Is it hot in here?

True or False?

At any given point in time, there are at least two opposite points on Earth with exactly the same temperature.

Assume that temperature at a point is a continuous function of the coordinates.

True False

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13 solutions

João Areias
Jul 26, 2018

Relevant wiki: Intermediate Value Theorem

Consider taking the temperature T 1 T_1 and T 2 T_2 at two opposite points, and consider T = T 1 T 2 T = T_1 - T_2 .

Now consider taking the temperatures at their opposite points like so:

The difference of the temperature in the second case will be T -T . Since the temperatures and their difference varies continuously as you move the points through the surface of the earth, and it went from T T to T -T , it must have crossed zero somewhere, which would mean T 1 T 2 = 0 T_1 - T_2 = 0 so T 1 = T 2 T_1 = T_2 .


Edit: A lot of people seemed to be uncomfortable with the previous proof which is fine since my oversimplification can leave out some details, but I thought I'd go a bit more in-depth on what I was doing up there. When I say, take T = T 1 T 2 T = T_1 - T_2 , what I'm actually saying is, define T ( x , y , z ) = T 1 ( x , y , z ) T 2 ( x , y , z ) T(x, y, z) = T_1(x, y, z) - T_2(x, y, z) where T 1 T_1 is the temperature at the point ( x , y , z ) (x, y, z) and T 2 T_2 is the temperature at the antipode of the point ( x , y , z ) (x, y, z) . Now, the temperature at a given space will be continuous, so, T 1 ( x , y , z ) T_1(x, y, z) and T 2 ( x , y , z ) T_2(x, y, z) are continuous too and since they are a continuous function T ( x , y , z ) T(x, y, z) is continuous too.

Now, let's consider what happens if we evaluate the function at a given point ( x 0 , y 0 , z 0 ) (x_0, y_0, z_0) , we get T ( x 0 , y 0 , z 0 ) = α T(x_0, y_0, z_0) = \alpha now consider what happens if we evaluate the function at this point's antipode ( x 1 , y 1 , z 1 ) (x_1, y_1, z_1) , well we defined T 1 T_1 as the temperature at the point and T 2 T_2 as the temperature at the point's antipode, so T 1 ( x 1 , y 1 , z 1 ) = T 2 ( x 0 , y 0 , z 0 ) T_1(x_1, y_1, z_1) = T_2(x_0, y_0, z_0) and T 2 ( x 1 , y 1 , z 1 ) = T 1 ( x 0 , y 0 , z 0 ) T_2(x_1, y_1, z_1) = T_1(x_0, y_0, z_0) which means that T ( x 1 , y 1 , z 1 ) = T 1 ( x 1 , y 1 , z 1 ) T 2 ( x 1 , y 1 , z 1 ) = T 2 ( x 0 , y 0 , z 0 ) T 1 ( x 0 , y 0 , z 0 ) = α T(x_1, y_1, z_1) = T_1(x_1, y_1, z_1) - T_2(x_1, y_1, z_1) = T_2(x_0, y_0, z_0) - T_1(x_0, y_0, z_0) = -\alpha . Now we have a continuous function that goes from a negative value to a positive one or vice versa, so we can apply the intermediate value theorem and see that the function has a zero. What does it mean for this function to have a zero? It means that the temperature at that point must be equal to the temperature at its antipode.


Edit 2: For people that are still having a hard time understanding this problem, I'd like to first link to a few resources that explain it in a more visual form which helps, those are great resources and definitely worth checking, and second, I decided to go over one example here of the solution.

Now for the example:

We called T ( p ) = T 1 ( p ) T 2 ( p ) T(p) = T_1(p) - T_2(p) with T 1 T_1 being the temperature at the point p p and T 2 T_2 the temperature at the point opposite to the point p p . So let's take point a a with a temperature of 20 ° \ang{20} and point b b opposite to point a a with a temperature of 10 ° \ang{10} . If we evaluate those functions we get that T ( a ) = 20 10 = 10 T(a) = 20 - 10 = 10 and T ( b ) = 10 20 = 10 T(b) = 10 - 20 = -10 , now consider moving around and evaluating this function on the points from a a to b b , no matter the shape of the function, or the shape of the earth, or where the sunlight is hitting, we have a continuous function that goes from 10 10 to 10 -10 and since it can't jump over the x-axis, it must cross 0 0 to do so, which means that, somewhere between a a and b b , and we don't know where, there is a point c c where T ( c ) = 0 T(c) = 0 .

Bonus: Show that at any given time there are two opposite points with the same air pressure and air humidity.

This is a special case of the Borsuk-Ulam Theorem .

Brian Charlesworth - 2 years, 10 months ago

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And unprovable it seems?

Kevin Luddy - 2 years, 10 months ago

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What's unprovable?

Richard Desper - 2 years, 10 months ago

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@Richard Desper unprovable because earth conditions are constantly changing through time as the link Brian provided states.

Roger AB - 2 years, 10 months ago

This is Cool. Really cool

wonde nicholson - 2 years, 10 months ago

I’m familiar with the theorem, but not sure this proof makes sense. Your proof seems to be an attempt to claim that any two arbitrary points opposite on a sphere have the same temperature which is not true.

