bad physical approximation" is appropriate or inappropriate depending on the context, obviously, and, for many purposes, the assumption of continuity renders feasible many calculations and understandings with few errors or other problems.

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).
3.15.2006 9:21pm
Dick King:
Some cliffs are overhung [so when you climb them you are holding yourself on with your hands]. In that case, there is a particular latitude and longitude that has three or more points.

-dk
3.15.2006 9:56pm
Aaron Bergman (mail):
Siona: By connected, I meant path connected. Sorry.

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.
3.15.2006 9:57pm
Siona Sthrunch (mail):
Bergman writes:


It's easy to see that r is continuous and that r(0) = r(1), so it is a circle


It is not true that r must be a circle. For example, r could be an ellipse, correct? Or any closed antipodally symmetric loop.
3.15.2006 10:06pm
SenatorX (mail):
From the "problem" :"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)."

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.
3.15.2006 10:07pm
Aaron Bergman (mail):
Sorry. I was speaking loosely. Read "loop" where ever you see "circle".
3.15.2006 10:37pm
Lev:
Antipodes, islands, New Zealand - rocky uninhabited islands, 24 sq mi (62 sq km), South Pacific, c.550 mi (885 km) SE of New Zealand, to which they belong. Explored by British seamen in 1800, the Antipodes are so named because they are diametrically opposite Greenwich, England.

Prove to me that Greenwich and The Antipodes never have the same temperature.
3.15.2006 11:19pm
exfizz:
Lev: "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 180o out of phase), I'd wager that TG(t)=TA(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 TG(t)=TA(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
3.16.2006 2:47am
Daryl Herbert (www):
I agree with Kevan that James' explanation was best, and I think Kevan's illustration is best.

"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 :-)
3.16.2006 4:21am
BobGo:
So much nonsense is inspired by asking for a mathematical proof. Only Philistine seemed to appreciate that day/night and winter/summer real-world variations make it improbable, though not impossible, for temperatures at antipodes to be equal.

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.
3.16.2006 6:33am
AppSocRes (mail):
By definition the "antipode" of a point on a sphere is the point defined by the intersection of a line through the original point and the center of the sphere and the surface of the sphere and that is not the original point. All arguments using the Intermediate Value Theorem, Mean Value Theorem, Rolle's Theorem or any variation just show that at least two points on a circumference of a sphere have the same value for any continuous function mapping the surface of the sphere to the real number line. None of these proofs demonstrates that these two points are antipodes except in very special cases. The property everyone is attempting to prove is a global property of the sphere and dependent on the sphere's topology. What everyone's trying to prove must be a corollary of the Brouwer Fixed Point Theorem: probably a proof by contradiction involving a constructed pull-back function. I don't have time to work out the details. By the way, there is an elementary introduction to category theory -- a mathematically inclined high school student could handle it -- that contains a proof of the fixed point theorem using only category theory. Very neat stuff. I didn't see a proof of Brouwer's theorem till my junior year at university.
3.16.2006 8:25am
Aaron Bergman (mail):
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.
3.16.2006 10:59am
AppSocRes (mail):
I was wrong. Aaron Bergman has convinced me the Mean Value proofs work.
3.16.2006 2:03pm
Robert Ayers:
Aaron Bergman et all have the fine proof that involves
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
3.16.2006 2:33pm
Jam (mail):
The Earth is tilted, so that the poles are in complete darkness or complete light at certain times of the year. The only times that the poles may have the same temperatures is when the Earth's tilt is tangental to the Sun -- this happens twice a year.
3.16.2006 9:24pm
Siona Sthrunch (mail):
Bergmann wrote:

Sorry. I was speaking loosely. Read "loop" where ever you see "circle".


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.
3.16.2006 11:13pm
Siona Sthrunch (mail):
Also, Bergmann, I am only picking on you because you seem to be one of only about 5% of the posters on this thread who do not clearly have no idea whatsoever what mathematics or a mathematical proof even is; the remaining 95% of the posts cannot even be discussed.
3.16.2006 11:17pm
Siona Sthrunch (mail):
AppSocRes wrote:

By the way, there is an elementary introduction to category theory -- a mathematically inclined high school student could handle it -- that contains a proof of the fixed point theorem using only category theory.

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.
3.16.2006 11:21pm
Oren (mail):
I had no idea so many other math people read the VC!

/feels a little better now
3.17.2006 4:13am
BobGo:
Exfizz has nicely shown, with his temperature range comparisons for NZ and UK, that there are overlaps in late spring and early fall, with the likelihood of equal temperatures twice daily during those seasons. Not very likely during the rest of the year, however. And even during the overlap seasons, not at just any moment; you have to wait for the varying times when the temperature lines cross. There are a few other antipodal places on Earth where such seasonal temperature range overlaps could occur, but antipodal land areas are quite sparse.

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.
3.17.2006 5:56am
Very Anonymous Coward:
Siona Sthrunch:

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.
3.17.2006 6:51am
Very Anonymous Coward:
AppSocRes:

Yes, that was my first impression as well. But I agree now; the IVT is all you really need.
3.17.2006 6:53am
Some Guy:
BobGo--

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.
3.17.2006 2:31pm
PT (mail):
What's "IVT" ?
3.17.2006 2:57pm
John Bouvier (mail):
Since this does require continuity of both the temperature and the surface of the earth, wouldn't a discontinuity allow for the conjecture to fail?

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??
3.17.2006 6:00pm
Oren (mail):
PT - http://en.wikipedia.org/wiki/IVT (jeez, minimum of effort).

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).
3.18.2006 11:55am