 Posers and Puzzles

1. 15 Jan '07 09:49
This one is almost the same as the buddhist monk, ascending and decending a hill in . It goes like this:

Is there two different points of the surface of earth that have exactly the same temperature *and* exactly the same air pressure?
Prove or disprove.
2. 15 Jan '07 13:16
yes because you didnt say how far apart they had to be so one point could be less than 1 mm away from the other. both being so close together that no difference occurs
3. 15 Jan '07 13:19
Originally posted by battery123
so close together that no difference occurs
There would still be a difference, albeit infinitesimally small 🙂
4. 15 Jan '07 15:411 edit
OK, I think this is sound (though there may be some gaps to fill)...

Assume the pressure varies (if it doesn't the problem is trivial). Choose a pressure between the minimum and maximum values. Plot the points where the pressure has this value. There will be at least one closed loop (because pressure varies continuously).

Now plot a graph of how temperature varies as we follow that loop. If it's constant we can pick any two points, so assume it isn't. It must start and end with the same value (since the end point is the same as the start point), and again must vary continuously. So choose a temperature between the maximum and minimum value on that loop. There must be at least two points on the loop that have that temperature.

If so, we've found our two points with the same temperature and pressure.
5. 15 Jan '07 16:251 edit
Sounds ok. Maybe you could use the words isobar, isotherm and intersection to increase legibility though 😀

Come to think of it, there may be some work on the actual existence of two intersections. One might imagine all iso-lines touch or something 🙂
6. 15 Jan '07 16:341 edit
Originally posted by piderman
Sounds ok. Maybe you could use the words isobar, isotherm and intersection to increase legibility though 😀

Come to think of it, there may be some work on the actual existence of two intersections. One might imagine all iso-lines touch or something 🙂
I don't think my proof relies on assuming isobars and isotherms intersect. It doesn't use isotherms, it looks at the variation of temperature along an isobar.

(Happy now? 🙂)
7. 17 Jan '07 16:022 edits
My proof rely heavy on intersecting closed loops. On isotherm and one isobar.

An isotherm is a line that closes itself in a loop. On the one side of the loop temp is lower, on the other side temp is higher. On the loop itself the temp is exactly the same everywhere all the loop around.
An isobar is a line that closes itself in a loop. On the one side of the loop pressure is lower, on the other side pressure is higher. On the loop itself the pressure is exactly the same everywhere all the loop around.

Take a arbitrary point on the surface on the Earth. (Like where I am.) Find the isotherm loop and the isobar loop for this point. If they intersect at this point it must intersect each other at another point also. Why? Because they are closed loops.

This is my proof.
8. 17 Jan '07 16:39
Originally posted by FabianFnas
My proof rely heavy on intersecting closed loops. On isotherm and one isobar.

An isotherm is a line that closes itself in a loop. On the one side of the loop temp is lower, on the other side temp is higher. On the loop itself the temp is exactly the same everywhere all the loop around.
An isobar is a line that closes itself in a loop. On the one side ...[text shortened]... rsect each other at another point also. Why? Because they are closed loops.

This is my proof.
Nice work! Since the closed loops for isobars and isotherms are constantly shifting, the second point of intersection (relative to you at the first point) would change with time. There would always be two, but not necessarily the same two.
9. 17 Jan '07 17:23
Originally posted by HandyAndy
Nice work! Since the closed loops for isobars and isotherms are constantly shifting, the second point of intersection (relative to you at the first point) would change with time. There would always be two, but not necessarily the same two.
If I am at one point, the other point will move, perhaps rather violently, I don't know. But at any on point of time there is always two points (at least) on the surface on the Earth that have exactly the same presssure and themperature.

This is only a proof of existance, not to prove the exact location at any particular or changing time.