Sean McCloskey - 2 years, 10 months ago

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I think the concept behind your proof makes sense. What you probably meant to say was if traveler A leaves from T1 to T2 in one direction and traveler B leaves from T2 to T1 in the opposite direction simultaneously at the same speed and measure the difference between the temperatures between A and B, at some time before they reach their destinations their temperatures will agree. This is actually a more strict proof than the problem because it proves that such points exist on any great circle.

Sean McCloskey - 2 years, 10 months ago

Thanks for your comment. So, the point I was trying to make, and I'm not sure I was able to make it, is that when you take T1 - T2 you actually create a function that takes a temperature T1 at a given point a and a temperature T2 at a's antipode. You may think of it as travelers going at the same speed, or any other form, me I like to think as a line crossing the center, but the fact is, once you pick a point a, the other point is not arbitrary, it's a's antipode. The other important realizations are that if you evaluate the function at a and get a value α \alpha , evaluating at a's antipode will give a value α -\alpha and that temperatures vary continuously so their difference also varies continuously. With those, it's enough to claim by the intermediate value theorem that there will be a point where if you evaluate the function there, the result is 0 which means that a and it's antipode are at the same temperature. Maybe you can help me elaborate on the proof so it's more clear for future readers, how do you think I could restate this without getting convoluted? My intention was to write an approachable problem for beginners.

João Areias - 2 years, 10 months ago

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I wrote my second comment below before your reply to my first comment was posted. But as I said in my second comment, I think your explanation makes sense, and I probably just misinterpreted it. If there is anything you could have said to make it more clear, I would say probably defining the points f(alpha) and f(-alpha) as antipodal points along a great circle between T1 and T2, and saying that at some point on the path f(alpha) = f(-alpha) = 0, (rather than T1 = T2 which to me seemed to imply that the starting points were equal).

Sean McCloskey - 2 years, 10 months ago

You define the function f:S^2 -> R defined by f(x,y,z) = T(x,y,z) - T(-x,-y,-z). Note that f(x,y,z) = 0 iff T has the same value at (x,y,z) and at its antipode. Now suppose f is non-vanishing. Then it is a continuous function from the sphere to the real numbers that takes on both positive and negative values without taking on a value of zero. This violates the fact that the continuous image of connected set must be connected. (A generalization of the Intermediate Value Theorem that holds in general for all metric spaces.)

Richard Desper - 2 years, 10 months ago

There is a subtlety of the understanding of the proof in that it is not that there are two arbitrary points that have this function, rather there exists a specific pair of points that have this function. I don't see that the proof used claims this at all. Part of the problem in the understanding of this proof is in realising (and accepting) that proving this to be true is simple, but finding the location where it is true is nearly impossible.

David Finch - 2 years, 10 months ago

New theorem to me, learned something, thanks sir!

Kelvin Hong - 2 years, 10 months ago

yeah but where is the part that justifies that the two points have to be directly opposite to each other....people who think this is right have not thought about your answer hard enough. you merely claim that there exists two points such that t1 = t2 but not the positions

William G. - 2 years, 10 months ago

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The thing is, I defined them to be like that and then showed that there are such points, they don't need to be the only points, neither do I need to know where they are, this proof only shows that it exists. I suggest you to check Sean McCloskey's reply to my proof, where Richard Desper and I went more in depth on the proof.

João Areias - 2 years, 10 months ago

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i thought about it a little more and it makes sense

William G. - 2 years, 10 months ago

I disagree. What situation when we consider point on Equator with the maximal temperature? In this case there is on Equator not any second point with the same temperature!

Vratislav Hlubucek - 2 years, 10 months ago

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This method tells that there are such pair of points, not where they are. Not every point will have a corespondent point with the same temperature but there exists a pair of points somewhere right now that are opposite on the globe and yet are at exactly the same temperature.

João Areias - 2 years, 10 months ago

I disagree: Imagine a world where the Southern hemisphere is at 60 degrees, and the Northern one at 70. There is no point with an opposite with the same temperature. If you're uncomfortable with the splitting line (the equator), you can say that the temperature is rising evenly from 60.1 to 69.9 in micro increments (as small as you define a "point"). You could even extend it to the entire globe: start somewhere at 0°F and set the temperature of one adjacent "point" to be that plus 0.000000000000001 °F. Repeat until the whole globe is covered. No two points will have the same temperature.

Noémie Durand - 2 years, 10 months ago

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Your temperature gradient isn't continuous.

José Torspecken Lobo de Miranda - 2 years, 10 months ago

The scenario you suggest is flawed. Firstly the whole Southern hemisphere cannot be 60 and the whole Northern 70 without there being a gradient point between them where the temperature moves from 60 to 70, as it is not possible for it to just switch from one to the other at the equator...that's just straight physics. Secondly...if you assume that the whole Southern hemisphere is colder than the northern hemisphere (which is what I assume you are suggesting), that still leaves the equator to have at least one pair of opposite points with the same temperature...so the stated case is true. Check the James Grime video referred to above, as it explains it quite simply.

David Finch - 2 years, 10 months ago

IF the question had been about 'a theoretical earth sized sphere of homogeneous composition and linear temperature variations over distance' I MIGHT buy it. As soon as you said 'Earth' its a whole new game. Ask any meteorologist. (edit: I understand the math now and agree that at least one pair of such points must exist.)

Garrett Roy - 2 years, 10 months ago

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There is no need for the earth to be spherical, it can be blob-shaped and this would still work, as long as the function is continuous if you go from negative to positive you must cross zero

João Areias - 2 years, 10 months ago

Is there any assumption on symmetrical order needed that I missed? I explicitly don't see the answer for the following example: What if I draw a spiral from south pole to north pole with each point of the spiral having a different temperature. How can there be opposite points with same temperatures then?

