# Poincare's Universe: simple Calculus question

Printable View

• April 29th 2010, 04:36 PM
onecentrob
Poincare's Universe: simple Calculus question
I have been reading "Prelude to Mathematics" by W.W. Sawyer. In it he describes Poincare's hypothetical universe.

This universe is contained in the interior of a circle. The temperature of this universe is high in the centre and gets cooler as you move towards the circumference. The law for the temperature is: If a is the radius of the circle, at the distance r from the centre of the circle the temperature is T= a² - r².

The size of objects inside this universe are affected by the variation in temperature. The length of any object varies in proportion to the temperature T. At the circumference of the circle, where r=a and T=0, the length of an object will shrink to zero.

Inhabitants of this universe are unaware of the changing temperature as they have no sensitivity to it. They are also unaware of their change in size, as anything they use to measure their size also changes its size.

The people/creatures inside this universe can never get to the boundary because the nearer they get to the boundary the more rapidly they shrink and the smaller their steps become. So although the universe is finite and bounded to us, to the inhabitants it is infinite.

This all makes perfect sense to me. The only part I do not understand is Sawyer's proof using simple calculus that the inhabitants can never reach the boundary (or circumference) of their universe. I studied basic calculus 20 years ago but I am a bit rusty and cannot follow Sawyer's calculation pasted below. Please can someone explain the last part of this calculation to me? Is it possible to break it down into simpler steps, or at least explain how the last calculation is arrived at? I would be most grateful for your help. Rob

http://www.curiouser.co.uk/poincare_crop.jpg
• April 29th 2010, 06:03 PM
Gusbob
Quote:

Originally Posted by onecentrob
I have been reading "Prelude to Mathematics" by W.W. Sawyer. In it he describes Poincare's hypothetical universe.

This universe is contained in the interior of a circle. The temperature of this universe is high in the centre and gets cooler as you move towards the circumference. The law for the temperature is: If a is the radius of the circle, at the distance r from the centre of the circle the temperature is T= a² - r².

The size of objects inside this universe are affected by the variation in temperature. The length of any object varies in proportion to the temperature T. At the circumference of the circle, where r=a and T=0, the length of an object will shrink to zero.

Inhabitants of this universe are unaware of the changing temperature as they have no sensitivity to it. They are also unaware of their change in size, as anything they use to measure their size also changes its size.

The people/creatures inside this universe can never get to the boundary because the nearer they get to the boundary the more rapidly they shrink and the smaller their steps become. So although the universe is finite and bounded to us, to the inhabitants it is infinite.

This all makes perfect sense to me. The only part I do not understand is Sawyer's proof using simple calculus that the inhabitants can never reach the boundary (or circumference) of their universe. I studied basic calculus 20 years ago but I am a bit rusty and cannot follow Sawyer's calculation pasted below. Please can someone explain the last part of this calculation to me? Is it possible to break it down into simpler steps, or at least explain how the last calculation is arrived at? I would be most grateful for your help. Rob

http://www.curiouser.co.uk/poincare_crop.jpg

Interesting. That integral should actually read

length $s = \int_0^a \frac{a^2}{a^2-r^2}~dr$

In which case we get:

$= a^2 \int_0^a \frac{1}{a^2-r^2} dr$

Now using the table of standard integrals

$= a \left[tanh^{-1} \left(\frac{r}{a}\right) \right ]_0^a$

$= a \times \infty - 0 = \infty$

If you want to I can give more information on why $\int \frac{1}{a^2 - r^2} dr =\frac{1}{a} tanh^{-1} \frac{r}{a}$
• April 30th 2010, 06:19 AM
onecentrob
Many thanks for your reply. I wish these forums had existed when I was at school. As I said, I am a bit (ok, very!) rusty.

1) I don't understand why $dr = \frac{a^2 - r^2}{a^2} ds$
Why isn't is just $dr = ({a^2 - r^2}) ds$ ?

2) I'm not 100% sure why integrating $ds$ gives the creature's estimate of the distance from the centre to the edge of the circle. Is this effectively the sum of all the lengths $ds$ as $ds$ moves from the centre to the circumference of the circle?

3)
Quote:

Originally Posted by Gusbob

If you want to I can give more information on why $\int \frac{1}{a^2 - r^2} dr = tanh^{-1} \frac{r}{a}$

Yes please!
• April 30th 2010, 06:29 PM
Gusbob
Quote:

Originally Posted by onecentrob
Many thanks for your reply. I wish these forums had existed when I was at school. As I said, I am a bit (ok, very!) rusty.

1) I don't understand why $dr = \frac{a^2 - r^2}{a^2} ds$
Why isn't is just $dr = ({a^2 - r^2}) ds$ ?

The temperature at r distance from centre is $T = a^2 - r^2$

So when a small length ds is moved to from the centre, it will vary in proportion to the temperature T.

The new length $dr \propto T ds \rightarrow dr = k(a^2 - r^2)ds$ for some constant k.

