Challenge 10a: Temperatures of the Earth

  • Context: Undergrad 
  • Thread starter Thread starter Office_Shredder
  • Start date Start date
  • Tags Tags
    Challenge Earth
Click For Summary

Discussion Overview

The discussion revolves around a mathematical challenge regarding the temperatures at two points on opposite sides of the Earth. Participants explore the conditions under which these temperatures can be shown to be equal, considering theoretical assumptions about the Earth's shape and temperature distribution.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests assuming a continuous and differentiable model of the Earth for the challenge.
  • Another participant proposes a parameterization of a great circle to analyze temperature differences using the intermediate value theorem.
  • A different participant references the Borsuk-Ulam theorem as a basis for the result, though not providing a formal proof.
  • One participant describes a heuristic approach involving two individuals moving along paths on the Earth's surface, recording temperatures, and using graphical analysis to find intersections of temperature readings.
  • Another participant questions the validity of the heuristic approach, specifically regarding the requirement for the points to be "exactly" opposite each other.
  • A participant acknowledges a misunderstanding in their previous argument and refines their approach to ensure the two individuals are always at opposite points on a diameter while recording temperatures.
  • One participant notes that their solution closely resembles an earlier one but with less mathematical notation.

Areas of Agreement / Disagreement

Participants express various approaches and interpretations of the challenge, indicating that multiple competing views remain. There is no consensus on a single solution or method, and some participants refine their arguments in response to critiques.

Contextual Notes

Some arguments rely on assumptions about the continuity and differentiability of temperature functions, and there are unresolved questions about the implications of the "exactly opposite" requirement in the problem.

Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
Messages
5,706
Reaction score
1,592
For the sake of this challenge, you can assume that the world behaves nicely (everything is continuous, differentiable etc, the world is a sphere blah blah blah) but figurative bonus points if you assume fewer things. If you're looking for a bigger challenge check out Challenge 10b.

The Challenge: Show that at any point in time there exist two points on exact opposite sides of the Earth which have the exact same temperature.
 
Physics news on Phys.org
If I've seen a very similar problem, should I still post an answer?
 
Apologies for any typesetting issues as I'm attempting an answer from my phone.

Take any great circle and choose a point on the circle to be the origin. Parameterize the circle uniformly from ##x=0## to ##x=1## where the origin is at both endpoints.

Let ##f:[0,1]\rightarrow \mathbb{R}## be the temperature function. Define ##d(x) = f(x+1/2) - f(x)##. If ##f(1/2)=f(0)## then we're done. Otherwise ##d(0)## and ##d(1/2)## have opposite signs , so (assuming continuity of ##f##) the intermediate value theorem gives us a point ##x## with ##d(x) =0##, hence ##f(x+1/2) = f(x)##.
 
HS-Scientist said:
If I've seen a very similar problem, should I still post an answer?

I would lean towards saying that you can give answers as long as you aren't just copying it from someone else.

Nice solution jbunniiii, anyone have an alternate way of approaching the problem?
 
Not a proof but...this results follows from Borsuk-Ulam theorem.
 
An alternative heuristic approach: See attached image. Let ADBC be a great circle on Earth surface.
AB diameter. At A, temperature is t1, at B temperature is t2. Let us imagine two men moving at same speed from A to B along paths ACB and ADB respectively. The distance from A to B =d. Now the men are recording temperatures continuously. We draw the graphs of the temperatures vs distance on the same axes.
Now if we draw a straight line corresponding to any temperature t=t0, (between t1 and t2) parallel to distance axis. Both the temperature curves shall cut this line at least once (intermediate value theorem).
This two intersection points are our answer.
[/PLAIN]
 

Attachments

  • nysr4o.jpg
    nysr4o.jpg
    24.5 KB · Views: 153
Last edited by a moderator:
ssd - Maybe I am misunderstanding your argument, but I don't see why the two intersection points ##(d_1, t_0)## and ##(d_2, t_0)## correspond to points on "exact opposite sides of the Earth" as the problem requires. In other words, what guarantees that ##d_1 + d_2## equals the arc length from ##A## to ##B##?
 
Last edited:
Oh, that is right. My answer is wrong. I missed the "exact" opposite part.

Well, let us consider that the first man is moving from A to B via C and the second man is moving from B to A via D. As before the are recording temperature continuously. They start at same time and move at same constant speed. This implies that they are always exactly on opposite points of a diameter.


Consider the graph of the difference in the temperature readings (of same instant) of two men vs the distance. Let, t1<t2. The graph starts with -ve value and ends with +ve value. According to the given conditions it must intersect the distance axis (that is, will attain value 0) somewhere between distance 0 to d. That point is the answer. If t1=t2 then the start and end point is one answer.
 
Last edited:
ssd said:
Consider the graph of the difference in the temperature readings (of same instant) of two men vs the distance. Let, t1<t2. The graph starts with -ve value and ends with +ve value. According to the given conditions it must intersect the distance axis (that is, will attain value 0) somewhere between distance 0 to d. That point is the answer. If t1=t2 then the start and end point is one answer.

As far as I can tell this is the solution that jbunniii gave, just with a lot less notation.
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 43 ·
2
Replies
43
Views
13K
  • · Replies 42 ·
2
Replies
42
Views
9K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 86 ·
3
Replies
86
Views
24K
  • · Replies 48 ·
2
Replies
48
Views
12K
  • · Replies 67 ·
3
Replies
67
Views
17K
  • · Replies 48 ·
2
Replies
48
Views
13K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 105 ·
4
Replies
105
Views
15K