Philipp Mielkau - 2 years, 10 months ago

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this is exactly what i was thinking

Vahid Fozi - 2 years, 10 months ago

Your function sill goes from a negative to a positive, so you still must cross 0, don't matter what path you take. Now, the proof I posted doesn't make it obvious but there is not just a single point, there is actually a continuous band of points like that which you have to cross in order to invert the points. You may prove this by contradiction, imagine you only have a point, or you have multiple points but not continuous, then you can draw a path where your continuous function can go from negative to positive without crossing zero, which goes against the intermediate value theorem , so, there is a continuous band that you must cross in order to go from a negative to a positive. You can also see this using the Borsuk-Ulam Theorem which, if you check my answer, there is a video from Vsause that explains it very well.

João Areias - 2 years, 10 months ago

And this holds true even though Earth isn't a sphere? Not only is it oblate, it has a rough surface.

Samuel Penn - 2 years, 10 months ago

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Yep, that's the magic of calculus

João Areias - 2 years, 10 months ago

This hold true for anything topologically equivalent to a sphere, for example a cube.

CHIN KEE HAW - 2 years, 10 months ago

Here is a great video by ThreeBlueOneBrown explaining and using the Borsuk Ulam theorem, highly related to this problem https://youtu.be/FhSFkLhDANA.

Victor Leq - 2 years, 10 months ago

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Nice, great video, I will link it with the other ones up there

João Areias - 2 years, 10 months ago

Let's look at this from a qualitative standpoint. Since we are at points on the equator, the sun's rays or the ability of the ground to hold the heat should be able to produce the same temperature . One point is day and the other is night could be a factor. Parts of the Eastern Hemisphere stay hot in the evening and African deserts can pretty hot during the day.

Robert Eligman - 2 years, 10 months ago

The positioning is the part that makes this completely false. "At any given point in time, there are at least two OPPOSITE points on Earth with exactly the same temperature." REAL temperatures are NOT linear progressions in one direction but vary with local conditions, wind speed, humidity, elevation, cloud cover, water etc. Florida is NOT always hotter than Georgia, New York, Canada, or the North Pole. IF you asked, GIVEN any point AND a "line" including that point AND circling the earth if there is at least one point on THAT circle that is the exact same temperature? Than SURE it is True. BUT OPPOSITE, meaning end point of line going THROUGH the center of the Earth AND the starting point, then the answer is CLEARLY FALSE. Poorly worded question with Theory and not Practicality in mind.

Jason Heffernan - 2 years, 10 months ago

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100% agreed! They messed this one up

Rafæl Couto - 2 years, 10 months ago

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I second that.

Ali Adams - 2 years, 10 months ago

I made the same mistake. However, as warmer points cool and cooler points warm at some point in time they are bound to cross paths.

Wm Schwartz - 2 years, 10 months ago

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@Wm Schwartz However, this approach is only applicable in theory and cannot be shown practically as both the temperature and pressure at a given point on Earth keep on changing continuously at variable rates.

Rafæl Couto - 2 years, 10 months ago

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@Rafæl Couto You keep making the mistake I was making. The Earth is in motion. Over time, the warm spots are cooling and the cool spots are warming if for no other reason than different parts of the Earth are being exposed to direct sunlight at different times of the day.

Earth's temperatures stay within a fixed range. There are currently absolute maximums and absolute minimums that are possible. While some points may never be the same, "At any point in time" some spot that is cooling (because it is night or cloudy or raining) is crossing paths with some other spot that is warming due to its solar orientation and phenomenon like the Jet Stream. As warmer spots cool and cooler spots warm, at some moment at least two will have the same temperature. We must consider the fact that the Earth is spinning and temperature changes are occurring over time. At some point in time a point that is cooling and a point that is warming will cross paths and have the same temperature.

Wm Schwartz - 2 years, 10 months ago

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@Wm Schwartz If you consider that it happens "at any point in time", you can forget the motion and take a single snapshot of any point in time. In this snapshot, there must be two opposite points with the same temperature. I see that it will always be the case now, because it's continuous and even with an uneven distribution, the lines on the temperature graph will cross at least once. Thanks for trying to make me understand Schwartz! Cheers

Rafæl Couto - 2 years, 10 months ago

It doesn't need to be linear, it just need to be continuous, nature doesn't really have many situations with infinite slopes or discontinuities and temperature is sure not one of them.

João Areias - 2 years, 10 months ago

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Correct, it doesn't need to be linear. Linear vs Continuous is not the issue however. I was trying to express a simplistic example to prove the point that this problem is False. One negative example proves false. Linear IS continuous. Yes, If you think of two graphs being continuous, let's say MANY waves, and thinking of them crossing as one goes from high to low and the other from low to high, it WOULD cross at some point, probably MANY points. HOWEVER, there is the second criteria that makes the question asked unlikely, those crossing points must ALSO MATCH with a third graph which is a graph of distance from the center of the earth to the line passing through those two points, which must equal ZERO. If all 3 graphs work THEN you would get True for JUST THAT case. NOT all the time. Distance off of Center = Zero & Temp point 1 = Temp point 2 both must be met. All the theories quoted leave out the Distance off of Center, 2D versus 3D thinking. Practice each and find the pros and cons for both if you want to increase your creativity and predictive skills.