When you get the Aha-sense, as you get with this problem and the same with the monk problem, when a problem of which you've not a clue suddenly converts itself to be chrystal clear in a moment, gives you a chill along your spine. This kind of problems I like a lot.
10. 17 Jan '07 17:44
Don't get me wrong, I understand the answer and I think it has been presented elegantly by the posters above, but they are hand-waving arguments and not proofs. You can't say "and they must intersect - why? Because they are closed loops", even though that is the case, and then say you've proven it. It really needs something else, like a proof by contradiction: "If you assume that the loops do not cross, then you end up with the following contradiction, therefore they must cross."
11. 17 Jan '07 21:501 edit
Originally posted by FabianFnas
Take a arbitrary point on the surface on the Earth. (Like where I am.) Find the isotherm loop and the isobar loop for this point. If they intersect at this point it must intersect each other at another point also. Why? Because they are closed loops.
Not quite enough. If the isotherm is tangential to the isobar at that point, there won't necessarily be another intersection.

To take this approach, you need to add a bit more:

- make sure you pick a point on the isotherm where the isobar is not tangential
- if this doesn't exist - i.e. isobars are tangential to the isotherm at all points on the isotherm - then the isobar and isotherm must be exactly the same curves. In which case you can pick any two points on the curves.
12. 17 Jan '07 21:56
Originally posted by PBE6
Don't get me wrong, I understand the answer and I think it has been presented elegantly by the posters above, but they are hand-waving arguments and not proofs. You can't say "and they must intersect - why? Because they are closed loops", even though that is the case, and then say you've proven it. It really needs something else, like a proof by contradiction ...[text shortened]... not cross, then you end up with the following contradiction, therefore they must cross."
Well, yes, I'm sure a full proof would lead to several pages and would contain a lot of maths. I've no inclination to try and produce that, and I don't suppose anyone want to read it if I did. These are always going to be sketch proofs, with the assumption that they are built upon other proofs. It only really matters if someone wants to start disputing one of those foundations.
13. 18 Jan '07 00:44
Originally posted by mtthw
Well, yes, I'm sure a full proof would lead to several pages and would contain a lot of maths. I've no inclination to try and produce that, and I don't suppose anyone want to read it if I did. These are always going to be sketch proofs, with the assumption that they are built upon other proofs. It only really matters if someone wants to start disputing one of those foundations.
indeed.

the proof is relatively simple if the isobar forms a closed interval [a,b]. We can simply invoke the Extreme Value Theorem of continuous functions combined with the fact that for the continuous temperature function f f(a) = f(b)

Now let f(x) = K be the highest value of f. (if x=a then let f(x) be the lowest value of f and reverse the signs, if x is still a then we have f(x) = K for all x and QED).

Now pick a J < K. and look at the interval [a,x] according to the intermediate value theorem there must exist a point x' in [a,x] for which f(x'😉 = J. Equally when we look at the interval [x,b] there must exist a point x'' in [x,b] for which f(x''😉 = J

Therefore we have f(x'😉 = f(x''😉 for some x' /= x'' and QED

The proof is a little more complex if the isobar is a half open interval [a,b) (it is still true we just need to take a closed subset of [a,b) call it [a,c] and show that if there doesn't exist a x' /= x'' for which f(x'😉 = f{x''😉 then we can always pick another c -> b which does)
14. 18 Jan '07 09:161 edit
Points taken, dear friends. I used a rather low grade definition of the word proof. Most of us here can't follow a rigorous proof so I didn't work of a water tight proof. I just presented the general idea of a proof.

But, guys, do we really have to prove the problems given in a water proof way? Then when do we stop?, at the axioms themselves? Then, where is the fun of it?

This problem is in two dimensions only, even on the curvature of a sphere. Can anyone formulate another similar problem in three dimensions?
15. 18 Jan '07 11:32
Originally posted by FabianFnas
Can anyone formulate another similar problem in three dimensions?
A classical example of the Kakutani theorem is stirring a coffee cup. At least one molecule ends up in the same place it began.

(note: here molecule = point, so we suppose a continuity that doesn't really exist)