Long Live π!

Poll - Total Votes: 15

The only π I like is apple π.

Stick with your books, old nerd!

You may be a math nerd, but you're sexy as hell!

Show Results

You can only vote on one answer.

(This is an old post but I thought it might be a good idea to resurrect it today...)
Happy birthday π !

Today (3/14) is π day.

(1) π is equal to the ratio of the circumference of a circle and its diameter.

(2) π is a transcendental number. Briefly, this means that we do not know its value, and we never will, haha. π, you’ll always be a mystery.

(3) π is (most likely) a normal number: the distribution probability of digits within π is random. If we assign a pair of consecutive digits to a character (such as "65"="A", cf. ASCII code table) we will find the contents of each book that has ever been written (even bad books, those behind the front row on our shelves) somewhere along the digits of π, and also any book that has not been written, and all variations thereof. There will be a version of Hamlet where Ophelia is called Helen.

(4) The value of π is more fundamental than the physical constants. If the universe did not exist, the physical constants would loose their meaning, but π would remain the same, indicating that spirit is more fundamental than matter. We, as humans, can change a lot of things, but we can’t change π. Even God cannot change π, sorry Sir, You may have created the world but You did not create π. It’s always been there. And you don't know the value of π either.

(5) π is considered to be one of the five fundamental numbers: i, e, π, 1, and 0. These numbers appear in the famous equation e^(i*π) + 1 = 0 (Gauss? Euler? Don’t remember.)

Happy birthday π!
You’re one of a kind, and we are big fans!

We love you π !

This page is a permanent link to the reply below and its nested replies. See all post replies »

JoyfulSilence · 51-55, M

Pi describes circles in flat, Euclidean space.

But what about "circles" on abstract, smooth, 2D manifolds?

That is, pick a point P on the manifold, then plot all nearby points X whose distance from P is equal to a fixed value R, which I will call the "radius".

This assumes, of course, that the distance between P and a candidate point X is well defined. I suppose one could plot all possible paths between P and X, and sum the total distance travelled in the metric indicated by the metric tensor for the manifold. Then take the minimum path length as the "distance". There may be more than one such path, and even an infinite number, but I presume one minimum distance since everything is compact. Such paths, if I not mistaken, are the geodesics (I think a geodesic is defined as the shortest path).

I presume, since things are smooth, that the set of all such X form some sort of curvy loop. At least locally. But would it self-intersect, occasionally? Would it be knotty, or not just a 1D object? Suppose I call the length C of this loop the "circumference".

I suppose one could calculate Pi(P,R) for various combinations of P (center) and "radius" R. Where Pi(P,R) = C/(2R). Assuming C is well defined. I think the smoothness of things may ensure C is. Even with self-intersections.

I do know that as R approaches zero, the manifold approaches Euclidean space (since it is smooth), so Pi(P,R) will approach the normal Pi.

Fun stuff to think about. I have not yet searched for any literature. I have never thought about this before.

ElwoodBlues · M

@JoyfulSilence Good question!! My sources inform me that
pi appears in non-Euclidean geometry (smooth manifolds) not as the ratio of a circle's circumference to its diameter, but within formulas for area, arc length, and trigonometric angles.

For example, differential geometry introduces the notion of a geodesic -- a shortest path between two points on a manifold -- and pi will come into play calculating the length of the geodesic, taking into account the local curvature of the space.