Cheers,

Jason Heffernan - 2 years, 10 months ago

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@Jason Heffernan It's not 2D vs 3D thinking, in fact for the Borsuk Ulam approach to this problem (which was also mentioned) it requires it to be 3D. I recommend you to go back to the solution and think a little deeper on what was said, the second and third approach may help so will the videos which can be of great help, because when I took those points I was defining them to be opposite to one another. I'm not taking two random points but the temperature at a point and it's antipode.

João Areias - 2 years, 10 months ago

@Jason Heffernan Thought experiment. If you take the bottom point say Temp 0 and on left hemisphere you increase 1 degree per degree you go around the ball. And on the other hemisphere you add 2 degrees per degree as you go around the other side. Near where the 2 hemispheres meet make a slight slope to have them stay continuous, if needed start one slope a little farther back than the other to make sure the Equator slopes don't match. Then when you get to the top have the same small slope to match the high temps together. In this there is no opposite equal values.

Jason Heffernan - 2 years, 10 months ago

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@Jason Heffernan You are thinking of taking 2 separate points moving around, but as I said, when I pick one, the other is already chosen automatically, again, I encourage you to go through the second explanation and the example I posted.

João Areias - 2 years, 10 months ago

@Jason Heffernan Jason... what about T0 + 120 ? At 120 degrees offset on the left hemisphere and at 60 degrees offset on the right. These 2 points should be opposite, and have the same temperature, don't they?

C . - 2 years, 9 months ago

You should have said two opposite points on a sphere rather than on Earth where geography varies. I would like to see temperature sensors placed on the Earth that produce the stated result other than by coincidence.

Wm Schwartz - 2 years, 10 months ago

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It doesn't really matter the shape, the theorem holds true, if you go from positive to negative you must cross 0

João Areias - 2 years, 10 months ago

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The word "opposite" threw me (as it did others) as did thoughts of microclimates resulting from variations in the local geography. In short, focusing on the wrong issues.

As to whether at least two points somewhere on Earth have the same temperature at the same time, there probably are. As warmer points cool and cooler points warm at some point in time they are bound to cross paths.

Wm Schwartz - 2 years, 10 months ago

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@Wm Schwartz I know, but that's the beauty of this problem, not that it's hard or meant to throw off people, but that it shows in a very intuitive manner something that is completely counter intuitive, I had the same sensation when I first came across it but for me this is one of the problems that shows the magic of mathematics.

João Areias - 2 years, 10 months ago

It took me quite a while but thanks to this great community I now know: CONTINUITY is the key! Temperature can't drop from one value to another over distance without having all "temperatures" in between. Thank you, João!

Philipp Mielkau - 2 years, 10 months ago

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You're welcome I'm glad I could help

João Areias - 2 years, 10 months ago

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This is false.

David Turner - 2 years, 10 months ago

False. Temperature need not be continuous. A flame against water ice need not have molecules at every temperature between the fire's heat and icy cold. The Earth need not have continuous temperatures and this false assumption negates your conclusion, despite how elegant and correct the math itself may be.

Dave Toms - 2 years, 10 months ago

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The question assumes the temperature is continuous so this question is faulty.

CHIN KEE HAW - 2 years, 10 months ago

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instead of "faulty" i would use "a simplified model of reality". Or even "theoretical".

C . - 2 years, 9 months ago

Unless you go down to the quantum level it's a pretty safe assumption, that's why it was stated on the problem

João Areias - 2 years, 10 months ago

Is it a fact that Temperature is a continuous function?

Ran Fuchs - 2 years, 10 months ago

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Depends how you measure it. But at the quantum level measuring individual particles energy then it most certainly isnt.

David Turner - 2 years, 10 months ago

Unless you go down to the quantum level it's a pretty safe assumption, that's why it was stated on the problem.

João Areias - 2 years, 10 months ago

Very elegant solution

Soumil Baksi - 2 years, 10 months ago

I think those that believe that a mathematical proof applies to reality are mistaken. For instance, I just learned that there is a mathematical proof that a (mathematical) object can be broken into seven equal pieces of the same size as the original piece. Obviously this mathematical object has no relationship with our current understanding of reality.

Shannon Lieb - 2 years, 10 months ago

this question is surely very easy to see, but i wonder how you prove temperature IS continuous? is it possible not so?

Cherry Lee - 2 years, 10 months ago

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Well, think of the implications of the temperature not being continuous, first, case scenario would be that some molecules would not have temperature, not that they would be at 0 degrees but the concept of temperature would not make sense for them, now the other case is if you had a instantaneous jump (or fall) in temperature, which, by the equation q = k Δ T q = -k\Delta T would require infinite energy.

João Areias - 2 years, 10 months ago

Hello, nice solution but I have something to ask. I consider that T is a continuous function of three variables x, y and z as you defined them and I called A = { (x, y, z), T 1(x, y, z) = T 2(x, y, z)} with T 1 and T 2 as you defined them too. So, because of the amount of way to exchange T 1 and T 2 (as you did) isn't countable, we can conclude that A is an infinite set. And that is something that is hard for me to accept, so if you can make something to solve my trouble ...​

Adrien Kaczmarek - 2 years, 10 months ago

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Yes, you are right, it's not countable, but in all of the ways you can do this, you must cross 0. Considering moving around the points to exchange T1 and T2, there are infinite paths you can take for that, but all of them goes from a positive value to a negative, thus, they must cross 0. This start to imply that there's not only a single pair of points at the same temperature but actually a continuous band where points in there are at the same temperature which is key for the Borsuk-Ulam Theorem approach. The Vsauce video explores just that if you want to check it out.

