# B Ratio of circumference to diameter for infinitely large circ

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1. Jul 23, 2016

### Darkmisc

If you divided the circumference of an infinitely large circle by its diameter, would the result be pi?

2. Jul 23, 2016

### andrewkirk

There is no such thing as an infinitely large circle, even in theory.

3. Jul 23, 2016

### Staff: Mentor

There is no such thing as an infinitely large circle. For every circle growing bigger and bigger it stays true.

Edit: at least the same wording

4. Jul 23, 2016

### Darkmisc

Cheers. Thanks.

5. Jul 24, 2016

### micromass

There is actually. It's a straight line. Although we often call lines generalized circles instead of circles. https://en.wikipedia.org/wiki/Generalised_circle
Although I don't know how to give those generalized circles a diameter and circumference in general. Perhaps by going to the Riemann sphere model...

6. Jul 24, 2016

### Staff: Mentor

I have thought about a solution of this kind, but couldn't imagine a sensible way to define diameter and circumference. Also projective spaces seem to be of little help. I'm quite sure that the passage to the limit as phrased in the OP would only lead to something like $\frac{\infty}{\infty} = \pi$. But interesting to know about generalized circles. (Where the h... do you know all these exceptional and exotic stuff from? I never even came close to it.)

A thought of mine has been another generalization: What happens in higher dimensions?
If we consider $n-$spheres, then the volume (of the surface) becomes $V(S_r^n) = c(n) \cdot r^{n}$ for some function $c(n)$.
Then $c(n) = 2 \pi^{\frac{n+1}{2}} \Gamma(\frac{n+1}{2})$ and all the magic about the definition of $\pi$ is camouflaged by this function $c$.
In this case we would have driven research on $c$ and $\pi$ would have been simply $\frac{1}{2} c(1)$.
Of course the magic will return by the vast number of occurrences of $\frac{1}{2} c(1)$ and we might would have named it $\pi$. However, the question in the OP would look rather exotic from this point of view: What happens to $\frac{V(S_r^1)}{c(1)}$ if $r$ is infinitely large?
The $2-$dimensional world is a rather special one and so is $\pi$. A metric at infinity appears edgy to me.

7. Jul 24, 2016

### micromass

You mention some interesting stuff. Circles with centers at infinity do indeed exist in projective geometry: they consist of a normal line together with the line at infinity. I won't go into detail, but the related theory is very exciting: see "perspectives on projectie gometry" by Richter-Gebert.

I'll only say this (which is the top of the iceberg really). A circle with center $(a,b,c)$ (homogeneous coordinates and through a point $(x,y,z)$ can be described by a matrix $M$. A point $p=(q,r,s)$ lies on the circle iff $pMp^T = 0$. The matrix $M$ can seen to be:
$$\left(\begin{array}{cc} c^2 & 0 & -ac\\ 0 & c^2 & -bc\\ -ac & -bc & 2axc + 2byc - c^2 (x^2 + y^2) \end{array}\right)$$

Extracting the factor $c$ and canceling it by homogenization, and then setting $c=0$ gives us
$$\left(\begin{array}{cc} 0 & 0 & -a\\ 0 & 0& -b\\ -a & -b & 2axc + 2byc \end{array}\right)$$
which is a conic consisting of the line through infinity and the line at infinity.

Using the right limit process, it doesn't seem outrageous to find that the "circumference" divided by the "diameter" is again $\pi$. But everything depends on the definition of these terms. Defining the circumference as $\pi d$, the result is rather trivial. Using another definition for circumference might make things less trivial.