But if you think about it, when you're at the centre w/ r = 0, you'll get $dr = ka^2 ds$ which is true iff $k = \frac{1}{ a^2 }$

Therefore $dr = \frac{a^2-r^2}{a^2} ~ds$
__________________________________________________ _______
A more intuitive way to think of it is using the ratio of lengths (L).

We have a ratio $\frac{dr}{ds}$ .

Because the lengths themselves are proportional to the specific temperature at the radius they are at,

$\frac{dr}{ds} = \frac{T_r}{T_s} = \frac{a^2 - r_2}{a^2 - r_1}$

where $r_1, r_2$ are the distance from center the lengths dr and ds are at.

At $L = ds, r = r_1 = 0, T_r = a^2$
At $L = dr, r = r_2, T_s = a^2 - r_2^2$

So we have $\frac{dr}{ds} = \frac{a^2-r_2^2} {a^2} \Rightarrow dr = \frac{a^2-r^2}{a^2} ds$

Quote:

2) I'm not 100% sure why integrating $ds$ gives the creature's estimate of the distance from the centre to the edge of the circle. Is this effectively the sum of all the lengths $ds$ as $ds$ moves from the centre to the circumference of the circle?
Yes. That's another way of saying this is the value of all the sums of dr. Because dr changes with distance from radius, we sum all the small values of dr at each distance to make the radius. If we make dr sufficiently small, as with integration, it will be a good estimation for the radius.

Quote:

Yes please!
You should probably read up a bit (wiki helps) on the hyperbolic trigonometric functions first. At least look at the relationships between those functions.

Then read this:
Derivative of the inverse hyperbolic tangent function, atanh, arctanh, tanh^{-1} Digital explorations
• May 1st 2010, 02:49 AM
HallsofIvy
That is, by the way, Poincare's "disk" model for hyperbolic geometry. I don't believe Poincare talked about "temperature"- that is a cute interpretation.

In the disk model, straight lines are modeled by either a line through the center of the disk or a circle meeting the circle bounding the disk at right angles.

All the axioms- and therefore all the theorems of hyperbolic geometry can be modeled in the Poincare disk. The important point of this is that it shows that Euclid's parallel postulate cannot be proved from the other postulates. If there were such a proof, we could duplicate it in the disk model, which is itself in Euclidean geometry, getting a contradiction in Euclidean geometry.
• May 1st 2010, 04:32 AM
Deadstar
what happens if these folks get to the center of the disk? Do they get infinitely large (and thus larger than the disk..?)
• May 1st 2010, 10:27 AM
onecentrob
Quote:

Originally Posted by Gusbob
The temperature at r distance from centre is $T = a^2 - r^2$

So when a small length ds is moved to from the centre, it will vary in proportion to the temperature T.

The new length $dr \propto T ds \rightarrow dr = k(a^2 - r^2)ds$ for some constant k.

But if you think about it, when you're at the centre w/ r = 0, you'll get $dr = ka^2 ds$ which is true iff $k = \frac{1}{ a^2 }$

Therefore $dr = \frac{a^2-r^2}{a^2} ~ds$
__________________________________________________ _______
A more intuitive way to think of it is using the ratio of lengths (L).

We have a ratio $\frac{dr}{ds}$ .

Because the lengths themselves are proportional to the specific temperature at the radius they are at,

$\frac{dr}{ds} = \frac{T_r}{T_s} = \frac{a^2 - r_2}{a^2 - r_1}$

where $r_1, r_2$ are the distance from center the lengths dr and ds are at.

At $L = ds, r = r_1 = 0, T_r = a^2$
At $L = dr, r = r_2, T_s = a^2 - r_2^2$

So we have $\frac{dr}{ds} = \frac{a^2-r_2^2} {a^2} \Rightarrow dr = \frac{a^2-r^2}{a^2} ds$

Yes. That's another way of saying this is the value of all the sums of dr. Because dr changes with distance from radius, we sum all the small values of dr at each distance to make the radius. If we make dr sufficiently small, as with integration, it will be a good estimation for the radius.

Thank you very much indeed for a very clear explanation. That all makes perfect sense.

Quote:

Originally Posted by Gusbob
You should probably read up a bit (wiki helps) on the hyperbolic trigonometric functions first. At least look at the relationships between those functions.

Then read this:
Derivative of the inverse hyperbolic tangent function, atanh, arctanh, tanh^{-1} Digital explorations

So, presumably, strictly speaking:

$
\int \frac{1}{a^2 - r^2} dr = \frac{1}{a} tanh^{-1} \frac{r}{a}
$
+ C?
• May 1st 2010, 10:37 AM
onecentrob
Quote:

Originally Posted by HallsofIvy
The important point of this is that it shows that Euclid's parallel postulate cannot be proved from the other postulates. If there were such a proof, we could duplicate it in the disk model, which is itself in Euclidean geometry, getting a contradiction in Euclidean geometry.

Yes, the section of Sawyer's the book in which this concept appears is all about non-Euclidean geometry. It is all very well explained. It was just the calculus proof which I struggled to follow. Thanks for your input.