João Areias - 2 years, 10 months ago

This is most certainly false. Why must the crossover point be exactly the opposite side of the planet? T1 and T2 could cross over 20 times traversing Earth's circumference but there's no guarantee it's on opposite points.

Mustafa Eroglu - 2 years, 10 months ago

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I think you may be misinterpreting the solution. I suggest you watch the videos, the visuals make it more clear.

João Areias - 2 years, 10 months ago

Very nice! Actually this works as a foundation to prove that there is not only a pair but an infinite set of pairs of opposite points with same temperature. The argument is that you can always find two opposite points P1 and P2 with T = T1 - T2 > 0 (otherwise you get the trivial case of T = 0 everywhere) and clearly you have an infinite set of "exclusive" paths to go from P1 to P2.

Mauricio Álvarez - 2 years, 10 months ago

Love the original.

I just guessed and got it using probability, but you explain so well and provide a scientific answer and extra sources What can I say except...

Perfection.

Zoe Codrington - 2 years, 9 months ago
Johanan Paul
Aug 5, 2018

Consider the opposite points on the globe and let's call them A and B.

If you move the two points so that A is now in B's initial position and B is on A's initial position, and were to plot the temperature of their path, the two graphs have to intersect at least once.

This works on any arbitrary path, including one that goes on every point on the globe.

Yuup I agree. It's basically Rolles theorem.

dipesh bhatia - 2 years, 10 months ago

"and were to plot the temperature of their path, the two graphs have to intersect at least once." Sorry I don't understand this part could you try and help me understand it, please?

Roman Smith - 2 years, 10 months ago

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Let's say A is at 30°C and B is at 50°C. If you move A to B, the temperature would look like a curve moving up from 30°C to 50°C. Same reasoning applies for B, but from 50°C to 30°C. Since the graphs are continuous, they will have to cross.

Johanan Paul - 2 years, 10 months ago

To be more precise, let the x axis be path length from A to B or B to A and let the y axis be the temperature. Then the paths can wander without intersecting so long as one starts with T-A and goes to T-B and the other goes from T-B to T-A. Intersection is not the issue but at every temperature between T-A and T-B, a horizontal line can be drawn for that temperature that intersects both curves. I have a drawing but can't figure out how to post it.

Paul Palmer - 2 years, 10 months ago

This is more clearly explained than the first two.

Albert Kirsch - 2 years, 10 months ago

Continuity of the function does not imply continuity of the slope of the function; therefore the same temperature does not have to be the same at the antipodes.

Bob Lord - 2 years, 10 months ago

Why would the intersection be on opposite sides of the globe? What if the rate of temperature change is not the same from A to B as from B to A?

Glo Gonzales - 2 years, 10 months ago
Steven Andrew
Aug 6, 2018

The Borsuk-Ulam theorem states that every continuous function from an n-sphere into Euclidean n-space maps some pair of antipodal points to the same point.

Let's think of that mapping process as the following:

Taking every point on Earth and plotting a scatter graph of Temperature vs. Pressure. Since both of these functions are continuous, there must be at least 1 pair of antipodal points with the same temperature and pressure.

Hence, there must be at least two opposite points on Earth with exactly the same temperature.

Nice, I love the Borsuk-Ulam approach too since you can show that there's not only points with the same temperature but you can show 2 variables like temperature and pressure or pressure and air humidity

João Areias - 2 years, 10 months ago

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Practically this doesnt make any sense whatsover to me (I am not that great at theoretical math, so bear with me please). Temperature is not a constant function of distance. If you move one point north the antipode automatically moves south. Its very possible given the way temperature works that the temperature rises for both directions (eg. since they both are moving towards the equator), but also that one drops and one stays the same, because youre moving into a desert or a river, or a city or whatever (Besides that, antipodes would most cases mean that one is during the day and the other during the night and one during winter and the other during summer). Mathematically speaking Id say its pretty random. Obviously its extremely likely that some points coincendetally will have the same temperature, although I dont think this can be absolutely certain. I figure this also depends on how exact you measure it (Id say its more likely that two points are 9 degrees than it is that two points will be 9.679273492953 degrees).

  • Edit - Ok, I think I should regard more or less as a limit. Because from any point you can move in an endless amount of directions, one should always approach the correct temperature eventually. But I still wonder what would happen if its winter in the northern hemisphere and its colder than 18 degrees everywhere and summer in the southern hemispehere and its warmer than 19 degrees everyhwere (ok, I realize this is practically impossible).

Leen Verkade - 2 years, 10 months ago

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Vsauce posted a video that goes more in depth about this, with nice visualizations and all here is the link if you want to check it out, I'd also suggest for you to check the Intermediate value theorem . The idea is that heat will vary continuously along the surface of the earth, you may get some pretty step rises and falls but nature usually doesn't have many infinite slope type of situations. Since the temperature varies continuously, if I take the temperature at two points, and move those points around, the difference in temperature will also vary continuously so for this problem I took two points opposite to one another and moved those two points around and found that if I vary them until they switch sides, the difference in their temperature will go from a positive to negative (or negative to positive) which, with the information that this is continuous, it's enough for me to say that the difference in their temperature crossed a 0 somewhere on their path.

João Areias - 2 years, 10 months ago

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@João Areias Hi Joao, I appreciate taking the time to explain it. I get the math side of things now. But I still think it wouldnt work on earth. Say you have two opposite points of 19 and 20 degrees and can move them the same distance and opposite direction, where theyd flip to 20 and 19 (I think that already would be a pretty rare occurence), even if the temperature only varries between 19 and 20 in that distance (which is also a pretty forgiving assumption), there is no way for sure that they had the same temperature at the same point along the line. I think the theorem only work for functions.

Leen Verkade - 2 years, 10 months ago

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@Leen Verkade The idea is, let's say we have a point a a at 20 degrees and a point b b at 19 degrees, if I evaluate my function at point a a , the one I mentioned in my answer, I will get T ( a ) = 1 T(a) = 1 , if I now evaluate the same function at the same time at point b b I get T ( b ) = 1 T(b) = -1 , now, I know my function is continuous, so if I keep evaluating this functions on the points between a a and b b there must be a point c c where, when I evaluate it there T ( c ) = 0 T(c) = 0 . And you also don't need to assume that it varies only between 19 and 20, it can go to any temperature, and so can their difference, but in order to go from positive to negative, it must cross 0 somewhere.

João Areias - 2 years, 10 months ago

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@João Areias I think this whole conversation highlights the difference between theoretical and practical problem-solving, echoing some Ancient Greek stuff. Zeno v Archimedes. Of course it is possible to argue the point about the temperature being equal from a philosophical standpoint, but anyone with practical common sense knows it can't be true in a real world sense. It's like a parlour game, a little trick to amuse people. The practical people commenting in this thread have pointed out that antipodean points have different seasons and times of day, so of course the temperature is highly unlikely to be the same. The philosophers 'get around' this practical consideration by drawing theoretical Ines to connect the points, and reversing temperatures - it's like a sleight of hand, but for the mind. None of this works in the real world. Zeno could prove all sorts of nonsense like this. Frogs and lily pads!

Marlin Helliwell - 2 years, 10 months ago

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@Marlin Helliwell You may miss the most important point - the temperature is continuous. The theoretical approach is correct in the real world if, and only if, the temperature in real life is continuous aswell. We could discuss for hours if the temperature proceeds continuously in the microscopic world, but given our temperature measurement devices work in the macroscopic world and considering thermodynamic laws, the assumption that temperature is continuous and therefore the theoretical approach works, is indeed fine.

EDIT: Differing seasons and daytimes are unimportant. If you could stop the time and do the experiment you will see a line of point with their antipodes with equal temperatures. This all comes from the fact, that the temperature is continuous and if you have two workers chaning their point on the earth, the temperature they experience has to flip and therefore have to be equal at some point.

Heribert Plum - 2 years, 10 months ago

@Marlin Helliwell While there are some honest problems regarding applying mathematical theory to reality, your mention of Zeno and time/seasons betrays your lack of understanding of both these problems and their resolutions.

Mathematics itself doesn't prescribe reality, but we can fit mathematical models to reality. In a broad sense, this is the essential difference between ancient philosophy and science. E.g. The question of how to fit the mathematical sphere to the globe has some real concerns, but we can deal with them. The concept of "temperature at a point on the Earth" is a technically fuzzy concept, but again, we can deal with it.

All your comment about the time/seasons means is that at 4 5 45^\circ latitude and 1 PM 1 \text{ PM} in July is incredibly unlikely to yield your pair of antipodal points. They are much more likely to be near the equator, during sunrise/sunset in the more temperate months, or with one of the points in the middle of the summer night. Luckily, every great circle of antipodal points must cross the equator, so we are very likely to get our pair of antipodal points "near" the equator.

Brian Moehring - 2 years, 10 months ago

@João Areias I think the math is very interesting, but feel it only applies to mathematical functions. Does it necessarily apply to real-world data? As anyone who has piloted a small aircraft will tell you, the temperature at any given point above the earth varies in a very irregular fashion. On a summer's day you get varying uplift as you fly over wooded ares and lakes due to rapidly changing temperature, pressure and air velocity fluctuations. Temperature is obviously continuous if we measure it on a fine enough grid(albeit with some pretty steep gradients between any two points), but I wouldn't like to try to describe the variation mathematically. So I think it would be a valid comparison to replace temperature in your problem with the height of the land at any given x,y coordinate on the Earth. If we replace your T1 and T2 with H1 and H2 for the respective heights at each point, and move your pode/anti-pode around the world until they have reversed positions, they will clearly have each changed values millions of times. At many points they are very likely to be equal, but is that really significant or even interesting mathematically - Can you prove that 2 opposite points on the Earth have the same height? If you take one point from the peak of Everest to the lowest point in the marianas trench, you obviously transition between every possible height on the Earth.It's still not clear to me that there physically has to be two opposite points of equal height along this path. I think you are saying that if you could describe the height of the land as an (almost infinite) n-th degree polynomial , you could prove that you have at least one opposite pair of points equal. I think your original solution is assuming a very simple function forT(x,y,z) - you appear to want a monotonic function f(theta) if you describe theta as the angle of rotation of your points around the normal to the great circle that passes through the points. I think most of the reactions are essentially "But that's not even close to a realistic assumption". I can't think of a real, parameter that varies in any simple mathematical way as f(x,y,z). You suggest that the math also applies to pressure. Perhaps it does, but can you say this without finding a mathematical function that accurately describes f(x,y,z)?

Gary Scarr - 2 years, 10 months ago

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@Gary Scarr Most functions in the real world have no nice concise form. This doesn't stop us from saying something about their properties.

For instance, however complicated the temperature is, it is by definition an averaging of the kinetic energies of the atoms, so any discontinuity that might happen should disappear instantaneously (basically by a local version of the laws of thermodynamics). That is, in real life we should expect the temperature to vary continuously almost surely. The only problem this invites is how to define the temperature at a point, but it hopefully makes sense to define this as the limit of the temperatures averaged over smaller and smaller regions.

The only property needed is continuity, nothing like monotonicity or polynomial or even analyticity matters. And importantly, as long as we have a continuous definition for a temperature at a point and a way of defining the antipode of a point continuously, the property we want follows.

(Incidentally, defining the "height" at a point on the globe invites more problems than does the temperature)

Brian Moehring - 2 years, 10 months ago

@Gary Scarr The trick here is not to think about what function is that we are using, which would overcomplicate the problem, but realize that there is such a function and try to figure out the implications of that. You can show that it is, in fact, a function because you don't get 2 temperatures for the same point, and, assuming you don't go as low as the quantum level, you can show that it is continuous because you don't get points with an undefined temperature and if you take Fourier's Equation q = k Δ T q = -k\Delta T an infinite temperature gradient would mean infinite energy. So you have a continuous function, that we don't care how it looks like, it could be any shape and doesn't need to be described necessarily by a polynomial, but the theorems in Calculus still hold, so we can still analyze this function.

João Areias - 2 years, 10 months ago

I still don't get it but I really want to! Is there any assumption on symmetrical order needed that I missed? I explicitly don't see the answer for the following example: What if I draw a spiral from south pole to north pole with each point of the spiral having a different temperature. How can there be opposite points with same temperatures then?

Philipp Mielkau - 2 years, 10 months ago

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i’m with philipp on this one!

Vahid Fozi - 2 years, 10 months ago

Vsause has a great video on that, which makes this very intuitive, check out here

João Areias - 2 years, 10 months ago

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Skip to 9:30

Brandon Parker - 2 years, 10 months ago

Isn't there a minimum temperature? Why would there be two places with the same minimum temperature? Sorry still don't see the explanation.

Robert Lurie - 2 years, 10 months ago

You require a loop. A spiral from south to north pole on it's own is not enough....because the whole point (pun not intended) is that it requires a closed system to have an set of opposite points to work with .

David Finch - 2 years, 10 months ago

It has been my experience that mother Earth is not a sphere. Never heard of the Borsuk-Ulam theorem so I don't know how it treats almost spheres, but I suspect that the theorem may almost point to the same point.

Robert Winkler - 2 years, 10 months ago

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I'm not really sure if this argument is valid 'cause I'm not that good in topology, so if people want to correct me fell free to do so, but I believe that the theorem still applies, if I'm not mistaken, since you are allowed to deform the sphere, topologically speaking the earth still a sphere. But even if it didn't the idea behind it still holds as Michel presented in Vsauce's video.

João Areias - 2 years, 10 months ago

The only possible concern with "almost spheres" in the geometric sense is how you define "opposite/antipodal points". As long as you're able to define that concept, the rest of the problem is actually purely topological, and the earth, at least on the large scale, can topologically be considered a sphere (the concept of "almost spheres" only makes sense geometrically... topologically, they're just spheres).

Brian Moehring - 2 years, 10 months ago

Its seems then that coldest or hottest records would always be shared by at least two opposite locations

Max Planck - 2 years, 10 months ago

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Not at all. The theorem doesn't let you specify what the common temperature is at your antipodal points.

Unless I misunderstood and you meant the record coldest temperatures and the record highest temperatures were your two functions. In that case, the answer is "not necessarily" depending on what you mean by the records. If you take the record temperature over a continuum time interval, the result can be discontinuous in which case the theorem wouldn't apply. On the other hand, if you are taking the record temperatures measured at each second in UTC time for the last hundred years, then yes, it would work! The only technical problem you may have to deal with is the Earth changes (e.g. due to plate tectonics)

Brian Moehring - 2 years, 10 months ago

Sorry, but this problem isn't up to the usual Brilliant standard. The concept is very poorly laid out -- what is "opposite"? I suspect something got lost in translation.

Owen Barnes - 2 years, 10 months ago

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The globe is modeled as a sphere. Draw any line through the center of the sphere. The two points the line hits on the surface of the sphere are opposites of one another.

Even though the globe isn't a sphere, we can still use longitude/latitude coordinates as if it were a sphere to identify opposite points.

Brian Moehring - 2 years, 10 months ago

I am quite aware of Brouwer's Fixed Point Theorem, as it applies to CONTINUOUS functions. But there's nothing in the statement of this problem that explicitly says we're supposed to assume a continuous function. And if one allows discontinuous functions, then the problem is false.

David Grossman - 2 years, 10 months ago

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Borsuk-Ulam is strictly more general than Brouwer, so I don't know why you mentioned it here. Maybe the statement also follows by Brouwer? I certainly don't see how right now.

The function taking points on the globe to the temperature will be continuous almost surely with respect to both space and time. You can see this from various physical laws.

Besides which, from the statement of the problem:

Assume that temperature at a point is a continuous function of the coordinates.

Brian Moehring - 2 years, 10 months ago
Abhishek Krishna
Aug 6, 2018

Earth's temperature depends on many factors and location of earth with respect to sun is the most important factor. So , imagine that there are two balls facing each other. One ball is earth and other is sun. As their centres lies on the same plane so earth's temperature will be symmetrical in both of it's halves due to sun. Hence , there must be atleast two points on earth symmetrically placed and opposite to each other which have the same temperature.

Shawn Ligocki
Aug 8, 2018

Paint the globe in the following way. Paint a point Red if it is hotter than it's opposite point. Paint it Blue if it's colder. White if it is equal. This question is equivalent to asking "Are there any White points on the globe?"

Visually, consider how this looks, there will be Red regions and Blue regions, but (assuming temperature varies continuously) there must always be a White line/region separating the adjacent Red and Blue regions. (As others have pointed out, the Intermediate Value Theorem is the rigorous way to prove this).

Nice approach, very creative!

João Areias - 2 years, 10 months ago
Blan Morrison
Aug 7, 2018

For those who think it is false and can still not see why, try creating a scenario where this is true. You will find that it is impossible.

Here are two scenarios: Imagine a world where the Southern hemisphere is at 60 degrees F, and the Northern one at 70. There is no point with an opposite with the same temperature. If you're uncomfortable with the splitting line (the equator), you can say that the temperature is rising evenly from 60.1 to 69.9 in micro increments (as small as you define a "point"). You could even extend it to the entire globe: start somewhere at 0°F and set the temperature of one adjacent "point" to be that plus 0.000000000000001 °F. Repeat until the whole globe is covered. No two points will have the same temperature.

Noémie Durand - 2 years, 10 months ago

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Except you are introducing a discontinuity that does not exist in nature.

Robert Kelleher - 2 years, 10 months ago

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With climate change, who knows how "nature" will look like in 1000 years. Besides, the earth is far from a perfect sphere. But ok, I see now the "continuous function of the coordinates" condition and indeed, my model does not fit that, I guess.

Noémie Durand - 2 years, 10 months ago

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@Noémie Durand By "nature" he means the laws of physics.

Blan Morrison - 2 years, 10 months ago

In the first scenario, if you pick 2 opposite points on the equator, they will be the same temperature (65°). The second scenario defies the Intermediate Value Theorem. Please note that we are dealing with irrational numbers, which means there is always a temperature gradient.

Blan Morrison - 2 years, 10 months ago

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You're right, my 0.00000001 degree example was a beginner's mistake. Let's say that the temperature on the equator is equal to the angle from an arbitrary starting point, so temperature will be between 0 and 360 (if we take degrees), with no 2 points having the same temperature. We can also divide that by 40 and add 60.5 to have temperatures between 60.5 and 69.5 for more real life values (in °F).

Noémie Durand - 2 years, 10 months ago

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Consider that arbitrary starting point. When you walk around the equator clockwise from that point x x meters, the temperature is x x °. When you walk to the counterclockwise an infinitesimally small amount, you go from 0° to 360°, which defies the Intermediate Value Theorem. That was a creative solution, though.

Blan Morrison - 2 years, 10 months ago
Corwin Silverman
Aug 6, 2018

There must be opposite points on the equator.

We will assume that f, where f(a) is the temperature at point a, is a continuous function from the unit circle to the real numbers. By the Borsuk-Ulam theorem, there exist antipodal points a and -a on the unit circle such that f(a)=f(-a). It follows that I points a and -a have the same temperature.

Jörg Straube
Aug 9, 2018

Not a proof but pragmatic: I visit the South pole (Antarctica) and my brother visits the North pole (Arctic ice pack). I'm pretty sure that we can wander around to find a spot so we are opposite each other and have the same freezing temperature. And this at any given point in time.

I think you forgot that half of the year the one pole is pointing away from the sun while the other pole points toward the sun. The other half of the year the poles switch positions relative to the sun. Using these two locations (not points in the mathematical sense) at any time of the year has virtually no possible solution to this problem.

Shannon Lieb - 2 years, 10 months ago
Peter Lawrence
Aug 8, 2018

Following on from J Areias' solution, are there not an infinite number of such twin pairs of antipodes, since all possible great circles must contain at least one? And further, is there a path (not straight) joining these points, so that one could in theory travel along a temperature contour back to one's starting point?

David Turner
Aug 8, 2018

On the surface of a sphere, there is an infinite number of points. Particle speed is however discreet(particles jump from one speed to another instantly at the quantum level). therefore temperature scale is a smaller infinity than the number of infinitesimal points of a sphere. (the numbers can not be mapped to each other). Therefore there is an infinite number of opposite points on the earth that are the same temperature.

Ervyn Manuyag
Aug 7, 2018

Yes bcuz when f=c, they both equal to 40.

Ben Mitchell
Aug 7, 2018

This seems to further imply that there is a connected path about the world for which the temp is the same on the other side.

And if you continue with this line of though you will see how this is just a special case to the Borsuk-Ulam Theorem and that you can actually map 2 variables like that, instead of only one, like temperature and pressure or pressure and humidity.

João Areias - 2 years, 10 months ago
Roger Ab
Aug 7, 2018

There is a Vsauce video about this, the topic starts at minute 11.

https://youtu.be/csInNn6pfT4

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